Frame Flex on GCN

[etu]

Interesting new video on GCN with a nice illustration on how frame flex might or might not make a frame better.

https://www.youtube.com/watch?v=BH_AL4rxrp8

[Kontact]

I've been saying the same thing for years. Bikes are springs, and springs are actually very efficient.

Too stiff a frame and you are going to have friction losses because the tires are going to scrub to an extent instead of conforming to the road surface because of the uneven loading that pedaling causes.

[ultraman6970]

Look back in the day was saying something similar and designed the KG86 to spring back helping the rider in the next pedaling, basically creating harmonics in the BB area, ask lemond he used that thing to win a tour :p

There is limit for everything, you have a stiff bike put stiff wheels or you get nowhere, stiff wheels in a no stiff frame and you have loss too IMO.

[Kontact]

While there is obviously a minimum and maximum stiffness that you'd want in a drivetrain, springy materials like those used in bikes are so efficient at storing and re-transmitting force that you can bend them nearly 90° and get enormous return:

https://i.imgur.com/vYwEYUb.gif

Can you imagine any other unpowered device that would allow a man to jump 20 feet vertically with nothing more than the energy of a running start? The poles are just fiberglass, like a Look KG56.

[ergott]

A spring is only a benefit if its return to original position is propelling the rider and bike forward. It also assumes your legs are rigid enough to too. If your legs simply absorb that energy it won't matter.

I haven't seen any complete studies of this that are very conclusive either way.

Sent from my SM-G950U using Tapatalk

[Kontact]

A spring is only a benefit if it's return to original position is propelling the rider and bike forward. It also assumes your legs are rigid enough to too. If your legs simply absorb that energy it won't matter.

I haven't seen any complete studies of this that are very conclusive either way.

Sent from my SM-G950U using Tapatalk

If your legs aren't rigid enough, you couldn't pedal with them.

If the drivetrain returning to normal isn't in line with propelling the bike forward, it will never be in line to propel the bike.

[Look585]

Francois Pervis, Flying 200m on a NJS steel track frame. (https://m.facebook.com/plugins/video.php?href=https%3A%2F%2Fwww.facebook.com%2FFr ancoisPERVIS1%2Fvideos%2Fvb.1422272264687513%2F147 0646453183427%2F%3Ftype%3D3&show_text=0&width=560&hc_location=ufi)

I don't know if this clarifies anything about energy recapture from flexy frames, but it does show what a guy that can do a 9.3s 200M does to a steel frame. (I'm pretty sure his Look did not flex as much.)

[ultraman6970]

Cant see the darn video but since long time ago the the guys do pretty much the same times they are doing now but in steel frames. And besides being light, a funny bike from the 80s probably is as fast as the best aero pinarello TT machine.

Marketing do magnificent things to sell stuff sometimes.

There is a video from trek with LA, they were saying that he was getting a bunch of frames custom made in the R&D stage of the products and who knows what else. happened now that in his cast for the tour he said stragith forward that his bikes were the same cr_P you can buy all over the place but with another paint job. Great news for the ones that bought those 7000 bucks replicas :D

[Peter P.]

All this discussion is what the term "planing" is all about.

If the spring-like characteristics of the frame match the power output and pedaling style of the rider, then wonderful things happen with regard to speed and bike feel.

Stiffness is over-rated.

[Kontact]

All this discussion is what the term "planing" is all about.

If the spring-like characteristics of the frame match the power output and pedaling style of the rider, then wonderful things happen with regard to speed and bike feel.

Stiffness is over-rated.

Planing may or may not be any old flex - I imagine there is a certain type of flex and frequency that feels especially ergonomic and in tune with pedaling. Flex that is of greater than ideal magnitude for "the plane" will be felt, but the question is if it is inefficient even if it is annoying.

[Bantamben]

I had a xl Merlin titanium bike that I could sprint up 2-3 minute climbs like no other bike I ever had it just seemed like a spring. But it was terrifying descending down the other side of the hill anything over 35mph was speed wobble city

[andrewsuzuki]

No mention of Jan Heine! or BQ!

[andrewsuzuki]

I know flexy frames aren't less efficient, but I still don't understand how flex enables more power (besides flexy bikes subjectively feeling smoother and thus enticing the rider to put more power into them).

Though today I was riding some 38mm tires and noticed they squirmed quite a bit when I did hard out-of-saddle efforts up a hill at relatively low cadence. First, when a tire "squirms", is is a relatively large hysteretic loss (as in > 5 watts)? Second, is it possible that lateral frame flex could store enough energy on each hard pedal stroke to reduce the flex in the tire?

Another theory I have is that since the rear wheel sees a smoother power input, and therefore the entire bike-rider system accelerates less on each pedal stroke. How big are these micro-accelerations between pedal strokes and does it amount to any nontrivial loss?

I'm most skeptical of the claim that flex lengthens the power stroke / reduces the dead spot at the top/bottom of the crank. How would this be any different than elliptical chainrings? Last I checked, there's really no consensus on elliptical chainrings enabling more power, and certainly no claims that they enable 12% more power on a hard effort.

[etu]

All this discussion is what the term "planing" is all about.

If the spring-like characteristics of the frame match the power output and pedaling style of the rider, then wonderful things happen with regard to speed and bike feel.

Stiffness is over-rated.

Or can we as riders adapt to the a specific bike without giving up comfort and efficiency?

[Kontact]

I know flexy frames aren't less efficient, but I still don't understand how flex enables more power (besides flexy bikes subjectively feeling smoother and thus enticing the rider to put more power into them).

Though today I was riding some 38mm tires and noticed they squirmed quite a bit when I did hard out-of-saddle efforts up a hill at relatively low cadence. First, when a tire "squirms", is is a relatively large hysteretic loss (as in > 5 watts)? Second, is it possible that lateral frame flex could store enough energy on each hard pedal stroke to reduce the flex in the tire?

Another theory I have is that since the rear wheel sees a smoother power input, and therefore the entire bike-rider system accelerates less on each pedal stroke. How big are these micro-accelerations between pedal strokes and does it amount to any nontrivial loss?

I'm most skeptical of the claim that flex lengthens the power stroke / reduces the dead spot at the top/bottom of the crank. How would this be any different than elliptical chainrings? Last I checked, there's really no consensus on elliptical chainrings enabling more power, and certainly no claims that they enable 12% more power on a hard effort.

I don't think this is any different than elliptical chainrings. Q rings have increased gearing on the power stroke, Biopace had decreased gearing on the power stroke, and round rings are right between. All of them seem to work about the same once you get used to them, despite all doing different things to the way power is transmitted, yet all having claims about the way they increase output through greater efficiency of the pedal stroke.

And they may be pretty much like too stiff, too springy and just right.

[cachagua]

A spring is only a benefit if its return to original position is propelling the rider and bike forward. If your legs simply absorb that energy it won't matter...


Correct. Strain energy in the frame cannot "spring back" as acceleration.

I'm not arguing for maximum stiffness in frames, I very much prefer springier ones. But their advantage does not come from returning energy they store as deflection. When force you exert with your legs flexes the frame, the frame is held steady at the other end by its (and your) resistance to acceleration -- and when your legs lessen their pressure, that resistance stays the same. This means the returned energy will never go in that direction.

If it goes back into your legs, can you store it there and get it back again? Mammalian connective tissue has a great deal of stretch to it and is used for energy storage in a number of settings: the strides of kangaroos and of wolves involve stretching tendons and allowing them to spring back, and tendon was used in bows and catapults since Classical times. I don't exactly see how this would work for pedaling a bike, but I'm open to anybody's theory -- I'd like to understand it better.

[93KgBike]

Interesting new video on GCN with a nice illustration on how frame flex might or might not make a frame better.

https://www.youtube.com/watch?v=BH_AL4rxrp8

Retrogrouches everywhere are grumbling, "No sh*t."

So will the price of steel go up or down now?

[Kontact]

So will the price of steel go up or down now?

Carbon springs great, as does Ti - as long as the diameter isn't gigantic.

Aluminum is the wet blanket.

[ergott]

Correct. Strain energy in the frame cannot "spring back" as acceleration.

I should know better than to type replies on my phone. My typing/writing skills fall apart!

[Kontact]

Correct. Strain energy in the frame cannot "spring back" as acceleration.

I'm not arguing for maximum stiffness in frames, I very much prefer springier ones. But their advantage does not come from returning energy they store as deflection. When force you exert with your legs flexes the frame, the frame is held steady at the other end by its (and your) resistance to acceleration -- and when your legs lessen their pressure, that resistance stays the same. This means the returned energy will never go in that direction.

If it goes back into your legs, can you store it there and get it back again? Mammalian connective tissue has a great deal of stretch to it and is used for energy storage in a number of settings: the strides of kangaroos and of wolves involve stretching tendons and allowing them to spring back, and tendon was used in bows and catapults since Classical times. I don't exactly see how this would work for pedaling a bike, but I'm open to anybody's theory -- I'd like to understand it better.


I think you're treating this like a simple pogo stick situation, where the cyclist is pushing a spring, and the spring goes slack at times. But what's really happening is that the bicycle is storing energy by twisting - the chainstays go up and down and essentially get shorter. And then, without the pedaling force ever going to zero, they go to neutral and twist the other way. That wouldn't happen if we pedaled with our legs together like a rowing machine, but it happens because our input is 180° out of phase. So the stored energy of the twisted stays simply feed into the alternate pedal stroke, not because we are removing force, but because the direction of force changes.

[fa63]

After watching the video, the impression I got was not that a "more springy" frame stores and returns energy into the rider in a meaningful way, but that frame stiffness does not matter as much as people think it does.

[andrewsuzuki]

Correct. Strain energy in the frame cannot "spring back" as acceleration.

The whole GCN video implies that it actually does (although I'm not sure their test on the trainer is completing accurate for demonstrating it). Jan/BQ's been saying that it does for years. And I remember looking over some physics professor's finite element analysis showing that all flex is transferred into forward movement but I can't find it anymore...anyone have a link?

So the stored energy of the twisted stays simply feed into the alternate pedal stroke, not because we are removing force, but because the direction of force changes.

I think it might happen even sooner. The return is so quick that the frame probably "tracks" the current downstroke-ing foot as the power (and therefore lateral displacement) tapers off in the bottom half of the stroke. So when the foot reaches the 6 o'clock position, the frame is already in equilibrium.

[andrewsuzuki]

After watching the video, the impression I got was not that a "more springy" frame stores and returns energy into the rider

Not exactly -- it returns energy directly into the drivetrain which becomes lossless forward movement.

but that frame stiffness does not matter as much as people think it does.

indeed. stiffness=efficiency is the intuitive conclusion which is why it's so easy to market, which in turn further inflates its importance

[fa63]

I think I was editing my comment as you were quoting me; I meant to say return energy into the rider in a meaningful way.

[Mark McM]

Let's say that maybe some type of spring compliance to store and release energy during the pedal stroke can help deliver more of the rider's power to the rear wheel. But since much of the frame flex is lateral and torsional, which is off-axis from the drive torque, maybe there is a better way to do it. For example, the Interdrive crank, with built-in spring energy storage:


http://pardo.net/bike/pic/mobi/d.interdrive-crank/farm7.staticflickr.com/6156/6231286452_105fa3d08a_o_d.jpg



I wonder why these cranks never became popular? Maybe because there was no obvious power gain?

[cachagua]

The chainstays go up and down and essentially get shorter. And then, without the pedaling force ever going to zero, they go to neutral and twist the other way...

Mmmm... so the frame springing back sums with your next pedal stroke, and the combined force moves you forward more than your pedal stroke would alone?

Maybe you could flesh that out a little bit? I'm not quite seeing it yet.

[andrewsuzuki]

Mmmm... so the frame springing back sums with your next pedal stroke

Basically yes, although as I noted in #22 it might happen a bit sooner than the next pedal stroke.

and the combined force moves you forward more than your pedal stroke would alone?

No because the energy from the spring returning was already taken from the original pedal downstroke.

Maybe you could flesh that out a little bit? I'm not quite seeing it yet.

See 4:35 to 6:00 in the GCN video

[rain dogs]

There is no free energy. It defies the laws of physics. 100W in is 100W out.

Certainly, there is a line of thinking that looks very close to: "all flex represents total loss and that no smaller amount is 'coming back' into the frame", which is incorrect, but what's 'coming back' is less than if the frame didn't flex at all in the first place (at least in a lab environment)

The stiffer frame will always lose less energy to flex. Always. But it may be a difference that is immeasureably small and has no real world influence.

The other factor is that too stiff of a frame may be possible and totally uncomfortable, but bigger tires and lower pressure takes care of that easily.

[mt2u77]

Rest assured, even the stiffest frame out there still deflects and acts like a spring. It's "spring constant" is just greater. It will still store and return just as much energy-- e.g. twice as much force to move it half the distance.

I don't see how more flex is better unless it somehow acts as a mental cue to help our pedal cycles stay on point, or indirectly as enabling some other attribute such as lighter weight, better aerodynamics, comfort, etc.

Likewise, more stiffness is not better as long as it's above some design minimum. It's not worth trading off other desirable traits to get more stiffness than is necessary. How much is necessary? Depends on the rider/use, but somewhere north of stuff rubbing together that shouldn't. I'm guessing the lower bound has a lot more to do with psychology and confidence in the bike than it has to do with performance.

[Kontact]

Mmmm... so the frame springing back sums with your next pedal stroke, and the combined force moves you forward more than your pedal stroke would alone?

Maybe you could flesh that out a little bit? I'm not quite seeing it yet.
No. I'm saying that the net force delivered should be the same, but delivered in different ways because the springy frame is storing the peak torques and delivering them later in the pedal stroke. Much like Biopace.

That said:

Rest assured, even the stiffest frame out there still deflects and acts like a spring. It's "spring constant" is just greater. It will still store and return just as much energy-- e.g. twice as much force to move it half the distance.

I don't see how more flex is better unless it somehow acts as a mental cue to help our pedal cycles stay on point, or indirectly as enabling some other attribute such as lighter weight, better aerodynamics, comfort, etc.


There may be a point when the frame resists flex to such an extent that the asymmetric forces of pedaling overwhelm the system and go somewhere else that is energy wasting - like sidewall flex or a lateral loss of traction.


I really like climbing on my old Cannondale 3.0. The short, stiff stays delivered the peak force of my standing pedal stroke directly to forward motion. However, the bike tolerated sprinting uphill less than other bikes because the tires broke traction more easily if you weren't careful. So while it felt efficient, it limited when and how much power I could put through it without wasting it. My Merlin Extralight doesn't feel like it climbs with as much authority because it feels 'soft', but I would be shocked if I was actually expending more energy, and I know the tires don't break traction in the same circumstances as the Cannondale.

Just an illustration from my experience, not saying the above is definitive. We all like the way a stiff bike responds to input, but we don't have a somatic way to measure efficiency.

And that's ignoring the chainstays' role in distributing fatiguing bumps.

[cachagua]

See 4:35 to 6:00 in the GCN video...

I didn't want to respond to the video here in our discussion because the ideas in it are embarrassingly ill-thought-out.

Here's what's happening in the video: he pushes the pedal down to flex the frame, using the rear brake to simulate the resistance to acceleration. Then he maintains that pressure on the pedal, while releasing the brake -- and snap, the frame relaxes, and the wheel spins.

This would correspond to your maintaining a perfectly unvarying pressure on the pedals through all 360 degrees, and your bike becoming easier to accelerate for a tiny moment at the bottom of each pedal stroke. But the former, while of course we try, isn't possible, and as for the latter -- I'm afraid Isaac Newton outlawed it a long time ago. The demonstration in the video is not an accurate model of riding your bike.

The bike-and-rider system's resistance to acceleration does not ever decrease. When strain energy comes out of the frame -- when you're actually riding, not dickin' around with a trainer -- there will always be something easier for it to push than you, forward.

[Kontact]

The bike-and-rider system's resistance to acceleration does not ever decrease. When strain energy comes out of the frame -- when you're actually riding, not dickin' around with a trainer -- there will always be something easier for it to push than you, forward.

And that would be what?

[cachagua]

And that would be what?

Two answers to that: first, it doesn't matter -- not to the point I'm making, which is that frame flex does not rebound in a way that contributes to moving you up the road.

But second, let me offer this guess: as the frame releases its strain energy, rather than increasing your forward motion, it slows your pedaling.

Here's how I imagine that happening: when you push on the pedals of a hypothetical 100% efficient bike, the resistance your legs encounter is the inertia of making you go forward. When you push against a real bike, some of your push makes you go, and some flexes the bike. All the time you're pushing, the bike is pushing back -- the energy in its flex wants to come back out, you might say. And when you push a little softer, between the peaks in your output, your legs will go a little shorter distance around the pedal circle as the frame straightens.

What other way could the system behave? Frames flex when you pedal, none of us questions that. When you ease up on the pedals, and the frame springs back, does the energy go to where it couldn't go before, i.e. your inertia? It's still just as hard to move you up the road as it's been all this time. Or instead, does it go toward where it's just become easier for it to go, now that you're not pushing so hard?

That's how I picture things.

[Kontact]

Two answers to that: first, it doesn't matter -- not to the point I'm making, which is that frame flex does not rebound in a way that contributes to moving you up the road.

But second, let me offer this guess: as the frame releases its strain energy, rather than increasing your forward motion, it slows your pedaling.

Here's how I imagine that happening: when you push on the pedals of a hypothetical 100% efficient bike, the resistance your legs encounter is the inertia of making you go forward. When you push against a real bike, some of your push makes you go, and some flexes the bike. All the time you're pushing, the bike is pushing back -- the energy in its flex wants to come back out, you might say. And when you push a little softer, between the peaks in your output, your legs will go a little shorter distance around the pedal circle as the frame straightens.

What other way could the system behave? Frames flex when you pedal, none of us questions that. When you ease up on the pedals, and the frame springs back, does the energy go to where it couldn't go before, i.e. your inertia? It's still just as hard to move you up the road as it's been all this time. Or instead, does it go toward where it's just become easier for it to go, now that you're not pushing so hard?

That's how I picture things.

I don't know why pushing against the bike would a variable thing where your legs do it sometimes but not others. Why would imposing a spring in the transmission of energy cause your legs to work differently than they do when the transmission is rigid?

That sounds very arbitrary. Why would your legs "know" to absorb feedback from stored spring energy, but not absorb feedback from fighting inertia?
Both are feeding back into your legs.

[Kontact]

I think this animation of a handcar might be instructive:

https://upload.wikimedia.org/wikipedia/commons/1/15/Hand_Car_Animation_with_men.gif

Imagine that the blue drive shaft is a piston with a spring in it. The spring can store energy by compressing or stretching. When Mr. Lefty pushes down the spring shaft compresses, storing energy. When Mr. Righty starts to push down, the motion is reversed, so his effort is pulling the compressed shaft up. Before that motion can stretch the spring, it first gets a free boost from Mr. Left's spring compression, so he has an easier time getting his lever moving. Compression gets traded for stretch, and when it is Lefty's turn again the drive shaft is preloaded to assist the start of his push.

Both men are keeping tension on their levers, so the stored spring energy is never going to just sproing away as loss. As long as the spring is efficient (and real springs are quite efficient), loss of energy from elastic deformation is low. And that small loss may be balance out (or exceeded) by the increased efficiency from a decrease in input amplitude.

[Mark McM]

I think this animation of a handcar might be instructive:

https://upload.wikimedia.org/wikipedia/commons/1/15/Hand_Car_Animation_with_men.gif

Imagine that the blue drive shaft is a piston with a spring in it. The spring can store energy by compressing or stretching. When Mr. Lefty pushes down the spring shaft compresses, storing energy. When Mr. Righty starts to push down, the motion is reversed, so his effort is pulling the compressed shaft up. Before that motion can stretch the spring, it first gets a free boost from Mr. Left's spring compression, so he has an easier time getting his lever moving. Compression gets traded for stretch, and when it is Lefty's turn again the drive shaft is preloaded to assist the start of his push.

Both men are keeping tension on their levers, so the stored spring energy is never going to just sproing away as loss. As long as the spring is efficient (and real springs are quite efficient), loss of energy from elastic deformation is low. And that small loss may be balance out (or exceeded) by the increased efficiency from a decrease in input amplitude.

Ah, that explains why all reciprocating crank drive systems (such as internal combustion piston engines) use elastic connecting rods!

Oh wait ... they don't.

If the connecting compresses when Mr. Lefty pushes down, Mr. Left has to exert more energy, because he has to push the lever further than if the rod were rigid (the extra energy is stored in the rod). Mr. Righty might get a little bit boost as the rod rebounds, but then he has to re-exert the same amount of energy to stretch the rod as he pushes down. In the end, there is no net energy gained or lost as power is transmitted through the crank. But there very well might be a loss as power is applied to the lever, since both Mr. Lefty and Mr. Righty have to move the lever further in both directions.

As shown in the spring drive crank above, there have been several attempts to add 'energy storage' into the drive system. All have failed to produce more efficient drive systems.

[andrewsuzuki]

Finally found an archive to the site describing how the strain energy is stored and returned to the drivetrain, with a finite element analysis.

https://web.archive.org/web/20060214082906/http://bikethink.com/Frameflex.htm

Additionally a graph of power going in and out of the frame

https://web.archive.org/web/20060214135954/http://bikethink.com:80/Power_output.htm

[fa63]

From the study:

"Conclusion:

Having concluded that frame flex does not waste energy, I do not believe that frame stiffness is irrelevant. You could say that a stiff frame feels more responsive. A stiffer frame can give the rider more confidence especially in a sprint."

[William]

When I started racing I was on a frame that flexed quite a bit under pressure in the standing sprint, on an uphill sprint I could actually get it to ghost shift when really laying down power. That said I won a lot of races on that bike and honestly, I don't think the flex really added power directly back to the drive train. What I think was happening was the flex actually caused it to "snap" back fractions faster than stiffer frames I was on later. That would get the frame back faster for the top of the stroke on the other side and vice versa. Translating into a faster sprint rhythm and quicker pedal stroke....harmonics. Probably only fractions of a second on each side but that adds up over the length of the sprint.

I've often wondered why subsequent frames, production and custom didn't quite have the feel of that frame in the sprint, but then I was under the impression that I needed to go stiffer due to my height and strength. Too much flex isn't good either though...

Sprints...
https://www.youtube.com/watch?v=nqWANNRlqpo

Science of the Sprint (sharing for the overhead of the sprint bike throw from side to side)...
https://www.youtube.com/watch?v=SIruLOp-PrM


Keep in mind I'm not a scientist and I don't play one on TV either.:)






William

[Kontact]

Ah, that explains why all reciprocating crank drive systems (such as internal combustion piston engines) use elastic connecting rods!

Oh wait ... they don't.

If the connecting compresses when Mr. Lefty pushes down, Mr. Left has to exert more energy, because he has to push the lever further than if the rod were rigid (the extra energy is stored in the rod). Mr. Righty might get a little bit boost as the rod rebounds, but then he has to re-exert the same amount of energy to stretch the rod as he pushes down. In the end, there is no net energy gained or lost as power is transmitted through the crank. But there very well might be a loss as power is applied to the lever, since both Mr. Lefty and Mr. Righty have to move the lever further in both directions.

As shown in the spring drive crank above, there have been several attempts to add 'energy storage' into the drive system. All have failed to produce more efficient drive systems.
There is no net energy loss. Correct.

How are Lefty and Righty going to move the crank further in both directions? The crank is a fixed length.

But on a bicycle with BB flex, the load path of the crank arm isn't a perfect circle because the flex will make it slightly ovalized.


No one is saying springy-ness adds any energy to the system. The point is that sprung parts don't actually give energy away - it is still trapped in the drive train between the motor and the road, and it can't go out side that or back upstream.

[Mark McM]

There is no net energy loss. Correct.

No net energy loss in the connecting rod or crank. But due to the extra travel of the lever, there may be extra energy loss at the lever.

How are Lefty and Righty going to move the crank further in both directions? The crank is a fixed length.

I didn't say the crank moves further, I said the lever (operated by Messrs. Lefty and Righty) moves further. Since the lever moves the connecting rod, if the connecting rod contracts/expands, the lever will have to move an extra distance corresponding to the change in length of the connecting rod. And that means that Messrs. Lefty and Right have to move their arms further to move the lever further, likely requiring them to expend more energy.

No one is saying springy-ness adds any energy to the system. The point is that sprung parts don't actually give energy away - it is still trapped in the drive train between the motor and the road, and it can't go out side that or back upstream.

There may not be energy lost in the cranks/frame, but if the rider has move their legs through a longer distance, there may be energy lost in the rider's legs.

[Kontact]

No net energy loss in the connecting rod or crank. But due to the extra travel of the lever, there may be extra energy loss at the lever.


There may not be energy lost in the cranks/frame, but if the rider has move their legs through a longer distance, there may be energy lost in the rider's legs.

Why does more distance become more energy? Are you saying long crank arms waste energy somehow?

[Black Dog]

I think part of the issue that we are missing here is the energy expended by a rider that does not generate muscle contraction and limb motion. Think of pushing against a wall with as much force as you can generate. Since the wall does not move there is no work being done. However, you are burning calories and exhausting muscles. This is why we have gears on bikes so that we can generate movement when we apply force to the peddles under a wide range of conditions (wind, hills etc..). To illustrate: try starting from a dead stop on a hill in a 53-11.

If a frame has some flex and the energy is returned then it is less likely that a rider would be generating none motive force and thus would be more efficient.

[Kontact]

I think part of the issue that we are missing here is the energy expended by a rider that does not generate muscle contraction and limb motion. Think of pushing against a wall with as much force as you can generate. Since the wall does not move there is no work being done. However, you are burning calories and exhausting muscles. This is why we have gears on bikes so that we can generate movement when we apply force to the peddles under a wide range of conditions (wind, hills etc..). To illustrate: try starting from a dead stop on a hill in a 53-11.

If a frame has some flex and the energy is returned then it is less likely that a rider would be generating none motive force and thus would be more efficient.

I honestly don't understand any of this, and I think your last sentence must be missing some words. But if you push against a wall your muscles are definitely contracting.

[cachagua]

Both men are keeping tension on their levers...

If in that diagram the men are maintaining unvarying pressure on their levers, that's exactly what doesn't happen on a bike. If it did, we'd be having a very different conversation, but it does not -- I don't think you're seriously trying to assert that.

Here's another way of thinking about it. Remember that video? The rider pushes the pedal down onto the block, the frame flexes, he lets the brake loose and the energy stored in the frame's flex turns the wheel.

Again, for clarity: the release of the frame flex turns the rear wheel when what's been resisting its rotation stops resisting, and the thing that's been pushing it -- the rider's foot on the pedal -- keeps pushing just as hard.

But this is not what happens when you ride. What happens when you ride would be illustrated by the rider's pushing down the pedal, flexing the frame, and then NOT releasing the brake but instead easing up on the pedal. Bike + rider's inertia stays the same, pedaling force decreases.

What would happen then? The pedal would rise back up to where it started, pushed by the strain energy being released from the frame.

Compression gets traded for stretch...

In other words, after the frame flexes from one leg's pedal stroke, that flex, when released, adds its energy to the next stroke? I don't think it does.

Let's look at the bike on the trainer again. You keep the brake squeezed, but ease up on the cranks, the right pedal rises up from the block -- but at the same time, the left pedal also makes a counter-rotating motion. The release of the frame flex does not advance the other pedal but moves it backwards, and thus doesn't increase the force of the next stroke, but decreases it, too.

[Kontact]

If in that diagram the men are maintaining unvarying pressure on their levers, that's exactly what doesn't happen on a bike. If it did, we'd be having a very different conversation, but it does not -- I don't think you're seriously trying to assert that.

Here's another way of thinking about it. Remember that video? The rider pushes the pedal down onto the block, the frame flexes, he lets the brake loose and the energy stored in the frame's flex turns the wheel.

Again, for clarity: the release of the frame flex turns the rear wheel when what's been resisting its rotation stops resisting, and the thing that's been pushing it -- the rider's foot on the pedal -- keeps pushing just as hard.

But this is not what happens when you ride. What happens when you ride would be illustrated by the rider's pushing down the pedal, flexing the frame, and then NOT releasing the brake but instead easing up on the pedal. Bike + rider's inertia stays the same, pedaling force decreases.

What would happen then? The pedal would rise back up to where it started, pushed by the strain energy being released from the frame.



In other words, after the frame flexes from one leg's pedal stroke, that flex, when released, adds its energy to the next stroke? I don't think it does.

Let's look at the bike on the trainer again. You keep the brake squeezed, but ease up on the cranks, the right pedal rises up from the block -- but at the same time, the left pedal also makes a counter-rotating motion. The release of the frame flex does not advance the other pedal but moves it backwards, and thus doesn't increase the force of the next stroke, but decreases it, too.
At no point when you are pedaling do you remove all tension from the pedals. If you believe something different something different there's nothing to discuss.

[Black Dog]

I honestly don't understand any of this, and I think your last sentence must be missing some words. But if you push against a wall your muscles are definitely contracting.

Ok. Let me try to be more articulate. I am not saying that muscles are not contracting, they are. I am saying that the contractions are not generating movement of the limbs because they can not overcome the resistance. I am speculating that a frame with some flex could absorb some of the force that would otherwise be wasted not turning the cranks due to excessive resistance and then return it to the system. This could also reduce peak muscle loads and reduce fatigue. I hope this is clearer than my previous post.

[Kontact]

Ok. Let me try to be more articulate. I am not saying that muscles are not contracting, they are. I am saying that the contractions are not generating movement of the limbs because they can not overcome the resistance. I am speculating that a frame with some flex could absorb some of the force that would otherwise be wasted not turning the cranks due to excessive resistance and then return it to the system. This could also reduce peak muscle loads and reduce fatigue. I hope this is clearer than my previous post.
I agree. The flex reduces the peaks and delivers that force later.

[cachagua]

At no point when you are pedaling do you remove all tension from the pedals...

Of course not. You don't have to reduce your pedaling force to zero, you just have to reduce it, for what I describe to happen. When you're putting forth maximum force, it's equal to the riding inertia you're overcoming plus the frame flex, and when you back off even the tiniest bit the frame flex springs back correspondingly. Back to the bike on the trainer: if 150 pounds will push the pedal down onto the block, then 75 would let the pedal back up part way -- but still would not turn the back wheel, if the brake remained locked.

The flex reduces the peaks and delivers that force later.

Yes, please say more. Reduces the peaks by how much and in what way, and delivers the force where and how?

[Kontact]

Of course not. You don't have to reduce your pedaling force to zero, you just have to reduce it, for what I describe to happen. When you're putting forth maximum force, it's equal to the riding inertia you're overcoming plus the frame flex, and when you back off even the tiniest bit the frame flex springs back correspondingly. Back to the bike on the trainer: if 150 pounds will push the pedal down onto the block, then 75 would let the pedal back up part way -- but still would not turn the back wheel, if the brake remained locked.



Yes, please say more. Reduces the peaks by how much and in what way, and delivers the force where and how?
In much the same way that Biopace does - by diverting some of the force into storage then expending it later in the pedal stroke. A small part of peak force gets turned into the frame flexing, and when the shift is made to the opposite crank arm, it is expended as forward motion of the bicycle as the chainstays extend back to normal length.

[msl819]

This may have been asked but if the bike in the video didn’t have a chain would you get the same results of rear tire spin? Is the energy being stored in the frame and returned through the rear wheel or is the chain being flexed under pressure and once the rear brake is released it turns the wheel over.

[Kontact]

This may have been asked but if the bike in the video didn’t have a chain would you get the same results of rear tire spin? Is the energy being stored in the frame and returned through the rear wheel or is the chain being flexed under pressure and once the rear brake is released it turns the wheel over.

The resistance of stepping down on the pedal is from the chain in tension. No chain, no resistance. No chain, nothing to pull the wheel when the brake is released.

[martl]

I see it that way: it only recently has been widely acknowledged that "frame response" actually plays a part in how a bike *feels*. Whether it plays a part in how the bike *performs* and if so, in what way, is still very much up to debate - as this thread shows.

So, the GCN vid is to be welcomed as a contribution to the debate. A definite truth it won't and can't tell.

[msl819]

The resistance of stepping down on the pedal is from the chain in tension. No chain, no resistance. No chain, nothing to pull the wheel when the brake is released.

Right, what I am asking is if the energy that is stored and transferred is more in the chain or frame. Obviously, no frame flex no stored energy in the chain and vice versa. But when we are loading our frame in a real world situation and the rear wheel isn’t locked to build the tension do the physics function in the same way given different variables. Maybe the question I am asking is does the chain function the same way when it isn’t held in tension with the rear brake as it does in the video and if not does that alter the experiment?

[andrewsuzuki]

what I am asking is if the energy that is stored and transferred is more in the chain or frame

The FEA (https://web.archive.org/web/20060214082906/http://bikethink.com/Frameflex.htm) suggests ~1% (https://web.archive.org/web/20060214135954/http://bikethink.com:80/Power_output.htm) of rider power goes towards flexing the frame (a steel frame of decent quality). Not sure about chain elasticity however.

I see it that way: it only recently has been widely acknowledged that "frame response" actually plays a part in how a bike *feels*.

Well, people looking for performance steel bikes have long known to look at tubing specs. Reynolds 531? must be a good bike! Triple-butted? must be a good bike! Whether or not they actually reasoned out why 531 or triple-butted tubes tend to make a good bike.

[Kontact]

Right, what I am asking is if the energy that is stored and transferred is more in the chain or frame. Obviously, no frame flex no stored energy in the chain and vice versa. But when we are loading our frame in a real world situation and the rear wheel isn’t locked to build the tension do the physics function in the same way given different variables. Maybe the question I am asking is does the chain function the same way when it isn’t held in tension with the rear brake as it does in the video and if not does that alter the experiment?

While everything stretches or compresses to an extent, the tension shown in the video is from the frame changing shape - specifically it is twisting. Does that frame twist that much while riding? Yes, but mainly when sprinting or climbing. But we can easily see frames twisting under load, and that twisting is going to decrease the distance from the rear cog to the chainring.

[Mark McM]

In much the same way that Biopace does - by diverting some of the force into storage then expending it later in the pedal stroke. A small part of peak force gets turned into the frame flexing, and when the shift is made to the opposite crank arm, it is expended as forward motion of the bicycle as the chainstays extend back to normal length.

While there may be some energy storage/return in the frame, the two mechanism described above don't work in that regard:

1) Biopace chainrings don't store energy. They merely change effective leverage ratio (gear ratio) of the drivetrain.

2) Chainstays are far too stiff in tension/compression to store any meaningful energy in compression, so their is essentially no energy "expended as forward motion of the bicycle as the chainstays extend back to normal length". Instead, the primary mode of chainstay flex is bending/torsion. This produces a frame deflections (and deflections at the crank) which are orthogonal to the direction of chain travel. This is the root of the whole quandary about how lateral/torsional elastic deflection can result in forward motion of the bicycle.

[Kontact]

While there may be some energy storage/return in the frame, the two mechanism described above don't work in that regard:

1) Biopace chainrings don't store energy. They merely change effective leverage ratio (gear ratio) of the drivetrain.

2) Chainstays are far too stiff in tension/compression to store any meaningful energy in compression, so their is essentially no energy "expended as forward motion of the bicycle as the chainstays extend back to normal length". Instead, the primary mode of chainstay flex is bending/torsion. This produces a frame deflections (and deflections at the crank) which are orthogonal to the direction of chain travel. This is the root of the whole quandary about how lateral/torsional elastic deflection can result in forward motion of the bicycle.

Biopace rings "store" energy by keeping crank speed higher going into the dead zone in exchange for lower input force at 3 o'clock.


I didn't say the chainstay is compressing. I said that the twisting of the chainstays compress the chainstay distance. This is just like what a coil spring does - the change in deflection of the coils 'compresses' the overall length of the spring.

[Mark McM]

Biopace rings "store" energy by keeping crank speed higher going into the dead zone in exchange for lower input force at 3 o'clock.

If anything, just the opposite occurs. The asymmetry of Biopace chainrings is orthogonal to most non-round chainrings, and the cranks slow down during the "dead zone", and speed up during the power stroke.


I didn't say the chainstay is compressing. I said that the twisting of the chainstays compress the chainstay distance. This is just like what a coil spring does - the change in deflection of the coils 'compresses' the overall length of the spring.

Chainstays are not anything like a coil spring. Chainstays are not coiled. Chainstays are oriented in-line with the drive force whereas in a coil spring (or any other torsion spring) the flexural member is oriented orthogonally to the force. If chainstays are suppose to act like coil springs, they are completely mis-oriented.

[Kontact]

If anything, just the opposite occurs. The asymmetry of Biopace chainrings is orthogonal to most non-round chainrings, and the cranks slow down during the "dead zone", and speed up during the power stroke.
Correct. They speed up in the power stroke because less energy is tapped off to make bicycle velocity, and they slow down in the dead zone as that 'stored' crank velocity is used up making power later. You are not disagreeing with me.




Chainstays are not anything like a coil spring. Chainstays are not coiled. Chainstays are oriented in-line with the drive force whereas in a coil spring (or any other torsion spring) the flexural member is oriented orthogonally to the force. If chainstays are suppose to act like coil springs, they are completely mis-oriented.

They aren't "supposed to" act like anything but structural members, but we are talking about what they actually do on bicycles, and what they do is twist around their shared centerline. And when they twist, the distance between the BB and the rear hub decreases. When the energy used to twist them is released, the rear center elongates. This elongation could be said to push the bike forward from the rear hub, or it could be said to pull the chain forward, driving the wheel. These aren't actually any different - just "feet push down on the earth, the earth pushes up on your feet" kind of semantics.

Each individual chainstay is not compressing, but the 'rear center' assembly is compressing. And I used a coil spring as a comparison because the wire composing the spring does not compress, but the spring as a whole does because the wires flex. This is all just an explanation for something that was implied by MSL819's question - that the chain might stretch or the individual chainstay might compress. The chainstays change shape, which decreases their length like a bow gets shorter when you pull the string.

[rain dogs]

Pretty much every component on a bicycle is designed to operate on a very linear plane. The more a frame flexes, every bearing, bushing and connection/contact point is being loaded and operated outside of this ideal plane. That is friction and loss in the same way chainline losses occur. The stiffer the frame the less loss to these tangential loads.

Again, you can argue that this loss is miniscule and insignificant in the real world (I'm guessing you don't invest in oversize pulleys and ceramic bearings) but it is watts. It cannot not be.

The only possible argument for frame flex leading to a more efficient bicycle is rider comfort and fatigue factors on the rider. But 2-5mm wider supple tires and 5-10psi less tire pressure will do magnitudes more to increase comfort than a flexy frame. This conversation is like the the ontological argument for the existence of ____... the sasquatch. I get it.... it's not so much loss to frame flex, but it sure as heck is no gains.

So hey, if you like the flexy frame.... ride the hell out of it!

[Ungaro]

Just from my humble and non-professional experience:
1) Stiffness is a fine line.
2) I've compared two bikes side by side on a computerized diagnostic set up - one was a fillet brazed number made of Columbus Foco tubing (anyone remember that light weight stuff?), the other was a new 2007 Orbea Orca. The Orca was more efficient compared to the Foco tubed bike, but please note that the Foco tubed frame was shot. No really, I had ridden the snot out of that thing for 8 years, and say what you will, a lugged steel frame will hast longer than a non-lugged one.
3) Carbon can be VERY stiff. So stiff, in fact that over a road course, the accumulated micro vibrations can soak into you and wear you slap out.
4) Carbon is it's own worst enemy - What I mean is that it's a great ride for the first few years, but that stiffness helps to also degrade it over time. In the end, it's way less efficient.
5) What happens to plastic when you leave it in the sun? Think what happens to your carbon bike.
6) Last year I reconditioned a 1971 Raleigh Pro for Eroica California. I road that almost exclusively for 2 months. At the end of that time, I got back on the Orbea (now 10 years old) and felt invigorated...like "yeah, this beast is fast!"....until I checked Strava afterward. I was worn out and exhausted and it wasn't even a fast pace. Two days later, I repeated the same course with the 1971 Raleigh. Result was a faster time and I felt refreshingly great at the end of 30 miles. I repeated this little experiment a few times to see if it wasn't a fluke. It wasn't.
7) Sold the Orbea frame and purchased a new lugged steel frame (Bottecchia Leggendaria), and have had incredible rides ever since.

So, for me, yes....a little flex is good. Too much flex is not.
FWIW....just my experience.

[Kontact]

Pretty much every component on a bicycle is designed to operate on a very linear plane. The more a frame flexes, every bearing, bushing and connection/contact point is being loaded and operated outside of this ideal plane. That is friction and loss in the same way chainline losses occur. The stiffer the frame the less loss to these tangential loads.

Again, you can argue that this loss is miniscule and insignificant in the real world (I'm guessing you don't invest in oversize pulleys and ceramic bearings) but it is watts. It cannot not be.

The only possible argument for frame flex leading to a more efficient bicycle is rider comfort and fatigue factors on the rider. But 2-5mm wider supple tires and 5-10psi less tire pressure will do magnitudes more to increase comfort than a flexy frame. This conversation is like the the ontological argument for the existence of ____... the sasquatch. I get it.... it's not so much loss to frame flex, but it sure as heck is no gains.

So hey, if you like the flexy frame.... ride the hell out of it!

I could see this argument if bikes had shaft drives, but the only component that really has to deal with the flex is the chain, and we are already quite happy operating that outside of the ideal plane by using multispeed gear trains.

I disagree that bigger tires are a better solution than flexible frames. The frame flex appears to be energy we can recover, but sidewall flex of larger rubber tires is heat loss. A balance between the two would be more efficient than a stiff frame and big tire.

[rain dogs]

I could see this argument if bikes had shaft drives, but the only component that really has to deal with the flex is the chain, and we are already quite happy operating that outside of the ideal plane by using multispeed gear trains.

I disagree that bigger tires are a better solution than flexible frames. The frame flex appears to be energy we can recover, but sidewall flex of larger rubber tires is heat loss. A balance between the two would be more efficient than a stiff frame and big tire.

1. Squeaking and creaking is friction, every bearing is dealing with frame flex
2. Bigger tires are faster and more efficient that has been proven time and time again.
3. You cannot recover 100% of anything. It's better to lose as little of it in the first place.

[Mark McM]

Correct. They speed up in the power stroke because less energy is tapped off to make bicycle velocity, and they slow down in the dead zone as that 'stored' crank velocity is used up making power later. You are not disagreeing with me.

Well which is it? Do they speed up in the dead zone to store energy, or do they slow down in the dead zone to store energy? It sounds like you can't make up your mind.

They aren't "supposed to" act like anything but structural members, but we are talking about what they actually do on bicycles, and what they do is twist around their shared centerline. And when they twist, the distance between the BB and the rear hub decreases. When the energy used to twist them is released, the rear center elongates. This elongation could be said to push the bike forward from the rear hub, or it could be said to pull the chain forward, driving the wheel. These aren't actually any different - just "feet push down on the earth, the earth pushes up on your feet" kind of semantics.

You're envisioning things that don't happen. I challenge you to measure any meaningful change in the distance between BB and hub on any rigid (non-articulated) rear triangle. We can easily measure lateral deflection at the BB and we can easily measure rotational deflection at the BB (relative to the rear hub), but there is no meaningful change in actual distance between BB and hub.

Energy stored in a spring is inversely proportional to the spring constant (stiffness). Since the linear stiffness between the BB and hub is so very high, there is virtually no energy stored in this deflection. Instead, there is far more energy stored in lateral and torsional deflection at the BB - but since these deflections are orthogonal to the chain force, you'll have to come up with a mechanism to redirect these forces/energies.

[William]

A simple exercise: Standing my bike up and leaning it like it would be in a spring, I apply pressure to the crank arm (at the bottom of the stroke) and the frame flexes. The quicker and with more pressure I do it and let off, the quicker it snaps back. In a sprint under hard torque where does that rebound go?

Interesting to note how much flex this guy is getting in the handle bars and built up stationary bike frame when he is applying power. He can't rock the bike as much as in the real world but consider if the guy in the GCN video could flex that frame by just standing on the pedal and holding the brake, how much flex will there be in a full on standing sprint when he mashing and torquing the pedals and handle bars back and forth? Quite a bit I would imagine.

https://www.youtube.com/watch?v=kiBw8mSgjRU







William

[William]

If you look at that guys related video with the push down/pull up exercise, there is a lot of torque being generated.





William

[JStonebarger]

There is no free energy. It defies the laws of physics. 100W in is 100W out.

Certainly, there is a line of thinking that looks very close to: "all flex represents total loss and that no smaller amount is 'coming back' into the frame", which is incorrect, but what's 'coming back' is less than if the frame didn't flex at all in the first place (at least in a lab environment)

The stiffer frame will always lose less energy to flex. Always. But it may be a difference that is immeasureably small and has no real world influence...

Where does that lost energy go? If energy loss is significant at all shouldn't my whippy steel frame heat up as the frame flexes? Funny, I've never noticed that on even the longest climbs.

[Kontact]

Well which is it? Do they speed up in the dead zone to store energy, or do they slow down in the dead zone to store energy? It sounds like you can't make up your mind.

As I said, the cranks gather velocity in the power zone and that velocity is carried into the dead zone where it is used up. The "storage" is just between power and dead.

I don't know why you don't understand what I mean.


You're envisioning things that don't happen. I challenge you to measure any meaningful change in the distance between BB and hub on any rigid (non-articulated) rear triangle. We can easily measure lateral deflection at the BB and we can easily measure rotational deflection at the BB (relative to the rear hub), but there is no meaningful change in actual distance between BB and hub.

I don't have to measure it - the GCN video quite clearly shows the "expanding" stays pulling on the chain. Instead of letting the wheel go completely, they could have let the brake out slowly and demonstrated exactly how much distance the chain is pulled for the amount of crank deflection.

[Kontact]

A simple exercise: Standing my bike up and leaning it like it would be in a spring, I apply pressure to the crank arm (at the bottom of the stroke) and the frame flexes. The quicker and with more pressure I do it and let off, the quicker it snaps back. In a sprint under hard torque where does that rebound go?


It goes into the drivetrain. None of the flex in bicycle is ever immediately released like a mousetrap because the rider is always pushing on the drivetrain and the tire is always pushing against the road. For the energy to be released in a snap you would have to break you leg or have the rear wheel come off the road.

I was a pole vaulter in high school. Sprint, plant the pole, pole smoothly flexes, you change direction, pole expands to full length and you let it go way up in the air. But one time the pole broke just at the beginning of bending, and a tremendous amount of energy was released all at once into my arms and I just fell forward into the mat. It was awful, but demonstrated how that controlled release of energy is useful and immediate release is not.

[andrewsuzuki]

Where does that lost energy go? If energy loss is significant at all shouldn't my whippy steel frame heat up as the frame flexes? Funny, I've never noticed that on even the longest climbs.

Pretty much every component on a bicycle is designed to operate on a very linear plane. The more a frame flexes, every bearing, bushing and connection/contact point is being loaded and operated outside of this ideal plane. That is friction and loss in the same way chainline losses occur. The stiffer the frame the less loss to these tangential loads.

In a sprint under hard torque where does that rebound go?

I'll post this link once more with the explanation yall:

https://web.archive.org/web/20060214082906/http://bikethink.com/Frameflex.htm

As for heat loss, which yes exists, I've been looking over this paper Elastic Hysteresis of Steel (http://rspa.royalsocietypublishing.org/content/royprsa/89/614/528.full.pdf), which happens to use drawn tubing almost identical in stiffness to a 25.4 steel tube with .5mm wall thickness (although admittedly it isn't the exact composition or manufacturing process as bicycle tubing). Given the results on page 535 and some quick "broscience" calculations I did, the heat loss under very hard pedaling (1-2cm lateral flex) is something around the magnitude of 10^-4 watts which doesn't even put it on the same magnitude as oversized pulley wheels. Also, the damping ratio of composite and aluminum is in the same order of magnitude of steel.

But 2-5mm wider supple tires and 5-10psi less tire pressure will do magnitudes more to increase comfort than a flexy frame

The argument for lateral flex doesn't have much to do with vertical compliance. In fact, I'd almost argue that bikes that are more laterally flexy tend to be less vertically compliant, since flexy bikes tend to correlate with horizontal top tubes, meaning longer seat tubes, meaning less exposed seatpost (which is the only thing that really flexes vertically on the rear end of a bike).

[fa63]

Where does that lost energy go? If energy loss is significant at all shouldn't my whippy steel frame heat up as the frame flexes? Funny, I've never noticed that on even the longest climbs.


If there was any heat generated due to frame flex, it would be so small that you would never notice it by touch.

[JStonebarger]

If there was any heat generated due to frame flex, it would be so small that you would never notice it by touch.

Sounds insignificant to me. Meanwhile, I suspect the role of frame stiffness has mostly to do with ad copy/sales...

[ultraman6970]

I relate with this post a lot.

Just from my humble and non-professional experience:
1) Stiffness is a fine line.
2) I've compared two bikes side by side on a computerized diagnostic set up - one was a fillet brazed number made of Columbus Foco tubing (anyone remember that light weight stuff?), the other was a new 2007 Orbea Orca. The Orca was more efficient compared to the Foco tubed bike, but please note that the Foco tubed frame was shot. No really, I had ridden the snot out of that thing for 8 years, and say what you will, a lugged steel frame will hast longer than a non-lugged one.
3) Carbon can be VERY stiff. So stiff, in fact that over a road course, the accumulated micro vibrations can soak into you and wear you slap out.
4) Carbon is it's own worst enemy - What I mean is that it's a great ride for the first few years, but that stiffness helps to also degrade it over time. In the end, it's way less efficient.
5) What happens to plastic when you leave it in the sun? Think what happens to your carbon bike.
6) Last year I reconditioned a 1971 Raleigh Pro for Eroica California. I road that almost exclusively for 2 months. At the end of that time, I got back on the Orbea (now 10 years old) and felt invigorated...like "yeah, this beast is fast!"....until I checked Strava afterward. I was worn out and exhausted and it wasn't even a fast pace. Two days later, I repeated the same course with the 1971 Raleigh. Result was a faster time and I felt refreshingly great at the end of 30 miles. I repeated this little experiment a few times to see if it wasn't a fluke. It wasn't.
7) Sold the Orbea frame and purchased a new lugged steel frame (Bottecchia Leggendaria), and have had incredible rides ever since.

So, for me, yes....a little flex is good. Too much flex is not.
FWIW....just my experience.

[Kontact]

1. Squeaking and creaking is friction, every bearing is dealing with frame flex
2. Bigger tires are faster and more efficient that has been proven time and time again.
3. You cannot recover 100% of anything. It's better to lose as little of it in the first place.

1. Flexible bikes don't squeak any more than rigid bikes. If you account for the all the PF30 bikes, it would appear that not flexing is more of a problem. Having a BB shell flex with the cranking motion is going to put less side load on the bearings, not more.

2. Bigger tires are faster if we are talking 20s vs 23s or 25s. Above 25s the rolling resistance goes up when you are riding the correct pressure for your weight. The only graphs that show decreasing rolling resistance with size are when all the tires are at the same pressure. No one is riding 28s at 110psi.

3. That's correct, but you can choose where you lose it. Which is faster: A four wheel drive sports car or a mechanically more efficient two wheel drive? If (and this is an if) a flexy bike aligns forces with bearing better and keeps tire contact better, than the flex loss would be worth it for increased bearing and tire efficiency.

[etu]

a bit neutral on whether flexible frames are more efficient, but I thought the idea was that more you're in sync with your bike, the more motivated you are to put out more power. also there is the thought that being limited by our cardiovascular limits means we're closer to our potential power output than being limited by burning in our legs/butts. would love to hear other's thoughts on this latter claim.

[Mark McM]

As I said, the cranks gather velocity in the power zone and that velocity is carried into the dead zone where it is used up. The "storage" is just between power and dead.

I don't know why you don't understand what I mean.

Because it is clear that you do not understand what you mean.
If you understood what was going on, then you'd realize that the reciprocating motion of the legs already incorporates some amount of energy storage (as the cranks rise) and release (as the crank drops) without needing some type eccentricity. In fact, the rotary velocity changes with Biopace cranks probably results in extra losses in leg motion compared to constant velocity rotation.




I don't have to measure it - the GCN video quite clearly shows the "expanding" stays pulling on the chain. Instead of letting the wheel go completely, they could have let the brake out slowly and demonstrated exactly how much distance the chain is pulled for the amount of crank deflection.

You must be looking at a different video, because that is not what is shown at all. The GCM video does not show "expanding" chainstays, nor do they even claim that they do. Instead, the video shows that the frame flexes torsionally; the drive force causes the drive side of the BB to drop as the frame twists. When the rear brake is released, the frame untwists, causing the drive side of the BB to rise (return to its original position). Because the pedal is fixed in position with the crank still primarily horizontal, the rising of the BB causes a rotation of the cranks, which is what cause the wheel to rotate.

Of course, this demonstration is completely rigged, and does not reflect reality. When riding, the pedal is not stopped when the crank is horizontal - instead, the full force remains on the pedal, and the frame remains flexed, until the pedal reaches the bottom of the stroke. So in a real pedaling system, the BB is not unloaded (and the "spring" energy is not released) until the cranks are vertical. When the crank is vertical, the rising of the BB does not result in crank rotation (in fact if the BB rises when the crank is past BDC, the rising BB acts to try to rotate the crank backward). Imagine if they tried the experiment, except instead stopping the pedal on the down stroke, they stopped it at the very bottom of the stroke. If they did this, then the rise in the BB when unloaded would not result in any crank rotation, and the wheel would not spin. This gets the crux of the problem I noted earlier - the frame deflection is orthogonal to the drive force, so you need a mechanism to convert/re-direct the spring energy.

[rain dogs]

1. Flexible bikes don't squeak any more than rigid bikes. If you account for the all the PF30 bikes, it would appear that not flexing is more of a problem. Having a BB shell flex with the cranking motion is going to put less side load on the bearings, not more.

2. Bigger tires are faster if we are talking 20s vs 23s or 25s. Above 25s the rolling resistance goes up when you are riding the correct pressure for your weight.


1. First, this isn't binary. There aren't "flexy" bikes and "rigid" bikes. All bikes flex. The same things you are arguing for your "flexy" bikes... if true, would occur on ALL frames, just less so on a more stiff frame. It's not like a flexy steel bike obeys certain laws of physics and a less flexy carbon or aluminum or steel bikes obeys different laws of physics.

So, if flexy bike A absorbs 20watts and returns 80% of that (16watts) you're losing 4 watts. If stiff (read: less flexy) bike absorbs 10 watts and returns 80% of that (8 watts) you're losing 2 watts. And if very stiff (read: much less flexy) bike absorbs 5 watts and returns 80% (4 watts) you're losing 1 watt.

Numbers are "imaginary" for a simple example.

2. The lowest rolling resistance has been tested in tires that are 1.5" wide (38mm) when all other things are equal (I'll have to find the link when I have more time). AND In regard to your incorrect claims about 28's:

"according to figures from another tyre brand, Continental, a 20mm tyre with 160psi, a 23mm tyre at 123psi, a 25mm tyre at 94psi and a 28mm tyre at 80psi all have the same rolling resistance."

"Similarly, the Zipp Tangente Speed clinchers were faster in the 28mm (31.3w) than the 25mm (33.2w) on the rough-surface drum." <The fastest tire in this test was 28.5mm the Schwalbe Pro One Tubeless.

And regarding pressure for weight. The wider tire the lower pressure you can ride for your weight.

[Kontact]

Because it is clear that you do not understand what you mean.
If you understood what was going on, then you'd realize that the reciprocating motion of the legs already incorporates some amount of energy storage (as the cranks rise) and release (as the crank drops) without needing some type eccentricity. In fact, the rotary velocity changes with Biopace cranks probably results in extra losses in leg motion compared to constant velocity rotation.

Mark, you didn't understand initially when I said "going into the dead zone", which means "has additional velocity before the dead zone" and you have been confused ever since.

The comparison I made to Biopace, which your first "correction" also reflects, is that Biopace (like a frame that flexes more) does not put as much peak power to the wheel compared to a round ring, and then causes that peak energy to be carried as additional crank velocity (again, compared to round) going into the dead zone where it is then converted to power at the wheel. All of this is only in comparison the a round ring, and none of it is a net increase or decrease in total power, but a change in the amplitude and distribution of power coming down the chain throughout the cycle. Lower peaks, but higher valleys.

I'm guessing that you got lost in "crank speeds up" vs. chain speed which is why you thought we were talking about two different things, or you got confused between "going into" and "in the dead zone".




You must be looking at a different video, because that is not what is shown at all. The GCM video does not show "expanding" chainstays, nor do they even claim that they do. Instead, the video shows that the frame flexes torsionally; the drive force causes the drive side of the BB to drop as the frame twists. When the rear brake is released, the frame untwists, causing the drive side of the BB to rise (return to its original position). Because the pedal is fixed in position with the crank still primarily horizontal, the rising of the BB causes a rotation of the cranks, which is what cause the wheel to rotate.

Of course, this demonstration is completely rigged, and does not reflect reality. When riding, the pedal is not stopped when the crank is horizontal - instead, the full force remains on the pedal, and the frame remains flexed, until the pedal reaches the bottom of the stroke. So in a real pedaling system, the BB is not unloaded (and the "spring" energy is not released) until the cranks are vertical. When the crank is vertical, the rising of the BB does not result in crank rotation (in fact if the BB rises when the crank is past BDC, the rising BB acts to try to rotate the crank backward). Imagine if they tried the experiment, except instead stopping the pedal on the down stroke, they stopped it at the very bottom of the stroke. If they did this, then the rise in the BB when unloaded would not result in any crank rotation, and the wheel would not spin. This gets the crux of the problem I noted earlier - the frame deflection is orthogonal to the drive force, so you need a mechanism to convert/re-direct the spring energy.

The GCN example loads the crank where the crank actually gets loaded during cycling - the power stroke. If we attempted to ride bicycle by only pushing down at 6 o'clock the cranks wouldn't turn - power has to be applied at something other than parallel to the orientation of the crank arm. But if you apply power at 90° to the crankarm anywhere in the pedal stroke you will get some sort of frame flex and motile power. But in the case of an upright bicycle that power is largely applied 90° from 3 o'clock of each crank, and applying power at that position will cause the frame to flex in a particular way, and that way is really the only one worth discussing. If we could pull up on the back of the stroke as hard as we can push down then the BB wouldn't flex and we wouldn't be talking about it.

But the BB does flex - bringing the chainrings down during right pedal power and up on left pedal power. And it is flexing that way not because we are standing on a pedal that is at 6 o'clock, but because we are pushing down at 3 o'clock which isn't an ideal position for pushing the BB down, but happens because the crank is being restrained from flexing in other directions by the tension of the chain.

That's the crux of all of this - the flex we see is the product of the opposition of pedal force and chain resistance, not just bending that is coincidental to the force applied to the chain.


Here's another experiment GCN could have done: Hold the crank at 3 o'clock with body weight, and push the bike backwards to simulate the resistance of applying power to the road. If the right pedal is blocked, the tension in the chain will cause the right side of the BB to drop. When the force holding the bike backwards is removed, the pedal won't move, the crank won't turn but the bike will move forward as the flex comes out of the BB. And when you do it with the left pedal the right side of the BB will rise, yet all the same things will happen.

[Kontact]

1. First, this isn't binary. There aren't "flexy" bikes and "rigid" bikes. All bikes flex. The same things you are arguing for your "flexy" bikes... if true, would occur on ALL frames, just less so on a more stiff frame. It's not like a flexy steel bike obeys certain laws of physics and a less flexy carbon or aluminum or steel bikes obeys different laws of physics.

So, if flexy bike A absorbs 20watts and returns 80% of that (16watts) you're losing 4 watts. If stiff (read: less flexy) bike absorbs 10 watts and returns 80% of that (8 watts) you're losing 2 watts. And if very stiff (read: much less flexy) bike absorbs 5 watts and returns 80% (4 watts) you're losing 1 watt.

Numbers are "imaginary" for a simple example.

2. The lowest rolling resistance has been tested in tires that are 1.5" wide (38mm) when all other things are equal (I'll have to find the link when I have more time). AND In regard to your incorrect claims about 28's:

"according to figures from another tyre brand, Continental, a 20mm tyre with 160psi, a 23mm tyre at 123psi, a 25mm tyre at 94psi and a 28mm tyre at 80psi all have the same rolling resistance."

"Similarly, the Zipp Tangente Speed clinchers were faster in the 28mm (31.3w) than the 25mm (33.2w) on the rough-surface drum." <The fastest tire in this test was 28.5mm the Schwalbe Pro One Tubeless.

And regarding pressure for weight. The wider tire the lower pressure you can ride for your weight.

I didn't say anything is binary flex/no flex, but if you are using two discreet bikes to test, that is binary. And I was speaking of the relative results of testing two obviously different bikes.



As far as rolling resistance goes, take the chart of rolling resistance of tires vs psi from the chart in this article:
https://www.bicyclerollingresistance.com/specials/conti-gp4000s-ii-23-25-28

And plug in the recommended PSI from this chart:
http://www.biketinker.com/wp-content/uploads/2010/08/BQ_berto_inflationgraph_2.jpg

So I picked 88 lbs as the input number, which gives a recommendation of 64psi for the 28c and 91psi for the 23c. When you plug those pressures into the rolling resistance chart, you get about 13.6 rolling resistance for the 23@91psi and about 13.8 for the 28@64psi.

If we take the bike weight = tire pressure thing seriously, those are the tire pressures to use to compare rolling resistance. You can certainly pump the 28 up to 70psi to lower its rolling resistance down to where the 23 is, but then you should be increasing the 23 to be fair.

[cachagua]

How do you know she's a witch?

[rain dogs]

So I picked 88 lbs as the input number, which gives a recommendation of 64psi for the 28c and 91psi for the 23c. When you plug those pressures into the rolling resistance chart, you get about 13.6 rolling resistance for the 23@91psi and about 13.8 for the 28@64psi.


I was being very conservative when I said 2-5mm wider and 5-10psi less pressure and you disagreed implying you'd need to be putting 110psi in a 28mm tire. And now you've posted a response where you wrote that 27psi less! and 5mm wider results in a difference of 13.6 and 13.8... so 0.2 Watts and nearly 30psi less!

Now, you imagine the difference of comfort of 25-27psi less. Heck even 20psi less and a 5mm wider tire. Not to mention the better traction and road feel.

So I guess your own post/data completely backs up my original point. Thanks.

So again, a less flexy bike, wider tires at the appropriate psi will give you less loss to frame flex (no matter how small), more comfort and traction (less of your claimed 'skidding under power' :confused:), and a lower rolling resistance. In other words greater efficiency, no?

[dddd]

The pressure drop appropriate to the use of a wider tire would seem to me to perhaps be larger than what is intuitively assumed.
Wider tires put proportionally more tension stress into their casing at equivalent pressure, which hardens the casing of the tire in terms of it's ability to yield to roughness of the road surface, so the pressure drop needs to firstly overcome this increased casing tension before we even start to see any increase in the tire's ability to conform to the road surface. So the wider tire widths really should be tested at quite-lower pressures before we start looking at rolling resistance data.


As for the issue of chainstay compression, of course this does occur, but this is part of the greater load path which encompasses all of the flexible elements that contribute to drivetrain elasticity outside of lateral flex at the bottom bracket.
This load path adds up to the sum total of:
--Chain elasticity
--Rear axle and freehub body flex and including bearings
--Chainstay elastic compression
--Bottom bracket spindle flex and including bearings
--Chainring flex in torsion along axis of bottom bracket
--Chainring+cog lateral flex in response to chain tension and cross-chain angle

Note that this entire net drivetrain load path flexing/elasticity constitutes windup flex peaking at the origin of force at the crankset, expressed as degrees rotation per unit torque delivered through the selected chainring.
Further, this entire load path loading and flex is dependent, quantitatively, on the inverse of the chainring size to the second power(!). So a 10% smaller chainring imposes an increase of a larger 23.4% increase in flex felt at the pedal eye in response to pedaling force. And this using the same overall gear ratio (as when for theoretical example where the rider changes from a 50/25t to a 34/17t combination).
And with a most-common 50-34t chainset, the flex felt at the pedal eye increases by a whopping 116% after changing from the big ring to the small ring (excluding rear hub/cassette/wheel windup, which is one place where torque isn't affected by changes in chainring size assuming same overall gear ratio).

So again, it is the entire drivetrain load path which is significant and not just chainstay compression by itself.

[earlfoss]

All that math won't save you when you're facing a selection in a race, going over the crest of the hill at max effort for 15-30 sec. I can almost guarantee you the dude on a flexy frame is going to be working much harder to keep up with those on stiffer framesets. However, after he gets dropped he'll have ample time to consider the superior comfort and dampening abilities of his rig. He'll feel fresher after the race is over, too.

The ti Hampsten I raced for a season was an awesome bike. I loved it, but when the hammer was down, the little bit of bb flex was a liability. I didn't have ghost shifting or anything that extreme. Going from that to a Giant Propel, the difference was night and day. I'd say that for modern racing, to be competitive you need to have a bike that is at least as stiff as anything the average guy is riding in your event if that makes any sense.

[dddd]

All that math won't save you when you're facing a selection in a race, going over the crest of the hill at max effort for 15-30 sec. I can almost guarantee you the dude on a flexy frame is going to be working much harder to keep up with those on stiffer framesets. However, after he gets dropped he'll have ample time to consider the superior comfort and dampening abilities of his rig. He'll feel fresher after the race is over, too.

The ti Hampsten I raced for a season was an awesome bike. I loved it, but when the hammer was down, the little bit of bb flex was a liability. I didn't have ghost shifting or anything that extreme. Going from that to a Giant Propel, the difference was night and day. I'd say that for modern racing, to be competitive you need to have a bike that is at least as stiff as anything the average guy is riding in your event if that makes any sense.


I would say that it depends on whether the "you" needing a bike "that is at least as stiff as anything the average guy is riding" is close to the weight/strength of the other riders, and whether sprinting is "your"/their thing.

I agree though that at times, say during a short/intense uphill sprint, that I (even at my 140lbs) have felt that I would be accelerating faster with a stiffer entire bike. But whether I really would be going significantly faster (especially after having had to ride and pedal the earlier part of the ride or race on that same stiffer bike) at the end of that sprint remains a point of curiosity, since I don't really know.
I do feel that for shorter efforts, when the legs are fresh, that yes, I probably could accelerate faster on a stiffer bike, but wouldn't the higher peak forces be more likely to bring on the onset of cramping during such a sprint toward the end of a ride or race? Again, I can't say for sure, though having enough solid resistance to push against would seem to improve the muscles' generating more overall power with that resistance force not being delayed and softened within the pedal stroke.
Perhaps a frame (or just drivetrain) can be developed with a more-constant deflection curve at the bottom bracket in response to pedaling effort, either through electro-mechanical gadgetry or some sort of material property akin to thixotrophy or viscoelasticity for rough analogy here(?).
Does that take us back to BioPace? Ohhhhh-Nooooo.:eek:
(Though I confess to doing almost all of my Road, XC and CX racing over the last 20+ years using BioPace II and similar BioPace SG chainrings!!!)

[andrewsuzuki]

All that math won't save you when you're facing a selection in a race, going over the crest of the hill at max effort for 15-30 sec. I can almost guarantee you the dude on a flexy frame is going to be working much harder to keep up with those on stiffer framesets. However, after he gets dropped he'll have ample time to consider the superior comfort and dampening abilities of his rig. He'll feel fresher after the race is over, too.

The ti Hampsten I raced for a season was an awesome bike. I loved it, but when the hammer was down, the little bit of bb flex was a liability. I didn't have ghost shifting or anything that extreme. Going from that to a Giant Propel, the difference was night and day. I'd say that for modern racing, to be competitive you need to have a bike that is at least as stiff as anything the average guy is riding in your event if that makes any sense.

Substitute "flexy frame" with "25mm tires" and "stiffer frame" with "19mm tires". You can easily imagine saying something like this ten years ago, just like 19mm tires "feel" faster

[earlfoss]

Say what you want, but at least at the P12 level you're going to want a stiffer frame these days. At the end of the day, race what you brung though.

[Kontact]

I was being very conservative when I said 2-5mm wider and 5-10psi less pressure and you disagreed implying you'd need to be putting 110psi in a 28mm tire. And now you've posted a response where you wrote that 27psi less! and 5mm wider results in a difference of 13.6 and 13.8... so 0.2 Watts and nearly 30psi less!

Now, you imagine the difference of comfort of 25-27psi less. Heck even 20psi less and a 5mm wider tire. Not to mention the better traction and road feel.

So I guess your own post/data completely backs up my original point. Thanks.

So again, a less flexy bike, wider tires at the appropriate psi will give you less loss to frame flex (no matter how small), more comfort and traction (less of your claimed 'skidding under power' :confused:), and a lower rolling resistance. In other words greater efficiency, no?
What an odd response. Pick any point you want on the graph using any starting weight you want. The Rollingresistance guys started at 120psi and worked their way down from there. The point is that there needs to be comparable psi for both tires, and the point of comparison is load placed on them by the bicycle.

Your response isn't math. Use the math if you want to make a mathematical statement about rolling resistance.

[Ungaro]

How do you know she's a witch?


I know! She if she floats!!!!:banana:

[Mzilliox]

How do you know she's a witch?

throw her into a pond and see if she floats?

[Bonesbrigade]

All that math won't save you when you're facing a selection in a race, going over the crest of the hill at max effort for 15-30 sec. I can almost guarantee you the dude on a flexy frame is going to be working much harder to keep up with those on stiffer framesets. However, after he gets dropped he'll have ample time to consider the superior comfort and dampening abilities of his rig. He'll feel fresher after the race is over, too.

The ti Hampsten I raced for a season was an awesome bike. I loved it, but when the hammer was down, the little bit of bb flex was a liability. I didn't have ghost shifting or anything that extreme. Going from that to a Giant Propel, the difference was night and day. I'd say that for modern racing, to be competitive you need to have a bike that is at least as stiff as anything the average guy is riding in your event if that makes any sense.

Damon Rinard would disagree with you. For anyone interested - here is an excellent podcast on the subject:

https://cyclingtips.com/2017/06/cyclingtips-podcast-does-frame-stiffness-matter/

[skouri1]

its psychological.
which does actually count. :)
contador refused aero stuff for a long time. Preferred lighter, shallower wheels. Emonda over aero bike etc.

Chances are though--if you felt bad on the hampsten, you would also have felt bad on whatever uber bike you were on. toasted is toasted, and as long as you have the same position and at least 9 gears, you should be competitive.

[earlfoss]

Well, duh, finding a bike with the right stiffness for you is important. I'm not saying that everyone should race on something that's as stiff as a Propel. It's not a binary thing. There's a reason why the aero road setups aren't commonplace in the pro peloton. I bet they beat you to hell. There's a medium to strike for everyone, true, but in competition I think that medium will naturally fall on the stiffer side of things. You're not going to want to race on something that visibly wags the bottom bracket around when the watts are high. Maybe you do want to though and that's ok because who cares anyway? If you enjoy riding it, whatever it is, then do it.

[William]

Well, duh, finding a bike with the right stiffness for you is important. I'm not saying that everyone should race on something that's as stiff as a Propel. It's not a binary thing. There's a reason why the aero road setups aren't commonplace in the pro peloton. I bet they beat you to hell. There's a medium to strike for everyone, true, but in competition I think that medium will naturally fall on the stiffer side of things. You're not going to want to race on something that visibly wags the bottom bracket around when the watts are high. Maybe you do want to though and that's ok because who cares anyway? If you enjoy riding it, whatever it is, then do it.



Just got to find that sweet spot that feels good for your weight and the right mix of flex and stiffness. :banana:




William

[William]

I went back through this thread and took all the physics principles, scientific explanations, and graphs and compressed, collated, and calculated them together and came up with this graph that explains it all...






:banana::banana::banana:
William

[fa63]

That is awesome :D I could argue that flex is the lack of stiffness (so one is the inverse of the other), and that rider weight and riding style should be separated, but your diagram captures the essence beautifully :)

[Fivethumbs]

I watched the video and this is what I saw:


When the rider stepped on the pedal with the brake applied, the first thing that happened is that the tension was applied to the chain trying to rotate the wheel in the forward direction while the brake prevented the wheel from turning.

Next the chain stretched until it could stretch no further at which time the frame flexed to the side until the pedal contacted the block.

When the brake was released the tension that was applied to the chain now caused the wheel to turn. Then the frame move back to its normal position.


Wouldn't the same thing have happened if the frame did not flex at all and the chain stretched enough to allow the pedal to touch the block? In that case frame flex would be ruled out yet the wheel would still turn.

[Kontact]

I watched the video and this is what I saw:


When the rider stepped on the pedal with the brake applied, the first thing that happened is that the tension was applied to the chain trying to rotate the wheel in the forward direction while the brake prevented the wheel from turning.

Next the chain stretched until it could stretch no further at which time the frame flexed to the side until the pedal contacted the block.

When the brake was released the tension that was applied to the chain now caused the wheel to turn. Then the frame move back to its normal position.


Wouldn't the same thing have happened if the frame did not flex at all and the chain stretched enough to allow the pedal to touch the block? In that case frame flex would be ruled out yet the wheel would still turn.

Chains don't stretch by very much. Some frames don't flex very much, either. But if you are talking about a frame of somewhat normal flex you get this sort of dramatic result where a notable amount of the pedaling force temporarily gets used to change the shape of the frame rather than going all into the chain.

[Fivethumbs]

My question is: What caused the wheel to turn, the pre-loaded tension on the chain or the frame moving back to normal?

[Kontact]

My question is: What caused the wheel to turn, the pre-loaded tension on the chain or the frame moving back to normal?

Compared to a bike where the BB didn't move, I would say the frame. But someone would object, so consider this:

You take a super stiff bike and the GCN bike and you lock your leg to hold the pedal at 3 o'clock. Your helper grabs the handlebars and pushes you straight back simulating a load on the drivetrain, like during climbing.

On the stiff bike, he won't be able to push the bike back as far before the chain, spokes, frame and everything else wind up enough to stop him.

On the flexy bike, he will be able to push you back further because the BB shell will twist and allow even more chain to be pulled over the cassette. He is pushing with the same force, but getting more flex distance out of the drivetrain as the stiff bike.


When he releases the stiff bike, it will scoot forward a little bit to its starting point.

When he releases the flexy bike it will jump forward more, but also end up where it started.

That extra jump is the flex of the BB that was trapped between your leg and the load on the rear wheel.

[Fivethumbs]

I guess the point I was trying to make was that the pedal started at point A and moved forward to point B. The wheel turned. That would happen whether the frame flexed or not.

To me their experiment showed that if the pedal is moving forward the rear wheel will turn and while frame flex may delay the movement momentarily, it won't stop it. That is so long as the frame returns to its original point.

If people are propositioning that frame flex by itself can propel a bike forward I don't see that. If people are propositioning that frame flex will not rob you of forward motion so long as the pedals are turning, then I can see that.

[Kontact]

I guess the point I was trying to make was that the pedal started at point A and moved forward to point B. The wheel turned. That would happen whether the frame flexed or not.

To me their experiment showed that if the pedal is moving forward the rear wheel will turn and while frame flex may delay the movement momentarily, it won't stop it. That is so long as the frame returns to its original point.

If people are propositioning that frame flex by itself can propel a bike forward I don't see that. If people are propositioning that frame flex will not rob you of forward motion so long as the pedals are turning, then I can see that.

If the frame was stiff enough, the pedal wouldn't have moved to point B.

[Fivethumbs]

If the frame was stiff enough, the pedal wouldn't have moved to point B.

But it did. And the wheel turned.

[Kontact]

But it did. And the wheel turned.

Did you watch a different GCN video with a super stiff frame?

[cachagua]

You lock your leg to hold the pedal at 3 o'clock...

Even if you could hold your leg that still, it wouldn't prove what you're trying to make it prove. The problem here is the same flaw that invalidates both the video and the FEA: they depend on the resistance at the rear wheel decreasing. That doesn't happen when you're riding.

If people are [suggesting] that frame flex will not rob you of forward motion so long as the pedals are turning, then I can see that...

Yes, people are suggesting that, but it's a misconception.

The flex in your frame does not ever make you go forward, period. That's because for this to happen, a lesser force would have to overcome a greater one. It is that simple.

As you're riding along, your legs vary how much force they're exerting between a maximum equal to (overcoming your inertia)+(frame flex), and some amount less than that. Now here comes the crucial point: your inertia never lessens.

The only time when the frame can un-flex is when your legs ease off, and the suggestion that when that happens, the flex energy is going to overcome your inertia -- the greater force, now that your legs are applying a lesser force -- is incorrect.

[Kontact]

Even if you could hold your leg that still, it wouldn't prove what you're trying to make it prove. The problem here is the same flaw that invalidates both the video and the FEA: they depend on the resistance at the rear wheel decreasing. That doesn't happen when you're riding.



Yes, people are suggesting that, but it's a misconception.

The flex in your frame does not ever make you go forward, period. That's because for this to happen, a lesser force would have to overcome a greater one. It is that simple.

As you're riding along, your legs vary how much force they're exerting between a maximum equal to (overcoming your inertia)+(frame flex), and some amount less than that. Now here comes the crucial point: your inertia never lessens.

The only time when the frame can un-flex is when your legs ease off, and the suggestion that when that happens, the flex energy is going to overcome your inertia -- the greater force, now that your legs are applying a lesser force -- is incorrect.

That is even vaguely true. The frame un-flexes when the load shifts from right to left. And that happens without the pedal forces going to zero.

What's getting left out of your analysis is that there are actually two ways for the frame to store some pedaling energy - by lowering the chainring (right pedal load), or raising the chainring (left pedal load). Moving between them is when the stored energy is pushed down the stays.


All the various static examples are demonstrating is that the flex is still in line with the drivetrain, and isn't just flying out into space. Then you take that understanding and apply it to the fact that the tension on the chain never goes to zero and the BB flex automatically alternates from right-to-neutral-to-left.

[johnniecakes]

But what about the crank arm and pedal spindle? Wouldn't they have also flexed or deflected and contributed in some way?

[kramnnim]

But it did. And the wheel turned.

To me, it looks like the trainer is flexing more than either frame.

[cachagua]

That is even vaguely true.

I apologize; I intended to be anything but vague.

The frame un-flexes when the load shifts from right to left. And that happens without the pedal forces going to zero.

Who said anything about the pedal force going to zero? All I said was it has to drop from its maximum.

What's getting left out of your analysis is that there are actually two ways for the frame to store some pedaling energy - by lowering the chainring (right pedal load), or raising the chainring (left pedal load). Moving between them is when the stored energy is pushed down the stays.

In other words, what we see when the BB "sways" from one side to the other and back? Again: the inertia at the rear wheel is pushing harder than your pedals, because your pedaling force has decreased in relation. In response to the decreased force at the pedals, the frame discharges some of its strain energy as we see, but there ain't no way that can get out the back wheel and move you along the road. The softer push is not going to overcome the harder one. Do you dispute that?

I don't know how I can make this simpler. If you and I push our hands together and push equally hard, our hands will stay still. If you relax a little and push softer, and I keep on pushing just as hard, will our hands move toward me as a result?

All the various static examples are demonstrating is that the flex is still in line with the drivetrain, and isn't just flying out into space. Then you take that understanding and apply it to the fact that the tension on the chain never goes to zero and the BB flex automatically alternates from right-to-neutral-to-left.

Flex is in line with the drivetrain -- not sure what you mean. Does it matter?
Tension on the chain never goes to zero -- doesn't need to; not part of my account.
BB flex alternates from right to neutral to left -- okay, if it goes back to neutral in the middle then manifestly there's nothing carried over to the next pedal stroke.

[Kontact]

I don't know how I can make this simpler. If you and I push our hands together and push equally hard, our hands will stay still. If you relax a little and push softer, and I keep on pushing just as hard, will our hands move toward me as a result?

No, but that isn't what the drivetrain is doing. You are describing something that would happen in a straight line. The drivetrain is twisting, and the untwisting is what is adding to the forward motion. If downforce on the pedal was constant, that flex would never go away and the stored flex energy would never get used. But on bicycles the flex oscillates from left to right, putting force back into the drive line every time it goes through the center.



Flex is in line with the drivetrain -- not sure what you mean. Does it matter?
Tension on the chain never goes to zero -- doesn't need to; not part of my account.
BB flex alternates from right to neutral to left -- okay, if it goes back to neutral in the middle then manifestly there's nothing carried over to the next pedal stroke.

Nothing is carried to the next pedal stroke - that's the point. The right side flex is used up by the BB returning to center, and then new left side flex is produced, then it is used up as the BB returns to center for the right pedal stroke.

It isn't like a spring that is simply being allowed to uncompress. Pedaling actively untwists the drivetrain, putting the work used to flex the frame to use. And that would be true even if the frame had a pivoting BB that required no real force to "flex". If the distance from chainring to cassette varied throughout the pedal stroke because the crank pivoted laterally on a bearing, then you'd still get the effect of force going through the chain from pedaling through neutral.

[ergott]

So hypothetically the twisting of the chainstays could be isolated and provide forward motion by themselves? A jig could be made to illustrate that movement and demonstrate how it could propel a bike or at the very least rotate the rear wheel.

[Kontact]

So hypothetically the twisting of the chainstays could be isolated and provide forward motion by themselves? A jig could be made to illustrate that movement and demonstrate how it could propel a bike or at the very least rotate the rear wheel.

Yes. I'd have to think about it awhile to figure out how to do it, but the twisting of the stays and BB act very much like a step cycle in miniature:

https://i.pinimg.com/736x/e6/a2/04/e6a2045ad3aaa6be08ee8769a6f38deb--body-fitness-stand-up.jpg

[cachagua]

The twisting of the stays and BB act very much like a step cycle in miniature...

We've covered that already:

Let's look at the bike on the trainer again. You keep the brake squeezed, but ease up on the cranks, the right pedal rises up from the block -- but at the same time, the left pedal also makes a counter-rotating motion. The release of the frame flex does not advance the other pedal but moves it backwards, and thus doesn't increase the force of the next stroke, but decreases it, too.

I think I see what you're driving at -- that when the frame releases the strain energy that one pedal stroke put into it, that could move the other pedal a little farther around, or make it easier for it to rotate. It's a beautiful idea! It would be great if it were true! But as we see, what retards one pedal retards the other equally.

[Kontact]

We've covered that already:



I think I see what you're driving at -- that when the frame releases the strain energy that one pedal stroke put into it, that could move the other pedal a little farther around, or make it easier for it to rotate. It's a beautiful idea! It would be great if it were true! But as we see, what retards one pedal retards the other equally.

It has nothing to do with the pedals or advancing them. It is that the stays get shorter from twisting when the BB sways, and when the sway goes through the center point the distance between chainring and cassette gets longer. When that happens either the crank is going to stop your legs or the chain will pull the wheel. Since we know your legs don't hiccup at the bottom of the pedal stroke, the elongation goes toward moving the bike.

Again, this isn't because chain tension has relaxed, it is because the lateral displacement of the BB has relaxed into a position that requires more chain. And even if the frame didn't want to relax to that position, the other half of your pedal stroke doesn't give it any choice.

Right pedal, neutral, left pedal, neutral.
Short stays, long stays, short stays, long stays.
Less chain, more chain, less chain, more chain.

[ergott]

That movement wouldn't be enough to advance the bike even one click of the freehub even on the finest of engagement mechanisms. The bike would just be sitting there flexing in both directions. If the bike were a fixed gear, it would still just rock back forth. It's not just energy output, it's energy over time and there's no way you could propel a bike with that tiny of a movement by itself. The movement of the BB is over the time of a riders cadence so it's too gradual.

A spring can launch something in the air, but if you slow down the release of that energy (energy over time) you could have a car spring that couldn't even launch a penny.

Seems more like trying to row a boat leaving the oars in the water the whole time.

[Black Dog]

That movement wouldn't be enough to advance the bike even one click of the freehub even on the finest of engagement mechanisms. The bike would just be sitting there flexing in both directions. If the bike were a fixed gear, it would still just rock back forth. It's not just energy output, it's energy over time and there's no way you could propel a bike with that tiny of a movement by itself. The movement of the BB is over the time of a riders cadence so it's too gradual.

A spring can launch something in the air, but if you slow down the release of that energy (energy over time) you could have a car spring that couldn't even launch a penny.

Seems more like trying to row a boat leaving the oars in the water the whole time.


This is the million dollar question. Where exactly is the spring's energy going? Like your penny example, slowing down the spring will transfer the energy to the device that is slowing down the spring and not the penny. With hub crank and pedal based power meters it would not be too hard to do some empirical testing with frames of various stiffness.

[ergott]

With hub crank and pedal based power meters it would not be too hard to do some empirical testing with frames of various stiffness.

I'm willing to bet that anything going in here is way under the accuracy tolerances of the best power meters available (±1%).

[fa63]

This is the million dollar question. Where exactly is the spring's energy going? Like your penny example, slowing down the spring will transfer the energy to the device that is slowing down the spring and not the penny. With hub crank and pedal based power meters it would not be too hard to do some empirical testing with frames of various stiffness.

If you listen to the Cycling Tips podcast which was linked earlier in the thread, Damon Rinard (Road Engineering Manager at Cannondale) talks about how he was unable to detect any loss when looking at simultaneous power measured at both the crank and the hub.

Using a simpler approach (strain energy), he also mentions in the podcast that he calculated about 40W for power which goes into flexing a frame under an all-out sprint (I think it was for a 1500-2000W effort). This was meant as an upper-bound value, but even if it is true, it is about 2% of the power input, which is more or less the accuracy of most strain-gage based power meters on the market today.

[Kontact]

That movement wouldn't be enough to advance the bike even one click of the freehub even on the finest of engagement mechanisms. The bike would just be sitting there flexing in both directions. If the bike were a fixed gear, it would still just rock back forth. It's not just energy output, it's energy over time and there's no way you could propel a bike with that tiny of a movement by itself. The movement of the BB is over the time of a riders cadence so it's too gradual.

A spring can launch something in the air, but if you slow down the release of that energy (energy over time) you could have a car spring that couldn't even launch a penny.

Seems more like trying to row a boat leaving the oars in the water the whole time.

If the movement is so little to power the bike, then it is also so little to be a loss.

That's the only real point here. Either the energy stays in the drivetrain either way, or there is no energy difference.

[johnniecakes]

Me wonders how much unrecoverable energy has been spent on this subject.

[Kontact]

Me wonders how much unrecoverable energy has been spent on this subject.

The forum should probably be closed so no energy is wasted on bike discussion any more.

[cachagua]

When the sway goes through the center point... either the crank is going to stop your legs or the chain will pull the wheel.

The former is close to what happens. The crank doesn't stop your legs entirely, but retards their motion. As I've mentioned, I'm not talking about the cranks stopping, or your legs' force dropping to zero -- instead of a hiccup, your legs are slowed down at the same gradual rate at which you reduce your pedaling force.

Right pedal, neutral, left pedal, neutral.
Short stays, long stays, short stays, long stays.
Less chain, more chain, less chain, more chain.

Good. Second and fourth in each of those sequences, the strain energy is emptied out of the frame -- however fast or slow it happens, wherever in your pedal stroke, whichever zodiac sign you happen to be in -- and nothing remains which would help the next stroke. You say, "the energy stays in the drivetrain" -- and that's exactly right: it cannot ever get out of the rear wheel to become forward motion.


How much unrecoverable energy has been spent on this subject...

Are you kidding? There is no more thrilling item in the forum just now! The energy may be unrecoverable, but when was the last time a couple of blind men stood around arguing whether an elephant is like a rope, or like a tree, for the entertainment of the onlookers? Uh, since the election, that is.

Nine pages, look at us go!

[Kontact]

Good. Second and fourth in each of those sequences, the strain energy is emptied out of the frame -- however fast or slow it happens, wherever in your pedal stroke, whichever zodiac sign you happen to be in -- and nothing remains which would help the next stroke. You say, "the energy stays in the drivetrain" -- and that's exactly right: it cannot ever get out of the rear wheel to become forward motion.

It has nothing to do with "helping the next stroke". I'm saying that the energy not used in the power stroke that is stored as BB sway is used later in that stroke, not the next one. And that's why I referenced Biopace eariler, which is another way that a drivetrain can delay the transmission of force to the rear wheel while still delivering all of it.

[cachagua]

The energy not used in the power stroke that is stored as BB sway is used later in that stroke, not the next one....

Nope, we've covered that too. That's the lesser-force-can't-overcome-the-greater-one thing. You ease off on the pedals, but the rear wheel doesn't ease off on the inertia.

Try this: if I offer you $5000 for your car, and someone else offers you $10,000, who's gonna end up with your car?

[Kontact]

Nope, we've covered that too. That's the lesser-force-can't-overcome-the-greater-one thing. You ease off on the pedals, but the rear wheel doesn't ease off on the inertia.

Try this: if I offer you $5000 for your car, and someone else offers you $10,000, who's gonna end up with your car?

We've covered this - the mass of the bike is not greater than your ability to pedal. That's why we don't come to a stop on steep climbs just because you pedal through the dead zone.

[cachagua]

I think you know that's not what I'm talking about.

[Kontact]

I think you know that's not what I'm talking about.

Despite you keep saying "we've covered this", I don't know what you're talking about. When you pedal the drivetrain, even a flexy drivetrain, does not feed back upstream. Even in the dead spots, which aren't really dead.

[cachagua]

I don't know what you're talking about.

I think you're right.

And maybe this is the most telling difference between our positions. I know what you're talking about, I just think it's incorrect.

However if I can't get any picture of what I'm thinking about across to you, then our fellow Paceliners will have to look elsewhere for entertainment, won't they? But we've had a good run. I thank you. And I wish you safe passage over those damned trolley tracks.

[kramnnim]

To me, it seems obvious that the energy stored in the flexy frame would simply result in the frame bending back and forth as the force to the pedal relaxes and switches over to the opposite pedal.

[Mark McM]

To me, it seems obvious that the energy stored in the flexy frame would simply result in the frame bending back and forth as the force to the pedal relaxes and switches over to the opposite pedal.

As pointed out earlier, the energy that goes into frame flex is just a small percentage of the total energy transmitted through the drive train. The frame is highly elastic and has virtually no damping, so the energy is returned in some form when the frame unflexes - but as cachagua has pointed out, there is no mechanism for this energy to be transferred into the drivetrain to directly drive the bike forward. Instead, I think the most likely action is that the small amount of energy stored in the frame during the downward push of the forward leg is returned by giving a little help raising the rear leg on the back of the pedal stroke.

It is also interesting to take note of how the rider applies forces to the pedals. Instrumented pedals have shown that rider typically maintain large downward forces on the pedals all the way to the bottom of the pedal circle, and don't unweight the pedals until the pedal is rising. Here's a typical example of the pedal force vectors:

http://forums.roadbikereview.com/attachments/racing-training-nutrition-triathlons/310628d1447372208-pioneer-power-meter-system-review-zone2.jpg

Energy stored in the frame from the downward pedal force can not be returned until the force decreases. Since this happens at the back of the back of the pedal circle (as the pedal is rising), it can't be transferred into the drivetrain to drive the bike forward. Instead, the action of the returned energy is simply to raise the rear leg. Of course, this is useful in its own right, as it decreases the energy the rider needs to expend to raise their leg.

So in the end, energy expended by the rider to flex the frame on the down stroke may be returned to the rider on the up stroke, so the net energy expended by the rider may be same with or without frame flex. But it is clear that energy stored in the frame does not directly drive the bike forward.

[Kontact]

I think you're right.

And maybe this is the most telling difference between our positions. I know what you're talking about, I just think it's incorrect.

However if I can't get any picture of what I'm thinking about across to you, then our fellow Paceliners will have to look elsewhere for entertainment, won't they? But we've had a good run. I thank you. And I wish you safe passage over those damned trolley tracks.

Don't be insulting. I follow your argument, I just don't see where you get this idea that we "ease off the pedals" when that isn't how pedaling works.

[Kontact]

As pointed out earlier, the energy that goes into frame flex is just a small percentage of the total energy transmitted through the drive train. The frame is highly elastic and has virtually no damping, so the energy is returned in some form when the frame unflexes - but as cachagua has pointed out, there is no mechanism for this energy to be transferred into the drivetrain to directly drive the bike forward. Instead, I think the most likely action is that the small amount of energy stored in the frame during the downward push of the forward leg is returned by giving a little help raising the rear leg on the back of the pedal stroke.

It is also interesting to take note of how the rider applies forces to the pedals. Instrumented pedals have shown that rider typically maintain large downward forces on the pedals all the way to the bottom of the pedal circle, and don't unweight the pedals until the pedal is rising. Here's a typical example of the pedal force vectors:

http://forums.roadbikereview.com/attachments/racing-training-nutrition-triathlons/310628d1447372208-pioneer-power-meter-system-review-zone2.jpg

Energy stored in the frame from the downward pedal force can not be returned until the force decreases. Since this happens at the back of the back of the pedal circle (as the pedal is rising), it can't be transferred into the drivetrain to drive the bike forward. Instead, the action of the returned energy is simply to raise the rear leg. Of course, this is useful in its own right, as it decreases the energy the rider needs to expend to raise their leg.

So in the end, energy expended by the rider to flex the frame on the down stroke may be returned to the rider on the up stroke, so the net energy expended by the rider may be same with or without frame flex. But it is clear that energy stored in the frame does not directly drive the bike forward.

What made you decide to ignore the mechanism I outlined for returning flex energy to the drivetrain via rear center elongation?

[dave thompson]

Tranquilós amigos, tranquiló.

[kramnnim]

Don't be insulting. I follow your argument, I just don't see where you get this idea that we "ease off the pedals" when that isn't how pedaling works.

But it is how it works, as seen in the screenshot from the pedaling efficiency graph from a Pioneer power meter.

[Kontact]

But it is how it works, as seen in the screenshot from the pedaling efficiency graph from a Pioneer power meter.

The graph is for one pedal, and shows that the one pedal is actually pushing against the crank rotation at times. If the other pedal truly went to zero at any point where the first pedal was anti-rotation, the pedals would stop momentarily, the chain would get slack and there would be a clank as the pedals started to re-engage the freewheel pawls.

The reality is that, despite the highs and lows, we provide varying but continuous power to the rear wheel. The tension in the chain, on the spider and through your legs never goes to zero.

If you were to take that pedaling vector diagram and sum it with the other pedal you would find no spots where there isn't net pro-rotation force.

[Mark McM]

What made you decide to ignore the mechanism I outlined for returning flex energy to the drivetrain via rear center elongation?

Didn't we already go over this? There is no meaningful change in the rear center distance (the GCN video didn't show any, and they didn't claim any either). If there is no significant deflection, there is no significant energy storage. You're grasping for straws.

[kramnnim]

No. The graph shows both pedals. Marked L and R.

When the crank arms are vertical, there is, at most, only slight rotational pressure being applied to the pedals. Not enough pressure to flex a frame.

Also, very little of the flex is stored in the drivetrain, ready to be put in to rotational force. This whole discussion is based around the frame flexing, the BB area moving left/right in relation to the rest of the frame. That force truly does go to zero as your legs alternate pressure.

The graph is for one pedal, and shows that the one pedal is actually pushing against the crank rotation at times. If the other pedal truly went to zero at any point where the first pedal was anti-rotation, the pedals would stop momentarily, the chain would get slack and there would be a clank as the pedals started to re-engage the freewheel pawls.

The reality is that, despite the highs and lows, we provide varying but continuous power to the rear wheel. The tension in the chain, on the spider and through your legs never goes to zero.

If you were to take that pedaling vector diagram and sum it with the other pedal you would find no spots where there isn't net pro-rotation force.

[Kontact]

No. The graph shows both pedals. Marked L and R.

When the crank arms are vertical, there is, at most, only slight rotational pressure being applied to the pedals. Not enough pressure to flex a frame.

Also, very little of the flex is stored in the drivetrain, ready to be put in to rotational force. This whole discussion is based around the frame flexing, the BB area moving left/right in relation to the rest of the frame. That force truly does go to zero as your legs alternate pressure.

The two diagrams are not in phase. If they were in phase and you added them, you would see that there is always force applied to the chainring.

And you are right that there isn't force enough to flex the pedals when the cranks are vertical, but that flex is the product of previous pedal input and the resistance of the rear wheel through the chain. That tug of war doesn't stop when the pedals are vertical, it just decreases to the point that the BB straightens.

The question is: If force through the cranks and against the resistance of the chain deflects the BB to the side, where does that deflection force go when the deflection is taken out? The frame doesn't SPROING! back to center, so it isn't just released. It comes out about at the same rate it went in, and if constant pressure is applied to the pedals, where can it go?


Some folks seem to think that the BB flex is separate from the force applied to the chain, but it comes from the tension between the chain, pedal and spindle. Otherwise it wouldn't occur at 3 o'clock.

[cachagua]

I follow your argument, I just don't see where you get this idea that we "ease off the pedals" when that isn't how pedaling works.

We do not apply constant, unvarying pressure on the pedals. We ease off, what else would you like to call it?

Not stop, not go backwards, not apply zero pressure, but apply a pressure less than the maximum we apply in some other o'clock of the pedal rotation.

http://forums.roadbikereview.com/attachments/racing-training-nutrition-triathlons/310628d1447372208-pioneer-power-meter-system-review-zone2.jpg

In the lower right, compare the four o'clock position and the five o'clock position. Four is maybe about the maximum, in that pedal stroke, and five is --
less. That's all I'm talking about when I say "easing off on the pedals", and according to those strain gauges, at least, it is indeed how pedaling works.

Mark has a suggestion that, as the frame un-flexes, it can help the opposite leg through the back part of its stroke, but given that the un-flexing retards the motion of the leg we're looking at, and the BB spindle connects the two cranks, I think we have to rule that out too.

[William]

So the forces in a sprint and a standing climb never have the bike completely vertical through the pedal stroke. If you look at the attached video of Cavendish sprinting (and not even going full out) the frame is always at an angle as the forces are applied, this seems to me that it would promote more possible flex to the BB area? More so than a bike sitting upright in a trainer as referenced in the original video.

https://www.youtube.com/watch?v=Kb67p8Cb7v0





William

PS: Apologies If I missed it in the running commentary.

[Kontact]

We do not apply constant, unvarying pressure on the pedals. We ease off, what else would you like to call it?

Not stop, not go backwards, not apply zero pressure, but apply a pressure less than the maximum we apply in some other o'clock of the pedal rotation.

http://forums.roadbikereview.com/attachments/racing-training-nutrition-triathlons/310628d1447372208-pioneer-power-meter-system-review-zone2.jpg

In the lower right, compare the four o'clock position and the five o'clock position. Four is maybe about the maximum, in that pedal stroke, and five is --
less. That's all I'm talking about when I say "easing off on the pedals", and according to those strain gauges, at least, it is indeed how pedaling works.

Mark has a suggestion that, as the frame un-flexes, it can help the opposite leg through the back part of its stroke, but given that the un-flexing retards the motion of the leg we're looking at, and the BB spindle connects the two cranks, I think we have to rule that out too.

The point I was making, which you countered with the "ease off" thing, was that the power coming off the chainring is never zero and the motion is never zero. So there is no reason to think that the frame flex is just disappearing up our legs and we don't notice. We can feel it when we put that flex in, and we would feel it feeding back as it came out if it pushed against our legs.

[Mark McM]

Mark has a suggestion that, as the frame un-flexes, it can help the opposite leg through the back part of its stroke, but given that the un-flexing retards the motion of the leg we're looking at, and the BB spindle connects the two cranks, I think we have to rule that out too.

I'm not sure how you conclude that. Take, for example, the cases where the cranks are vertical (one pedal is at the top of the stroke, and one pedal is at the bottom). In this position, the differential between the right/left down forces on the pedals is maximum, which results in the maximum (torsional) frame flex. Due to the torsional rotation at the BB, the bottom pedal is at the maximum downward deflection, and the top pedal is at the maximum upward deflection. As the pedals continue to rotate through the next 180 degrees, the right/left force differential reverses, and the deflections also reverse. In other words, as the pedal rises at the back of the pedal circle, the reversal of the deflection acts to raise the pedal. Until the frame returns to its neutral, unflexed position, the stored energy in the frame is indeed helping to lift the rear leg.

Curiously, the torsional flex at the BB has an interesting affect on the shape of the pedal 'circle'. Instead of moving in a perfect circle, the upward and downward deflections cause the shape to be stretched into an ellipse. I wonder what affect this has on muscle utilization?

[Kontact]

I'm not sure how you conclude that. Take, for example, the cases where the cranks are vertical (one pedal is at the top of the stroke, and one pedal is at the bottom). In this position, the differential between the right/left down forces on the pedals is maximum, which results in the maximum (torsional) frame flex. Due to the torsional rotation at the BB, the bottom pedal is at the maximum downward deflection, and the top pedal is at the maximum upward deflection. As the pedals continue to rotate through the next 180 degrees, the right/left force differential reverses, and the deflections also reverse. In other words, as the pedal rises at the back of the pedal circle, the reversal of the deflection acts to raise the pedal. Until the frame returns to its neutral, unflexed position, the stored energy in the frame is indeed helping to lift the rear leg.

Curiously, the torsional flex at the BB has an interesting affect on the shape of the pedal 'circle'. Instead of moving in a perfect circle, the upward and downward deflections cause the shape to be stretched into an ellipse. I wonder what affect this has on muscle utilization?

The problem is, this isn't true. The maximum frame deflection is during the power stroke, somewhere around 3 o'clock. The frame isn't flexing because we are stepping on it, it if flexing because we are trying to move the rear wheel so hard that tension in the chain is enough to pull the BB off center.

[kramnnim]

The frame isn't flexing because we are stepping on it, it if flexing because we are trying to move the rear wheel so hard that tension in the chain is enough to pull the BB off center.

Seems like most of the flex is the BB area moving left/right in relation to the centerline of the bike/frame because of the downward force of the pedals...not because of the rotational force...

[Kontact]

Seems like most of the flex is the BB area moving left/right in relation to the centerline of the bike/frame because of the downward force of the pedals...not because of the rotational force...

Those are the same forces at 3 o'clock. But the BB flexes most under the most drivetrain force - which is during the power stroke, not at 6 o'clock.

[kramnnim]

The question is: If force through the cranks and against the resistance of the chain deflects the BB to the side, where does that deflection force go when the deflection is taken out? The frame doesn't SPROING! back to center, so it isn't just released. It comes out about at the same rate it went in, and if constant pressure is applied to the pedals, where can it go?


Constant, even pressure is not applied to the pedals.

I don't think the frame would spring back at the same rate because the force against the pedals varies throughout the pedal stroke. You can stretch a rubber band quickly, and release it slowly... Maybe the peak force on the pedal is at 3 oclock, maybe it is 100 newton meters, and maybe the frame flexes 6mm at 100nm. At the bottom of the pedal stroke, the pressure would have gradually dropped from 100 back to 0-10nm, and the frame would gradually flex back to straight before flexing the opposite direction as the opposite leg begins to apply pressure. The frame isn't flexing enough to be all that noticeable. It's not like bobbing around on a full suspension bike with no lockout.

[kramnnim]

Those are the same forces at 3 o'clock. But the BB flexes most under the most drivetrain force - which is during the power stroke, not at 6 o'clock.

The BB is flexing because of the downward pressure from the rider, not the rotational force. If you stuck a motor in the seatpost, the BB would not flex left/right very much, if any.

[Kontact]

The BB is flexing because of the downward pressure from the rider, not the rotational force. If you stuck a motor in the seatpost, the BB would not flex left/right very much, if any.

But the BB would flex in the same direction if you pulled up on the opposite crank. The flex isn't a result of a simple stepping on the crank, but because that's the only direction the BB can go when it is trapped between the power stroke and the chain.

If it was just downward force, the BB would flex most at 6 o'clock, not 3 o'clock.

[Kontact]

Constant, even pressure is not applied to the pedals.

I don't think the frame would spring back at the same rate because the force against the pedals varies throughout the pedal stroke. You can stretch a rubber band quickly, and release it slowly... Maybe the peak force on the pedal is at 3 oclock, maybe it is 100 newton meters, and maybe the frame flexes 6mm at 100nm. At the bottom of the pedal stroke, the pressure would have gradually dropped from 100 back to 0-10nm, and the frame would gradually flex back to straight before flexing the opposite direction as the opposite leg begins to apply pressure. The frame isn't flexing enough to be all that noticeable. It's not like bobbing around on a full suspension bike with no lockout.

I didn't say "even pressure". You inserted that.

If you watch this video of what the BB is doing under heavy load, you'll see that the most BB sway is right during the power stroke and is largely gone by the time the pedal is at the bottom. If this was just an issue of putting weight on the pedals, the flex would be at the bottom to match Mark's graphic. Instead, that's when it is going away.

[kramnnim]

But the BB would flex in the same direction if you pulled up on the opposite crank. The flex isn't a result of a simple stepping on the crank, but because that's the only direction the BB can go when it is trapped between the power stroke and the chain.

If it was just downward force, the BB would flex most at 6 o'clock, not 3 o'clock.

It flexes the most at 3 oclock because that's when the leg is applying the most force

[kramnnim]

I didn't say "even pressure". You inserted that.

If you watch this video of what the BB is doing under heavy load, you'll see that the most BB sway is right during the power stroke and is largely gone by the time the pedal is at the bottom. If this was just an issue of putting weight on the pedals, the flex would be at the bottom to match Mark's graphic. Instead, that's when it is going away.

Yes, we agree on when the BB is flexing and when it goes away. But I don't see how the flexing converts to rotational force and propels the bike.

[Kontact]

Yes, we agree on when the BB is flexing and when it goes away. But I don't see how the flexing converts to rotational force and propels the bike.

The easiest way I've found to thing about it is that the rear center gets shorter when the stays twist along with the BB sway. When the BB straightens out the rear center elongates. Since the chain is already under tension it has to get longer too by yanking on the cassette.

[Kontact]

It flexes the most at 3 oclock because that's when the leg is applying the most force

Right, but it isn't vectored to bend the crank down and over like it is at 6 o'clock. The BB bends at 3 because of the chain tension. At 6 there is still plenty of down force, yet it is unbending. The thing that is causing the bend is chain, and the chain moves the bike. You can't separate the BB flex from max chain tension.

[kramnnim]

Right, but it isn't vectored to bend the crank down and over like it is at 6 o'clock. The BB bends at 3 because of the chain tension. At 6 there is still plenty of down force, yet it is unbending. The thing that is causing the bend is chain, and the chain moves the bike. You can't separate the BB flex from max chain tension.

This doesn't make sense to me. The BB flexes because of the leverage from the crank arm. If your BB was frozen with the arms at 3 and 6, the BB would flex if you apply pressure on either arm, no chain needed.

[Kontact]

This doesn't make sense to me. The BB flexes because of the leverage from the crank arm. If your BB was frozen with the arms at 3 and 6, the BB would flex if you apply pressure on either arm, no chain needed.

But they aren't frozen, and if you were coasting there is no way you could make the BB flex from 3 and 9.

[kramnnim]

But they aren't frozen, and if you were coasting there is no way you could make the BB flex from 3 and 9.

You could at 6 while coasting and none of that flex would propel you forward.

[kramnnim]

The easiest way I've found to thing about it is that the rear center gets shorter when the stays twist along with the BB sway. When the BB straightens out the rear center elongates. Since the chain is already under tension it has to get longer too by yanking on the cassette.

This doesn't sound like any frame I have ridden.

[Kontact]

You could at 6 while coasting and none of that flex would propel you forward.

But we aren't talking about what happens at 6 because, as you noted in the video, the BB doesn't sway at 6.

And I doubt it would sway like it does at 3 if the BB was frozen.



This doesn't sound like any frame I have ridden.

How would you know?

[William]

Came across Dave Kirk's thoughts on frame flex: Three types of flex in the bike...

http://kirkframeworks.com/resources/technical/frame-flex/







William

[William]

Good example of the lateral/torsional flex...

https://www.youtube.com/watch?v=8fbpnn7TUSE





William

[William]

Short but shows bottom bracket flex...

https://www.youtube.com/watch?v=73inseR2zwI





William

[Mark McM]

The problem is, this isn't true. The maximum frame deflection is during the power stroke, somewhere around 3 o'clock. The frame isn't flexing because we are stepping on it, it if flexing because we are trying to move the rear wheel so hard that tension in the chain is enough to pull the BB off center.

Well, that's clearly not true. The frame is flexing precisely because we are stepping on it (or rather, because we are stepping on the pedal). If we step on the pedal when the crank is forward of the BB, then we generate a torque on the crank to drive the front wheel. But if we step on the pedal when the crank is vertical, no drive torque is generated, but we still flex the frame just the same.

[etu]

came across dave kirk's thoughts on frame flex: Three types of flex in the bike...

http://kirkframeworks.com/resources/technical/frame-flex/







william

1+

[Mark McM]

The easiest way I've found to thing about it is that the rear center gets shorter when the stays twist along with the BB sway. When the BB straightens out the rear center elongates. Since the chain is already under tension it has to get longer too by yanking on the cassette.

Maybe you think about this way, but you're thinking is incorrect. You keep claiming that the stays get shorter and longer, but unfortunately there is no evidence that they do to any meaningful extant. So there can be virtually no energy storage due to this type of deflection. The deflection modes of the frame were modelled and analyzed in a web page that was reference on the 3rd page of this discussion (see link below). This analysis shows that the primary energy storage in the frame is torsional deflection around the the longitudinal axis. In other words, the cranks/pedals are going up and down due to a torsional twisting of the frame, causing a rotation of the BB shell as the frame twists.

https://web.archive.org/web/20060214082906/http://bikethink.com/Frameflex.htm

[Kontact]

Good example of the lateral/torsional flex...

https://www.youtube.com/watch?v=8fbpnn7TUSE



I honestly can't tell if his frame is flexing or if his wheels and fork are. I can't see the HT and ST going out of plane with each other.

[Mark McM]

Okay, so here's a quick summary of what's been discussed: I think we can all agree to the following:

The frame deflects under vertical pedal forces;

The primary deflection mode of the frame is a torsional rotation of the BB shell around the longitudinal axis;

The torsional rotation of the BB shell causes the crankarms/pedals to rise or fall, depending on which direction of rotational flex;

As the forces on the pedals continue to change during the crank rotation, the frame un-flexes, on its way to reversing its direction of flex;

Energy stored in frame flex is not lost, but is restored in some way to the system when the frame unflexes;

So the key question is, just where does this energy go?

It appears that the answer to this question is largely dependant on where in the crank rotation the frame unflexes, which is itself dependant on how the pedal forces vary during the crank rotation. That sounds simple, but in reality, there can be a lot of variation in how the pedal forces vary during crank rotation.

The amount of torsion flex at the BB will be dependent on the right/left pedal force differential. In the pedal force vector diagram below, the right/left pedal differential is maximized around 6 o'clock. This means the downward flex at a pedal increases throughout the downstroke, and then decreases on the upstroke. So the flexural energy is restored as the pedal rises, acting to lift the rearward leg. None of the restored energy goes directly into the drivetrain, but instead helps the rider to move their legs.

http://3.bp.blogspot.com/_urSQl6wUA5g/S7MGtK-9gSI/AAAAAAAAH6s/WRXTRJS-NIU/s400/ClockPlot_ForceProfile_CF_Standing1-1024x571.JPG

But in this second pedal force vector example just below, the right/left pedal differential (and therefore the maximal downward flex at the pedal) is maximized around 4 o'clock. This means that much of the unflexing occurs while the downward moving crank is still forward of the BB. The unflexing of the frame acts to raise the BB while the pedal is still descending. This results in forward rotation of the crank. In this case, some of the energy stored in the frame goes directly into generating crank torque, and therefore directly into the drivetrain. (Note: most of the frame unflexing in this case still occurs in a portion of the crank rotation where the 'spring' force/restored energy primarily acts to raise the bottom leg).

http://i32.photobucket.com/albums/d7/k4drd/Bicycles/PedalForces.jpg

In both cases, it is a case of a zero sum gain. The energy stored in the frame is resturned, and helps to drive the bike forward either indirectly or directly: Indirectly, in that it reduces the amount of energy the rider needs to expend to bring their leg back up for the next power stroke; or directly, in that the energy goes into the drivetrain during the power stroke. Or, most likely in many cases, a little of both.

[William]

I honestly can't tell if his frame is flexing or if his wheels and fork are. I can't see the HT and ST going out of plane with each other.

If you look at the faceplate on the stem when he first starts torquing it (before the camera pulls right) it looks like the head tube and seat tube are moving out of plane with each other. How much is frame and how much is wheel/fork is hard to tell for sure.





William

[Kontact]

Well, that's clearly not true. The frame is flexing precisely because we are stepping on it (or rather, because we are stepping on the pedal). If we step on the pedal when the crank is forward of the BB, then we generate a torque on the crank to drive the front wheel. But if we step on the pedal when the crank is vertical, no drive torque is generated, but we still flex the frame just the same.

And yet, the BB flex goes away before we pedal through 6 where you've shown that there is still plenty of down force on the pedal - down force that is much more in line with BB flex than it even is at 3 o'clock.

Maybe you think about this way, but you're thinking is incorrect. You keep claiming that the stays get shorter and longer, but unfortunately there is no evidence that they do to any meaningful extant. So there can be virtually no energy storage due to this type of deflection. The deflection modes of the frame were modelled and analyzed in a web page that was reference on the 3rd page of this discussion (see link below). This analysis shows that the primary energy storage in the frame is torsional deflection around the the longitudinal axis. In other words, the cranks/pedals are going up and down due to a torsional twisting of the frame, causing a rotation of the BB shell as the frame twists.

What is that, a royal fiat?


Did you read the article you linked to?

For a final part of this analysis, I have used an FEA model to determine the response of a frame to the four loads described earlier. It is somewhat trivial to look at how much energy goes in and out of the frame since we know that it gets released into the drive train. But it is interesting to understand about how much energy gets temporarily stored in the frame.
....

Having concluded that frame flex does not waste energy, I do not believe that frame stiffness is irrelevant.


I have offered the "twisting stays" model as a basic way of thinking about what happens when you have a plane between two lines (chain stays between hub and BB shell), and you move one of those points. I'm not talking about where the energy is specifically stored, but how that energy is redelivered to the drivetrain. Obviously, the energy is stored in every component that is bending - especially the down tube, seat tube and chain stays.

You might not like that way of looking at it, but frankly the ability to model the actual geometry of all of this is a bit much for anyone to do in their head, so we have to talk about this with simplified versions. So here's a very simple example of the problem:

If I take a flat strip of spring steel and twist the ends in opposite directions, would you not agree that the overall length of the strip gets shorter?

When I release that twist, does the strip not push itself back to full length?

And if I cut a slot down the center of the strip like the space between the chainstays, does any of that change?

[kramnnim]

So Kirk is saying that the flexed frame, returning back to neutral as the rider is still applying pressure, is directing the energy to the rear wheel... If the frame moves 5mm at the peak of the pedal stroke, and was able to flex back those 5mm while the rider was still applying pressure, the crank arm would rotate slightly. (5mm?)

That does make sense...as long as the frame is flexing back against the pedal stroke. If it can't flex back until 6 oclock, the energy would not go back through the chain to the rear wheel and would be wasted.

[Kontact]

If you look at the faceplate on the stem when he first starts torquing it (before the camera pulls right) it looks like the head tube and seat tube are moving out of plane with each other. How much is frame and how much is wheel/fork is hard to tell for sure.



Not saying it isn't happening, I just can't tell if what I'm looking at is twist or just what happens when something leans out of plane and one part is closer to the observer than the others. There is a perspective issue with vanishing points that makes it very hard to observe if the head tube and seat tube aren't actually crossing in view.

[Kontact]

None of the restored energy goes directly into the drivetrain, but instead helps the rider to move their legs.

This is one of those statements like "the man's feet don't push against the earth, the earth pushed up on his feet."

You can't argue that "helping the rider move their legs" is different than "helping the rider move the crank" in a system where net force never goes to zero. Until force goes to zero, the bike and the legs are all one mechanical linkage and you don't get to pick where the force is acting within it.

[Kontact]

That does make sense...as long as the frame is flexing back against the pedal stroke. If it can't flex back until 6 oclock, the energy would not go back through the chain to the rear wheel and would be wasted.

Why not? Did the the rider input go to zero and the system went completely slack?

Under normal pedaling do you ever feel the freewheel disengage, even momentarily, at any part of the pedal stroke?

[cachagua]

The bike and the legs are all one mechanical linkage and you don't get to pick where the force is acting within it.

No, we don't get to choose. You're right about that.

I still think the force will act where it is resisted less. It sure would put my mind at rest if someone could say how it would not do that, but instead would act on the thing that's resisting it more...?

[Kontact]

No, we don't get to choose. You're right about that.

I still think the force will act where it is resisted less. It sure would put my mind at rest if someone could say how it would not do that, but instead would act on the thing that's resisting it more...?

It's a fair question. I found this rather scholarly article that is more about how to get results than about the results themselves, but it does show that pedaling is always net positive torque value. See figure 6 on page 7:

https://www.uni-konstanz.de/mmsp/pubsys/publishedFiles/QuDaSa15.pdf

It might be useful to know just how much torque it takes to notably flex a frame, because that number might compare usefully to the minimum pedaling torque to suggest how much un-flexing could feed back against pedaling. It would also be good to know when the unflexing happens - at minimum torque or somewhere before or after it.

[cachagua]

It's a fair question. I found this rather scholarly article that is more about how to get results than about the results themselves, but it does show that pedaling is always net positive torque value...

I don't quite see the connection with what I was asking. Or is that point related to something else?

It would also be good to know when the unflexing happens - at minimum torque or somewhere before or after it.

Is "when it happens" really the way to think about it? It certainly doesn't occur all at once -- we could say it begins to happen when your power decreases below its maximum, and continues to happen more as you decrease more -- and then, on into the next stroke when you begin to push harder again, the frame begins to flex again (in the direction appropriate to that stroke, of course). Insofar as the flex is a response to the power you're putting out, it will vary in tandem with that.

And clarifying things this far is making me begin to wonder if the BB swaying Mark describes as a result of non-torque pedal pressure may have to be dealt with separately. Can we conveniently separate frame flex that's a result of torque, and frame flex that's a result of non-torque forces?

[kramnnim]

Why not? Did the the rider input go to zero and the system went completely slack?

Under normal pedaling do you ever feel the freewheel disengage, even momentarily, at any part of the pedal stroke?

Indoors at low power and low RPM, sitting up, hand off the bars...yes, I hear several clicks of the freewheel at the 12/6 dead spot.

No, it isn't normal pedaling, but it exaggerates just how dead the dead spot is.

If we compare the motion of the BB flexing back against the pedal stroke to when you are up out of the saddle, rocking the bike back and forth, using your arms to rock the bike left as your right foot pushes down...that movement does nothing useful at 12/6.

That type of rocking would be one of the times when a frame flexes the most, and it seems like the flexed frame would be rebounding right around 11/5 to 12/6. So it would be wasted just like bobbing around on a full suspension MTB. Or are you going to say that the energy stored in the compressed suspension is also pushed back into the drivetrain...

[Kontact]

I don't quite see the connection with what I was asking. Or is that point related to something else?



Is "when it happens" really the way to think about it? It certainly doesn't occur all at once -- we could say it begins to happen when your power decreases below its maximum, and continues to happen more as you decrease more -- and then, on into the next stroke when you begin to push harder again, the frame begins to flex again (in the direction appropriate to that stroke, of course). Insofar as the flex is a response to the power you're putting out, it will vary in tandem with that.

And clarifying things this far is making me begin to wonder if the BB swaying Mark describes as a result of non-torque pedal pressure may have to be dealt with separately. Can we conveniently separate frame flex that's a result of torque, and frame flex that's a result of non-torque forces?

The point I was making is that torque never comes off the chain and that torque has a minimum level. Despite that minimum level the BB flex comes out, in part because it is pulled to the other side by the alternating pedaling rather than just the drop in pedal torque. So when the BB starts to move back to center and then gets to center, how much tension was on the chain, when was it on it and how much of BB recentering is just decreases in torque and how much is the other leg doing to tug it that way?

[cachagua]

When the BB... gets to center, how much tension was on the chain?

Less than some other place in the pedal stroke. That's all that's necessary for what I'm suggesting.

How much of BB recentering is just decrease in torque and how much is the other leg doing to tug it that way?

That's kind of what I was asking above -- can we separate torque effects from non-torque effects? But I'm starting to think it doesn't matter. I'm not sure whether all the flex comes out of the frame at any phase of a crank revolution, now that we're looking at a number of different flex modes, but that's not necessary either -- what does come out still looks to me like it doesn't speed the bike up, but rather slows your legs down.

I think that was what we were originally trying to figure out?

[Mark McM]

And yet, the BB flex goes away before we pedal through 6 where you've shown that there is still plenty of down force on the pedal - down force that is much more in line with BB flex than it even is at 3 o'clock.

And what bit of magic effect do think does that? If there is differential in downward force at the pedals, there will be torque applied to the BB, and the frame will be flexed. All the pedal force vector diagrams show that there is still a downward force on the bottom pedal (which is greater than the force on the top pedal), so the frame will be flexed. That's just simple physics.



What is that, a royal fiat?


Did you read the article you linked to?


Yes, but I don't think you did. There is no royal fiat, just empirical evidence and numerical analysis. This page (https://web.archive.org/web/20060214140009/http://bikethink.com:80/Strain_energy.htm)shows the strain energies from the different reaction forces. From the graph, the strain energy due to stays compressing (labeled the "Horizontal force" in the graph) is very small, just a few percent of the total strain energy in the frame.



If I take a flat strip of spring steel and twist the ends in opposite directions, would you not agree that the overall length of the strip gets shorter?

When I release that twist, does the strip not push itself back to full length?

You are really grasping. Yes, if the twist is very large, there will be very small change in length. But in this case, the twist is small enough that the change in length is essentially zero. This is covered in this text chapter on torsion reaction in structural members (http://ocw.nthu.edu.tw/ocw/upload/8/252/Chapter_3-98.pdf). Instead, the primary deflection is the rotation of the BB shell, resulting in vertical deflection at the pedals, and the primary strain energy is stored in this torsional twisting along an axis perpendicular to the crank torque.

[Mark McM]

This is one of those statements like "the man's feet don't push against the earth, the earth pushed up on his feet."

You can't argue that "helping the rider move their legs" is different than "helping the rider move the crank" in a system where net force never goes to zero. Until force goes to zero, the bike and the legs are all one mechanical linkage and you don't get to pick where the force is acting within it.

Nope, you're wrong again. Let's consider one crank revolution, starting with the leg at the top of the stroke: As the rider pushes down, the frame flexes. Some of the rider's energy goes directly into the drivetrain, and some goes into the frame in the form of strain energy. At the bottom of the pedal stroke, the rider still exerts a significant downward force on the pedal, so the frame is still flexed. As the rider continues the crank rotation, he has to expend energy to raise his leg. But as the frame un-flexes, the energy released by the frame helps the rider to raise his leg, so the rider's net energy is the same as if the frame had not flexed. But none of the strain energy went into the drivetrain - all the energy that went into the drivetrain occured during the downstroke. If a rider began from a standing start and only made one crank revolution, then it is very clear that the strain energy restoring the rider's leg position made no contribution to drive the bicycle. And it is made particularly clear if there is a time delay between the rider pushing all the way to the bottom of the stroke, and re-starting the revolution to bring their leg back up.

[cachagua]

If there is differential in downward force at the pedals, there will be torque applied to the BB, and the frame will be flexed...

We've got a couple of different torques here that we shouldn't confuse. One's the swinging of the BB that we'd see looking fore-to-aft along the bike, and the other's the rotation of the crank. The BB swinging isn't going to contribute to the bike's forward motion, but of course the rotation of the crank is exactly what makes the bike go forward.

However I think it there is another flex mode, not the BB swinging, that does effectively "shorten" the stays. The video we started with demonstrates it. If there were no strain released in that mode when the rider lets go of the rear brake -- no "re-lengthening" of the effective stay length -- then the wheel wouldn't move.

Would anyone like to try this at home? Duplicate the set-up in the video and do exactly the same thing, only let the rear brake off very gradually, so the wheel doesn't spin freely but just rotates a tiny, tiny bit when the frame relaxes. A degree? Two? It would be very little. But that's going to correspond to what we're calling "compression" of the stays.

[Kontact]

Nope, you're wrong again. Let's consider one crank revolution, starting with the leg at the top of the stroke: As the rider pushes down, the frame flexes. Some of the rider's energy goes directly into the drivetrain, and some goes into the frame in the form of strain energy. At the bottom of the pedal stroke, the rider still exerts a significant downward force on the pedal, so the frame is still flexed. As the rider continues the crank rotation, he has to expend energy to raise his leg. But as the frame un-flexes, the energy released by the frame helps the rider to raise his leg, so the rider's net energy is the same as if the frame had not flexed. But none of the strain energy went into the drivetrain - all the energy that went into the drivetrain occured during the downstroke. If a rider began from a standing start and only made one crank revolution, then it is very clear that the strain energy restoring the rider's leg position made no contribution to drive the bicycle. And it is made particularly clear if there is a time delay between the rider pushing all the way to the bottom of the stroke, and re-starting the revolution to bring their leg back up.

The rider's legs, the way you are describing it, are part of the drivetrain.

No one is claiming that there is extra energy coming out of a flexy frame. I'm claiming that there is no real loss because the flex is simply used later in the pedal stroke, which is what the article you posted claims and it also seems to be what you are claiming, too.

The only conflict appears to be that you believe there is a difference between "helps the rider to raise the leg" and any other pro-forward motion input. I used the stays to illustrate another way stored energy could contribute to forward motion, but I don't think there is any real difference between "helps push the bike" or "helps turn the pedals" - those are the same thing. You are "pushing against the foot" and I'm saying they are the same thing.



But the author, you and I are all taking a very different stance from that typified by cachagua who believes that the release of strain energy drags the riders leg against pedaling motion, causing an anti-forward motion input, slowing the bike overall.

[cachagua]

And does that released strain energy, the energy that rotates the rear wheel in the video, contribute to your going forward, or does it not? That, I think, is our original, central question.

And we've been getting rather far afield.

[cachagua]

The author, you and I are all taking a very different stance from that typified by cachagua who believes that the release of strain energy drags the riders leg against pedaling motion, causing an anti-forward motion input, slowing the bike overall.

Yes thank you, exactly right, except that "slowing the bike" perhaps overemphasizes the effect I'm envisioning -- I'd call it "doesn't contribute to your going forward", which physics-wise is about the same thing, but more accurately reflects our experience of riding.

[Kontact]

We've got a couple of different torques here that we shouldn't confuse. One's the swinging of the BB that we'd see looking fore-to-aft along the bike, and the other's the rotation of the crank. The BB swinging isn't going to contribute to the bike's forward motion, but of course the rotation of the crank is exactly what makes the bike go forward.

However I think it there is another flex mode, not the BB swinging, that does effectively "shorten" the stays. The video we started with demonstrates it. If there were no strain released in that mode when the rider lets go of the rear brake -- no "re-lengthening" of the effective stay length -- then the wheel wouldn't move.

Would anyone like to try this at home? Duplicate the set-up in the video and do exactly the same thing, only let the rear brake off very gradually, so the wheel doesn't spin freely but just rotates a tiny, tiny bit when the frame relaxes. A degree? Two? It would be very little. But that's going to correspond to what we're calling "compression" of the stays.

The flex in the video comes from the tension between the pedal going down and the chain pulling back. The frame can't compress to the rear, so it ends up flexing into the lateral plane. And that seems to be what makes people's heads explode - this idea that flex energy could be stored outside the plane that we are comfortable thinking about.


If you let the brake out slowly the wheel will move. The fact that they let the wheel move enough to cause it to coast isn't really different than letting it move the amount of distance dictated by distance of the flex. Those are just two different illustrations of the same thing:

If you release the brake all at once, the resulting velocity of the wheel is a measure of released energy - the energy it took to accelerate the wheel from zero to whatever rotational speed it achieves. That demonstrates a work per second rate - Watts.

If you release the brake slowly, you are releasing a lot of that energy into the brake pads, so what you are demonstrating is no longer a transfer of Watts, but a transfer of work without reference to time - Joules. The drivetrain flexing will provide enough work to move the wheel from A to B. But since we took time out of it we are no longer talking about energy the way cyclists normally do - in Watts.


So you can demonstrate whichever you want, but they are demonstrations of different ways of understanding energy, not two separate things.

[Mark McM]

The rider's legs, the way you are describing it, are part of the drivetrain.

No one is claiming that there is extra energy coming out of a flexy frame. I'm claiming that there is no real loss because the flex is simply used later in the pedal stroke, which is what the article you posted claims and it also seems to be what you are claiming, too.

The only conflict appears to be that you believe there is a difference between "helps the rider to raise the leg" and any other pro-forward motion input. I used the stays to illustrate another way stored energy could contribute to forward motion, but I don't think there is any real difference between "helps push the bike" or "helps turn the pedals" - those are the same thing. You are "pushing against the foot" and I'm saying they are the same thing.

I don't think most people would define the rider (or their legs) as part of the drivetrain. I think most people would see it as the "engine". So most would separately classify the energy paths through them. But putting aside the definitions of energy paths, in the end there appears to be no net loss of energy in frame flex - the strain energy transferred into the frame is not lost, but is returned in a useful way (through one mechanism or another). Tests and rider performances appear to bear this out.

However .. I think the mechanism of the energy return actually is important at least in one regard. As we've been discussing, the changing force vectors on the pedals through the crank revolution will not only affect how much strain energy goes into the frame, but also at what points in the crank rotation that it goes in and comes back out. This in turn affects the mechanisms by which the returned energy is utilized. The reason this is important is because riders often vary their pedaling styles to match different situations (for example, I don't think anyone will claim that they pedal the same way between standing or seated, or between steady continuous effort and short full-power sprints). I think it is highly likely that riders will vary there pedaling styles (if only subtly) in response to frame stiffness, and with practice will adapt their pedaling to optimize their pedal motions/forces with the frame response.

[kramnnim]

I don't think the original GCN video is an accurate representation of what happens while riding because the trainer is flexing a lot.

But, even if it was accurate- yes, wheel will be propelled when the brake is released if the crank remains at 3 oclock.

I don't think the tension is released until around 6 oclock.

So to see what the effect would be, you would have to flex the frame+trainer at 3 oclock, hold it there while you allow the crank to move down to 6 oclock, and then release. I don't think the wheel would move nearly as much, If at all.

[Kontact]

I don't think most people would define the rider (or their legs) as part of the drivetrain. I think most people would see it as the "engine". So most would separately classify the energy paths through them. But putting aside the definitions of energy paths, in the end there appears to be no net loss of energy in frame flex - the strain energy transferred into the frame is not lost, but is returned in a useful way (through one mechanism or another). Tests and rider performances appear to bear this out.

However .. I think the mechanism of the energy return actually is important at least in one regard. As we've been discussing, the changing force vectors on the pedals through the crank revolution will not only affect how much strain energy goes into the frame, but also at what points in the crank rotation that it goes in and comes back out. This in turn affects the mechanisms by which the returned energy is utilized. The reason this is important is because riders often vary their pedaling styles to match different situations (for example, I don't think anyone will claim that they pedal the same way between standing or seated, or between steady continuous effort and short full-power sprints). I think it is highly likely that riders will vary there pedaling styles (if only subtly) in response to frame stiffness, and with practice will adapt their pedaling to optimize their pedal motions/forces with the frame response.

I don't think a anyone who broke a driveshaft from too much torque on a muscle car would consider the motor separate from the drivetrain when looking for causes. If you are examining how power is distributed from leg muscles to road it is a bit arbitrary to say that flesh doesn't count.


As to adapting to how the energy goes in and comes out, that sounds like something your brain figures out in about 4 pedal strokes and is little different than the learning curve of walking with new shoes. Why would it be important?


It seems like the only important thing that these discussions really have to offer is convincing people that, despite how it might feel, you aren't actually throwing away energy if your frame flexes.

[Kontact]

I don't think the original GCN video is an accurate representation of what happens while riding because the trainer is flexing a lot.

But, even if it was accurate- yes, wheel will be propelled when the brake is released if the crank remains at 3 oclock.

I don't think the tension is released until around 6 oclock.

So to see what the effect would be, you would have to flex the frame+trainer at 3 oclock, hold it there while you allow the crank to move down to 6 oclock, and then release. I don't think the wheel would move nearly as much, If at all.

The GCN thing isn't supposed to be an accurate model of riding. It is just supposed to show that lateral displacement of the BB is still "in line" with power transmission. It is a separate question how and when that energy gets to the rear wheel during actual pedaling.

[cachagua]

If you release the brake all at once, the resulting velocity of the wheel is a measure of released energy...

If you release the brake slowly, what you are demonstrating is a transfer of work without reference to time...

So you can demonstrate whichever you want, but they are demonstrations of different ways of understanding energy, not two separate things.

Could you go over what you thought I was trying to say? I'm not quite able to read the above as a response. Or if it's a response to something else -- my mistake.


It seems like the only important thing that these discussions really have to offer is convincing people that, despite how it might feel, you aren't actually throwing away energy if your frame flexes.

Naw, I think there's some other value besides that to this conversation -- coupla things, actually.

[Kontact]

Could you go over what you thought I was trying to say? I'm not quite able to read the above as a response. Or if it's a response to something else -- my mistake.


No, it was a response to your suggestion of letting off the brake very slowly. Both demonstrate the same trapped energy, but in different ways. I was illustrating why your suggestion is different than the GCN way - because it subtracts time so it doesn't show Watts of energy, just Joules of work.

Your suggestion is like the difference between slowly carrying a book upstairs and throwing it.

[cachagua]

No, it was a response to your suggestion of letting off the brake very slowly. Both demonstrate the same trapped energy, but in different ways. I was illustrating why your suggestion is different than the GCN way - because it subtracts time so it doesn't show Watts of energy, just Joules of work.

Your suggestion is like the difference between slowly carrying a book upstairs and throwing it.

I guess so, thanks. What I hoped to throw some light on is that there is a component of frame flex that puts the rear axle closer to the BB (since there had been some divergence of opinion on that). The distinction between letting the released strain energy spin the wheel, and using the brakes to keep it from freewheeling and only rotate as far as the chain actually pulls it, is (for my purpose) only useful to show how little the wheel rotates in direct response to the un-flexing.

And I wanted to highlight that because I think that's the component of flex, or the main one, that we've been trying to look at.

I'm totally curious about this but I don't have a trainer, or else I'd do the experiment myself. I'll bet the wheel moves just a degree or two at most. I could be wrong but I think it must be the tiniest, tiniest bit. But some. Definitely some.

[Kontact]

I guess so, thanks. What I hoped to throw some light on is that there is a component of frame flex that puts the rear axle closer to the BB (since there had been some divergence of opinion on that). The distinction between letting the released strain energy spin the wheel, and using the brakes to keep it from freewheeling and only rotate as far as the chain actually pulls it, is (for my purpose) only useful to show how little the wheel rotates in direct response to the un-flexing.

And I wanted to highlight that because I think that's the component of flex, or the main one, that we've been trying to look at.

I'm totally curious about this but I don't have a trainer, or else I'd do the experiment myself. I'll bet the wheel moves just a degree or two at most. I could be wrong but I think it must be the tiniest, tiniest bit. But some. Definitely some.

It isn't a great deal of energy. But if you told a Strava guy you were going to take away that amount of energy on every pedal stroke he'd blow a gasket.

[cachagua]

YAAAAH-ha-ha-ha-ha-ha! I would never, ever breathe a word of this to a Strava guy.

[cachagua]

There appears to be no net loss of energy in frame flex - the strain energy transferred into the frame is not lost, but is returned in a useful way (through one mechanism or another)...

Can you elaborate? Which mechanism returns strain energy in a useful way? I'm still waiting for an explanation of how a lesser force can overcome a greater one. Or if you're thinking about the release of BB sway "lifting" the pedal on its upward stroke -- that gets the pedal farther off the ground but I don't think it rotates the cranks forward. It tips the cranks back towards "vertical" (with respect to the bike) but it doesn't turn them.

Tests and rider performances appear to bear this out.

Not sure what you mean here. The tests we've been looking at, and rider performance (if we believe the riders) couldn't be more contradictory. The video and the FEA both claim you get all your pedaling energy back out of the frame (note however that they're both based on the same incorrect assumption) but riders generally agree that a stiffer frame wastes less energy, i.e. that frame flex is lost energy.

The changing force vectors on the pedals through the crank revolution will not only affect how much strain energy goes into the frame, but also at what points in the crank rotation that it goes in and comes back out. This in turn affects the mechanisms by which the returned energy is utilized. The reason this is important is because...

Now, you got to stop talking like that. Baffling bulls**t is what I do, it's my specialty and my trademark, and if you start sounding like that I call plagiarism! :)

I think it is highly likely that riders will vary their pedaling styles (if only subtly) in response to frame stiffness, and with practice will adapt their pedaling to optimize their pedal motions/forces with the frame response.

I guess you mean, in response to the different stiffnesses of different frames? Interesting... might that mean that stiffer frames of recent times aren't really faster than old ones, but today's riders think they are because they don't know how to ride the old ones? That's actually kind of tasty... but again, we'd better make sure not to say it to Strava-guy.

[Mark McM]

I guess so, thanks. What I hoped to throw some light on is that there is a component of frame flex that puts the rear axle closer to the BB (since there had been some divergence of opinion on that).

Yes, there is a component of frame flex that puts the rear axle closer to the BB. But of all the flex components, this is one of the smallest. Since strain energy is inversely proportional to stiffness, this flex component will be a very small part of the total strain energy in the frame. While the GCN video may not be an entirely realistic simulation of pedalling, it clearly shows that the majority of the strain energy in the frame is from the horizontal twisting of the BB, not from the (linear) compression of the rear triangle. Anyone who wants to claim otherwise, will need to come up with direct evidence to support their argument.

[Mark McM]

Can you elaborate? Which mechanism returns strain energy in a useful way? I'm still waiting for an explanation of how a lesser force can overcome a greater one. Or if you're thinking about the release of BB sway "lifting" the pedal on its upward stroke -- that gets the pedal farther off the ground but I don't think it rotates the cranks forward. It tips the cranks back towards "vertical" (with respect to the bike) but it doesn't turn them.

There doesn't need to be lesser force overcoming a greater one for strain energy to be released to the drivetrain. Even a drive force less than the drag on the bicycle acts to propel the bike. Here's why: Pedaling drive force is periodic - it oscillates larger and smaller during the crank revolution. But as long as the average drive force equals the drag force, the bike maintains a constant average speed. During the portion of the crank revolution where the drive force exceeds the drag force, the bike accelerates. During the portions of the crank revolution where the drive force is less than the drag force, the bike decelerates. But even when the drive force is less than the drag force, it acts to propel the bike, because it causes the bike to decelerate more slowly than if there was no drive force. So anytime there is a drive force (regardless of its magnitude), energy is transferred into the drivetrain.

As you know, as the rider presses down on the pedal during the down stroke, the frame flexes (by twisting at the BB), and both the pedal and the BB drop in response. The peak downward force in fact exceeds the bicycle drag force during this phase. Towards the end of the down stroke, the rider may ease off on the down force. This will allow the frame to unflex, because the flex force then exceeds to the pedal force. The pedal's downward speed may also decelerate, as the BB starts to rise - but as long as enough force is applied to the pedal so the pedal doesn't rise as fast as the BB, the the upward motion of the BB will cause the crank to rotate forward. And the energy that caused this extra crank rotation (and thus extra energy applied to the drivetrain) came from the strain energy stored in the frame. The drivetrain force due to the release of the strain force may not exceed the drag force, but it is results in the strain energy enterring the drivetrain.



Not sure what you mean here. The tests we've been looking at, and rider performance (if we believe the riders) couldn't be more contradictory. The video and the FEA both claim you get all your pedaling energy back out of the frame (note however that they're both based on the same incorrect assumption) but riders generally agree that a stiffer frame wastes less energy, i.e. that frame flex is lost energy.

Laboratory tests (like the Damon Rinard test) have yet to find energy lost due to frame flex. And rider performance isn't measured by how the rider feels (or believes) - it is measured with speedometers and stop watches. While riders may feel that a flex bike robs them of power, it largely has not shown up on the stop watch. Plenty of races have been won on flexy frames, and lost on stiff frames.



I guess you mean, in response to the different stiffnesses of different frames? Interesting... might that mean that stiffer frames of recent times aren't really faster than old ones, but today's riders think they are because they don't know how to ride the old ones? That's actually kind of tasty... but again, we'd better make sure not to say it to Strava-guy.

Lots of things have changed over the past few decades, not just frame stiffness. The biggest difference is probably the effectiveness (and prevalence) of PEDs, as demonstrated in how race speeds dropped dramatically after the Reasoned Decision, practically to pre-EPO era speeds(despite the use of "stiffer" frames). Other factors include: Better training methods, better road surfaces, more aerodynamic bikes and clothing, lower rolling resistance tires, etc. There are too many confounding factors to attribute changes in race speeds to frame stiffness.

[benb]

Laboratory tests (like the Damon Rinard test) have yet to find energy lost due to frame flex. And rider performance isn't measured by how the rider feels (or believes) - it is measured with speedometers and stop watches. While riders may feel that a flex bike robs them of power, it largely has not shown up on the stop watch. Plenty of races have been won on flexy frames, and lost on stiff frames.

For sure. I'd imagine there is never going to be any test that proves the flex returns energy either.

I think the thing is rider placebo is a valuable effect. If a rider feels more efficient on a stiff frame, that is good for them, if they feel better on a flexier frame, that's good for them.

No different than superstitions in baseball.

[Black Dog]

There doesn't need to be lesser force overcoming a greater one for strain energy to be released to the drivetrain. Even a drive force less than the drag on the bicycle acts to propel the bike. Here's why: Pedaling drive force is periodic - it oscillates larger and smaller during the crank revolution. But as long as the average drive force equals the drag force, the bike maintains a constant average speed. During the portion of the crank revolution where the drive force exceeds the drag force, the bike accelerates. During the portions of the crank revolution where the drive force is less than the drag force, the bike decelerates. But even when the drive force is less than the drag force, it acts to propel the bike, because it causes the bike to decelerate more slowly than if there was no drive force. So anytime there is a drive force (regardless of its magnitude), energy is transferred into the drivetrain.

As you know, as the rider presses down on the pedal during the down stroke, the frame flexes (by twisting at the BB), and both the pedal and the BB drop in response. The peak downward force in fact exceeds the bicycle drag force during this phase. Towards the end of the down stroke, the rider may ease off on the down force. This will allow the frame to unflex, because the flex force then exceeds to the pedal force. The pedal's downward speed may also decelerate, as the BB starts to rise - but as long as enough force is applied to the pedal so the pedal doesn't rise as fast as the BB, the the upward motion of the BB will cause the crank to rotate forward. And the energy that caused this extra crank rotation (and thus extra energy applied to the drivetrain) came from the strain energy stored in the frame. The drivetrain force due to the release of the strain force may not exceed the drag force, but it is results in the strain energy enterring the drivetrain.





Laboratory tests (like the Damon Rinard test) have yet to find energy lost due to frame flex. And rider performance isn't measured by how the rider feels (or believes) - it is measured with speedometers and stop watches. While riders may feel that a flex bike robs them of power, it largely has not shown up on the stop watch. Plenty of races have been won on flexy frames, and lost on stiff frames.





Lots of things have changed over the past few decades, not just frame stiffness. The biggest difference is probably the effectiveness (and prevalence) of PEDs, as demonstrated in how race speeds dropped dramatically after the Reasoned Decision, practically to pre-EPO era speeds(despite the use of "stiffer" frames). Other factors include: Better training methods, better road surfaces, more aerodynamic bikes and clothing, lower rolling resistance tires, etc. There are too many confounding factors to attribute changes in race speeds to frame stiffness.

This sums up so much. Our little sport is rife with "feelings", "beliefs", and assumptions that are untested or even more so are completely contrary to actual data. Human perception is simply not a reliable way to asses so many things. This has been shown time and time again, yet, without any evidence, we seem so sure that we can tell exactly what is going on in complex and dynamic systems based on feel and intuition.

[Kontact]

For sure. I'd imagine there is never going to be any test that proves the flex returns energy either.


How is a test that shows no loss due to flex not exactly the same as proving that flex energy is returned?

It takes energy to flex a frame. And it isn't hard to measure how much energy. If that isn't a measurable loss and the energy at the wheel is the same, doesn't that prove that the flex returned its energy? What else could have made up for the loss?

[kramnnim]

Towards the end of the down stroke, the rider may ease off on the down force. This will allow the frame to unflex, because the flex force then exceeds to the pedal force. The pedal's downward speed may also decelerate, as the BB starts to rise - but as long as enough force is applied to the pedal so the pedal doesn't rise as fast as the BB, the the upward motion of the BB will cause the crank to rotate forward.

This all makes sense to me...except I think the BB may be rising when the crank is at 6 oclock, and the rise would not help with crank rotation.

[Kontact]

This all makes sense to me...except I think the BB may be rising when the crank is at 6 oclock, and the rise would not help with crank rotation.

You are referring to one location on a structure that is bending and twisting, and then referring to the movement of that location in a single plane. But the outside of the BB is rising, twisting, swinging out and sliding forward. Meanwhile the chainstay attached to it is untwisting, expanding, curving, rising and swinging out.

I realize I'm repeating myself, but why do these analyses alway attempt to reduce a complex three dimensional movement to just what is observed perpendicular to the plane of the chainline? Is that even vaguely realistic when much of what is happening to the bike is like the bending of a bow from the middle of downtube all the way to the rear hub?

[benb]

How is a test that shows no loss due to flex not exactly the same as proving that flex energy is returned?

It takes energy to flex a frame. And it isn't hard to measure how much energy. If that isn't a measurable loss and the energy at the wheel is the same, doesn't that prove that the flex returned its energy? What else could have made up for the loss?

I know you've been arguing about this for pages and pages. What I meant no one is going to prove that the flex doesn't make you faster, or slower.

But if it (a certain degree of flex or lack of flex) makes you feel better, ride it.

[Mark McM]

This all makes sense to me...except I think the BB may be rising when the crank is at 6 oclock, and the rise would not help with crank rotation.

As I mentioned, when the energy is returned depends on pedaling style. In an earlier post, I included 2 diagrams of 2 different pedaling styles. The first diagram showed an example of standing while pedaling, in which the maximum R/L pedal force differential occurred at 6 o'clock (i.e., the rider is standing with all their weight on the bottom pedal at the bottom of the pedal stroke). In this case, the BB doesn't rises while the pedal is already moving upward, so it doesn't directly help to rotate the crank. Instead, it helps raise the rider's foot/leg. (If you've ever stood and sprinted on a very flexy bike, you may liken it to jumping up and down on trampoline, with bike springing back up on the back of the pedal stroke.)

In the second diagram, the maximum R/L pedal force differential occurs at about 4 o'clock, so the BB begins its rise after 4 o'clock - when the rising BB does help with crank rotation.

[Mark McM]

You are referring to one location on a structure that is bending and twisting, and then referring to the movement of that location in a single plane. But the outside of the BB is rising, twisting, swinging out and sliding forward. Meanwhile the chainstay attached to it is untwisting, expanding, curving, rising and swinging out.

I realize I'm repeating myself, but why do these analyses alway attempt to reduce a complex three dimensional movement to just what is observed perpendicular to the plane of the chainline? Is that even vaguely realistic when much of what is happening to the bike is like the bending of a bow from the middle of downtube all the way to the rear hub?

The Bike Think web page referenced earlier did analyze the pedal stroke (and the flex in the frame) in 3 dimensions. This analysis showed that by far, the largest flex (and strain energy storage) is from the twisting of the frame/BB around the longitudinal axis. The other modes of flex contributed very little to the stored strain energy - and in particular, the linear deflection between axle and BB accounted for no more than a few percent of the total strain energy.

The conclusions of the analysis can easily be confirmed by eye - the twisting of the frame/BB around the longitudinal axis is large and pronounced, and can be easily seen. At the same time, any shortening of the distance between the axle and BB can not be seen by the naked eye. Why do you keep concentrating on something which is so obviously inconsequential?

[Kontact]

I know you've been arguing about this for pages and pages. What I meant no one is going to prove that the flex doesn't make you faster, or slower.

But if it (a certain degree of flex or lack of flex) makes you feel better, ride it.

I think you could demonstrate the a certain amount of flex is ideal, just as science has changed how we used to think rolling resistance works. It just requires isolating the variables caused by people to be accounted for.


And I'm not really advocating for flex, just somewhat against the marketing of drive train stiffness - especially when it leads us to sacrifice ergonomics like Q or ankle clearance to provide a quality that we don't benefit from. Dealing with problematically weirdo stuff like BBRight for several years will make you want to talk about the Emperor's Clothes.

My personal experience has been that stiff bikes are fun to ride, and don't necessarily have to be unpleasant. But I did still see with those bikes that they were on the edge of ideal tire traction at times and could "waste" energy doing what they were supposed to be ideal for - sprinting up hills.

I have also had the experience of riding a flexible aluminum Vitus that just seemed slow - as if the relatively poor spring qualities of narrow aluminum tubing was maybe turning energy into the work hardening that more naturally springy materials didn't suffer from. So it is not hard to see why people love stiffness if they have had similar experiences.



So my interest isn't obsessive advocacy for one design philosophy or another, but just getting cyclists to see that the problems of frame design are actually subtle and even counter-intuitive because pedaling is not nearly as simple as a 2D textbook illustration would suggest.

It sounds like enough tests have been performed to suggest that relatively average BB flex is not less efficient, and the reasonable discussion at this point is really just about the mechanism of efficiency, rather then whether it is true or not.

It is a fun and interesting conversation.

[Kontact]

The Bike Think web page referenced earlier did analyze the pedal stroke (and the flex in the frame) in 3 dimensions. This analysis showed that by far, the largest flex (and strain energy storage) is from the twisting of the frame/BB around the longitudinal axis. The other modes of flex contributed very little to the stored strain energy - and in particular, the linear deflection between axle and BB accounted for no more than a few percent of the total strain energy.

The conclusions of the analysis can easily be confirmed by eye - the twisting of the frame/BB around the longitudinal axis is large and pronounced, and can be easily seen. At the same time, any shortening of the distance between the axle and BB can not be seen by the naked eye. Why do you keep concentrating on something which is so obviously inconsequential?

That's all good. I actually wasn't trying to say "the energy is stored here", but just that the flex is not to be seen and measured in one plane from one point in that plane.

Like I said earlier, there are also two separate flex issues - where the strain energy is stored, and how its release actually feeds back into the drivetrain.

[kramnnim]

In this case, the BB doesn't rises while the pedal is already moving upward, so it doesn't directly help to rotate the crank. Instead, it helps raise the rider's foot/leg. (If you've ever stood and sprinted on a very flexy bike, you may liken it to jumping up and down on trampoline, with bike springing back up on the back of the pedal stroke.)



I get what you're saying, but the rising BB is only giving back what it gave away...and it may be giving it back when you don't need it?

[Kontact]

I get what you're saying, but the rising BB is only giving back what it gave away...and it may be giving it back when you don't need it?

If there is constant, positive force going through the drivetrain at all times that you are pedaling, when would you not benefit from it?

[kramnnim]

If there is constant, positive force going through the drivetrain at all times that you are pedaling, when would you not benefit from it?

Sigh. Take your bike outside and climb out of the saddle at 60rpm, rocking the bike with your arms. Take note of how much rotational force is going to the chain at 6 oclock.

[kramnnim]

Also, stand to the right of your bike, facing it. Drive side at 6oclock. Hold bars and saddle, push down on the drive side pedal, frame will flex. Release. Does rear wheel move forward?

I think the frame flexes in a similar way at 3oclock (or from 1 to 5) and releases around 6 oclock. And the flexing back doesn't help rotate the crank arm.

[Kontact]

Sigh. Take your bike outside and climb out of the saddle at 60rpm, rocking the bike with your arms. Take note of how much rotational force is going to the chain at 6 oclock.

Why would I do that when we have sensitive electronics showing that there is rotational force at 6 o'clock?

Sigh.

Also, stand to the right of your bike, facing it. Drive side at 6oclock. Hold bars and saddle, push down on the drive side pedal, frame will flex. Release. Does rear wheel move forward?

I think the frame flexes in a similar way at 3oclock (or from 1 to 5) and releases around 6 oclock. And the flexing back doesn't help rotate the crank arm.

Why would you flex the frame at 6 when you are claiming that 6 is the point when the flex comes out? Are you purposely trying to compare reality to something that has little to do with pedaling?

[kramnnim]

Why would I do that when we have sensitive electronics showing that there is rotational force at 6 o'clock?



Do you have said electronics ready for use? I do. I don't really have a way to record what the head unit shows, though. Maybe cyclists with better pedaling technique have some rotational force at 6oclock at low rpm out of the saddle, I don't.

[kramnnim]

Why would you flex the frame at 6 when you are claiming that 6 is the point when the flex comes out? Are you purposely trying to compare reality to something that has little to do with pedaling?

Already typed this, but I cannot come up with a way to flex the frame with the arm at 3oclock, hold the flex in place, rotate the arm down to 6 oclock, then release. Which is what I think would be happening in low rpm out of saddle pedaling.

[Mark McM]

I get what you're saying, but the rising BB is only giving back what it gave away...and it may be giving it back when you don't need it?

Also, stand to the right of your bike, facing it. Drive side at 6oclock. Hold bars and saddle, push down on the drive side pedal, frame will flex. Release. Does rear wheel move forward?

I think the frame flexes in a similar way at 3oclock (or from 1 to 5) and releases around 6 oclock. And the flexing back doesn't help rotate the crank arm.

Well of course, the frame flex does not generate energy - it can only give back what was put in.

At what point does the rider not need the energy restored by the frame? After 6 o'clock, the rider needs to bring their leg back up to the top of the pedal circle, and that takes energy. If the frame is flexed downward, than it's upward unflexing helps raise the leg, thus reducing the energy the rider needs to expend to raise their leg. The rider expends extra energy on the downstroke to flex the frame, and then gets it back on the upstroke, so the net energy loss is zero.

[cachagua]

The rider expends extra energy on the downstroke to flex the frame, and then gets it back on the upstroke, so the net energy loss is zero.

Yes, exactly as you say, your legs get the energy back. And when they do, of course that energy doesn't move the bike forward. I don't think that's what we're trying to call zero net energy loss.

Put a balloon on a scale, and then put two five pound weights on the balloon. What's the scale read? Ten pounds, or close enough. Now remove one of the weights. Does the scale read seven and a half pounds? And does the expanding balloon raise the remaining weight only as high as if it indeed weighed seven and a half?

That is roughly what you're proposing, if you say released strain energy from the frame drives the bike.

When does the frame un-flex? When you reduce the pressure on the pedals. And does the resistance at the rear wheel change, then? Nope, stays the same. So, big push at one end of the frame's flex, little push at the other end -- where is the difference in push going to do some work?

It's important to distinguish frame flex induced by drive torque from that induced by other forces, but thinking about the paths your feet take as they go around, and the sideways BB sway summed with the kinking of the stays, is needless distraction. All those things happen when you ride, but they do not affect the way the system behaves. It is perfectly accurate to regard the system of frame flex induced by drive torque as two forces squeezing a spring, or weights on a balloon. And it's painfully obvious that such a system cannot behave in violation of simple physics: the lesser force cannot overcome the greater.

[Kontact]

Yes, exactly as you say, your legs get the energy back. And when they do, of course that energy doesn't move the bike forward. I don't think that's what we're trying to call zero net energy loss.

Put a balloon on a scale, and then put two five pound weights on the balloon. What's the scale read? Ten pounds, or close enough. Now remove one of the weights. Does the scale read seven and a half pounds? And does the expanding balloon raise the remaining weight only as high as if it indeed weighed seven and a half?

That is roughly what you're proposing, if you say released strain energy from the frame drives the bike.

When does the frame un-flex? When you reduce the pressure on the pedals. And does the resistance at the rear wheel change, then? Nope, stays the same. So, big push at one end of the frame's flex, little push at the other end -- where is the difference in push going to do some work?

It's important to distinguish frame flex induced by drive torque from that induced by other forces, but thinking about the paths your feet take as they go around, and the sideways BB sway summed with the kinking of the stays, is needless distraction. All those things happen when you ride, but they do not affect the way the system behaves. It is perfectly accurate to regard the system of frame flex induced by drive torque as two forces squeezing a spring, or weights on a balloon. And it's painfully obvious that such a system cannot behave in violation of simple physics: the lesser force cannot overcome the greater.
What's the greater force?


I think both of you are missing the forest for the trees: There is no difference between "helping the motor" and "powering the drivetrain".

[cachagua]

There is no difference between "helping the motor" and "powering the drivetrain".

Perhaps so, but the energy going back into your legs (in Mark's phrase) does not help them -- it pushes against them and retards them.

[kramnnim]

In the pedaling situation that I keep mentioning, where the rider is out of the saddle, rocking the bike with his or her arms, 60rpm or so- I think the frame flexes between 2 and 5 oclock, taking torque/rotational force from the rider. The frame flexes back at 6 oclock, and I don't think it helps the crank arm rotate. Sure, the riders foot may go upwards, but it took away a fraction of a degree of rotation, and did not give it back. So I think that energy was wasted.

...I tested this for myself yesterday with a Pioneer power meter+head unit and there's no force shown at 7 oclock. At 8, it shows me pulling up.

[kramnnim]

I've also mentioned the full suspension bike with no lockout several times. The suspension compressing can easily take away downward force that could go towards rotating the cranks. When the suspension rebounds, yes, the rider's foot will move upwards, but it will not help rotate the cranks.

[Kontact]

Perhaps so, but the energy going back into your legs (in Mark's phrase) does not help them -- it pushes against them and retards them.

If you're following Mark's logic, it moves your legs in a pro-pedaling direction.

For anything to retard them the release of energy would be in an anti-pedaling direction. What mechanism are you suggesting reverses the motion of the drivetrain, and if that mechanism exists, why wouldn't it just make the chain go slack?

[Kontact]

In the pedaling situation that I keep mentioning, where the rider is out of the saddle, rocking the bike with his or her arms, 60rpm or so- I think the frame flexes between 2 and 5 oclock, taking torque/rotational force from the rider. The frame flexes back at 6 oclock, and I don't think it helps the crank arm rotate. Sure, the riders foot may go upwards, but it took away a fraction of a degree of rotation, and did not give it back. So I think that energy was wasted.

...I tested this for myself yesterday with a Pioneer power meter+head unit and there's no force shown at 7 oclock. At 8, it shows me pulling up.
"Upwards" as in pro-spin or anti-spin. If it is pro-spin, then it is putting energy back into the drivetrain.

I've also mentioned the full suspension bike with no lockout several times. The suspension compressing can easily take away downward force that could go towards rotating the cranks. When the suspension rebounds, yes, the rider's foot will move upwards, but it will not help rotate the cranks.

Rear suspension has always been known to waste energy to a degree and most systems have tried to combat that by putting the pivot along the same line as the tensioned chain. But that has nothing to do with what a rigid rear end does when it twists - completely different beasts.

[kramnnim]

"Upwards" as in pro-spin or anti-spin. If it is pro-spin, then it is putting energy back into the drivetrain.




Neither. Put the crank at 6. Press down. Frame flexes. Release. Crank doesn't rotate. Already went over this. You countered, I responded, and I don't think you replied to my response.

[kramnnim]

Rear suspension has always been known to waste energy to a degree and most systems have tried to combat that by putting the pivot along the same line as the tensioned chain. But that has nothing to do with what a rigid rear end does when it twists - completely different beasts.

Both result in downward force being wasted.

[Kontact]

Neither. Put the crank at 6. Press down. Frame flexes. Release. Crank doesn't rotate. Already went over this. You countered, I responded, and I don't think you replied to my response.

You have said two contradictory things - that flex is coming out by 6 o'clock, and that flex happens at 6 o'clock because of down force.

When I get a sense for which one you believe is the case, I will respond. But the majority of BB flex happens closer to 3 o'clock. If doesn't really matter if your right foot is pushing straight down at 6 if it doesn't cause BB flex and your left foot is getting a boost from the flex coming out.

[Mark McM]

Yes, exactly as you say, your legs get the energy back. And when they do, of course that energy doesn't move the bike forward. I don't think that's what we're trying to call zero net energy loss.

All the energy to drive the bike ultimately comes from the rider. The rider expends energy when pushing the pedals down (most of this energy goes into the drive train) - and then has to expend some more energy to raise their legs back up. And of course, these are happening at the same time - as the rider pushes down with one leg, they are raising the other leg.

If the rider is producing a constant power output (energy delivered per unit time), any energy the rider does not have to expend to raise the leg on the upstroke is available to push down on the downstroke. Which is important, since a part of that downstroke energy goes into flexing the frame. As long as the rider gets back the same amount of frame flex energy on the upstroke as they expended on the downstroke, the energy (and power) going into the drivetrain is equal to the total energy (power) expended by the rider. Thus zero net energy loss.

[Kontact]

Both result in downward force being wasted.

Sitting on the seat also "wastes" downforce. But we aren't really talking about that, but flex causing waste.

[kramnnim]

You have said two contradictory things - that flex is coming out by 6 o'clock, and that flex happens at 6 o'clock because of down force.



No. I've said this:


I think the frame flexes in a similar way at 3oclock (or from 1 to 5) and releases around 6 oclock.

[kramnnim]

As long as the rider gets back the same amount of frame flex energy on the upstroke as they expended on the downstroke, the energy (and power) going into the drivetrain is equal to the total energy (power) expended by the rider. Thus zero net energy loss.

I don't think this happens. When the frame flexes on the downstroke, the rider loses a fraction of a degree of crank rotation. When the frame flexes back, the rider does not get crank rotation back. The foot might get some lift, but if the frame wouldn't have flexed, it wouldn't need that lift.

[Kontact]

No. I've said this:

But you keep talking about this, and I don't know why:

Neither. Put the crank at 6. Press down. Frame flexes. Release. Crank doesn't rotate.

[Kontact]

I don't think this happens. When the frame flexes on the downstroke, the rider loses a fraction of a degree of crank rotation. When the frame flexes back, the rider does not get crank rotation back. The foot might get some lift, but if the frame wouldn't have flexed, it wouldn't need that lift.

How can you have a mechanism that is one way with breaking the tension in the chain? If flex loses a fraction of a degree, how can it be permanently gone?

[cachagua]

If you're following Mark's logic, it moves your legs in a pro-pedaling direction.

For anything to retard them the release of energy would be in an anti-pedaling direction. What mechanism are you suggesting reverses the motion of the drivetrain, and if that mechanism exists, why wouldn't it just make the chain go slack?

Yes, Mark seems to suggest that returned flex energy helps your legs, but I don't think that's what happens. It "goes back into your legs", as he says, but opposite the force they're trying to apply, not in the same direction.

In your terms, the release of energy is in an anti-pedaling direction. However it need not and does not reverse the motion of the drivetrain. Of course the drivetrain doesn't pause, the pedals don't go backwards, and the chain doesn't goes slack. The release of energy is in an anti-pedaling direction; the location where it's applied continues to rotate in the pedaling direction.

Here's a simple way to put it: when the flex energy is released, as you reduce the pedal force that originally induced that flex, it's very very slightly harder to pedal than if there were no flex energy in the frame -- i.e. if your frame didn't flex at all.

[kramnnim]

But you keep talking about this, and I don't know why:

To demonstrate how the frame flexing back into the relaxed state does not result in crank rotation the way it does in the original video with the crank arm fixed at 3 oclock

[kramnnim]

How can you have a mechanism that is one way with breaking the tension in the chain? If flex loses a fraction of a degree, how can it be permanently gone?

It's not gone, it's just returned in a way that is not helpful to the rider.

If you think of the rider's legs to be like pistons...and the crank arm is the crankshaft. I think frame flex is comparable to connecting rods that get shorter in length during the compression stroke and return to original length on the upstroke.

[kramnnim]

The flexy frame would also be like rowing a boat with flexy oars... Part of your arm stroke would be used up bending the oar, and the oar would bend back straight once you lifted it out of the water...and would not help to propel the boat.

[Kontact]

To demonstrate how the frame flexing back into the relaxed state does not result in crank rotation the way it does in the original video with the crank arm fixed at 3 oclock

That isn't what it demonstrates. It demonstrates that you can flex a frame lots of different ways that have nothing to do with pedaling. You could flex the frame by laying it down and stepping on the seat tube - that also isn't like pedaling.

The flexy frame would also be like rowing a boat with flexy oars... Part of your arm stroke would be used up bending the oar, and the oar would bend back straight once you lifted it out of the water...and would not help to propel the boat.

Except with pedaling we don't take the oar out of the water. With normal, vigorous pedaling the net output from the crank never goes to zero. If it did the freewheel would disengage, and we would hear it clunking as we reapplied power on every new downstroke. It is like spoke tension, it varies but never goes to zero.

[kramnnim]

That isn't what it demonstrates. It demonstrates that you can flex a frame lots of different ways that have nothing to do with pedaling. You could flex the frame by laying it down and stepping on the seat tube - that also isn't like pedaling.



My position has been that energy is wasted with a flexy frame when pedaling out of the saddle, low rpm (60ish) and rocking the bike back and forth with your arms. Like...a less graceful version of Contador.

Based on what I saw on my Pioneer head unit while pedaling as described, my pedaling force is all straight down, from 2 oclock to 6. At 7 it is zero, and 8 is slight upward pressure. (meanwhile the other leg is pushing straight down) Which is why I disagree with you here, I think the flex at 6 oclock while pedaling as described is very similar to the flex of just pushing down while the bike is stationary.

[kramnnim]

Except with pedaling we don't take the oar out of the water. With normal, vigorous pedaling the net output from the crank never goes to zero. If it did the freewheel would disengage, and we would hear it clunking as we reapplied power on every new downstroke. It is like spoke tension, it varies but never goes to zero.

Based on what I see on the Pioneer head unit, I think the rotational pressure at 12/6oclock is pretty much zero when pedaling in the manner I have repeatedly described.

[cachagua]

You can flex a frame lots of different ways that have nothing to do with pedaling. You could flex the frame by laying it down and stepping on the seat tube - that also isn't like pedaling.

*like!*

Two kinds of frame flex: A) the kind that's a response to drive torque, and stores energy taken from drive torque, and B) any other kind.

When B is released back out of the frame, it cannot drive the bike forward, because the source of its energy wasn't going towards that to begin with. When A is released, it cannot drive the bike forward, it can only make your legs work harder.

[kramnnim]

When A is released, it cannot drive the bike forward, it can only make your legs work harder.

Right, because it is actively pushing back against the drive torque. Like if you had a compression spring between the pedal and the cleat.

[Mark McM]

My position has been that energy is wasted with a flexy frame when pedaling out of the saddle, low rpm (60ish) and rocking the bike back and forth with your arms. Like...a less graceful version of Contador.

Based on what I saw on my Pioneer head unit while pedaling as described, my pedaling force is all straight down, from 2 oclock to 6. At 7 it is zero, and 8 is slight upward pressure. (meanwhile the other leg is pushing straight down) Which is why I disagree with you here, I think the flex at 6 oclock while pedaling as described is very similar to the flex of just pushing down while the bike is stationary.

If, when out of the saddle, the maximum downward leg force on the pedal is at 6 o'clock, then the maximum frame deflection, the maximum 'spring force' of the frame, and maximum strain energy in the frame, occurs at this point. If the downward leg force on the pedal decreases after this, then the larger 'spring force' of the frame acts to raise the leg, and the frame un-flexes. Since the frame is exerting a force (the frame 'spring force'), over a distance (the amount it un-flexes), then energy in the frame is being returned, and this energy is used to help raise the leg.

As long as the downward leg force decreases (which it must if the leg is to be raised), the frame must un-flex upward. The frame also has its 'spring force' that is exerted upward while un-flexing. Since energy is a force exerted over a distance (E = F x d), it must be transferring energy to the pedal as it un-flexes upward. Of do you think the Laws of Thermodynamics are wrong, and energy just disappears?

[kramnnim]

Not saying the energy disappears, I don't think it performs anything very useful. It takes away downward force and at best, helps to lift your leg a small amount. If the frame would not have flexed to start with, your leg would not need to be lifted that extra small amount.

[Kontact]

Not saying the energy disappears, I don't think it performs anything very useful. It takes away downward force and at best, helps to lift your leg a small amount. If the frame would not have flexed to start with, your leg would not need to be lifted that extra small amount.

There is little point in talking about this if "lift your leg" is "not useful". That sort of 'logic' is going to make any sort of engineering explanation pointless. You simply don't believe it is possible and there is no point arguing with faith.

[kramnnim]

Yeah, it's quite clear after how many pages of the same thing going around in circles that we aren't getting anywhere.

[Mark McM]

Not saying the energy disappears, I don't think it performs anything very useful. It takes away downward force and at best, helps to lift your leg a small amount. If the frame would not have flexed to start with, your leg would not need to be lifted that extra small amount.

I'm not sure what you mean by "takes away downward force". The downward leg force during the downstroke is the same, whether the frame flexes or not. In the case the frame flexes, the pedal travels downward further, but under the same force. The extra downward travel requires extra energy from the leg, which is transferred into the frame as strain energy. When the downward force from the leg decreases, the pedal (and/or BB) travels upward as the frame unflexes, and the energy is returned. If the leg down force is decreased before 6 o'clock, then the unflexing raises the BB more than the pedal, thus helping to rotate the crank, and some of the strain energy goes directly into the drivetrain. After 6 o'clock, the returned strain energy can't act to rotate the crank forward, but it does help lift the leg.

[kramnnim]

It my mind, it's like if you were trying to lift an object with a class 1 lever. You can only move the lever down so far before you hit the ground/floor. If the lever flexes, your downward force is wasted.

I don't think the upward lift at 6 oclock is useful because it is only giving back what it gave away. So you lost some downward force when you needed it, and don't get anything back.

[Mark McM]

It my mind, it's like if you were trying to lift an object with a class 1 lever. You can only move the lever down so far before you hit the ground/floor. If the lever flexes, your downward force is wasted.

I don't think the upward lift at 6 oclock is useful because it is only giving back what it gave away. So you lost some downward force when you needed it, and don't get anything back.

Does the pedal hit ground when you step on it? No. If the lever doesn't hit the ground, does it transfer the same force, regardless of whether the lever flexes or not? Yes it does.

I think you need to go back and review the concepts of Force and Energy (Work).

[cachagua]

When the downward force from the leg decreases, the pedal (and/or BB) travels upward as the frame unflexes, and the energy is returned... The returned strain energy can't act to rotate the crank forward, but it does help lift the leg.


Lifts your leg, but can't rotate the crank forward. Exactly right.

[kramnnim]

Does the pedal hit ground when you step on it? No. If the lever doesn't hit the ground, does it transfer the same force, regardless of whether the lever flexes or not? Yes it does.

I think you need to go back and review the concepts of Force and Energy (Work).

At 6 oclock, the downward force on the pedal can no longer provide rotational force...and additional downward force just flexes the frame, which doesn't benefit the rider.

[Kontact]

At 6 oclock, the downward force on the pedal can no longer provide rotational force...and additional downward force just flexes the frame, which doesn't benefit the rider.

Nor does it actually happen. So why keep bringing it up as if it does? At 6 o'clock rainbows don't appear, either.