Frame Flex on GCN

[Keith A]

Okay guys and gals, let's play nice

[kramnnim]

Originally Posted by Kontact

I have. Your responses always indicate that you either don't understand or just want to argue, because any time your specific question is answered you change your question.

I'm still waiting for you to explain how lateral flex results in crank rotation.

[andrewsuzuki]

Originally Posted by kramnnim

Do feel free to (re)link to these tests, they've gotten buried. Maybe they can explain how lateral flex results in crank rotation.

Linked this a few times in the first few pages, here they are again.

https://web.archive.org/web/20060214.../Frameflex.htm

Esp. check out all the links near the bottom of the page under "FEA Model Quantifying Strain Energy" right before the Conclusion heading.

[Mark McM]

Originally Posted by kramnnim

I'm still waiting for you to explain how lateral flex results in crank rotation.

This is just basic physics.

Lateral flex doesn't occur in isolation. The only way a vertical force can result in a lateral flex, is if the lateral flex is a component of the total flex. For example, if you push down on the right pedal, the torsional flex of the frame will cause the pedal to deflect down and to the left. This means that when the frame un-flexes in torsion, it will un-flex back to the right and UP.

Also note that since the flexing force is vertical, the lateral deflection by itself neither stores nor releases energy - energy is force times distance, and since the vertical force and lateral deflection are orthogonal, there is no force in the direction of the lateral displacement, and therefore no energy exerted in the lateral direction. Instead, the energy is stored and released by the flex that directly resulted from the vertical force - in the case of the BB, it is the torsional flex of the frame. The lateral displacement is only an artifact of the torsion, and therefore neither absorbs nor releases energy.

Here's a simple example of this: Raising a weight increases the potential energy of the weight, equal to the gravitation force times the distance the weight is raised. But what if I raise a ball by rolling up an angled ramp, does the ball store energy due to the lateral movement? No. Since the gravitational force on the ball is vertical, the energy required to roll the ball up the ramp (and therefore the potential energy stored in the ball) is still proportional to only the vertical distance traveled.

[Kontact]

Originally Posted by kramnnim

I'm still waiting for you to explain how lateral flex results in crank rotation.

And I've explained it with vectors and several other ways. So you either are being argumentative, or you simply aren't able to understand.

The result is the same, so you and I are done.

[kramnnim]

Originally Posted by andrewsuzuki

Linked this a few times in the first few pages, here they are again.

https://web.archive.org/web/20060214.../Frameflex.htm

Esp. check out all the links near the bottom of the page under "FEA Model Quantifying Strain Energy" right before the Conclusion heading.

This basically shows what the GCN video shows. It only works when the test is done with the arms horizontal. We've all agreed that the force that is holding the spring energy in the flexed frame is released at around 7 oclock.

[kramnnim]

Originally Posted by Mark McM

This is just basic physics.

Lateral flex doesn't occur in isolation. The only way a vertical force can result in a lateral flex, is if the lateral flex is a component of the total flex. For example, if you push down on the right pedal, the torsional flex of the frame will cause the pedal to deflect down and to the left. This means that when the frame un-flexes in torsion, it will un-flex back to the right and UP.

Also note that since the flexing force is vertical, the lateral deflection by itself neither stores nor releases energy - energy is force times distance, and since the vertical force and lateral deflection are orthogonal, there is no force in the direction of the lateral displacement, and therefore no energy exerted in the lateral direction. Instead, the energy is stored and released by the flex that directly resulted from the vertical force - in the case of the BB, it is the torsional flex of the frame. The lateral displacement is only an artifact of the torsion, and therefore neither absorbs nor releases energy.

Here's a simple example of this: Raising a weight increases the potential energy of the weight, equal to the gravitation force times the distance the weight is raised. But what if I raise a ball by rolling up an angled ramp, does the ball store energy due to the lateral movement? No. Since the gravitational force on the ball is vertical, the energy required to roll the ball up the ramp (and therefore the potential energy stored in the ball) is still proportional to only the vertical distance traveled.

I understand that the BB/crank/pedal moves up and down as it moves laterally. But the crank rotation is perpendicular to this motion.

When the BB moves up at 7oclock, the other crank arm is at 1, and that side of the BB is moving down. That leg is also pushing down. The fulcrum point is moving down, which is the opposite of what you want...

Think about how you rock the bike back and forth when you ride out of the saddle. You rock the bike to the left when you push down with your right leg. Why? The rocking motion lifts the fulcrum (the BB) upwards against the downward force of your leg. It's not helpful for the BB to move in the direction of your pedal because you ideally trying to rotate the crank arm around it.

[cachagua]

Originally Posted by Kontact

Substitute "store" for "absorb" and it is a go.

Sure, "store". Because we're going to get it back out of the frame later. So, your legs push on the frame, the resistance is pushing at the other end, and the frame, flexed and carrying some stored energy, is pushing back, equally hard, against your legs and against the road resistance -- that part okay? That would bring us about as far as my first diagram.

[kramnnim]

Originally Posted by Kontact

And I've explained it with vectors and several other ways. So you either are being argumentative, or you simply aren't able to understand.

The result is the same, so you and I are done.

Your explanations have been incorrect.

Bye!

[mtechnica]

Can we begin by assuming that if a flexible frame is less efficient, and the tension of the chain is the only thing between the bottom bracket and the power going into the wheel, then it stands to reason that there would be less tension going through the chain in a flexy frame?

But since the bottom bracket sways perpendicularly to the chainring, the distance between the chainring and the wheel will not decrease when the bottom bracket flexes.

Therefore I believe that it should be possible to attach a scale either to a chain or a wheel, and put a weight on the pedal, then see if there is more or less force going into the chain depending on the frame.

In my opinion, I think the force will be the same, but since the frame will eventually reach a point where it stops bending from a given force, it might just take slightly longer for the full force to be transmitted to the wheel.

Which raises an interesting question, is that delay a source of inefficiency since power is your force being applied over time, or does it only make a difference in the way it feels to the rider?

[andrewsuzuki]

Originally Posted by kramnnim

This basically shows what the GCN video shows. It only works when the test is done with the arms horizontal. We've all agreed that the force that is holding the spring energy in the flexed frame is released at around 7 oclock.

This graph suggests the frame is at equilibrium at the 7 o'clock position (~210deg), and the majority is returned between 135deg and 180deg.

This is all using real rider power curves, "an average of 17 road riders at 350W and 90rpm"

https://web.archive.org/web/20060214...edal_loads.htm

[Kontact]

Originally Posted by mtechnica

Can we begin by assuming that if a flexible frame is less efficient, and the tension of the chain is the only thing between the bottom bracket and the power going into the wheel, then it stands to reason that there would be less tension going through the chain in a flexy frame?

But since the bottom bracket sways perpendicularly to the chainring, the distance between the chainring and the wheel will not decrease when the bottom bracket flexes.

Therefore I believe that it should be possible to attach a scale either to a chain or a wheel, and put a weight on the pedal, then see if there is more or less force going into the chain depending on the frame.

In my opinion, I think the force will be the same, but since the frame will eventually reach a point where it stops bending from a given force, it might just take slightly longer for the full force to be transmitted to the wheel.

Which raises an interesting question, is that delay a source of inefficiency since power is your force being applied over time, or does it only make a difference in the way it feels to the rider?

Actually, it doesn't work that way. Flexible frames flex until the tension on the chain is the same as a stiff frame. That happens because the amount of force that it takes to flex the frame increases with the distance it is flexed. At full flex a flexy frame is "stiff".

And the chainwheel and wheel do get closer. The BB can't swing off to one side without the chainstays following it, and being a fixed length the chainstays have to pull on the dropouts to follow the BB.

A lot of what people are struggling to understand is because we tend to think of these movements in isolation. But the reality is that the frame is like a bow and the chain is a bow string.


So I don't think your experiment makes sense given the reality of how the bike actually functions.

[mtechnica]

If the chain stay is a fixed length and the bottom bracket shell moves left and right, wouldn’t that mean that the distance between it and the hub actually increases under load since it would be the hypotenuse of the triangle it would make? The distance can never be less than in the unoaded state correct? So long as the chain stay isn’t under compression.

[mtechnica]

At any rate shouldn’t energy loss be accountable with a hub based power meter?

[cachagua]

Originally Posted by Kontact

The chainwheel and wheel do get closer... The frame is like a bow and the chain is a bow string.

Fantastic, fantastic image! That's one of the strongest analogies we've come across. It's a good way to see the distinction I was trying to make between frame flex caused by drive force, vs. flex caused by forces that don't move the bike.