Frame Flex on GCN

[Kontact]

Originally Posted by kramnnim

Tension from pressure on the other crank arm at 1oclock...(which is playing catch up because the BB is flexing away from it...)

Pedal movement in the direction of pedaling is not crank rotation if the fulcrum point is also moving in the same direction at the same speed.

It isn't moving at the same speed. The crank arm is much further away from the axis point than the BB. Is a hub flange moving at the same speed as the rim?

[kramnnim]

Originally Posted by Kontact

It isn't moving at the same speed. The crank arm is much further away from the axis point than the BB. Is a hub flange moving at the same speed as the rim?

It's still not rotating.

[kramnnim]

Here's a video that shows a flexy frame flexing when pedaling seated.

https://www.youtube.com/watch?v=bDb4oTkuaMw

And an exaggerated and very short clip...

https://www.youtube.com/watch?v=x2QuWy1PL8s

[Kontact]

Originally Posted by kramnnim

It's still not rotating.

It is moving the same speed!

Nope.

Oh, okay, I'll insist something different.

[cachagua]

Here's my conception of what happens to the energy in frame flex.

First, let's restrict our attention to drive torque and the frame flex it produces. When we take this step, there are just a handful of elements in the bike/rider/road system, and it can be thought of the same as a spring with something squeezing it at each end. And besides simplicity, the reason we leave out flex caused by non-drive-torque forces is that, even if there were no frame flex, those forces weren't going to move you forward. Hence the name.

The elements we have to account for, after making the above restriction, are the drive torque the rider's applying, the frame flex it induces, and the bike's and rider's resistance to motion. This might include inertia, wind drag, rolling resistance, going uphill -- basically, everything you have to overcome to ride.

Each pedal stroke, the force of your drive torque fluctuates between a maximum, and some amount less than its maximum. Let's call your maximum force X. At your maximum, wherever that happens around the pedals' revolution, you're pushing X, and per Newton's 3rd Law, the resistance is pushing X against you. The two forces are in equilibrium.

But remember that the frame is between you and the road, where the resistance is being applied against your force. You're actually pushing X against the frame, and the resistance is pushing X against it at the other end. The frame is squeezed between the two forces, and it flexes a little, putting some of the force that's squeezing it into strain energy.

And to complete our account of what's happening here, we have to say you're pushing X against the frame, and, 3rd Law again, the frame is pushing X back against you -- and likewise, the frame is pressing X against the resistance where the resistance is pressing X back. All in equilibrium, like this:

Road ---X---><---X---frame---X---><---X---legs

Kind of crude, but you get the idea. So let's pedal a little way around to where you're not able to apply your maximum force, X, to the pedals, but you're applying somewhat less. Let's call this force y. Here's what this does to our diagram:

Road ---X---><---X---frame---X---><---y---legs

Now there's less force on the frame, where your legs are pressing against it, and in response the frame will un-flex a little. Some of its strain energy will be released, and will be available to do some work. The forces between the other elements will drop too, so that everything's in equilibrium, but the resistance to moving your bike up the road hasn't changed meaningfully -- it's still at the same level it where it's been all along. So now the released strain energy has to do some work either against a greater force, X, the bike's resistance -- or against a lesser force, y, your legs. Inescapably it will do some work where your legs are, because the force balancing it there has decreased.

The effect of the released strain energy's work is that the pedals are retarded a little. They don't go backwards, they don't stop, but they rotate a little less than where they would have been, if the strain energy were still in the frame. You don't feel this as you're riding, because it's a very small amount, and because it happens as smoothly and gradually as you reduce your pedaling force. All you know is, it's hard to push the pedals... and there's no surprise in that! But ultimately what the strain energy in the frame flex has done is to move your legs back, and not to move the bike forward.

This line of thinking indicates to me that frame flex, as we've defined it here, cannot be returned as forward motion, and is, for purposes of getting somewhere, wasted.

[Kontact]

Originally Posted by cachagua

Here's my conception of what happens to the energy in frame flex.

First, let's restrict our attention to drive torque and the frame flex it produces. When we take this step, there are just a handful of elements in the bike/rider/road system, and it can be thought of the same as a spring with something squeezing it at each end. And besides simplicity, the reason we leave out flex caused by non-drive-torque forces is that, even if there were no frame flex, those forces weren't going to move you forward. Hence the name.

The elements we have to account for, after making the above restriction, are the drive torque the rider's applying, the frame flex it induces, and the bike's and rider's resistance to motion. This might include inertia, wind drag, rolling resistance, going uphill -- basically, everything you have to overcome to ride.

Each pedal stroke, the force of your drive torque fluctuates between a maximum, and some amount less than its maximum. Let's call your maximum force X. At your maximum, wherever that happens around the pedals' revolution, you're pushing X, and per Newton's 3rd Law, the resistance is pushing X against you. The two forces are in equilibrium.

But remember that the frame is between you and the road, where the resistance is being applied against your force. You're actually pushing X against the frame, and the resistance is pushing X against it at the other end. The frame is squeezed between the two forces, and it flexes a little, putting some of the force that's squeezing it into strain energy.

And to complete our account of what's happening here, we have to say you're pushing X against the frame, and, 3rd Law again, the frame is pushing X back against you -- and likewise, the frame is pressing X against the resistance where the resistance is pressing X back. All in equilibrium, like this:

Road ---X---><---X---frame---X---><---X---legs

Kind of crude, but you get the idea. So let's pedal a little way around to where you're not able to apply your maximum force, X, to the pedals, but you're applying somewhat less. Let's call this force y. Here's what this does to our diagram:

Road ---X---><---X---frame---X---><---y---legs

Now there's less force on the frame, where your legs are pressing against it, and in response the frame will un-flex a little. Some of its strain energy will be released, and will be available to do some work. The forces between the other elements will drop too, so that everything's in equilibrium, but the resistance to moving your bike up the road hasn't changed meaningfully -- it's still at the same level it where it's been all along. So now the released strain energy has to do some work either against a greater force, X, the bike's resistance -- or against a lesser force, y, your legs. Inescapably it will do some work where your legs are, because the force balancing it there has decreased.

The effect of the released strain energy's work is that the pedals are retarded a little. They don't go backwards, they don't stop, but they rotate a little less than where they would have been, if the strain energy were still in the frame. You don't feel this as you're riding, because it's a very small amount, and because it happens as smoothly and gradually as you reduce your pedaling force. All you know is, it's hard to push the pedals... and there's no surprise in that! But ultimately what the strain energy in the frame flex has done is to move your legs back, and not to move the bike forward.

This line of thinking indicates to me that frame flex, as we've defined it here, cannot be returned as forward motion, and is, for purposes of getting somewhere, wasted.

I don't get why you would think that if you put a spring between you and the work you are doing, the spring is going to have the power to take something away from your output.

Consider this thing:



You have two versions, one made of springy stuff, one made of incredibly stiff stuff.

You throw the ball with the stiff one. Your effort accelerates the ball and it goes flying off.

You use the springy one. Your peak effort partially flexes the shaft, then your arm goes past peak effort and the arm springs forward, then the ball is released.


In the springy example, at what point did the unflexing spring feed back into your arm and take away from your throw?

Did the ball actually go further with the stiff thrower? Or did your arm perform the same work and get the same result because a tiny unflexing of the arm isn't going to override the massive power of the arm that is powering it?




There is never an orientation of the drivetrain components on a bike that make the linear relationship of leg to crank to frame to wheel change to where the frame is controlling the leg instead of the other way around.

[kramnnim]

Originally Posted by Kontact

It is moving the same speed!

Nope.

Oh, okay, I'll insist something different.

As you can see in the video, the BB shell is flexing left/right, not fore/aft.

Since the crank arm is a lever, the pedal does move faster+farther left/right.

This left/right movement does not help to rotate the crank arms.

[kramnnim]

Originally Posted by Kontact

I don't get why you would think that if you put a spring between you and the work you are doing, the spring is going to have the power to take something away from your output.

Consider this thing:



You have two versions, one made of springy stuff, one made of incredibly stiff stuff.

You throw the ball with the stiff one. Your effort accelerates the ball and it goes flying off.

You use the springy one. Your peak effort partially flexes the shaft, then your arm goes past peak effort and the arm springs forward, then the ball is released.


In the springy example, at what point did the unflexing spring feed back into your arm and take away from your throw?

Did the ball actually go further with the stiff thrower? Or did your arm perform the same work and get the same result because a tiny unflexing of the arm isn't going to override the massive power of the arm that is powering it?




There is never an orientation of the drivetrain components on a bike that make the linear relationship of leg to crank to frame to wheel change to where the frame is controlling the leg instead of the other way around.

Nice theory, but the flex is lateral. Congrats to the flexy frame for flinging your heels out away from the bike, it's super helpful.

[Cloozoe]

Surely there is an actual physicist somewhere on this board that can un-obfuscate this a bit. Although I remember years ago various supposedly actual physicists were claiming a thrown baseball couldn’t curve, that it was an optical illusion. Which would be true on the moon, but here on earth with air resistance...

So maybe even a real physicist wouldn’t help.

[Kontact]

Originally Posted by Cloozoe

Surely there is an actual physicist somewhere on this board that can un-obfuscate this a bit. Although I remember years ago various supposedly actual physicists were claiming a thrown baseball couldn’t curve, that it was an optical illusion. Which would be true on the moon, but here on earth with air resistance...

So maybe even a real physicist wouldn’t help.

People hear and elsewhere have offered explanations that are physics and match the power numbers in tests.

But some other posters "think" that tests and physics don't make sense. They won't make sense when explained by a real physicist, either.

[kramnnim]

Originally Posted by Kontact

People hear and elsewhere have offered explanations that are physics and match the power numbers in tests.

But some other posters "think" that tests and physics don't make sense. They won't make sense when explained by a real physicist, either.

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

[Kontact]

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.

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.

[cachagua]

Originally Posted by Kontact

There is never an orientation of the drivetrain components on a bike that make [sic] the linear relationship of leg to crank to frame to wheel change to where the frame is controlling the leg instead of the other way around.

H'mm. Let me see if I can tell where we're getting hung up. Your legs apply some force to the frame, that force is resisted by the bike's inertia -- the opposing pair of forces squeezes the frame, it flexes a little, and in doing so absorbs some energy -- are we good so far?

[superbowlpats]

Wow 21 pages and growing.

I don't know Mark McM but I've read enough on this forum to know he's always right

[Kontact]

Originally Posted by cachagua

H'mm. Let me see if I can tell where we're getting hung up. Your legs apply some force to the frame, that force is resisted by the bike's inertia -- the opposing pair of forces squeezes the frame, it flexes a little, and in doing so absorbs some energy -- are we good so far?

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