Frame Flex on GCN

[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]

Originally Posted by kramnnim

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.

Originally Posted by 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.

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]

Originally Posted by Kontact

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]

Originally Posted by Kontact

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]

Originally Posted by kramnnim

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?

Originally Posted by 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.

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]

Originally Posted by cachagua

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]

Originally Posted by cachagua

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]

Originally Posted by 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.

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

Originally Posted by 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.

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]

Originally Posted by Kontact

"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]

Originally Posted by Kontact

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]

Originally Posted by kramnnim

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.