Simple puzzle: bicycle motion

[DaveC426913]

TL;DR

Most "puzzles on SocMed are dumb. This one was a least a tiny bit challenging.

Clarification: Assume the bicycle is supported from falling over, but not prevented from moving forward or backward.

 

[anuttarasammyak]

Forward. I am poor at finding the puzzle here.

 

[renormalize]

TL;DR: Most "puzzles on SocMed are dumb. This one was a least a tiny bit challenging.

SPOILER!
The direction of motion depends on the both the angle of pull and the gear ratio:

 

[DaveC426913]

At first, I thought it was simple, but one commenter pointed out a very non-intuitive effect.

I had to go do the experiment myself with my bike to be sure.

Spoiler

Yes, the bike moves backward.
Its weirder than that though. The bike moving backward engages the gears and chain, causing the pedal to rotate counterclockwise, even as it is moving backward (leftward).

I had to make a little diagram to explain it to commenters:

The pedal moves backward with the bike (blue line and arrow) even as it rotates CCW (red arrows).

Pretty cool.

 

[DaveC426913]

The direction of motion depends on the both the angle of pull

The angle of pull is optimally straight back. That's in the instructions, so not optional.

and the gear ratio:

It worked fine in my experiment at the low gear it was in. I cannot say for certain that there is real-world gear ratio that would change the result, but I am open to the possibility that it exists.

 

[Halc]

Backwards, in any gear available.

 

[Ken Fabian]

But this seems to occur because there is some looseness to take up before the drive chain fully engages.

Start with none of that looseness - with the drive chain pre-loaded - it should move forward. And in a real situation - unless you are one legged - there would be forward force from the opposite pedal, cancelling that effect. No physics defying effect is in play,

 

[renormalize]

But this seems to occur because there is some looseness to take up before the drive chain fully engages.
Start with none of that looseness - with the drive chain pre-loaded - it should move forward.

That claim disagrees with the experiment shown in post #3. Did you watch the video?

 

[A.T.]

But this seems to occur because there is some looseness to take up before the drive chain fully engages.

Start with none of that looseness - with the drive chain pre-loaded - it should move forward.

No, it has nothing to do with chain slack. It moves backwards without slipping even after the chain is tensioned.

It's the same principle as with the spool pulled by the string from underneath.



Interestingly it moves in the same direction as the string pulling it but faster than the string. So it's also the same principle that allows you to move directly downwind faster than the wind that is propelling you.

 

[A.T.]

I cannot say for certain that there is real-world gear ratio that would change the result,

Have you watched the video in post #3? When you start swapping sprockets between wheel and pedals, you can get there. But it won't happen for most regular bikes.

 

[wrobel]

Let us write down the equations. Assume that there is no slipping between the bicycle wheels and the floor. Then this is a system with ideal constraints and a single degree of freedom.

Let us fix a point $A$ on the bicycle frame and let $x$ be the coordinate of this point on the horizontal axis. This is the generalized coordinate of our system. We consider the axis to be directed from left to right. Let $\boldsymbol F$ denote the force acting on the pedal. This force is directed left horizontally. The corresponding generalized force is

$$Q=\left(\frac{\partial \boldsymbol {r}_{pedal }}{\partial x},\boldsymbol F\right).$$

And this is where the most interesting part lies. The answer to the problem depends on the sign of the function $Q$ and is determined solely by the kinematics of the system. Indeed, let us write Lagrange's equation:
$$\frac{d}{dt}\frac{\partial T}{\partial \dot x}-\frac{\partial T}{\partial x}=Q,\quad\dot x(0)=0.$$
Here, the kinetic energy of the system has the form
$$T=\frac{1}{2}a(x)\dot x^2,\quad a>0.$$
Lagrange's equation takes the form
$$\ddot x=\frac{1}{a}\left(Q-\frac{1}{2}a'\dot x^2\right).$$
Thus, the direction of motion of the bicycle at the initial moment of time is entirely determined by the sign of the bracketed term on the right-hand side of the equation, and this sign is determined by the sign of $Q$.

 

[wrobel]

To summarize the analysis, the solution to this problem can be stated as follows:

If rolling the bicycle slightly to the right causes the pedal to displace rightward from its initial position, applying a leftward pull via the string (as shown in the video) will move the bicycle to the left.

Conversely, if rolling the bicycle to the right causes the pedal to shift to the left relative to its starting point, pulling the pedal leftward will result in rightward motion of the the bicycle.

 

[A.T.]

The answer to the problem depends on the sign of the function Q and is determined solely by the kinematics of the system.

Yes, without slipping it becomes a purely kinematic/geometric problem. It can be generalized with the formula in the figure below which depends only on the gear ratio A. If 0<A<1 then V/W > 1, so the vehicle moves in the same direction as the propelling object (blue) but faster.

A regular bicycle is like the 3rd variant shown below, if you make t (back wheel radius) > T (pedal lever arm).

 

[DaveC426913]

Conversely, if rolling the bicycle to the right causes the pedal to shift to the left relative to its starting point, pulling the pedal leftward will result in rightward motion of the the bicycle.

Except: no.

Pulling the pedal to the left causes the bicycle to move left, even as the pedal rotates CCW.

 

[DaveC426913]

so the vehicle moves in the same direction as the propelling object (blue) but faster.

So ... pedal goes left, bike goes left?

 

[A.T.]

Except: no.

Pulling the pedal to the left causes the bicycle to move left, even as the pedal rotates CCW.

You should read the statement by @wrobel that you quoted more carefully, in particular the if-part, which doesn't apply to regular bikes.

And you should watch the video in post #3 where both cases described by @wrobel are demonstrated.

Case 1: Regular bike (A < 1)

If rolling the bicycle slightly to the right causes the pedal to displace rightward from its initial position, applying a leftward pull via the string (as shown in the video) will move the bicycle to the left.

Case 2: Custom bike with an ultra low gear (A > 1)

Conversely, if rolling the bicycle to the right causes the pedal to shift to the left relative to its starting point, pulling the pedal leftward will result in rightward motion of the the bicycle.

 

[A.T.]

So ... pedal goes left, bike goes left?

If A<1 then pulling the string to the left will move the bike to the left. The pedal then also moves to the left, relative to the ground.

Additionally, if 0<A<1 the bike will move faster than the string is moving.

 

[wrobel]

Except: no.

Forgive me, but I have this silly habit of trusting equations slightly more than your intuition.

 

[Swamp Thing]

Another example of analogous devices embodying related principles, like the the ones posted above...

 

[A.T.]

Another example of analogous devices embodying related principles, like the the ones posted above...

Yes, that is a practical application of the ultra low gear demonstrated by the custom made bike in the post #3 video.

Here is another one:
https://en.wikipedia.org/wiki/Ice_jigger

 

[DaveC426913]

Forgive me, but I have this silly habit of trusting equations slightly more than your intuition.

Intution? I have the habit of trusting empirical experiment over equations. As I said, I tried it.

My point of contention isn't whether or not you go it wrong, it's that, with all the equations, you didn't ultimately appear to state a clear conclusion. So I looked for what seemed to be a conclusion. Ultimately, you're probably I assume you are right about the general case; it's just very hard to tell.

mea culpa

You should read the statement by @wrobel that you quoted more carefully, in particular the if-part, which doesn't apply to regular bikes.

As above. I was reacting to the wall of equations, apparently without a clear conclusion.

I probably could have expressed that more diplomatically.

 

[A.T.]

I was reacting to the wall of equations, apparently without a clear conclusion.

The post that you were quoting (#12) is a clear conclusion. But you only quoted the 2nd half of it (which doesn't apply to most bikes), while ignoring the 1st half (which does apply to most bikes).

 

[bob012345]

What is the difference between pulling the bottom pedal to the left and pushing the top pedal to the right? They are connected. It’s been awhile since I was twelve but I definitely remember when I pushed the top pedal to the right as seen in the diagram above, the bike and I moved forward.

 

[DaveC426913]

What is the difference between pulling the bottom pedal to the left and pushing the top pedal to the right? They are connected. It’s been awhile since I was twelve but I definitely remember when I pushed the top pedal to the right as seen in the diagram above, the bike and I moved forward.

Clarification: You are standing on the ground beside the bike, not sitting on it.

Pushing the top/port pedal to the right is not the same as pushing the bottom/starboard pedal to the left. Your frame of reference is the ground, not the bike. So a push to the left makes the object move to the left (just as if you were pushing on any solid part of the bike).

As you push the bottom/starboard pedal to the left, the pedals will actually rotate counterclockwise. (because the rear wheel is going in reverse (counterclockwise), which engages the chain, which engages the pedals) i.e. the bottom/starboard pedal you are holding will move left even as it rotates rightward-and-upward.

Weird, ain't it?

 

[A.T.]

What is the difference between pulling the bottom pedal to the left and pushing the top pedal to the right?

If you sit on the bike, there is no difference. But if you stand on the ground, there can be a difference in the bike's motion direction. Think about what the external vs internal forces are, what directions they have, and if/how their magnitudes are related.

 

[A.T.]

Pushing the top/port pedal to the right is not the same as pushing the bottom/starboard pedal to the left. Your frame of reference is the ground, not the bike. So a push to the left makes the object move to the left (just as if you were pushing on any solid part of the bike).

It's not about "frame of reference", but what the external forces on the moving system are.

And it's not "just like pushing on any solid part of the bike".

When standing on the ground:
- applying a backwards force to the bike frame, will always move the bike backwards
- applying a backwards force to the bottom pedal, will move the bike in a direction determined by the gearing

Just because both scenarios above can result in backwards motion of the bike, doesn't mean they are mechanically the same:
- When pulling back the bike frame, the bike will move at the same speed your hand moves.
- When pulling back the bottom pedal, the bike will move at a speed determined by the gearing, which can be smaller than, equal to, or greater than the speed of the pulling hand.

 

[bob012345]

I’ve seen a lot of complicated word explanations but so far no free body diagrams!

 

[A.T.]

I’ve seen a lot of complicated word explanations but so far no free body diagrams!

Mainly because:

without slipping it becomes a purely kinematic/geometric problem.

So you don't really need to consider forces, if the question is purely about kinematics.

For kinematic diagrams see post #13.

 

[wrobel]

I’ve seen a lot of complicated word explanations but so far no free body diagrams!

Of course. That's because a bicycle is a system of several rigid bodies interacting in a rather complex manner. If all problems could be easily solved the way you expect, there would be no need to devise the Lagrangian formalism
and other advanced topics


.

 

[bob012345]

Of course. That's because a bicycle is a system of several rigid bodies interacting in a rather complex manner. If all problems could be easily solved the way you expect, there would be no need to devise the Lagrangian formalism
and other advanced topics.

Are you saying a free body diagram would not help for this problem?