Rotation of bike riding off cliff

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  • #1
steveZ
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When a cyclist rides off a ‘drop’ (an abrupt step in topography, ranging from a curb to a cliff), the front wheel starts falling before the back wheel, so that by the time the back wheel comes off the drop, the bike will not be horizontal. The front wheel will be lower than the back wheel by some amount that depends largely on the speed of the bike.

My question is: what happens to the attitude of the bike as it falls to the ground? Does it maintain the angle that it had reached when the rear wheel went off the drop? Or does it rotate because the front wheel is falling faster than the back wheel? In either case, can the rider level the bike, by shifting his position or by pulling on the handlebars?
 

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  • #2
Lnewqban
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Welcome!
Bike and biker should keep rotating together around a common center of mass during the fall.
 
  • #3
berkeman
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Welcome to PF. :smile:

Are you asking about bicycles or motorcycles? With dirtbikes, you use the power of the engine to help control the balance of the motorcycle as you leave dropoffs and other kinds of jumps. And you can use the engine and rear brake to alter the pitch of the dirtbike in the air a fair amount (depending on engine speed and rear wheel speed at takeoff).

If you are asking about bicycles (like mountain bikes), you have some control over your pitch at takeoff with body lean and pedal motion and pulling on the handlebars, but far less control versus a dirtbike.

1643641449718.png

https://www.bertsmegamall.com/blog/how-to-jump-a-dirt-bike-like-a-pro--26862

1643641516336.png

https://www.bikeradar.com/advice/fi...ike-a-pro-learn-to-dirt-jump-with-mike-smith/
 
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  • #4
berkeman
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When a cyclist rides off a ‘drop’
BTW, nice MTB single-track trail in your avatar! :smile:

1643642620684.png
 
  • #5
hutchphd
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And you can use the engine and rear brake to alter the pitch of the dirtbike in the air a fair amount
Of course this is also the way spacecraft (like Hubble and ISS) change their attitude without expending any rocket propellent. I think the term of art is "reaction wheel".
 
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  • #7
steveZ
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Welcome to PF. :smile:

Are you asking about bicycles or motorcycles? With dirtbikes, you use the power of the engine to help control the balance of the motorcycle as you leave dropoffs and other kinds of jumps. And you can use the engine and rear brake to alter the pitch of the dirtbike in the air a fair amount (depending on engine speed and rear wheel speed at takeoff).

If you are asking about bicycles (like mountain bikes), you have some control over your pitch at takeoff with body lean and pedal motion and pulling on the handlebars, but far less control versus a dirtbike.

View attachment 296317
https://www.bertsmegamall.com/blog/how-to-jump-a-dirt-bike-like-a-pro--26862

View attachment 296318
https://www.bikeradar.com/advice/fi...ike-a-pro-learn-to-dirt-jump-with-mike-smith/
Thanks for your reply. I would prefer to keep engines and brakes out of the equations, for now. I just want to know if the bike has an angular momentum, imparted by the the fact that the front wheel lost the support of the ground first, and whether that leads to continued rotation once the vehicle is fully in the air.
 
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  • #8
PeroK
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Thanks for your reply. I would prefer to keep engines and brakes out of the equations, for now. I just want to know if the bike has an angular momentum, imparted by the the fact that the front wheel lost the support of the ground first, and whether that leads to continued rotation once the vehicle is fully in the air.
It must. When the centre of mass is over the edge, there is an unbalanced torque from the ground to the back wheel. That said, as the bike accelerates during the fall, air resistance may act against and slow the rotation significantly.
 
  • #9
steveZ
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Thanks! If we ignore air resistance, would you say it is basically a Newton's First Law situation, where the angular velocity caused by that initial torque continues until another force changes it?
 
  • #10
PeroK
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Thanks! If we ignore air resistance, would you say it is basically a Newton's First Law situation, where the angular velocity caused by that initial torque continues until another force changes it?
Yes, Newton's laws lead directly to conservation of angular momentum for an isolated system.
 
  • #11
steveZ
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Welcome!
Bike and biker should keep rotating together around a common center of mass during the fall.
Thanks! Any thoughts on how to quantify the torque that causes the rotation?
 
  • #12
PeroK
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Thanks! Any thoughts on how to quantify the torque that causes the rotation?
My initial thought is that it is quite complicated. Especially for the irregular shape and mass distribution of a bicycle. And, speed is a critical factor.
 
  • #13
berkeman
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Thanks! Any thoughts on how to quantify the torque that causes the rotation?
Can you say a bit more about why you are asking this? Is it just to better understand ##\tau = I \alpha## in general, or are you planning some bicycle experiements or something?
 
  • #14
hutchphd
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Why would it be that complicated?
At (high) speed v there is an "unbalanced" force for a duration of $$\Delta t=\frac {d_{axle}} v$$ where ##{d_{axle}## is the distance between the axles The speed is high enough that there is little rotation during the edge event (ie impulse approximation).
Assuming most of the mass is a human of mass m this will induce a torque of size $$ \tau=mg d_{butt} $$ and angular speed of $$md_{butt}^2 \Delta \omega=\tau\Delta t=mg d_{butt}\frac {d_{axle}} v $$ so$$\Delta \omega=\frac {d_{butt}} {d_{axle}}\frac g v$$ If you assume your butt is halfway between the axles this gives $$\Delta \omega=\frac 1 2 \frac g v$$ At 20 mph this would give an angular speed of 0.5 rad/s or roughly 5 rpm. At 10 mph twice that. Seems about right.
 
  • #15
berkeman
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LOL ##d_{butt}## :smile:
 
  • #16
berkeman
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BTW, I think it should be more like ##d_{butt} cos( \theta )##, since the torque lever arm goes from the rear tire contact patch up to the pedal axle. That ##\theta## probably starts out at something like 30° and rotates down while the bike is rotating down.

Still, your numbers are probably a good approximation for an upper limit... :smile:
 
  • #17
PeroK
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It's even more complicated with a person on the bike. A bike has a mass of about 10kg. The rider would have a mass of about 75kg (and is also an extended body).
 
  • #18
hutchphd
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Slight oops. For completeness one should include the "parallel axis" contribution to the moment of inertia explicitly I suppose. Of course the factor for butt distance was a swag also. Mia culpa.
 
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  • #19
steveZ
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Can you say a bit more about why you are asking this? Is it just to better understand ##\tau = I \alpha## in general, or are you planning some bicycle experiements or something?
Two related reasons: (1) I am trying to learn to do drops and jumps better on my mountain bike; and, (2) I am writing some Matlab code to simulate these operations. Both motivations stem from snapping my collarbone a couple of years ago, before I learned to drop.
 
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  • #20
steveZ
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Why would it be that complicated?
At (high) speed v there is an "unbalanced" force for a duration of $$\Delta t=\frac {d_{axle}} v$$ where ##{d_{axle}## is the distance between the axles The speed is high enough that there is little rotation during the edge event (ie impulse approximation).
Assuming most of the mass is a human of mass m this will induce a torque of size $$ \tau=mg d_{butt} $$ and angular speed of $$md_{butt}^2 \Delta \omega=\tau\Delta t=mg d_{butt}\frac {d_{axle}} v $$ so$$\Delta \omega=\frac {d_{butt}} {d_{axle}}\frac g v$$ If you assume your butt is halfway between the axles this gives $$\Delta \omega=\frac 1 2 \frac g v$$ At 20 mph this would give an angular speed of 0.5 rad/s or roughly 5 rpm. At 10 mph twice that. Seems about right.
Very nice! Thanks, Dr Hutch.
 
  • #21
berkeman
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Both motivations stem from snapping my collarbone a couple of years ago, before I learned to drop.
Been there, done that (well, it was ribs for me). Have you learned to manual yet? That seems like a skill that would help you with dropoffs and other obstacles:

1643659849140.png

https://www.mbr.co.uk/how-to-2/how-to-manual-351507
 
  • #22
steveZ
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It's even more complicated with a person on the bike. A bike has a mass of about 10kg. The rider would have a mass of about 75kg (and is also an extended body).
Exactly. I think its an important point that on a bicycle the rider masses much more than the bike. So it may be easy to change CoM.
 
  • #23
steveZ
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Been there, done that (well, it was ribs for me). Have you learned to manual yet? That seems like a skill that would help you with dropoffs and other obstacles:

View attachment 296332
https://www.mbr.co.uk/how-to-2/how-to-manual-351507
Learning to manual is a long-term project for me. I built a so-called "manual machine" out of wood to help learn, after an unfortunate incident where I "looped out" and landed hard on my back. Many experts suggest that you should only manual off a drop if (a) you are going too slow to simply ride off the drop (i.e. dt is long enough that rotation is significant) and (b) you are really good at manuals. Similar for wheelies--you can wheelie off a drop if you are going slow but you better be 100% on your wheelie success rate.
 
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  • #24
sophiecentaur
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My initial thought is that it is quite complicated.
I was thinking that the torque from foot and crank would have the same value. If the crank radius is 0.2m and the rider pushes down with his/her own weight then the torque could be up to say
80X10X0.2 = 160Nm. (similar to doing a wheely) With skill, that would be reduced appropriately (and dynamically) to balance the torque of the rider's weight force about the rear wheel contact with the ground.
Some fiddly Integration would be needed for a best answer. Amazing how the brain 'jus does' that stuff.
 
  • #25
steveZ
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I was thinking that the torque from foot and crank would have the same value. If the crank radius is 0.2m and the rider pushes down with his/her own weight then the torque could be up to say
80X10X0.2 = 160Nm. (similar to doing a wheely) With skill, that would be reduced appropriately (and dynamically) to balance the torque of the rider's weight force about the rear wheel contact with the ground.
Some fiddly Integration would be needed for a best answer. Amazing how the brain 'jus does' that stuff.
I am not sure I understand your observations related to the cranks. I should have specified in the original formulation of the problem that the rider would not be pedaling as he goes off the drop—the pedals would be horizontal and the rider would be standing on them (probably crouching, really, but with butt not touching the saddle) such that his weight would be centered over the bottom bracket, which can then act as an axis of rotation, or hinge, if the rider pulls or pushes on the handle bars. Does that affect your ideas?
 
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  • #26
sophiecentaur
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Does that affect your ideas?
Yes, it's clear that a wheelie is not the same as stopping the front from dipping as you go over the edge for several reasons but zero force on the ground from the front wheel would be the same situation as staying level when the bike stays horizontal.

With a motorcycle, the engine is the only thing getting the front off the ground but you are saying that the pedal force is not the only thing at work with a bike ( or even has no relevance at all).

To keep the bike lever when only the back wheel is supported, there has to be, either an appropriate torque on the back wheel or some motion of the rider's mass to keep the CM of the whole system moving horizontally. That will reduce the upward force on the back wheel to near-zero. Effectively, the bike has to 'jump' before the front wheel gets to the edge so the whole thing is in free fall by then. Some serious skill needed.

So it must involve a jerk down and a pull up on the handlebars, which is the only way to get any torque that's not from the pedal and that will throw the bike upwards until the back wheel leaves over the edge. To stop the front dipping, there has to be some ('upwards') angular momentum of the rider which can be used to keep the total angular momentum at zero by the time the rear wheel gets over the edge.

Is there some 'last minute' pedalling just as the back wheel rolls over the edge?
I wonder how a physics - savvy expert on the technique would describe the process, from their personal experience.
 
  • #27
hutchphd
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And indeed how do you do all the calculations so quickly facing the looming drop! (I prefer pavement and the smoother the better... )
 
  • #28
steveZ
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Yes, it's clear that a wheelie is not the same as stopping the front from dipping as you go over the edge for several reasons but zero force on the ground from the front wheel would be the same situation as staying level when the bike stays horizontal.

With a motorcycle, the engine is the only thing getting the front off the ground but you are saying that the pedal force is not the only thing at work with a bike ( or even has no relevance at all).

To keep the bike lever when only the back wheel is supported, there has to be, either an appropriate torque on the back wheel or some motion of the rider's mass to keep the CM of the whole system moving horizontally. That will reduce the upward force on the back wheel to near-zero. Effectively, the bike has to 'jump' before the front wheel gets to the edge so the whole thing is in free fall by then. Some serious skill needed.

So it must involve a jerk down and a pull up on the handlebars, which is the only way to get any torque that's not from the pedal and that will throw the bike upwards until the back wheel leaves over the edge. To stop the front dipping, there has to be some ('upwards') angular momentum of the rider which can be used to keep the total angular momentum at zero by the time the rear wheel gets over the edge.

Is there some 'last minute' pedalling just as the back wheel rolls over the edge?
I wonder how a physics - savvy expert on the technique would describe the process, from their personal experience.
I think I can shed a little light on this. The technique you are describing is often used, and it is known as a "bunny hop". The rider springs upward just before the front wheel goes off the edge. The light bicycle comes along with him and the result is that both wheels are off the ground and the bike is level as it sails over the edge of the drop.

It is commonly done, especially in situations where your speed is low enough that there would otherwise be a dangerous amount of rotation. Some people recommend against it because: (a) it increases the vertical distance of your fall, so you hit harder, and (b) if you do not hop high enough, the bike tire will strike the edge of the drop as you fall, which can be catastrophic.

If you recall, though, my original question was not how to avoid rotation. It was whether the rotation that does occur during the Δt between the front wheel going off the drop and the back wheel going off the drop continues to increase during the remainder of the fall to the ground.

In other words, if an angle θ develops during Δt, does that angle stay fixed as the bike falls to the ground, or does rotation continue such that theta increases until front wheel strikes the ground? Several people on the forum have indicated that the rotation should continue unless something is done to stop it.
 
  • #29
sophiecentaur
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The technique you are describing is often used, and it is known as a "bunny hop".
That's the sort of thing I was referring to. My point about all this is that any period during which the rear wheel alone has a force ion it from the ground will produce finite 'forward rolling' angular momentum. To avoid tumbling forward, there must be some opposite net angular momentum introduced before the front wheel hits the edge. The bunny hop whilst traveling forward could keep the bike off the ground during this time a long jump). The only other ways to counteract a rotational impulse would be either to provide an impulse, 'driving' the back wheel downwards as it rolls over the edge or to push the rider's body away from the handlebars , again before the front wheel passes the edge, which would be a sort of half bunny hop, keeping the net angular momentum to zero when all contact is lost.

As with all aerial stunts, the angular velocity can be controlled by changing the moment of inertia so as to get the best angle at the point of landing but there's a limit to how much that can do.
 

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