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Physics question about turn-in

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Physics question about turn-in

Old 10-10-21, 08:24 PM
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smashndash
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Physics question about turn-in

Nerdy question for anyone who might know about this stuff. When going in a straight line, the bike has no angular momentum in the roll axis. Once you initiate a turn, you start "falling" into it and thus have angular momentum. Your gravitational potential energy is being converted into kinetic energy. This is all fine.

Obviously at some point, you stop falling. You "catch" yourself by turning the bars deeper into the turn to straighten up the bike. But what happens to that kinetic energy? That angular momentum?

I think I know what the answer is but I'm curious whether anyone else has a better answer.
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Old 10-10-21, 10:13 PM
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Will give it a shot. When leaning before turning the bars, you are starting to fall and would crash unless you turned the bars. Turning the bars initiates the turn which causes a frictional force towards the center of the turn arc, which is the friction of the tires on pavement (i.e., at ground level). 3-4 feet off the ground, linear momentum makes your body and bike want to go in a straight line unless acted upon by a force, so you feel the "centrifugal force" from that which acts as a torque outwards (which wants to straighten the bike upright), which perfectly counteracts the torque caused by gravity during a steady-state turn, so that outward torque is what stops your angular momentum towards falling over. (Note that "centrifugal force" is in quotes because it's not an actual force; it's just a way of describing the way it feels locally to the rider, The only actual forces here are from friction which is required to cause the rotation around the turn arc, and gravity which is required to maintain a steady lean angle). As for energy, I'm guessing you probably convert potential into kinetic (thus speeding up a bit) during initial turn-in from "falling" into the turn, but the reverse of that would happen on the exit of the turn when you straighten up.

Last edited by jayp410; 10-10-21 at 10:50 PM. Reason: fixed typo: centripetal -> centrifugal
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Old 10-11-21, 07:13 AM
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...except we don't turn the bars, at least not at typical road bike cruising speed; it's the lean that initiates the turn.
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Old 10-11-21, 07:16 AM
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Originally Posted by Bob Ross View Post
...except we don't turn the bars, at least not at typical road bike cruising speed; it's the lean that initiates the turn.
It is early for popcorn.....
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Old 10-11-21, 07:46 AM
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The OP should search for videos that explain how to steer a motorcycle. After taking a motorcycle training course so I could legally ride a motorcycle in Colorado, I understood the topic better, but only from a point of cause and effect. On a motorcycle, you push forward on the right side of the handlebar to turn right. If you quit pushing, the bike will return to it's normal straight line course. A bicycle works the same way, but the force required is so much smaller that it can be difficult to feel or demonstrate unless you have winding mountain descents to practice on, like I do. Drop bars on road bikes also make the demonstration more difficult. Flat bar bikes make the demonstration of turning forces easier.
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Old 10-11-21, 08:49 AM
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Originally Posted by jayp410 View Post
As for energy, I'm guessing you probably convert potential into kinetic (thus speeding up a bit) during initial turn-in from "falling" into the turn, but the reverse of that would happen on the exit of the turn when you straighten up.
So you think that "catching" the fall results in the bike going faster? What would be the mechanism for that? If anything, the front tire is applying a braking force (hence the term "scrubbing off speed"). I do understand that the torque applied on the bike-rider system by the front tire via friction is critical to this discussion, but that part is fairly obvious and covered extensively. What I really want to know is what happens to that energy. When you're maintaining static lean through the turn, that energy is simply gone, right? Unless it turns out that you do actually speed up from the catch.

Actually my reference to the term "scrubbing off speed" may have given away where I believe the energy goes, but I want to believe I'm wrong.

Originally Posted by DaveSSS View Post
The OP should search for videos that explain how to steer a motorcycle.
I've read/watched more content on car, motorcycle and tire dynamics than is probably considered healthy for a human being. I'm very familiar with the idea of counter-steering LOL.
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Old 10-11-21, 08:53 AM
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Originally Posted by Bob Ross View Post
...except we don't turn the bars, at least not at typical road bike cruising speed; it's the lean that initiates the turn.
you definitely have to turn the bars to "catch" yourself. I never claimed anything about having to turn the bars to initiate the dive aka counter-steering.
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Old 10-11-21, 10:33 AM
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Originally Posted by smashndash View Post
But what happens to that kinetic energy? That angular momentum?
I don't think there's a simple answer to that question, other than "they go somewhere else."
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Old 10-11-21, 11:22 AM
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I think post 2 is correct. Note the emphasis on friction. Friction produces heat so that's where that energy goes, probably also some additional kinetic energy. Heat death of the universe and all that. Disregarding heat, the kinetic energy gain in "falling" is equal to its loss in "standing up." Thus coasting through a long series of chicanes on the flat will cause you to lose speed more quickly than coasting the same distance of travel in a straight line.
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Old 10-11-21, 11:59 AM
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Originally Posted by tomato coupe View Post
I don't think there's a simple answer to that question, other than "they go somewhere else."
well... yeah. What makes you think I'm asking for a simple answer?
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Old 10-11-21, 12:07 PM
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Originally Posted by Carbonfiberboy View Post
I think post 2 is correct. Note the emphasis on friction. Friction produces heat so that's where that energy goes, probably also some additional kinetic energy. Heat death of the universe and all that. Disregarding heat, the kinetic energy gain in "falling" is equal to its loss in "standing up." Thus coasting through a long series of chicanes on the flat will cause you to lose speed more quickly than coasting the same distance of travel in a straight line.
Right. So that's my hypothesis as well. I think we all know that cutting really hard causes a bike to slow down. It's no surprise that it might kill your angular momentum if it kills your linear momentum.

Now comes the juicy part. The main way tires generate heat is through internal friction aka hysteresis. So one can imagine that a high-hysteresis tire would be more effective at sapping away your angular momentum than a low-hysteresis one.

Many bike tires (like the GP5000) are trying their hardest to reduce hysteresis as much as possible. So what would happen if you rode tires with nearly 0 hysteresis? Would you still be able to kill your angular momentum? If not, what would happen?

The reason I'm saying this is that my observations have led me to the idea that hysteresis might actually be necessary for tires to deliver that confidence we need to flick hard. Low hysteresis tires may be able to handle braking and static lean fine, but are they compromising our ability to "catch" our rotational kinetic energy?

Imagine 2 baseball gloves. One covered in steel springs and one made of memory foam. The memory foam is going to make it WAY easier to catch a baseball.
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Old 10-11-21, 12:17 PM
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Originally Posted by smashndash View Post
What makes you think I'm asking for a simple answer?
Because you're asking the question on a bike forum, not a physics forum.
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Old 10-11-21, 12:18 PM
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Originally Posted by tomato coupe View Post
Because you're asking the question on a bike forum, not a physics forum.
fair point. I forgot that physics forums are a thing. I guess I'll look there. Thanks for the tip!
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Old 10-11-21, 05:52 PM
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Although there is loss due to tire friction (scrubbing speed), that is a secondary thing that we are not talking about at this level. Consider a 100% lossless system, where you have a wheel on a perfect bearing on a lossless rail. The rider in such a system would still have to lean in to provide the centripetal force component towards the arc center required for the turn, to make the center of mass rotate around the arc. And by leaning in, you are definitely losing gravitational potential, but due to conservation of energy in this system, there is only one place for that energy to go, and that is into kinetic energy.

Now in the turn, you are increasing your lean inward, which decreases the effective radius of the turn (on your center of mass, not the contact patch). Consider the angular momentum that you have (in a top down view). You and the bike have an angular momentum around the center of the turn arc. By leaning in and decreasing the radius, you are decreasing the moment of inertia (which is M*R*R) which causes a corresponding increase in angular velocity to compensate, due to conservation of angular momentum. It's the same effect that a speed skater takes advantage of when they are in a spin and they draw their arms closer to the center of their body and their rotation speeds up. (Think of you and the bike as the speed skater's hands). Gravity provides the force that keeps your center of mass moving around the circle, and gravity is also the force that causes your moment of inertia to decrease and therefore increases your angular velocity. Like the speed skater, drawing his/her arms in while rotating, gravity is providing a similar force inwards during initial lean-in, so gravity is doing work there, adding kinetic energy to the system temporarily until you exit the turn and straighten up.

So the above explains my hypothesis both in terms of energy and momentum analysis as to why you would temporarily speed up slightly while in the turn.

Last edited by jayp410; 10-11-21 at 06:35 PM.
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Old 10-11-21, 06:57 PM
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So in terms of friction, there is static friction, and non-static friction. Static friction exerts a force but the tire sticks and doesn't slide, and therefore it does no work and causes no energy loss. A bike when cornering is almost entirely static friction for the duration of the turn, unless you get aggressive. There is also non-static friction, where the tire slips a little and that is where energy is lost to heat and tire wear, but in a nice easy turn-in, there is effectively none of that to speak of, for the purposes of this discussion. In my initial post, I was discussing static (lossless) friction as the primary force that causes the bike to turn.

In the previous post I meant figure skater rather than speed skater. Ooops.
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Old 10-11-21, 11:24 PM
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Originally Posted by smashndash View Post
Obviously at some point, you stop falling. You "catch" yourself by turning the bars deeper into the turn to straighten up the bike. But what happens to that kinetic energy? That angular momentum?

I think I know what the answer is but I'm curious whether anyone else has a better answer.

Ignoring all friction, the bike is going to increase forward speed.

Kinetic energy is redirected into the direction of the turn by the wheel's grip on the ground is how the forward speed is increased.

Initial "fall" is converted to angular momentum and angular momentum to kinetic energy when this fall is arrested by turning the front wheel.

Noting that kinetic energy is converted back to potential energy as soon as bike stops turning and resumes straight path. Forward speed is restored to the speed prior to the turn.


You can see proof by rolling a bicycle wheel (only the wheel) on the ground at 5 to 10 kph watch it gain speed as it leans into a turn.

Last edited by cubewheels; 10-11-21 at 11:28 PM.
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Old 10-12-21, 09:04 AM
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Originally Posted by smashndash View Post
Right. So that's my hypothesis as well. I think we all know that cutting really hard causes a bike to slow down. It's no surprise that it might kill your angular momentum if it kills your linear momentum.

Now comes the juicy part. The main way tires generate heat is through internal friction aka hysteresis. So one can imagine that a high-hysteresis tire would be more effective at sapping away your angular momentum than a low-hysteresis one.

Many bike tires (like the GP5000) are trying their hardest to reduce hysteresis as much as possible. So what would happen if you rode tires with nearly 0 hysteresis? Would you still be able to kill your angular momentum? If not, what would happen?

The reason I'm saying this is that my observations have led me to the idea that hysteresis might actually be necessary for tires to deliver that confidence we need to flick hard. Low hysteresis tires may be able to handle braking and static lean fine, but are they compromising our ability to "catch" our rotational kinetic energy?

Imagine 2 baseball gloves. One covered in steel springs and one made of memory foam. The memory foam is going to make it WAY easier to catch a baseball.
The grippy rubber tread on a tire like a GP5000 allows for confident turns. The tire flexes over larger bumps for a smooth, efficient ride, but the tread is interacting with the micro texture of the road, too. At least until I hit a wet, oily patch...

On my other bike, running smooth tread 38mm tires, the turning grip gave me a "glued to the road" feeling. Interesting.

(thanks for a post about tire physics other than bicycle countersteering. Those threads never work.)
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Old 10-12-21, 11:08 AM
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You may find this enlightening:

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Old 10-12-21, 05:56 PM
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Man, I just spent 10 minutes reading a thread I didn't much understand.

Makes me wonder if I actually know how to ride a bike!😳😳😳
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Old 10-12-21, 08:59 PM
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In a turn the angular momentum is an effect of the gravity acting on the bike, rider, and lean of the bike.

If you start with Newton’s second law for a body in motion being acted on by a force we know: F = dp/dt where p is momentum (m*v).

Now if we imagine a circle the rider is turning through and since angular momentum must always have a reference point we’ll use the Center of the circle and imagine a vector r from the rider to the Center of the turning circle.

Multiple both sides of the equation by r: r*F = r * dp/dt
Of course r* dp is the change in angular momentum or dL.
And r*F is torque.
Torque = dL/dt
Or changing your angular momentum always results in torque.

Then I think the torque, ignoring friction, in this case should be something like or close to: torque = m*g*sin(theta). Where m is the mass of the bike and rider, g is gravitational force of the Earth, and theta is the angle of the plane of the bike to its vertical plane. (at zero degrees offset to the vertical plane the angular momentum is zero.) As you straighten the bike back up to reduce the torque from the turn to zero you regain your gravitational potential energy which is where the torque originally came from.

Could be wrong though.
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Old 10-12-21, 09:16 PM
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Originally Posted by smashndash View Post
Right. So that's my hypothesis as well. I think we all know that cutting really hard causes a bike to slow down. It's no surprise that it might kill your angular momentum if it kills your linear momentum.

Now comes the juicy part. The main way tires generate heat is through internal friction aka hysteresis. So one can imagine that a high-hysteresis tire would be more effective at sapping away your angular momentum than a low-hysteresis one.

Many bike tires (like the GP5000) are trying their hardest to reduce hysteresis as much as possible. So what would happen if you rode tires with nearly 0 hysteresis? Would you still be able to kill your angular momentum? If not, what would happen?

The reason I'm saying this is that my observations have led me to the idea that hysteresis might actually be necessary for tires to deliver that confidence we need to flick hard. Low hysteresis tires may be able to handle braking and static lean fine, but are they compromising our ability to "catch" our rotational kinetic energy?

Imagine 2 baseball gloves. One covered in steel springs and one made of memory foam. The memory foam is going to make it WAY easier to catch a baseball.
I don't know if your hypothesis is correct but since I use my bike for road riding and not slaloms or gymkhanas, I will take the low Crr every time.
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