Centripetal force, action and reaction, motorcycles

This is long. Sorry :)

Intro:
As a motorcycle rider as well as an analytical thinker, I like to know how my bike works. A discussion was undertaken that involved:
1) The act of countersteering
2) A rider travelling in a straight line towards a corner who wishes to take that corner at a high speed without running wide and off the road
3) A rider who is already leaned over and balanced in a turn who wishes to turn tighter (say, because they are about to run off the outer edge of the road)

For the purpose of this discussion, I want to make the following assumptions. Please don't bother arguing against countersteering.

Given: Countersteering. For the purpose of this discussion, is defined as rider input of a torque at the handlebars which causes the bike to lean.
Eg, to turn LEFT:
- rotate the handlebars clockwise viewed from above, by pushing forward on the left bar and/or pulling backwards on the right.
- This causes the front tyre to turn outwards (rotate clockwise), and the contact patch to move to the right.
- The mass of the system has inertia going straight. The friction at the front contact patch prevents the front tyre from sliding straight ahead (it wants to keep rolling, to the right). The bike is therefore forced to topple "forwards", which is to the left when viewed down the line of the front tyre. The bike is now leaning to the left.
- The pressure on the bars is reduced. Steering geometry causes the front tyre to "flop" back leftwards. It will reach a stable position once it is pointing somewhat leftwards. The degree to which it points leftwards is often hard to see with the naked eye at reasonable speeds.
- The bike now tracks a stable course through a left turn.
- Countersteering can be performed in a hypothetical universe without precession, and while precession does play a role in this universe it is usually slight compared to other forces at work.

Given: An inertial reference frame. I would rather not hear any mention of "centrifugal force". If you answer using a rotating reference frame I will almost certainly not understand how things are working out.

Given: simple, ideal conditions. The bike has no (or perfect) suspension, the tyres are not at the limit of traction at any time, air resistance is negligible, etc.

I will now outline the secenarios and the questions that arise from them. Sometimes I will assume answers to the questions to continue the scenario. Some of the questions may seem obvious at first, but when they come up again things get more complex.

If people could use equations in their arguments...
F = ma
tangential velocity = angular velocity * radius
centripetal force = mass * tangential velocity**2 / radius
etc, it would be appreciated.

Scenario 1: A rider on a bike for a total mass of MT is approaching a left hand turn with initial velocity IV1. They are positioned centrally on the seat. They perform a countersteer, leaning the bike left to some lean angle LB1 from vertical. For now this lean angle of the bike is also the lean angle to the centre of mass since the rider is centered on the seat. The bike tracks around the turn at tangential velocity TV1. If there is no throttle input I believe this will be less than IV1 since the circumerence of the tyre at the contact patch is smaller as the tyre leans over. I do not know whether we can "safely" assume that throttle is used to keep TV1 the same as IV1, or whether that becoems critical in later scenarios. In any case, they make a turn radius of TR1 at an angular velocity of AV1, and a centripetal force of CF1 = MT * TV1**2 / TR1

Scenario 2: Same as scenario 1, but before making the countersteer the rider firsts shift their body leftwards by a distance D2 on the seat. To do this, they push against the bike. These next questions may be made easier by assuming a frictionless seat (or perhaps that changes the whole thing! I don't know).

Question 2.1: To move their mass left, the rider exerts a rightwards force against the bike. Does this cause the bike to rotate such that it leans to the right?
Question 2.2: Once they have moved their mass left a distance D2, they exert a leftwards force on the bike to stop moving. Does this cause the bike to rotate back left and stand up straight, or merely to stop rotating to the right?
Question 2.3: If there is a rotation then is the rotation around the center of mass, or around the contact patches, or around some other point, and does this matter?

Assuming some answers here, the bike is now travelling straight with the rider "hanging off" left and the bike leaning right with the center of mass of the combination unchanged. The "lean angle" of the system as a whole is still vertical. The rider now performs a countersteer action such that the *bike* leans left to lean angle LB1, the same as in scenario 1. The lean angle to the CoM of the system is greater since the rider has shifted it to the left.

Question 2.4: Does this requires "more" countersteer since the the bike is initially leaning right?
Question 2.5: Does this result in an angular velocity AV2 which is higher than AV1? Are they running tighter? Has centripetal force changed?

Scenario 3: The rider from scenario 1 is half way through their turn, with lean angle LA1, tangential velocity TV1, turn radius TR1, angular velocity AV1. They now want to turn tighter. Let's see if this can be done by shifting their weight to the left without performing any other action. This is where things start getting difficult. Asking the same questions from before...

Question 3.1: As the rider shifts left, does this cause the bike to stand up (lean less left)?
Question 3.2: Once the rider has reached their desired amount of shift, will their pulling on the bike cause it to return to the previous lean, or will it just stop rotating, such that it is steady but at less lean?
Question 3.3: After performing this shift, and performing no other actions, has the CoM changed?
Question 3.4: Is the bike now running tighter as a result? Or must the rider input a countersteer to return the bike to its initial lean angle LB1, but with the CoM now further inward a higher lean angle to the CoM, resulting in a tighter line?
Question 3.5: Has centripetal force changed?

Here are two thought experiments to consider:

1) "ideal" bike leaning left, turning left. Pop gun on bike with muzzle pointing left. Pop gun is fired so the cork flies left. What happens? Is a torque applied? Does the bike tend to stand up (lean left less)? Would the bike eventually fall over to the right if nothing else happened? The cork has flown off and fallen some amount. What happens when the string goes taught? (Assume that this all happens fast enough that the bike hasn't changed direction very much. ie, assume a turn with a very large radius).

2) ideal bike leaning left turning left. Pop gun on bike with muzzle pointing "up", perpendicular to the angle of lean. That is to say, extending the line of fire back down through the center of mass of the bike would result in intersecting a line drawn through the contact patches. The gun is fired. What happens? Is a torque applied now? The cork flies off in a parabola, gravity has slowed its vertical component, which means the cork is now to the left of its initial vector. The string goes taught. Do we now apply a torque to the bike?

Do these two thought experiments show that whether you tighten your line depends on how the rider moves their mass? Ie, horizontally into the turn in the first case, vs "up" perpendicular to the lean angle and then straight down perpendicular to the earth, resulting in a net sideways movement in the second case.

Here's the big question from all this. Is the bike less likely to lose traction at a given angular V if you hang off? Does hanging off "help you get around a corner", or does this only matter once the bike is unable to be leaned any further because parts of it (eg, footpegs) are beginning to scrape along the ground?
















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