Way the cornering forces are felt when riding a motorcycle

[Erunanethiel]

Everybody says when going around a corner on a motorcycle and leaning, you feel the g forces vertically (like pushing down on your spine) as opposed to horizontally when cornering in a car.

I can't understand how that can be the case, if the rider corners at 45 degrees, the gravitational force will perpendicular to the ground, and the centrifugal force will always be parallel to the ground, no matter the amount. So how can the rider feel it pushing him down to the seat?

As for the amount of force that is felt:

If the rider is cornering at 0.5g, does he only feel 0.5g on his body? The total force he feels is less than 1g which is what he feels when standing on still on the ground?

 

[lekh2003]

You don't feel the g-force. You feel a force very similar to gravity and kind of feels like gravity but isn't. This is centripetal force (centrifugal force doesn't exist, that is a myth). The centripetal force is acting on the bike towards to the point of rotation.

To answer your problem, think about which direction the bike is being pushed by the centripetal force. You are at 45 degrees and spinning in a circle. If you think about it, the centripetal force is acting at an angle. According to Newton's third law, the bike exerts a force on you and you feel the force on your spine.

The force you are feeling is not entirely the g-force. You might feel some of it, but most of it is a result of the centripetal force pulling the bike inwards and keeping you in rotation.

I'm not an engineer or mechanic so I might give a misconstrued opinion. Take a look at this link:

https://www.wired.com/2015/09/just-far-can-motorcycle-lean-turn/

 

[Erunanethiel]

You don't feel the g-force. You feel a force very similar to gravity and kind of feels like gravity but isn't. This is centripetal force (centrifugal force doesn't exist, that is a myth). The centripetal force is acting on the bike towards to the point of rotation.

To answer your problem, think about which direction the bike is being pushed by the centripetal force. You are at 45 degrees and spinning in a circle. If you think about it, the centripetal force is acting at an angle. According to Newton's third law, the bike exerts a force on you and you feel the force on your spine.

The force you are feeling is not entirely the g-force. You might feel some of it, but most of it is a result of the centripetal force pulling the bike inwards and keeping you in rotation.

I'm not an engineer or mechanic so I might give a misconstrued opinion. Take a look at this link:

https://www.wired.com/2015/09/just-far-can-motorcycle-lean-turn/

I am confused

The link doesn't work

 

[lekh2003]

I am confused

The link doesn't work

Try this articles if you can, seems to be more information:
https://www.wired.com/2015/08/motorcycles-lean-far-without-tipping/

Let me know if this one doesn't work either. I could try attaching a document with its contents.

 

[Erunanethiel]

Try this articles if you can, seems to be more information:
https://www.wired.com/2015/08/motorcycles-lean-far-without-tipping/

Let me know if this one doesn't work either. I could try attaching a document with its contents.

I searched "how do motorcycles lean so far without tipping over" on google and read the first link, hope it is the same one. Even though I read and understood it, I still don't get what force the rider feels when cornering.

 

[lekh2003]

I searched "how do motorcycles lean so far without tipping over" on google and read the first link, hope it is the same one. Even though I read and understood it, I still don't get what force the rider feels when cornering.

Yes, I think it is the same link.

When a rider is cornering, he leans over, so the article explains the physics behind that (leaning over). When you study physics you are introduced to the centrifugal force, but this is actually the centripetal force. The bike is being pulled to the center of rotation by the centripetal force. Friction, the banking of the bike and you are acting in opposition. If you were to suddenly jump off and the road lost friction, then the bike would rush towards the corner due to the centripetal force.

What I am giving as an explanation pales in comparison to the link. Make sure you understand all the concepts in the link, there is no more I can provide you with.

 

[Erunanethiel]

Yes, I think it is the same link.

When a rider is cornering, he leans over, so the article explains the physics behind that (leaning over). When you study physics you are introduced to the centrifugal force, but this is actually the centripetal force. The bike is being pulled to the center of rotation by the centripetal force. Friction, the banking of the bike and you are acting in opposition. If you were to suddenly jump off and the road lost friction, then the bike would rush towards the corner due to the centripetal force.

What I am giving as an explanation pales in comparison to the link. Make sure you understand all the concepts in the link, there is no more I can provide you with.

Doesn't the centripetal force push the system "out" while gravity is pushing the system "down" towards the ground?

 

[lekh2003]

Doesn't the centripetal force push the system "out" while gravity is pushing the system "down" towards the ground?

See, this is the misconception I am trying to resolve. You are still under the assumption that there is a force pushing you outwards. There is no such force acting in any way. Try swinging an object held on with a string. Let go of the string and the object maintains its velocity. You let go of the string which was the centripetal force and then there was no force on the object. If this magic force was actually there, then the objects would fly outwards, but it doesn't.

In this situation, there is a centripetal force which pulls the bike inwards. The equal and opposite force is the feeling that you are pushed down. Gravity is acting, but so is the centripetal force.

 

[Erunanethiel]

See, this is the misconception I am trying to resolve. You are still under the assumption that there is a force pushing you outwards. There is no such force acting in any way. Try swinging an object held on with a string. Let go of the string and the object maintains its velocity. You let go of the string which was the centripetal force and then there was no force on the object. If this magic force was actually there, then the objects would fly outwards, but it doesn't.

In this situation, there is a centripetal force which pulls the bike inwards. The equal and opposite force is the feeling that you are pushed down. Gravity is acting, but so is the centripetal force.

Isn't the centripetal force in the "swinging a ball attached to a rope" analogy provided by the string? And in the case of the motorcycle, it is provided by the tires, and if the tires lost friction, the motorcycle would continue in a direction tangent to it's previous path right? Rather than going towards the corner as you said. I have a feeling I am missing something, thanks for helping out

 

[lekh2003]

Isn't the centripetal force in the "swinging a ball attached to a rope" analogy provided by the string? And in the case of the motorcycle, it is provided by the tires, and if the tires lost friction, the motorcycle would continue in a direction tangent to it's previous path right? Rather than going towards the corner as you said. I have a feeling I am missing something, thanks for helping out

Yes, it would continue at a tangent, sorry, I made an error.

Again, just read the article. It explains everything better than I ever can.

 

[Erunanethiel]

Yes, it would continue at a tangent, sorry, I made an error.

Again, just read the article. It explains everything better than I ever can.

These are all very confusing subjects for us mere mortals, I wish I was Newton or something so I could think myself out of the questions I think about

 

[lekh2003]

These are all very confusing subjects for us mere mortals, I wish I was Newton or something so I could think myself out of the questions I think about

You don't need to be Newton to understand this. You just need to study physics from the ground up. You will understand these concepts much better if you start learning about motion, then Newton's laws, then rotation, uniform circular motion and then you will be able to solve your bike problem on your own.

I suggest that you look up some course-site and take a course in basic classical physics. It will help with understanding mechanics.

 

[Erunanethiel]

You don't need to be Newton to understand this. You just need to study physics from the ground up. You will understand these concepts much better if you start learning about motion, then Newton's laws, then rotation, uniform circular motion and then you will be able to solve your bike problem on your own.

I suggest that you look up some course-site and take a course in basic classical physics. It will help with understanding mechanics.

Thank you, I have big exam in 6 months, after that I will do soSo, the centripetal force is pushing us inwards at 1g, and at 1g the system must be leaning 45 degrees, so if the force the rider feels is the centripetal force, it must be parallel to the ground, but the rider feels the forces "head to toe" how does that happen?

 

[jbriggs444]

See, this is the misconception I am trying to resolve. You are still under the assumption that there is a force pushing you outwards. There is no such force acting in any way. Try swinging an object held on with a string. Let go of the string and the object maintains its velocity. You let go of the string which was the centripetal force and then there was no force on the object. If this magic force was actually there, then the objects would fly outwards, but it doesn't.

In this situation, there is a centripetal force which pulls the bike inwards. The equal and opposite force is the feeling that you are pushed down. Gravity is acting, but so is the centripetal force.

In the case of a motorcycle on a curve, the centripetal force on the motorcycle is provided by the road on the tires -- friction. The road supports the tires against the downward force of the motorcycle due to gravity. The road provides the inward force that provides the acceleration toward the center of the curve.

The total force from the road consists of those two components. One pointing inward and one pointing upward. Those forces add as vectors. The result is a single force pointing diagonally up and in.

The force you feel on the seat of your pants is very much the same. Diagonally upward and in. But the bike and rider are also leaning diagonally upward and in. The force is directly aligned "under" the seat of the pants. So the force feels like an ordinary upward force. It's just that "up" is leaning in.

 

[Erunanethiel]

In the case of a motorcycle on a curve, the centripetal force on the motorcycle is provided by the road on the tires -- friction. The road supports the tires against the downward force of the motorcycle due to gravity. The road provides the inward force that provides the acceleration toward the center of the curve.

The total force from the road consists of those two components. One pointing inward and one pointing upward. Those forces add as vectors. The result is a single force pointing diagonally up and in.

The force you feel on the seat of your pants is very much the same. Diagonally upward and in. But the bike and rider are also leaning diagonally upward and in. The force is directly aligned "under" the seat of the pants. So the force feels like an ordinary upward force. It's just that "up" is leaning in.

If the rider is cornering at 0.5g, does he only feel 0.5g on his body? The total force he feels is less than 1g which is what he feels when standing on still on the ground?

 

[jbriggs444]

If the rider is cornering at 0.5g, does he only feel 0.5g on his body? The total force he feels is less than 1g which is what he feels when standing on still on the ground?

If you add 0.5 g inward to 1.0 g upward using the Pythagorean theorem, what do you get? That's what's pushing on the seat of your pants.

 

[Erunanethiel]

If you add 0.5 g inward to 1.0 g upward using the Pythagorean theorem, what do you get? That's what's pushing on the seat of your pants.

I don't know

 

[jbriggs444]

I don't know

Do you know how to add vectors?

If you walk 1 mile north and 0.5 miles east you could have cut the corner and walked directly to the endpoint along the diagonal. Pythagoras gave us a formula for the length of that diagonal.

Can you tell us how long that diagonal would be?

 

[berkeman]

Everybody says when going around a corner on a motorcycle and leaning, you feel the g forces vertically (like pushing down on your spine) as opposed to horizontally when cornering in a car.

Wait a minute. You said in one of your other motorcycle threads that you ride. Why are you having to ask this question?

I do not live in the USA, and I do ride, I have a Ducati 1098 currently.

 

[Erunanethiel]

Wait a minute. You said in one of your other motorcycle threads that you ride. Why are you having to ask this question?

Because I don't feel any g forces what so ever when cornering, maybe because I don't lean enough [emoji28]

 

[Erunanethiel]

Do you know how to add vectors?

If you walk 1 mile north and 0.5 miles east you could have cut the corner and walked directly to the endpoint along the diagonal. Pythagoras gave us a formula for the length of that diagonal.

Can you tell us how long that diagonal would be?

Nope

 

[berkeman]

Because I don't feel any g forces what so ever when cornering, maybe because I don't lean enough [emoji28]

Well, cornering hard at a racetrack, you mostly feel the sideways force in the direction of the turn. The vertical force is pretty much unnoticed.

 

[lekh2003]

Nope

You would feel 1.12 g. This is more than the natural downward force from gravity. I suggest you learn a little math and physics before diving head first into what is essentially a basic math problem.

 

[Erunanethiel]

Well, cornering hard at a racetrack, you mostly feel the sideways force in the direction of the turn. The vertical force is pretty much unnoticed.

I have never done a trackday, and its very hard to lean further than 45 degrees on the street. Let me know if you come to Istanbul, and I will lose my track virginity with you [emoji28]

 

[Erunanethiel]

You would feel 1.12 g. This is more than the natural downward force from gravity. I suggest you learn a little math and physics before diving head first into what is essentially a basic math problem.

Is all of the 1.12 g pushing you down to the seat while cornering at 0.5g?

That is incredible god (or people who are better than me at physics) knows how much force motogp riders experience when leaning at over 60 degrees

 

[jbriggs444]

That is incredible god (or people who are better than me at physics) knows how much force motogp riders experience when leaning at over 60 degrees

60 degrees from the vertical? 30 degrees from the horizontal? That's an easy angle. 2 g's on the diagonal.

You can easily see that with an equilateral triangle cut in half. The diagonal is twice the length of the half side that remains after the cut. The angle of an uncut corner is 60 degrees. The angle of the corner that was cut is 30 degrees. The half side represents gravity. The diagonal represents the diagonal acceleration. The length of the cut represents the centripetal acceleration.

[Introduction to trigonometry]

 

[Erunanethiel]

60 degrees from the vertical? 30 degrees from the horizontal? That's an easy angle. 2 g's on the diagonal.

You can easily see that with an equilateral triangle cut in half. The diagonal is twice the length of the half side that remains after the cut. The angle of an uncut corner is 60 degrees. The angle of the corner that was cut is 30 degrees. The half side represents gravity. The diagonal represents the diagonal acceleration. The length of the cut represents the centripetal acceleration.

[Introduction to trigonometry]

Well, how much force is that they feel when leaning 64 degrees which is what equals to 2g I think? Sorry I have no idea how to do the trigonometry bit [emoji28]

 

[jbriggs444]

Well, how much force is that they feel when leaning 64 degrees which is what equals to 2g I think? Sorry I have no idea how to do the trigonometry bit [emoji28]

Trigonometry...

You start with a right triangle. i.e. one with an angle that is at 90 degrees. That is the angle between gravity (vertical) and centripetal acceleration (horizontal).

You pick one one of the two remaining angles. I am going to pick the 26 degree angle -- the angle with respect to the horizontal. [Darn, that's a heck of a lean]

The "sine" of that angle is the ratio of the side opposite (gravity) to the hypotenuse (the diagonal -- or the total diagonal acceleration).

We are after the inverse of that ratio -- the ratio of diagonal acceleration to gravity.

So we pick up a calculator and ask for the sine of 26 degrees -- 0.438. And then take the reciprocal of that -- 2.28.

Edit: you mention 2.0 g's being associated with 64 degrees. Could that be centripetal acceleration? Let's figure out the centripetal acceleration at 64 degrees.

I'm still anchored to my 26 degree angle. Now we are after the ratio of the side adjacent (centripetal force) to the side opposite (gravity). That's the cotangent. My calculator does not have a cotangent button. So let's use the 64 degree angle instead and look for the ratio of the side opposite (centripetal force) to the side adjacent (gravity). That's the tangent -- 2.05

You'd need 63.4 degrees to get a tangent of 2.00. On a calculator, the button for that operation is the "inverse tangent". The inverse tangent (or arc tangent) of 2.00 is 63.4 degrees.

 

[Erunanethiel]

Trigonometry...

You start with a right triangle. i.e. one with an angle that is at 90 degrees. That is the angle between gravity (vertical) and centripetal acceleration (horizontal).

You pick one one of the two remaining angles. I am going to pick the 26 degree angle -- the angle with respect to the horizontal. [Darn, that's a heck of a lean]

The "sine" of that angle is the ratio of the side opposite (gravity) to the hypotenuse (the diagonal -- or the total diagonal acceleration).

We are after the inverse of that ratio -- the ratio of diagonal acceleration to gravity.

So we pick up a calculator and ask for the sine of 26 degrees -- 0.438. And then take the reciprocal of that -- 2.28.

It must be incredible to lean 64 degrees and having over 2g playing tricks with the fluid in your ear and being out of sync with the visual orientation of the world whilst the road goes a few cm away from your head at god knows how many kilometers an hour must be unbelievably fun and terrifying

 

[Nik_2213]

Please, please, please keep such antics on official tracks. I've seen enough bikers scraped off roads to weep. One patch of mud, dung or gravel-wash, one slick of sloshed diesel, one dozy driver and they're gone.

Where I used to live, we'd see a stream of bikers come off each IOM ferry after TT Week. They'd zoom across the big A59 junction's fly-over, lay into run-out's adverse-camber bend, straighten out and be waved down for doing 60+ in well-signed 30 zone. IIRC, record was 25 in an hour...

 

[Erunanethiel]

Trigonometry...

You start with a right triangle. i.e. one with an angle that is at 90 degrees. That is the angle between gravity (vertical) and centripetal acceleration (horizontal).

You pick one one of the two remaining angles. I am going to pick the 26 degree angle -- the angle with respect to the horizontal. [Darn, that's a heck of a lean]

The "sine" of that angle is the ratio of the side opposite (gravity) to the hypotenuse (the diagonal -- or the total diagonal acceleration).

We are after the inverse of that ratio -- the ratio of diagonal acceleration to gravity.

So we pick up a calculator and ask for the sine of 26 degrees -- 0.438. And then take the reciprocal of that -- 2.28.

Edit: you mention 2.0 g's being associated with 64 degrees. Could that be centripetal acceleration? Let's figure out the centripetal acceleration at 64 degrees.

I'm still anchored to my 26 degree angle. Now we are after the ratio of the side adjacent (centripetal force) to the side opposite (gravity). That's the cotangent. My calculator does not have a cotangent button. So let's use the 64 degree angle instead and look for the ratio of the side opposite (centripetal force) to the side adjacent (gravity). That's the tangent -- 2.05

You'd need 63.4 degrees to get a tangent of 2.00. On a calculator, the button for that operation is the "inverse tangent". The inverse tangent (or arc tangent) of 2.00 is 63.4 degrees.

Yes it was the centripetal acceleration I was talking about.

Isn't the number 2.28 still correct for the amount of force the rider feels at 64 degrees?

 

[jbriggs444]

Yes it was the centripetal acceleration I was talking about.

Isn't the number 2.28 still correct for the amount of force the rider feels at 64 degrees?

2.28 g's at 64 degrees. Somewhat less at the 63.4 degrees calculated above.

At 2.0 g's centripetal force, one can calculate the felt acceleration in either of two ways:

$$a=\sqrt{a_{centripetal}^2+a_{gravity}^2}=\sqrt{2^2+1^2}=\sqrt{5}= 2.24$$
$$a=\frac{1}{cos(arctan(\frac{a_{centripetal}}{a_{gravity}}))}=\frac{1}{cos({arctan(2)})}=2.24$$

 

[itfitmewelltoo]

The motorcycle leans. The combined forces on the rider are normal to the seat.

If course, if the rider hangs off of the bike, https://goo.gl/images/GYnzJn, the total center of gravity is shifted from that plane, the center of gravity of the rider is not aligned with the vehicle, the bike leans less, giving more clearance.

 

[berkeman]

the bike leans less, giving more clearance.

And the opposite is true as well, unless you have one of the new experimental flying MotoGP bikes...

http://photos.motogp.com/2015/01/29/146438.small.jpg

 

[itfitmewelltoo]

More interesting is what is called counter steering. The wheels, in particular, the front wheel, is a gyroscope. The angular momentum of the front wheel is significant. In order to turn left, the rider pushes forward on the left handle bar. The bike then leans to the left and the bike turns to the left. The effect is more pronounced as speed increases and there is a transition speed below which the steering is as expected and above which angular momentum dominates

 

[berkeman]

More interesting is what is called counter steering. The wheels, in particular, the front wheel, is a gyroscope. The angular momentum of the front wheel is significant. In order to turn left, the rider pushes forward on the left handle bar. The bike then leans to the left and the bike turns to the left. The effect is more pronounced as speed increases and there is a transition speed below which the steering is as expected and above which angular momentum dominates

You must have missed this recent (very long) thread: https://www.physicsforums.com/threads/motorcycle-physics-countersteering-and-bodysteering.927832/

 

[cjl]

More interesting is what is called counter steering. The wheels, in particular, the front wheel, is a gyroscope. The angular momentum of the front wheel is significant. In order to turn left, the rider pushes forward on the left handle bar. The bike then leans to the left and the bike turns to the left. The effect is more pronounced as speed increases and there is a transition speed below which the steering is as expected and above which angular momentum dominates

That's not a gyroscopic effect - rather, it comes from the fact that if you turn the wheel to the right (by pushing the left handlebar forward), the wheels will steer to the right, out from underneath the CG of the bike. Since the wheels are now to the right of the CG, the bike "falls" to the left, initiating the left lean needed for the left turn.

 

[rcgldr]

Everybody says when going around a corner on a motorcycle and leaning, you feel the g forces vertically (like pushing down on your spine) as opposed to horizontally when cornering in a car.

The rider feels a net force vertical with respect to the lean angle of the bike (if the rider is not leaning inwards or outward relative to the bike). If the bike is leaning at 45 degrees (and in a coordinated turn), then the rider and bike are leaning at 45 degrees, and the rider feels a net force that is also at 45 degrees, so the rider doesn't feel any sideways forces (if the rider is not leaning inwards or outward relative to the bike). This is essentially the same as a coordinated banked turn in an aircraft.

If the rider leans inwards on the bike, the rider would feel a torque and also a force that is not inline with the angle of the rider. The rider could lean to one side with the bike going straight (the bike leaned a bit the other way) and the feeling would be similar, just less force than if leaning inwards while cornering.

 

[itfitmewelltoo]

The minimum force and acceleration is when the cycle is parked (or moving at a constant velocity). The acceleration due to gravity and the cooresponding weight.

The total acceleration when the bike is cornering, accelerating or decelerating on a straight line will always be more, the vector sum of gravity and cornering.

 

[itfitmewelltoo]

I believe that what they are referring to when they say the rider experiences 0.5g is that which is not due to gravity.

 

[itfitmewelltoo]

That's not a gyroscopic effect - rather, it comes from the fact that if you turn the wheel to the right (by pushing the left handlebar forward), the wheels will steer to the right, out from underneath the CG of the bike. Since the wheels are now to the right of the CG, the bike "falls" to the left, initiating the left lean needed for the left turn.

https://en.m.wikipedia.org/wiki/Countersteering

 

[itfitmewelltoo]

That's not a gyroscopic effect - rather, it comes from the fact that if you turn the wheel to the right (by pushing the left handlebar forward), the wheels will steer to the right, out from underneath the CG of the bike. Since the wheels are now to the right of the CG, the bike "falls" to the left, initiating the left lean needed for the left turn.

Oh, reading further in that link, "One effect of turning the front wheel is a roll moment caused by gyroscopic precession. The magnitude of this moment is proportional to the moment of inertia of the front wheel, its spin rate (forward motion), the rate that the rider turns the front wheel by applying a torque to the handlebars, and the cosine of the angle between the steering axis and the vertical.[10]".

Note the term "of turning the front wheel is a roll moment caused by gyroscopic precession."

 

[berkeman]

https://en.m.wikipedia.org/wiki/Countersteering

As the link shows, you can countersteer at speeds slow enough so that gyroscopic effects are minimal. cjl is correct that the main effect is to move the front tire's contact patch out from under the COM, which causes the bike to tip in. There is a small gyroscopic effect as well at higher speeds, but only adds a bit in my experience.

 

[itfitmewelltoo]

That's not a gyroscopic effect - rather, it comes from the fact that if you turn the wheel to the right (by pushing the left handlebar forward), the wheels will steer to the right, out from underneath the CG of the bike. Since the wheels are now to the right of the CG, the bike "falls" to the left, initiating the left lean needed for the left turn.

Let's consider this intuitively. The classic presentation of the gyroscopic effect is a guy holding a spinning bicycle tire, sitting on a rotating chair. Compare the weigh of this bicycle tire and rotational velocity to the weight and rotational velocity of a motorcycle wheel at 40 mph. The bicycle pales by comparison. It can hardly be said that the precession of that motorcycle wheel is negligible. Clearly it is not.

And, the phenomena does not occur at low speed. There is a crossover speed at which the steering changes from as expected to counter steering. As speed increases, the effect becomes more pronounced. What has changed? What has changed is the rotational velocity of the wheel.

 

[berkeman]

Let's consider this intuitively. The classic presentation of the gyroscopic effect is a guy holding a spinning bicycle tire, sitting on a rotating chair. Compare the weigh of this bicycle tire and rotational velocity to the weight and rotational velocity of a motorcycle wheel at 40 mph. The bicycle pales by comparison. It can hardly be said that the precession of that motorcycle wheel is negligible. Clearly it is not.

And, the phenomena does not occur at low speed. There is a crossover speed at which the steering changes from as expected to counter steering. As speed increases, the effect becomes more pronounced. What has changed? What has changed is the rotational velocity of the wheel.

Do you ride? What do you ride?

 

[rcgldr]

Oh, reading further in that link, "One effect of turning the front wheel is a roll moment caused by gyroscopic precession. The magnitude of this moment is proportional to the moment of inertia of the front wheel, its spin rate (forward motion), the rate that the rider turns the front wheel by applying a torque to the handlebars, and the cosine of the angle between the steering axis and the vertical". Note the term "of turning the front wheel is a roll moment caused by gyroscopic precession."

The precession only exists while a torque is applied. Once the steering is completed, initially outwards to induce a lean, the precession stops since it's a reaction to torque, not angle.

There is a crossover speed at which the steering changes from as expected to counter steering.

There are two primary speed ranges, one below the speed of self-stability, one in the range of self-stability. For a mathematical model using razor thin tires, there is a third speed range (capsize speed), above which a bike would tend to fall inwards at an extremely slow rate, but with real tires, this speed range doesn't exist or only exists well beyond any speed that a bike could achieve.

Counter steering is in effect at all speeds. At slow speeds, it's much less noticeable as the steering input is very light. It's easier to see this effect in the case of unicycles, which don't have the self correction related to trail on a two wheeled bike.

 

[itfitmewelltoo]

The precession only exists while a torque is applied. Once the steering is completed, initially outwards to induce a lean, the precession stops since it's a reaction to torque, not angle.

There are two primary speed ranges, one below the speed of self-stability, one in the range of self-stability. For a mathematical model using razor thin tires, there is a third speed range (capsize speed), above which a bike would tend to fall inwards at an extremely slow rate, but with real tires, this speed range doesn't exist or only exists well beyond any speed that a bike could achieve.

Counter steering is in effect at all speeds. At slow speeds, it's much less noticeable as the steering input is very light. It's easier to see this effect in the case of unicycles, which don't have the self correction related to trail on a two wheeled bike.

"Counter steering is in effect at all speeds."

Yes. It is one of a number of effects that vary with things like wheel weight and radius, the bike's weight, angle of the front forks, distance from front to back wheels, and ... things I haven't considered. Some of these are in opposition to the counter steering affect, what we might call "normal steering.". It is apparent that the counter steering affect decreases as speed decreases, becoming negligible. My impression, from riding around a parking lot, is that at 20-25 mph, "normal steering" is predominant.

It begs the question if why is there no apparent speed at which counter and normal steering component simple cancel out and make the bike simply unsteerable?