Is Counter-Steering the Only Way to Change Direction on a Motorcycle?

In summary, the conversation discusses the possibility of changing the combined center of gravity of a bike and rider system by shifting weight without using counter-steering. It is suggested that by twisting the body and pushing with the feet, the center of gravity can be shifted and the bike can change direction. However, this may be more difficult on a motorcycle compared to a bicycle or tight rope. The conversation also explores the importance of the point of application of force on the motorcycle and its effect on the torque and movement of the system. It is ultimately concluded that a combination of torque and force is exerted on the bike and the pavement, causing the bike to lean and potentially steer in the opposite direction.
  • #1
Erunanethiel
On a frictionless road and tires, I know the rider wouldn't be able to change the combined center of gravity of the system by moving his weight around. If he pushes the bike to lean to the right, he would go to the left by the amount that would keep the combined cog the same.

But a bike with real tires and a real road under it is not a closed system, the tires wouldn't let the lower part of the bike left or right when trying to initiate lean. Does this mean it is possible to push with your bodies inertia to the bike, and the combined cog will be changed? I think if you do so above the cog of the bike, you could, but I am not sure.I know counter steering is the best way, but I would like to think it is possible to change direction without it, without the influence of handlebars
 
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  • #2
Yes. In the same way a trapeze artist keeps their balance on a tight rope. If you twist the body and push left with the feet on the rope (or contact patch) then the rope (or contact patch) must push rightward. This deflects your center of gravity rightward. It is easier if you have a long pole to use as an angular momentum sink. But the upper body can do in a pinch.

The idea of doing it on a motor bike rather than a light weight bicycle or tight rope is a bit daunting, however.
 
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  • #3
Lock the steering to see how tricky this. Use a bicycle not a motorcycle.
 
  • #4
jbriggs444 said:
Yes. In the same way a trapeze artist keeps their balance on a tight rope. If you twist the body and push left with the feet on the rope (or contact patch) then the rope (or contact patch) must push rightward. This deflects your center of gravity rightward. It is easier if you have a long pole to use as an angular momentum sink. But the upper body can do in a pinch.

The idea of doing it on a motor bike rather than a light weight bicycle or tight rope is a bit daunting, however.
You don't mean twisting like your upper body turning right and lower body turning left correct? Like this is not what you mean I think:
images+%2886%29.jpeg
 
  • #5
Erunanethiel said:
You don't mean twisting like your upper body turning right and lower body turning left correct? Like this is not what you mean I think:
Yes, that is not what I mean. I mean rather bending leftward or rightward at the waist. For example, shifting head and legs left while shifting waist right.
 
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  • #6
jbriggs444 said:
Yes. In the same way a trapeze artist keeps their balance on a tight rope. If you twist the body and push left with the feet on the rope (or contact patch) then the rope (or contact patch) must push rightward. This deflects your center of gravity rightward. It is easier if you have a long pole to use as an angular momentum sink. But the upper body can do in a pinch.

The idea of doing it on a motor bike rather than a light weight bicycle or tight rope is a bit daunting, however.
In the tight rope walker example there is nothing to counter his forces though, there is nothing that moves the other way to cancel your movement like motorcycle and the rider.

I also think that where I apply force to the motorcycle is important too, if I push below the center of mass of the motorcycle and closer to the ground, the more the tires oppose the force I applied to the motorcycle, since it produces less torque and less can it cancel my movement.

Thank you very much for your time
 
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  • #7
Erunanethiel said:
In the tight rope walker example there is nothing to counter his forces though
There is the rope, attached rigidly to the earth. That's a plenty big reaction mass.
 
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  • #8
jbriggs444 said:
There is the rope, attached rigidly to the earth. That's a plenty big reaction mass.
Thats true lol

Do you think I am right about the point of application making a difference? Like if I push with my knee to the motorcycle or I use my feet, I would cause different torques on the motorcycle, making my movement more effective?
 
  • #9
Unless the rider applies a force where the effective force vector passes through the rider's center of mass, the rider exerts a combination of torque and force on the bike. The net result is a chain of Newton third law pairs of forces, eventually the tire exerting a force onto the pavement, and the pavement exerting an opposing force equal in magnitude.

As for the original question, if the rider shifts weight, then since the bike and Earth have momentum and resist the shifting of the rider's weight, the center of mass of the system can be accelerated horizontally due to the eventual force that the pavement exerts onto the bike's tires. For example, if the rider shifts left, the pavement eventually exerts a leftwards force onto the tires, and the center of mass of bike and rider accelerate left. If the bike doesn't steer, then the system of bike and rider will just lean over and fall due to the torque related to gravity pulling down at the center of mass of the system, and the pavement pushing up on the tires. If the bike is steerable and has conventional steering geometry (trail), then the bike leans in the opposite direction of the rider's weight shifting, causing the front tire to steer "outwards", an indirect form of counter-steering.
 
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  • #10
rcgldr said:
If the bike is steerable and has conventional steering geometry (trail), then the bike leans in the opposite direction of the rider's weight shifting, causing the front tire to steer "outwards", an indirect form of counter-steering.

Can you explain this bit?
 
  • #11
Erunanethiel said:
Can you explain this bit?
 
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  • #12
rcgldr said:
Unless the rider applies a force where the effective force vector passes through the rider's center of mass, the rider exerts a combination of torque and force on the bike. The net result is a chain of Newton third law pairs of forces, eventually the tire exerting a force onto the pavement, and the pavement exerting an opposing force equal in magnitude.

As for the original question, if the rider shifts weight, then since the bike and Earth have momentum and resist the shifting of the rider's weight, the center of mass of the system can be accelerated horizontally due to the eventual force that the pavement exerts onto the bike's tires. If the bike doesn't steer, then the system of bike and rider will just lean over and fall due to the torque related to gravity pulling down at the center of mass of the system, and the pavement pushing up on the tires. If the bike is steerable and has conventional steering geometry (trail), then the bike leans in the opposite direction of the rider's weight shifting, causing the front tire to steer "outwards", an indirect form of counter-steering.
I think in the scenario where the bike doesn't steer, rider and the bike will fall to the side the rider leaned into (rider leans to right, motorcycle goes left, and they both eventually fall to the right)

In the scenario where the bike can steer, the tire turning adds even more force to falling, again to the side where the rider leaned to in the first place

Am I correct?
 
  • #13
Erunanethiel said:
In the scenario where the bike can steer, the tire turning adds even more force to falling, again to the side where the rider leaned to in the first place ...
With conventional steering geometry (trail), the front tire steers in the direction of bike lean. When a rider shifts weight, the bike initially leans and steers "outwards", but the steering then results in the bike leaning "inwards", eventually stabilizing into a steady lean at a steady turning radius, depending on speed. If the rider shifts weight left, then the bike initially leans and steers right, then the bike leans and steers left. At very high speeds, the gyroscopic related reactions resist any change in lean angle and shifting weight ceases to work (almost no response), and direct counter-steering is required.
 
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  • #14
Erunanethiel said:
I think in the scenario where the bike doesn't steer, rider and the bike will fall to the side the rider leaned into (rider leans to right, motorcycle goes left, and they both eventually fall to the right)
Rider leans right (by bending rightward at the waist, pushing the seat leftward and foot pegs rightward). Bike leans left. Bike wheels would be deflected rightward, but steering is locked, so they push rightward on pavement instead. Pavement pushes leftward on the wheels. The center of mass of the bike+rider assembly moves left and therefore tips left.

Edit: It's like wind-milling your arms. If you windmill forward, you can save yourself from a forward fall and fall backward instead. If you windmill rightward, you can save yourself from a rightward fall and fall leftward instead. The reflex to lean right when standing atop a cliffside at one's right and about to fall off is well learned and entirely natural.
 
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  • #15
jbriggs444 said:
Rider leans right (by bending rightward at the waist, pushing the seat leftward and foot pegs rightward). Bike leans left. Bike wheels would be deflected rightward, but steering is locked, so they push rightward on pavement instead. Pavement pushes leftward on the wheels. The center of mass of the bike+rider assembly moves left and therefore tips left.

Edit: It's like wind-milling your arms. If you windmill forward, you can save yourself from a forward fall and fall backward instead. If you windmill rightward, you can save yourself from a rightward fall and fall leftward instead. The reflex to lean right when standing atop a cliffside at one's right and about to fall off is well learned and entirely natural.
I don't think tires would push rightward on the pavement when the rider leans right, to have that kind of torque, you would need quite a long lever arm, or huge force. On normal circumstances with not huge torque, I think the tires would push to the left, and the ground would push to the right, and since the rider leaned to the right as well, the system would end up falling to the right

Am I missing something?
 
  • #16
rcgldr said:
With conventional steering geometry (trail), the front tire steers in the direction of bike lean. When a rider shifts weight, the bike initially leans and steers "outwards", but the steering then results in the bike leaning "inwards", eventually stabilizing into a steady lean at a steady turning radius, depending on speed. At very high speeds, the gyroscopic related reactions resist any change in lean angle and shifting weight ceases to work (almost no response), and direct counter-steering is required.
Can you change or edit your first reply to have "directions"? Like the rider leans to the "left"And wouldn't gyroscopic forces cancel the bikes lean when I initially push it, making it move less to the other side, and cancelling less of my movement the other way?
 
  • #17
Erunanethiel said:
wouldn't gyroscopic forces cancel the bikes lean when I initially push it, making it move less to the other side, and cancelling less of my movement the other way?
The gyroscopic reactions reduce (or nearly eliminate) the lean reaction of the bike, resulting in higher lateral force between pavement and tire in response to weight shifting.
 
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  • #18
rcgldr said:
The gyroscopic reactions reduce (or nearly eliminate) the lean reaction of the bike, resulting in higher lateral force between pavement and tire in response to weight shifting.
Can you edit your first post with directions please, it is hard to understand this way
 
  • #19
Erunanethiel said:
Can you edit your first post with directions please, it is hard to understand this way
It's updated now. I'll delete this reply later.
 
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  • #20
jbriggs444 said:
Rider leans right (by bending rightward at the waist, pushing the seat leftward and foot pegs rightward). Bike leans left. Bike wheels would be deflected rightward, but steering is locked, so they push rightward on pavement instead. Pavement pushes leftward on the wheels. The center of mass of the bike+rider assembly moves left and therefore tips left.

Edit: It's like wind-milling your arms. If you windmill forward, you can save yourself from a forward fall and fall backward instead. If you windmill rightward, you can save yourself from a rightward fall and fall leftward instead. The reflex to lean right when standing atop a cliffside at one's right and about to fall off is well learned and entirely natural.
You and Rcgldr are saying opposite things, if you check his first post which is now edited
 
  • #21
rcgldr said:
The gyroscopic reactions reduce (or nearly eliminate) the lean reaction of the bike, resulting in higher lateral force between pavement and tire in response to weight shifting.
Yes, so when the rider moves to the right, motorcycle pretty much won't go to the left, which shifts the systems center of mass to the right more than if there wasn't any gyroscopic forces right?
 
  • #22
Erunanethiel said:
Yes, so when the rider moves to the right, motorcycle pretty much won't go to the left, which shifts the systems center of mass to the right more than if there wasn't any gyroscopic forces right?
The motorcycle will lean left, and it's center of mass will move left due to the leaning. However the net force eventually exerted by the tires onto the pavement will be to the left, and the net force exerted by the pavement onto the tires will be to the right, so the system of bike plus rider accelerates to the right. The gyroscopic reactions reduce the amount of lean by the motorcycle.

Note that with conventional steering, the immediate reaction of the bike leaning left is for the bike to steer left, so that the initial force from the pavement is to the left, "steering" the tires outwards from under the bikes center of mass, causing the bike to lean right. Once the bike leans right, then the bike steers right, and the force from the pavement is to the right.
 
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  • #23
rcgldr said:
The motorcycle will lean left, and it's center of mass will move left due to the leaning. However the net force eventually exerted by the tires onto the pavement will be to the left, and the net force exerted by the pavement onto the tires will be to the right, so the system of bike plus rider accelerates to the right. The gyroscopic reactions reduce the amount of lean by the motorcycle.

Note that with conventional steering, the immediate reaction of the bike leaning left is for the bike to steer left, so that the initial force from the pavement is to the left, "steering" the tires outwards from under the bikes center of mass, causing the bike to lean right. Once the bike leans right, then the bike steers right, and the force from the pavement is to the right.
So, final question,

Does where I push on the motorcycle make a difference? Like get up and push with feet on the pegs or sit down and push with my knees on the tank?
 
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  • #24
Erunanethiel said:
Does where I push on the motorcycle make a difference? Like get up and push with feet on the pegs or sit down and push with my knees on the tank?
As mentioned in my prior post, the rider exerts both a force and a torque, except for the unusual (if not impossible) case where the force from the ride goes through the rider's center of mass. Say the rider exerts more downwards force on the left peg than the right peg. This results in the bike leaning left and the rider leaning right if the rider doesn't compensate for the peg related torque by resisting it with other contact with the bike. The rider could lift off the seat and exert a horizontal force on the handlebars and foot pegs in order to shift weight (relative to the bike), which would normally have more effect than trying to use downward force differential on the foot pegs.
 
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  • #25
rcgldr said:
As mentioned in my prior post, the rider exerts both a force and a torque, except for the unusual (if not impossible) case where the force from the ride goes through the rider's center of mass. Say the rider exerts more downwards force on the left peg than the right peg. This results in the bike leaning left and the rider leaning right if the rider doesn't compensate for the peg related torque by resisting it with other contact with the bike. The rider could lift off the seat and exert a horizontal force on the handlebars and foot pegs in order to shift weight (relative to the bike), which would normally have more effect than trying to use downward force differential on the foot pegs.
Applying horizontal force on the pegs and handlebars would have more effect as in the combined center of mass would tip to the side where the rider leaned into in the in the first place right? Since the bike doesn't move the other way as much as it would otherwise.Wouldn't applying horizontal force on the pegs cause my body to rotate other way since my feet are way below my center of gravity? Like if I push left on the pegs to move to the right relative to the bike and pegs push on my feet to the right, wouldn't my body try to tip to the left? Or is that just wrong thinking?

Thank you very much!
 
  • #26
Erunanethiel said:
Wouldn't applying horizontal force on the pegs cause my body to rotate other way since my feet are way below my center of gravity? Like if I push left on the pegs to move to the right relative to the bike and pegs push on my feet to the right, wouldn't my body try to tip to the left?
This gets back to my point about applying a force that doesn't pass through the riders center of mass. In order to generate a left force at the pegs, the rider would rotate counter-clockwise as viewed from behind, such as leaning left at the waist, and/or swinging the right leg to the right, ... . If the rider doesn't first shift his weight to the left, then the leftward force exerted by the rider at the pegs results in the rightward force from the pegs to the riders feet, causing the rider to shift to the right (and the bike to lean left). You can try this standing on the floor, lift the right leg and lean left, you lean left, but move right unless you shift your weight to the left first.
 
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  • #27
rcgldr said:
This gets back to my point about applying a force that doesn't pass through the riders center of mass. In order to generate a left force at the pegs, the rider would rotate counter-clockwise as viewed from behind, such as leaning left at the waist, and/or swinging the right leg to the right, ... . If the rider doesn't first shift his weight to the left, then the leftward force exerted by the rider at the pegs results in the rightward force from the pegs to the riders feet, causing the rider to shift to the right (and the bike to lean left). You can try this standing on the floor, lift the right leg and lean left, you lean left, but move right unless you shift your weight to the left first.

How does the rider shift his weight to the left at first?
 
  • #28
Erunanethiel said:
How does the rider shift his weight to the left at first?
By pushing right somewhere on the bike at first.

It should be noted that counter-steering is recommended over weight shifting when turning a motorcycle, even more so for racing motorcycles since weight shifting ceases to work at high speeds (100+ mph => 160+ kph). Racing motorcycle riders typically use a combination of counter-steering and weight shifting in order to keep a bike going straight while at the same time hanging off the bike (weight shifting) in preparation for a turn. So to hang off to the left, the rider counter-steers to the right just enough to keep the bike near vertical and prevent the bike from leaning right during approach on a straight before reaching the start of a turn. Between an S turn sequence, the rider will shift from one side to the other and if the straight is short, use more counter-steering input so that the bike leans (rolls) in the same direction as the direction of weight shift by the rider. For example, during the transition from a left to right turn sequence, the bike transitions from leaning left to leaning right while at the same time the rider shifts weight from the left side to the right side of the bike.
 
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  • #29
rcgldr said:
By pushing right somewhere on the bike at first.

It should be noted that counter-steering is recommended over weight shifting when turning a motorcycle, even more so for racing motorcycles since weight shifting ceases to work at high speeds (100+ mph => 160+ kph). Racing motorcycle riders typically use a combination of counter-steering and weight shifting in order to keep a bike going straight while at the same time hanging off the bike (weight shifting) in preparation for a turn. So to hang off to the left, the rider counter-steers to the right just enough to keep the bike near vertical and prevent the bike from leaning right during approach on a straight before reaching the start of a turn. Between an S turn sequence, the rider will shift from one side to the other and if the straight is short, use more counter-steering input so that the bike leans (rolls) in the same direction as the direction of weight shift by the rider. For example, during the transition from a left to right turn sequence, the bike transitions from leaning left to leaning right while at the same time the rider shifts weight from the left side to the right side of the bike.
Pushing right can be achieved either by standing up on the seat and applying a horizontal force or using the downward force differential on the pegs as you previously said I think. Correct? The reason body steering ceases to work in higher speeds is mostly due to the gyroscopic forces from the wheels increasing right?
 
  • #30
Erunanethiel said:
The reason body steering ceases to work in higher speeds is mostly due to the gyroscopic forces from the wheels increasing right?
At high speeds the self correction that steers a bike into the direction of lean goes away and instead the bike tends to hold a lean angle without counter-steering used to change the lean angle. The mathematical models call this "capsize" speed, where the modeled bike falls inwards at very slow rate, so slow that it's imperceptible. The mathematical models may not take into account that the contact patch shifts inwards due to deformation, which would counter the slow inwards falling. The actual effect reported by motorcycle racers is that once leaned, a bike at high speeds tends to hold the lean angle, and the rider has to counter-steer to make any change in lean angle, and that weight shifting has no perceptible effect. The amount of effort to counter-steer increases as speed increases, and at these high speeds, it takes a significant amount of counter-steering effort to change lean angle.
 
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  • #31
rcgldr said:
At high speeds the self correction that steers a bike into the direction of lean goes away and instead the bike tends to hold a lean angle without counter-steering used to change the lean angle. The mathematical models call this "capsize" speed, where the modeled bike falls inwards at very slow rate, so slow that it's imperceptible. The mathematical models may not take into account that the contact patch shifts inwards due to deformation, which would counter the slow inwards falling. The actual effect reported by motorcycle racers is that once leaned, a bike at high speeds tends to hold the lean angle, and the rider has to counter-steer to make any change in lean angle, and that weight shifting has no perceptible effect. The amount of effort to counter-steer increases as speed increases, and at these high speeds, it takes a significant amount of counter-steering effort to change lean angle.
Why does self correction that steers the bike go away at high speeds? Isn't it due to gyroscopic effects from the wheels? Or is it because of the camber thrust?
 
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  • #32
Erunanethiel said:
Why does self correction that steers the bike go away at high speeds? Isn't it due to gyroscopic effects from the wheels? Or is it because of the camber thrust?
It's due to gyroscopic effects becoming dominant at high speed. At "normal" speeds, the steering geometry results in self corrective steering that tends to return a bike to a vertical orientation, but at high speeds, the tendency is to hold the current lean angle. At "normal" speeds, a small amount of counter-steering is needed to overcome the steering geometry to hold a lean angle, while at high speeds, virtually no rider input is needed to hold a lean angle. At high speeds, a rider has to use counter-steering to make any change in lean angle.

Wiki has a section about the mathematical model. Start at

https://en.wikipedia.org/wiki/Bicycle_and_motorcycle_dynamics#Lateral_motion_theory

and skip down to the eigenvalues section. Note the graph which shows a "capsize" speed around 8 m /s. Turns out this was wrong for the bike being modeled as it was noted as being "very stable" at 8.33 m/s (30 kph) on a large treadmill test. The actual "capsize" speed would be higher than what the graph shows. The model uses "knife edge" tires, as opposed to the actual tires used on the bike, so the actual "capsize" speed is higher. The effect is real though. I only had one personal experience with high cornering speeds, and it's true that if you ease off the handlebars, nothing happens. At lower speeds, if I ease off the handlebars, the bike tends to return to a vertical orientation, which is what the steering geometry (trail) is supposed to accomplish. At lower speeds, I have to apply a small amount of counter-steering input in order to hold a lean angle and prevent the bike from returning to vertical orientation. I always counter-steer to control lean angle and never rely on weight shifting, and I didn't experiment with weight shifting at high speeds, but motorcycle racers report that weight shifting is useless at high speeds.

> Camber thrust

There's are conflicts in what camber thrust means. It's basically a relationship between contact patch deformation and the lateral force at the contact patch. This would be one of the differences with a mathematical model based on knife edge tires with no contact patch deformation and the real word. My understanding is the contact patch deformation increases the speed at which "capsize" mode occurs. Wiki article:

https://en.wikipedia.org/wiki/Camber_thrust
 
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  • #33
Is it possible to create another thread about this? All the people have answered greatly, but the way I asked the original question caused the thread not being able to yield the answer I was looking for.

If there are moderates that can inform me if I can create a new thread asking the same question but in a different and more specific way, please let me know. Thank you very much to all the people who have helped me in this thread and epically to Rcgldr who also helped me via messages.
 
  • #34
As a motorcycle rider you might start from the way that you initiate a turn. At low speeds, when the rotational inertia of the wheels is low, you turn the front wheel to the right to turn right. At high speeds, when the rotational inertia of the wheels dominates the behaviour of the motorcycle, you turn the front wheel to the left to turn right. When turning the rider moves on the motorcycle to lower the centre of gravity which it is suggested increases the traction (see the book by John Surtees) although this may have more to do with preventing the motorcycle from flipping outwards by reducing the turning moment about the tyre contact patch - watch what happens when the contact is lost and the motorcycle starts to slide versus being high sided.
 
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  • #35
GeorgeX45 said:
As a motorcycle rider you might start from the way that you initiate a turn. At low speeds, when the rotational inertia of the wheels is low, you turn the front wheel to the right to turn right. At high speeds, when the rotational inertia of the wheels dominates the behaviour of the motorcycle, you turn the front wheel to the left to turn right. When turning the rider moves on the motorcycle to lower the centre of gravity which it is suggested increases the traction (see the book by John Surtees) although this may have more to do with preventing the motorcycle from flipping outwards by reducing the turning moment about the tyre contact patch - watch what happens when the contact is lost and the motorcycle starts to slide versus being high sided.
A bicycle or motorcycle has to be leaned inwards in order to keep from falling outwards during a turn. Regardless of speed, counter-steering to setup the initial leaning motion is always needed. The unicycle guys understand this, since there's no steering reaction to lean as there is with a bike.



Getting back to the bicycle / motorcycle there are three speed ranges that affect how a bike behaves while turning.

There's a moderate speed where the bike is self stable, and the bike's steering geometry tends to return the bike to a vertical orientation if the bike is leaned. At moderate speeds, the rider has to apply a small amount of counter steering to hold a lean angle and prevent the bike from returning to vertical. On many bikes, braking while leaning causes an inwards steering torque on the front tire, and the rider has to counter-steer to keep the front tire from steering inwards.

At speeds below the self stable speed, a bike tends to fall inwards due to insufficient inwards steering, and the rider has to compensate by steering further inwards to maintain balance and lean angle.

At speeds above the self stable speed, called "capsize" mode, a bike with knife edge tires would tend to fall inwards at an extremely slow rate. With real tires, a bike tends to hold a lean angle with no perceptible tendency to fall inwards or straighten up. At these speeds, only counter-steering inputs can change lean angle.

Although motorcycle racers hang off a bike when turning for a variety of reasons, it's not needed when riding in a street situation.
 
<h2>1. What is counter-steering on a motorcycle?</h2><p>Counter-steering is the act of pushing the handlebars in the opposite direction of the turn you want to make in order to initiate the lean of the motorcycle.</p><h2>2. Is counter-steering the only way to change direction on a motorcycle?</h2><p>Yes, counter-steering is the most effective and efficient way to change direction on a motorcycle. It allows the rider to quickly and smoothly initiate a turn.</p><h2>3. Why is counter-steering necessary on a motorcycle?</h2><p>Counter-steering is necessary on a motorcycle because of the physics of how motorcycles turn. Unlike cars, motorcycles cannot turn by simply steering in the direction of the desired turn. Counter-steering allows the motorcycle to lean and turn in the desired direction.</p><h2>4. Can you explain the physics behind counter-steering?</h2><p>Counter-steering works by creating a gyroscopic effect on the front wheel of the motorcycle. When the rider pushes the handlebars in the opposite direction of the desired turn, the front wheel will momentarily turn in that direction, causing the motorcycle to lean in the opposite direction and initiate the turn.</p><h2>5. Do all riders use counter-steering?</h2><p>Yes, all experienced riders use counter-steering, whether they are aware of it or not. It is a natural and instinctive technique that becomes second nature with practice.</p>

1. What is counter-steering on a motorcycle?

Counter-steering is the act of pushing the handlebars in the opposite direction of the turn you want to make in order to initiate the lean of the motorcycle.

2. Is counter-steering the only way to change direction on a motorcycle?

Yes, counter-steering is the most effective and efficient way to change direction on a motorcycle. It allows the rider to quickly and smoothly initiate a turn.

3. Why is counter-steering necessary on a motorcycle?

Counter-steering is necessary on a motorcycle because of the physics of how motorcycles turn. Unlike cars, motorcycles cannot turn by simply steering in the direction of the desired turn. Counter-steering allows the motorcycle to lean and turn in the desired direction.

4. Can you explain the physics behind counter-steering?

Counter-steering works by creating a gyroscopic effect on the front wheel of the motorcycle. When the rider pushes the handlebars in the opposite direction of the desired turn, the front wheel will momentarily turn in that direction, causing the motorcycle to lean in the opposite direction and initiate the turn.

5. Do all riders use counter-steering?

Yes, all experienced riders use counter-steering, whether they are aware of it or not. It is a natural and instinctive technique that becomes second nature with practice.

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