Why countersteer causes leaning

It's been a long time since 1st year engineering physics. I wanted to really understand countersteering. Naturally, I went to Wikipedia and found a great graphic. I hope it displays or at least lets you click on it.

http://upload.wikimedia.org/wikipedia/commons/6/68/Gyroscopic_precession_256x256.png

A bike wheel has no gimbal pivot. But, the bottom of the wheel is held on the road and the top of the wheel can move, that is, lean.

I think this is the best illustration I have found, what do you think?

For very small steering head rotation, your explanation would not work. Did you look at the picture? It shows why the wheels lean.

The leaning comes first; outtracking will take a bit of time; outtracking could be done without any leaning, I think, for small angles, on an instantaneous basis.

The best description of motorcycle dynamics I've found is in John Robinson's book Motorcycle Tuning: Chassis. He uses classical Newtonian physics and explains the difference between slow speed turns. where you don't countersteer, and higher speed turns where you do.

The way I understand countersteering, is to realize the bike does not lean at the contact patches of the tires, as if the contact patches were the center line of a hinge. Imagine a centerline, running fore/aft through the bike, roughly just above the level of the cylinders, in the center of the bike. When the portion of the bike/rider combination goes in one direction left or right above the line, the rest of the bike/rider combination below that line has to goe the other way to balance.

Simple Newtonian physics, equal and opposite forces. Now, understanding that, if the front tire were slightly deflected off the center path to one side or the other (it being part of the bike/rider combination below the line), the upper portion of the bike/rider combination has to go the opposite way to maintain balance. Using that knowledge, it becomes very clear that in order to make a motorcycle lean accurately and quickly, the rider must steer the front tire off center of the "line". It also becomes very clear that new riders with no knowledge of motorcycle dynamics, easily crash when they try to "steer" a bike like a car and get the opposite result they thought they would get.

It also explains how we learned to ride a bicycle when very young. We had no concept of intentional countersteeing, and we wobbled/wiggled like crazy as we intuitively steered the bike like our eyes would direct us to do. But once the brain clicked onto the spatial idea of balance like it does when we learned to walk upright, then suddenly we gained balance and leaning control on a bicycle. Same principle on a motorcycle. Difference though, is the bike weighs MUCH more than the rider, so the rider must make direct intentional input, countersteering, to make the bike/rider combination lean as intended. It also explains why countersteering input is not needed for all riding, and why some people claim they can make the bike lean without countersteering.

To a point they are right, you can make a bike lean without DIRECT countersteering input. But ANY motorcycle lean is the result of either direct or indirect countersteering input. Indirect input is the result of the upper body "leaning into the turn" like some riders claim is their method to make a bike lean. Again, back to the "line", because the action of the rider leaning into the turn causes a force off center of the line that the bike reacts to by moving in the opposite direction. However, it is a slow method to make a bike lean. That is where the unknowing/untrained rider can get in trouble, when they expect the bike to lean accurately and quickly just by leaning the upper body into the turn. Then they get rigid at the grips and don't allow the front tire to move off center of the line, and the bike runs wide. Body lean countersteering is just not quick enough.

The "line" is like the horizontal frame of the gimbrel shown in the original post, and the "tire" rotates left and right around that frame. Except in the case of a bike in motion, we have the bike/rider in combination "rotating" around that line.

Actually in a low speed turn there is a momentary countersteer to initiate the turn, I didn't think so until I saw this video. http://www.youtube.com/watch?v=OLzB5oriblk . Once the bike is leaning the bars must be turned into the turn to get the wheels back under the centerline of the bike. At higher speeds another force comes into play where the shape of the contact patch of the tires wants to force the tires back under the bike. For lack of a better description of this force I refer to it a "scrub thrust" . This is probably the best video I found to demonstrate countersteering, http://www.youtube.com/watch?v=1_A8k58ysSw&feature=related

Some people call this push steering because you push the handlebar forward on the side to which you want to turn: right forward for right; left forward for left. See David L. Hough's Proficient Motorcycling for his explanation.

BMW Triumphant said:

snip outtracking could be done without any leaning, I think, for small angles, on an instantaneous basis.

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I have rethought this part of my post; It is incorrect. Let me try again.

The original post had a link to a diagram. I still like the diagram as it shows the reason for a wheel to lean due to countersteering. In the diagram, it would be a right turn situation.

A bit of countersteering to the left as in the diagram would cause a bit of lean to the right for front and back wheels. More countersteer, more lean. Less countersteer, less lean.

At the start of the turn, the front and rear wheels have the same track, assuming proper alignment.

With a bit of countersteer, the front wheel is no longer completely aligned with the rear. As the front wheel rotates, the track of this wheel will be more and more to the left (in a right turn). Hence, outracking is not instantaneous, but takes time to develop as the wheel rotates.

At some point, if the rider wishes to stop the turn and resume straight ahead travel, the countersteering must cease. As it does so, the leaning will lessen the turning will cease, and at the end, with neutral steering, front and rear tracks will be the same.

However, the bike may not be travelling in the original direction. That would require a symmetrical left turn as well.

When the bike is leaning, gravity can be said to act through the centre of mass which will be to the right of the contact patch in a right turn. This gives rise to a horizontal force acting on the contact patch to provide the centripetal force pointing right, to the centre of the turn radius at any given instant.

There, that is what I wanted to say ...

I believe countersteer effects do increase with rotational speed increase. There is no lower threshold as such, just a smooth and continuous increase as speed rises from dead slow to slow to faster.

Tire flex and suspension effects have been ignored, but they would have real impact on the above I'm sure.

But, once again, it was the diagram that was so appealing to me that it prompted the original post.

I think you are right. I have tired to counter-steer on a mountain bike. It doesn't work that well on mountain bikes that have 19" wheels. My guess is GS riders with their larger wheels might not have the same counter steer feel an RT rider would have. The RT has a smaller front wheel. In theory, you wouldn't have to go as fast to counter steer a bike with a smaller wheel. The smaller wheel spins faster and creates a greater gyroscopic force. Likewise, a mountain bike with a 19" wheel that is super light weight would have to go much faster to get a counter steer.

Motorcycles don't lean due to gyroscopic precession; they lean due to the lateral force at the contact patch due to tire slip angle and camber thrust.

Gyroscopic forces can be completely eliminated by some tricky engineering. For example, one engineer designed a motorcycle brake system where the front rotors rotate in the opposite direction from the front wheel at twice the speed (Link). This significantly reduced the gyroscopic effect of the front wheel, but the bike handles fine (I'm not sure I buy all the developer's claims, however).

Any helicopter mechanics or pilots out there? They could add to the discussion and explain why increasing the pitch of the rotor blades at certain points in their circle of travel will cause the copter to move in a direction that is 90 degrees away, in the direction of travel of the spinning blades, from the area that the pitch is changed...

We did this experiment 'back in the day' Hold a fast spinning bicycle wheel with both hands, like you are the forks of a bike. Hold it with your arms pointing straight down, rotate your arms like you are turning and watch which way the wheel wants to tip. Didn't believe it till I tried it!

kthutchinson said:

Motorcycles don't lean due to gyroscopic precession; they lean due to the lateral force at the contact patch due to tire slip angle and camber thrust.

(snip)

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Ok, I looked up slip angle and camber thrust in Wikipedia. Camber thrust was the clearest.

At http://en.wikipedia.org/wiki/Camber_thrust

I read, in part,

Camber thrust is generated when a point on the outer surface of a leaned and rotating tire, that would normally follow a path that is elliptical when projected onto the ground, is forced to follow a straight path due to friction with the ground. This deviation towards the direction of the lean causes a deformation in the tire tread and carcass that is transmitted to the vehicle as a force in the direction of the lean.

Ok, that makes sense and that force in the direction of the lean is the centripetal force that creates the turn.

However, I do believe that the "leaned and rotating tire" obtains the lean because of countersteering causing gyroscopic forces to lean the wheel. At that point, camber thrust operates as above.

I'm open to other ideas to how the wheel could obtain lean, if someone has something to offer...

BMW Triumphant said:

Ok, I looked up slip angle and camber thrust in Wikipedia. Camber thrust was the clearest.

At http://en.wikipedia.org/wiki/Camber_thrust

....
Ok, that makes sense and that force in the direction of the lean is the centripetal force that creates the turn.

However, I do believe that the "leaned and rotating tire" obtains the lean because of countersteering causing gyroscopic forces to lean the wheel. At that point, camber thrust operates as above.

I'm open to other ideas to how the wheel could obtain lean, if someone has something to offer...

Click to expand...

You are correct...with a fast rotating front wheel, with substantial mass, when you turn the handle bars to the left, that is the same as applying a force to right side of the leading edge of the tire, and a force to the left side of the trailing edge. This force is transmitted 90 degrees in the direction of the wheels rotation, causing the wheel to lean to the right, as if a force was pushing to the right at the top of the tire and pushing to the left at the bottom of the tire. Traction prevents the tire from slipping out from under the bike, so the force causes the bike to lean to the right. Then other forces come into play.

Im no physics expert, and could be wrong. Just conversing here. I'm a single dad so I enjoy an occasional conversation that doesn't revolve around XBox or Facebook!

Countersteering for idiots:

turning the handlebars at speed will cause bike to fall down

turning less will cause bike to fall down less, or "lean"

a leaning bike is a turning bike

Thinking about all this while riding may cause you to fall down

Thinking less may help

When the handle bars are turned left, a lateral force to the left is generated at the contact patch due to tire slip. This is the same force generated in your car when the steering wheel is turned left. This force creates a rolling moment to the right, causing the motorcycle to lean right.

Yeah it's all kind of "scientific," but there's a simpler way to think of it.

If your bike is going to lean left, the left handgrip is going to be lower, i.e. closer to the ground than the right one.

To get it that way, you're going to have to push it down.

Vice versa, of course.

You can "wheelbarrow" through a curve with the bike remaining upright/not leaning but you're not going to be able to do it at more than 1 mph or so. Probably less (haven't tested).

A very simple way to think of it is like a little kid pretending to be an airplane, "fly" the bars down the road like you would an airplane wing. Try it, it's kind of fun.

kthutchinson said:

When the handle bars are turned left, a lateral force to the left is generated at the contact patch due to tire slip. This is the same force generated in your car when the steering wheel is turned left. This force creates a rolling moment to the right, causing the motorcycle to lean right.

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I agree about the lateral force and believe this is the centripetal force that is required for the bike to turn.

I am not sure the direction is correct as turning bars to the left should cause a right turn and this requires a centripetal force to the right and this force has to occur at the tire patch...

I do not understand how the force causes a rolling moment which causes the bike to lean. When not leaning, the force of gravity is perpendicular to the road and the slip angle force is at right angles to that. Vectors at right angles can't have any components one upon the other, so they are independent.

I'm still of the mind that gyroscopic effects cause the lean and the lean plus the centripetal force caused by the slip angle and / or castor sources you mention are both required.

But, if you can clarify or add something, please do so.

Can you help me understand your model?

I may be able to add something...

Further research reminded me that a force on the tire at the road is lower than the centre of mass of the bike. This force is equivalent to a type of force called a couple and the couple would cause a rotation around the centre of mass.

The diagram at http://en.wikipedia.org/wiki/Couple_(mechanics) would seem to support a slip force to the right causing a rotation that would be a lean to the right.

So, at this point, I would say both the gyro forces and the slip forces contribute to the leaning of the bike. I am satisfied I have the directions correct and they both point to the centre of the turn, ie, provide the required centripetal force.

What I don't know is how big the forces are, one relative to the other.

Tires are pretty stiff and distorting them takes a lot of energy. But, wheels rotate quickly and might have appreciable gryroscopic forces. That would depend upon the wheel / tire system's Moment of Inertia, which could be calculated by someone whose calculus skills are a bit more recent than mine if they had the data.

So, unless there are more facts, I am going to say both of these forces contribute, but I don't know which is greater.

BMW Triumphant said:

I agree about the lateral force and believe this is the centripetal force that is required for the bike to turn.

I am not sure the direction is correct as turning bars to the left should cause a right turn and this requires a centripetal force to the right and this force has to occur at the tire patch...

I do not understand how the force causes a rolling moment which causes the bike to lean. When not leaning, the force of gravity is perpendicular to the road and the slip angle force is at right angles to that. Vectors at right angles can't have any components one upon the other, so they are independent.

I'm still of the mind that gyroscopic effects cause the lean and the lean plus the centripetal force caused by the slip angle and / or castor sources you mention are both required.

But, if you can clarify or add something, please do so.

Can you help me understand your model?

Click to expand...

BMW Triumphant,

My post #15 explains how an initial countersteering input to the left causes the bike to lean right. As the bike leans right, the countersteering input is removed and the bars are actually turned slightly to the right, into the turn. The lateral force at the contact patch is now to the right, and this is the centripetal force that causes the bike to turn.

Gravity will be trying to pull the bike farther over to the right, and the centripetal turning force will be trying to roll the bike upright to the left. When these rolling moments balance the bike will maintain a constant lean angle.

For another good technical explanation, see chapter 4 in Tony Foale's excellent book Motorcycle Handling and Chassis Design. He also gives an explanation of the role of gyroscopic effects in stability.

lkchris said:

Yeah it's all kind of "scientific," but there's a simpler way to think of it.

If your bike is going to lean left, the left handgrip is going to be lower, i.e. closer to the ground than the right one.

To get it that way, you're going to have to push it down.Vice versa, of course.

Click to expand...

Sorry, but that's exactly...wrong.

You don't push the left or right grip down, you push it forward.

And that's all you really have to know about countersteering: push right, go right, push left, go left.

Harry