Why does a bike turn

[wil780]

I would agree with everything except the term "slipping" sideways. I think a better word would be "rolling" sideways to the chassis. The slipping infers tread slip with the ground, which does happen but isn't the force that causes the bike to turn. It's the fact that the wheel is (even slightly) off axis to the bike that actually causes the bike's direction to change and turn.

Your preferred description is of course correct at low lean angles, when the turning force is so low that any sideways slippage is microscopic. However, as you corner harder it becomes significant and my understanding is that the front tire is working in the sliding friction range rather than with static friction. So whether the tire is "rolling sideways to the chassis" or slipping sideways becomes a chicken-and-egg discussion at moderate or greater lean angles.

I settled on using slip because anyone with car experience is familiar with the term slip angle, and because it's mainly the increasing slip sideways that tells you that the front tire is approaching the traction limit. (You get the sense that the bike is running wide of the line you're aiming at, and if you're not steering with the throttle that's because the front tire slips sideways more and more as you approach the limit. If you're sensitive to this and you approach the limit gradually you get some warning about losing the front end.) Anyways, it's the way I think about it but your description is valid and probably easier to understand.

 

[OldernYZer]

I read this whole thing, and there’s some truth here and there throughout, along with quite a bit of stuff that appears to have been made up on the fly because it sounds good. The post by Northwoods Snowman comes closest to painting the whole picture, but doesn’t quite cover the wall on the first coat. So I’m going to take a shot at it.

First, I want to be clear with anyone who bothers to read this on some terms I’m going to use. First there is the center of gravity, or “CG”. That speaks to the point in a 3 dimensional object around which it will rotate if it were to be tossed into the air and spun in any direction, and the point at which it would be balanced end for end, side to side, top to bottom if it could be perched on that single point. It’s found up in the middle of a motorcycle somewhere, and reflects the weight distribution in any of those directions. The “X axis” is the fore and aft center line running through the CG, and when the bike rotates on this axis, it is said to “roll”. The “Y axis” is a vertical line through the CG, around which the bike “yaws”. The “Z axis” runs from left to right through the CG, and the bike “pitches” around this one.

Vehicles with two inline wheels steer, or turn, using a combination of dynamics that blend to varying degrees, very well or poorly, depending on the overall chassis setup. First, since it’s been brought up so many times, let’s examine counter steering. Counter steering can be used to initiate a turn by encouraging the motorcycle, bicycle, or scooter to bank, or lean into a corner. Essentially, it is a way to force the bike to roll on the X axis in spite of the angular momentum (gyro effect) of the wheels which resists its doing so. The concept is rather like that used to balance a hammer or broom on one’s finger tip with the heavy end up. Balance is maintained by shifting the low end of the mass in any direction on the X-Z plane, causing the object to roll or pitch in such a way as to remain upright and counteract the tendency to fall one way or other. When a bike running straight and upright is counter steered to the left, the contact points of the wheels are moved to the left under the CG, and the bike rolls, or falls, toward the right. Note that it is also true that one can simply shift the rider weight to the right over the CG and cause the bike to roll right in response to that without counter steering, so counter steering is not an essential element of turning, just a more effective way of initiating a turn more quickly and with less motion than shifting the rider weight.

Gyroscopic precession does not initiate or maintain a turn at all, and in fact works specifically against any rotation of the axle in any direction from its position on the Z axis. It is a highly stabilizing influence on the chassis, owing to the mass of the rotating tire and rim assembly. The strength of the effect depends the mass involved, and on the diametrical location of the rotating mass, the farther from the center, the more effective it is. It resists leaning the bike in either direction, and that is the main reason that counter steering is used to more forcibly cause the lean than by shifting the weight of the rider.

Someone mentioned bicycles with very small wheels being ridable, which they are, but to anyone who suggests that the gyro effect doesn’t stabilize the bike, I would say they should try to ride a Razor scooter on a pair of the training rollers posted earlier. Such small wheeled bikes are balanced almost entirely through counter steering in the manner of balancing a hammer, as discussed above.

As to the whole tire contact patch thing, there may be some effect in that, but before you buy into that being a primary causal effect, try rolling a coin across a flat surface and watch what happens when it leans one way or other; it turns in that direction, and the more it leans, the tighter the turn is. Why? Think of the diametrical edge of the coin, wheel, etc., not as a continuous line, but as a series of points along the circle. When upright on edge, it rolls straight because, as it rolls off of one point of contact and onto the next, the next point that rolls down onto the ground is coming from a point directly in front of the current contact point, and the next one after that is also directly in front, and so on. When the coin leans left, the next point approaching the ground as the coin rolls is located to the left of vertical, and the next is farther left, and the next farther left than that, so the coin turns in a circle to the left. As the coin leans farther, the circle tightens until it finally looses traction and centripetal force makes it slip and fall. All that with a sharp edge as a contact patch. Any wheel and tire assembly will naturally roll in a curved line in the direction it is leaned over from vertical for this reason alone. The size of the circle will depend on the radius of the tire.

So part of the reason a motorcycle banks into a turn is to cause the wheels to roll in the direction of the lean, but also, there is the matter of countering the centripetal force that would otherwise cause the bike to fall over to the right when turned left. The Y axis through the CG must be placed in a balanced condition against this force in order to counter it.

Then there is the complex matter of “rake”, more accurately called steering axis, or steering head angle, and how it interacts with trail. As the previously posted link showed, trail is the distance between a line drawn through the steering axis and the point at which the tire touches the ground. Normally (and hopefully) this point is behind the intersection of the steering axis and the ground. Trail is the cause of most of the self centering force on the steering while riding in a straight line, caused by the drag of rolling friction pulling back along the X axis at the contact point, but it also has a very pronounced effect on the chassis behavior when the vehicle is banked or rolled from vertical, and this effect works in combination with the steering axis angle to give the bike much of its cornering and steering behavior.

Visualizing how all this ties together is aided by getting your hands on a bicycle, or a motorcycle light enough to be leaned over quite a way without touching the handlebars. If you hold a bike by its seat and lean it to the left, you see that the wheel will “fall in” to point to the left of the X axis. This happens because the force of gravity, which was pushing up in line with the Y axis, is now working at angle to that axis and is able to rotate the steering assembly by pushing against what amounts to an effective lever, called a force arm, that is the length of the trail measurement, turning the wheel toward the direction of lean. In the first few degrees of lean, the “fallover” of the front wheel in this static, non-rolling condition is more pronounced, and the wheel of the bicycle actually begins to turn back more in line with the chassis as it is leaned farther. This happens for two reasons. The first is that the bicycle is standing still, and there is no “caster” or self centering effect on the steering because there is no drag pulling back on the tire from the contact point. Remember this point as we continue.

The second reason the wheel begins to turn back toward straight as lean is increased, and the reason it doesn’t fall all the way 90 degrees to the side in the first place, is because the arc of the leading edge of the tire comes in contact with the ground. What happens here is that the contact patch of the tire has moved from a point behind the steering axis (trail measurement) to a point just about even with it, neutralizing the force of gravity that is trying to turn the wheel in by pushing up and over on the force arm created by trail. In motion, this effect can be problematic. Lots of us have had the experience of a bicycle front wheel suddenly turning sharply inward, or “tucking under” as a result of the wheel falling in too far, and the contact patch getting ahead of the steering axis, and then being launched over the front as we “ride over” the front wheel. At low speeds and with steep head angles, this can happen.

Head angle plays into this such that shallower head angles (more “rake”) reduce the tendency to tuck, and steeper ones encourage it. Picture a really extreme chopper, say with a nearly 90 degree steering axis. With such a setup, steering depends a great deal on the natural circular tracking of a leaned over wheel, and a lot more rotation of the fork in the steering head is necessary to produce the same amount of steering. There is very little tendency to tuck, but a very pronounced tendency to push the front end, or understeer. Conversely, with a steep head angle, the front tire can move the tire contact patch ahead of the steering axis very early in the lean-in maneuver, showing a severe tendency to tuck, but at the same time very little fork rotation is needed to steer the machine, and it will seem very precise and responsive to small steering inputs.

The balance that has to be struck at speed is between the caster, or centering effect of trail, and the tendency for the front wheel to turn in. Head angle has to be chosen based on the amount of stability required versus how responsive the steering should be. If too shallow, the bike will turn in a vague and sluggish manner and tend to push the front, but it will be less prone to speed wobbles. If too steep, it will be skittish and unstable at speed, and prone to wobble, but will turn seemingly with just a thought. Once the head angle is chosen to match the intended use of the bike and other factors, trail is chosen to compliment and modify the turn in behaviors.

So, a turn at speed is initiated most effectively by counter steering to start the bike leaning, at which point the contact patches of both the front and rear tires shift to the side of the tire center lines in and amount governed by the tire size and shape, and the combined influences of the tire’s circular form, steering head angle, and trail produce a more or less balanced natural amount of fork rotation, while centripetal force holds the bike from falling to the inside of the corner. How this complex interaction actually works will affect whether you find yourself applying force to the bars to steer the bike into the corner more, or pulling back toward straight as the corner proceeds.

When it all works right, it feels easy and natural. But it’s definitely not simple.

 

[Eagle Six]

My subjective impression from this test agrees with what scientists researching motorcycle dynamics have concluded, that gyroscopic forces account for roughly 30% of the inherent stability of a motorcycle.

Hi wil780, would you please post a reference link to the research report of these scientist, substantiating the "roughly 30%" results? Thank You.

I can't claim to have done an exhaustive search, but here are the highlights of what I've run across over the years in my amateur efforts to understand motorcycle dynamics.

An explicit claim that gyroscopic forces are not dominant: J Fajans, 2000. Steering in bicycles and motorcycles. Am. J. Phys. 68, 654 - 659. As far as I know, no one has pointed out any errors in the conclusions of this paper, despite the use of a simplified theory of motorcycle steering that allowed a specific focus on gyroscopic effects.

A report that stability does not require gyroscopic effects (because they were engineered to not exist): JDG Kooijman, JP Meijaard, JM Papadopoulos, A Ruina, AL Schwab, 2011. A bicycle can be self-stable without gyroscopic or caster effects. Science, March 20, 2011. (I got a preprint from the authors which doesn't have the full reference and don't have it handy. A quick search should find it for you if you need it.)

The most comprehensive theory currently available: JP Meijaard, JM Papadopoulos, A Ruina, AL Schwab, 2007. Linearized dynamics equations for the balance and steer of a bicycle: a benchmark and review. Proc. R. Soc. A. 463, 1955 - 1982. This shows that there are many effects that govern motorcycle stability, not just gyroscopic effects.

I've corresponded with the some of the authors of the second and third papers regarding the "cone effect" questioned in the original post of this thread (it's usually called camber thrust), and they confirmed my conclusion that it has almost nothing to do with turning a motorcycle. (Repeating what I posted early in this thread, it is one of many forces that try to turn the handlebars while a bike is established in a stable turn. These forces must balance if the bike is to go around a corner without needing constant pressure on the bars by the rider, but none of those forces cause the motorcycle to follow a curving path. That turning force comes from the front tire slipping slightly sideways. I say almost nothing, because I think that camber thrust from the rear wheel does apply a very small force that acts in the same direction as the slipping force from the front wheel. I have not run this idea past any experts.)

Thank You vil780 for the effort in your reply and references. I wasn't ignoring you, It just took me a few sessions to get through all the papers and references from the papers. I may have completely missed it, but I did not detect a direct reference to approximate 30% stability from gyroscopic force, although there is lots of reference to gyro effects/force. I assume the "approximate" may be due to the size and speed of the spinning mass (wheel & tire). There has been a lot of time, effort and mathematics spent on those 3 papers!!

Using your term "slipping force" I would assume this force has more effect at higher speeds and less effect at slower speeds. Although an interesting topic, I'm extremely happy I have the minimum knowledge and skills to effectively turn my bike on demand, without having to understand all the scientific principles, effects and the math to explain every aspect. So, this afternoon I ride!

 

[Fred W]

I do not think any actual significant slipping of the tire tread has anything to do with "normal" turning, right up until you are exceeding the point of tire adhesion (weight / thrust / friction).

While not a mental giant, I can (like all the other 6 year old kids in my neighborhood) manage to navigate a two wheel vehicle down the road without falling over. One thing I ponder occasionally is whether sometimes ignorance is bliss?

I know some people who have the nerve and kinetic awareness to ride their bikes very, very fast. I have the ignoble habit of attempting these feats occasionally , perhaps more often than would be prudent. But at the same time I suspect that many of these really fast hot shoes have no real awareness of the physical factors at play, and rather just push the limit until they fail (and then back off) in a state of oblivion?

Trial and error when you have only one life to submit is a somewhat harsh reality. It works somewhat better with sports like skiing or even car racing, where a crash is only a temporary setback, than it does with with on-road motorcycling, where your next bad judgement may very well be your last one.

Sorry, not quite sure where I was going with this. Just sharing some stream of consciousness thoughts with hopes others can relate.

Carry on with the physics.

 

[Eagle Six]

I do not think any actual significant slipping of the tire tread has anything to do with "normal" turning, right up until you are exceeding the point of tire adhesion (weight / thrust / friction).

I would agree with you Fred, dependant on the term "normal" probably being different for different riders. For me, normal would be 5-10 mph over. I think during normal operation there is more tire wear from the friction induced by forcing the tire forward in the direction of travel (quoting you again, "weight / thrust / friction"), with some additional small amount as the tire slips through the turn. None of which is close to exceeding the traction required to stabilize the bike. When we approach racing speeds, either on the track, a closed course or the street, then I think there is significant slipping at the friction point as the inside area of the contact patch fights with the outer area of the contact patch each area going at a different speed, and as front and rear tire traction fights the force of momentum, gravity and centrifugal forces. I think "slipping", as many other aspects, has an effect on turning, but more of a consequent of bikes turning rather than the reason bikes turn. But then again, what I know about geometry and physics could be balanced on a thimble!!

 

[wil780]

Eagle Six, when a bike is leaned at all, the outer edges of the contact patch must slip continuously because the part on the outside of the turn wants to rotate slower than the centre of the patch, and the part on the inside of the turn wants to rotate faster The rate of slip can be pretty substantial. Look at one of our rear tires and it appears to me that at 30 degrees of lean the radius of the wheel at the inner and outer parts of the contact patch can differ by an inch. Which means they want to rotate at speeds that differ by approximately 5%. So each edge is slipping at 2.5% of the road speed! (I think this is why it's possible to wear out the sides of the tires when even in the twisties you spend relatively little time in corners, and of course each corner uses only one side of the tire.) You might argue that the fact that the bike is turning would prevent the slip, since the inner edge is making a tighter turn than the outer edge. But that effect is insignificant since the radius of the turn the tire wants to take is defined by the point where the extended axle line touches the ground - just a few feet away at moderate lean angles.

Fred W, I think we agree on the key point. A bike goes around a corner because the front wheel is turned inwards, away from the direction it would point for the bike go in a straight line (assuming something like a line from a helicopter kept it from falling over). That generates the sideways force which deflects the bike from a straight path. What we're quibbling about is whether the tire slips sideways slightly as it generates that sideways force. I see the slip as a consequence of the tire being turned into the corner, and it was a mistake to emphasize the slip instead of the inward turn. Since all of the contact patch except the very centre is already slipping forward or backwards (because of the geometry of the contact patch relative to the axle), I think the tire is operating with sliding friction rather than static friction, so a bit of sideslip is required to generate the turning force. But that's really irrelevant to the issue of why a motorcycle turns, and I never should have brought it up. The front tire generates the turning force because it is turned inwards, away from the direction the bike is moving at any instant.

Fred W, it's probably true that some really fast racers don't think at all about what they do. But I suspect most of the truly great ones think very deeply about their riding. They just refuse to talk about technique to avoid giving away secrets. An excellent example is Cal Rayborn, probably the first roadracer to slide the front tire routinely. He refused to confirm or deny he did that. My all time hero, Kenny Roberts, constantly told anyone who would listen that you have to think yourself faster.

 

[wil780]

Thank You vil780 for the effort in your reply and references. I wasn't ignoring you, It just took me a few sessions to get through all the papers and references from the papers. I may have completely missed it, but I did not detect a direct reference to approximate 30% stability from gyroscopic force, although there is lots of reference to gyro effects/force. I assume the "approximate" may be due to the size and speed of the spinning mass (wheel & tire). There has been a lot of time, effort and mathematics spent on those 3 papers!!

No, Eagle Six, you didn't miss it. Because the same model describes bicycles and motorcycles which vary a great deal in the amount rotating mass and its distribution, you have to try to estimate where an FJR would fit. I meant that the experts seem to agree that gyroscopic effects provide appreciably less than half of the resistance to leaning for most motorcycles, and it appears to me that roughly 30% is what you would get if you had the data you need from the bike to plug into the equations, or just interpolate on the graphs.

 

[wil780]

Eagle Six, when a bike is leaned at all, the outer edges of the contact patch must slip continuously because the part on the outside of the turn wants to rotate slower than the centre of the patch, and the part on the inside of the turn wants to rotate faster The rate of slip can be pretty substantial. Look at one of our rear tires and it appears to me that at 30 degrees of lean the radius of the wheel at the inner and outer parts of the contact patch can differ by an inch. Which means they want to rotate at speeds that differ by approximately 5%. So each edge is slipping at 2.5% of the road speed! (I think this is why it's possible to wear out the sides of the tires when even in the twisties you spend relatively little time in corners, and of course each corner uses only one side of the tire.) You might argue that the fact that the bike is turning would prevent the slip, since the inner edge is making a tighter turn than the outer edge. But that effect is insignificant since the radius of the turn the tire wants to take is defined by the point where the extended axle line touches the ground - just a few feet away at moderate lean angles.

I wanted to edit the last line of this paragraph, but apparently once you log out you can't do that. So here's what the last line should say: But that effect is insignificant since the radius of the turn at which all parts of the contact patch naturally rotate at the same speed (no slipping forward or back) is defined by the point where the extended axle line touches the ground - just a few feet away at moderate lean angles.

 

[wil780]

OldernYZer wrote a long post starting with a claim that no one else understands the issues. Then he proceeded to write what would be a pretty good post (great if not for some genuine errors) if the title of the thread was "why does a bike feel the way it does while it is being leaned?" or "why does a bike feel the way it does while it is turning?" Nothing in his post tells us what force he thinks causes a bike to turn. Clearly he thinks no one else knows, so it would be great if he told us his opinion of why a bike turns.

I read this whole thing, and there’s some truth here and there throughout, along with quite a bit of stuff that appears to have been made up on the fly because it sounds good. The post by Northwoods Snowman comes closest to painting the whole picture, but doesn’t quite cover the wall on the first coat. So I’m going to take a shot at it.

 

[Eagle Six]

Thank You vil780 for the effort in your reply and references. I wasn't ignoring you, It just took me a few sessions to get through all the papers and references from the papers. I may have completely missed it, but I did not detect a direct reference to approximate 30% stability from gyroscopic force, although there is lots of reference to gyro effects/force. I assume the "approximate" may be due to the size and speed of the spinning mass (wheel & tire). There has been a lot of time, effort and mathematics spent on those 3 papers!!

No, Eagle Six, you didn't miss it. Because the same model describes bicycles and motorcycles which vary a great deal in the amount rotating mass and its distribution, you have to try to estimate where an FJR would fit. I meant that the experts seem to agree that gyroscopic effects provide appreciably less than half of the resistance to leaning for most motorcycles, and it appears to me that roughly 30% is what you would get if you had the data you need from the bike to plug into the equations, or just interpolate on the graphs.

OK, I understand. It appears you have made a conclusion estimate based on various reported factors and principles. I was thinking you may have read a specific research stated in relatively layman terms us mere mortals could comprehend. Thank You for the reply.

 

[Eagle Six]

First, I want to be clear with anyone who bothers to read this on some terms I’m going to use. First there is the center of gravity, or “CG”. That speaks to the point in a 3 dimensional object around which it will rotate if it were to be tossed into the air and spun in any direction, and the point at which it would be balanced end for end, side to side, top to bottom if it could be perched on that single point. It’s found up in the middle of a motorcycle somewhere, and reflects the weight distribution in any of those directions. The “X axis” is the fore and aft center line running through the CG, and when the bike rotates on this axis, it is said to “roll”. The “Y axis” is a vertical line through the CG, around which the bike “yaws”. The “Z axis” runs from left to right through the CG, and the bike “pitches” around this one.

OldernYZer, I think your 'first' point adds an unnecessary complexity to the 'Why does a bike turn' discussions, and was quite confusing to me as to why you wanted to clarify the terms of "X,Y, Z axis" as I think it is has more relation to a vehicle such as an aircraft suspended in air or submersible such as a submarine suspended in water. In reference to normal operation, a motorcycle is suspended at ground level with the entire weight distribution on two tires at their ground contact points. A motorcycle jumped off both wheels, or a wheelie or stoppie I think should be left out of the discussion as those are exceptions to normal operation. So, in normal operation.......

If both tires are contacting the ground, absorbing the full weight distribution of the bike, the 'X' axis is no where near the middle of the bike, rather at ground level. That is why we refer to the rotation as 'lean' rather than 'roll'. The bike will lean left and right from that ground contact point, not at the CG of a suspended bike.

Because of this tire contact on the surface, considering the surface is flat, as most of our streets, the bike is not going to rotate around the 'Y' axis, you reference as somewhere near the middle of the bike, rather not rotate at all, other than a small tire compression difference between front and rear dependant on the trust/brake loading. The frame, due to suspension would create more of a rotation as the front/rear loading changed, the the point of rotation would again be at ground level, not through the point of CG.

The only axis I think may be relative in the explanation is the 'Z' axis. This should be somewhere in the center, between the front and rear tire contact points, but will not change due to the CG loading. For example, the CG will change from solo riding to 2-up riding, position of the rider(s) sitting upright or leaning forward, or somewhat by the level of fuel on board, yet the axis point will never change as it is fixed by ground contact of the tires. OK, may be a very small bit dependant on the trail and angle of steering deflection.

Anyway, my response here isn't to drive off topic, rather I thought others may be confused as I was to the value of the relationship you used in this 'first' point. On the other hand, perhaps I'm still confused!!

I do know (at least I think I know!), if the steering head of our FJR's are locked, the bike will not turn and becomes unrideable. I know using counter steering is an expedient method of inducing a lean, which counters the centrifugal force of a turning bike. I also know the explanation often used such as simply, "push on the right grip to turn right" will almost always dump the bike! If we push forward on the right grip (to induce a lean to the right) and don't follow-up with turning the steering head into the turn, the bike will turn left, throw the weight to the right and without the rider moving his position adjusting the CG (works at parking lot speeds, not highway speeds), the bike will simply fall over to the ground.

Despite the relationship and effect of centrifugal force, gyro force, rake and trail, CG, contact patch, slipping and friction, etc., if I don't steer the front tire, after inducing the lean, back in the direction of the turn, I'm going down and not having a good day.

I'm not a scientist, nor engineer, and have never written and published a research paper, so I may fall into the category of, "along with quite a bit of stuff that appears to have been made up on the fly because it sounds good"........if so, I'm good with that.

 

[wil780]

I have a confession. There really are two different forces that cause a bike to turn. One operates at low lean angles and cornering forces, and works when the front wheel is pointing straight ahead (tangent to the curve the bike is following.) The other, the one I have been saying is the only turning force, is predominant at moderate and greater lean angles and turning forces, and arises because the front wheel is turned slightly into the corner and therefore has to be slipping slightly. Slipping causes a lateral frictional force that causes the front wheel to move to the side (which is the definition of a turning force).

I've been fibbing about this (Fred W noticed!) because until today I've found the first mechanism to be completely incomprehensible. I was hoping someone would come along and say, you dummy, there is this other way of generating a turning force and here's how it works. That didn't happen, but in thinking about one of the many bloopers in post #121 it occurred to me that the "coin effect" might be key to understanding the first mechanism. I went back to the book, and suddenly what had been mumbo-jumbo turned into something I think I understand, sort of. Further down I'll try to describe this other mechanism.

First, for you "I don't think, I just ride" types, please try to understand that today I am just as excited as I was the day of my first roadrace. Or, if not understand, at least try to accept that for me this is a worthwhile exercise.

Second, the "coin effect". Once a rolling coin leans over, it follows a curving path that looks similar to a motorcycle turn. Each contact point is displaced slightly from the previous one, relative to the direction the coin was moving while the previous point was touching the ground. The coin follows the curving path defined by the displacement (as long as traction allows it) because there's nothing to stop it (because the previous contact point lifts off the ground while the next one touches it). This can't work for a motorcycle, because there's the whole length of the contact patch behind the leading edge of the contact patch. It's absurd to think the front few thousandths of an inch of the contact patch could generate enough force to move the entire contact patch sideways.

Third, where is this coming from? My reference book is Motorcycle Dynamics by Vittore Cossalter, 2nd English Edition, 2006. ISBN 978-1-4303-0861-4. It was recommended to me by one of the authors of the last two papers I cited in post #117. (I corresponded with that group in regard to the "cone effect" that started this thread.) If you want to get some perspective on the stature of this group of scientists, read a brief biography of Jim Papadopoulos called "The bicycle problem that nearly broke mathematics" in Nature Volume 535, 338-341, Issue 7612, July 21, 2016.

The section of the book that describes why a motorcycle turns in only 3 pages (47 - 49). It consists of brief, terse verbal descriptions that probably have lost some meaning during translation and diagrams that should be helpful but make no sense unless you can figure out the right way to look at them. Here's my (possibly very limited) understanding of why a leaned motorcycle front tire that is pointing exactly in the direction the bike is moving can generate a lateral force that causes the bike to turn. Imagine the tire as a bunch of coins side-by-side and think about what happens to one coin. As the tire rotates, each point on the circumference of the coin passes in turn through the contact patch. The key point is imagining the shape of the line on the ground that is drawn by the coin as it passes through the contact patch. If there were only one coin the shape would be a curve, as noted by OldernWZer in post #121 and repeated above. However, the coin is not free to follow that curve, because as soon as a point on the coin reaches the front of the contact patch it is locked into the path the whole bike is following. So instead of drawing a curve on the road it draws a (nearly) straight line. Consequently the tire tread has to distort sideways within the contact patch. This distortion puts a lot of lateral force into the ground, and the reaction to that force is what moves the contact patch (and the front tire) sideways. And that's why a bike turns at low lean angles.

Yeah, even after I thought I finally understood the diagram in the book, I had to think this through dozens of times before it made a bit of sense. I would be delighted if someone could come up with a better way to describe this second turning force.

 

[FJRay]

I think I'm getting a headache. I'm not quite sure what or why about the turning but I'm damn sure glad it works.

 

[wil780]

Following up on my post #131, I think this tread distortion mechanism for turning explains something I've always wondered about. Why does the handling change so much when the front tire is nearly worn out? After all, the change in tire shape is really pretty small if you've been riding the twisties and the sides as well and the centre are worn past the wear lines. Maybe it's because there is very little tread left to distort, so this mechanism no longer works the same because it has to distort the carcass rather than the tread.

 

[OldernYZer]

Eagle Six,

I only brought up the standard layout of the three axes because I referenced them in the text, and I wanted anyone reading to be on the same page.

If both tires are contacting the ground, absorbing the full weight distribution of the bike, the 'X' axis is no where near the middle of the bike, rather at ground level. That is why we refer to the rotation as 'lean' rather than 'roll'. The bike will lean left and right from that ground contact point, not at the CG of a suspended bike.

The above is a common misconception. The bike still has an X axis, and it's still above ground. Taken as a static object, leaning a bike does cause it to swing in an arc centered over the center line of the tire contact, but the CG of the bike remains in the same place. This is actually why counter steering is more effective at initiating lean. In order to lean a static bike from one side to the other, the X axis through the CG must be set in motion and swung across the center of rotation like a heavy object on a hinge. But the bike is not static when it turns. By counter steering, one in effect moves the wheels out from under the X axis, rolling the bike on that axis, rather than having to move the center of mass itself.

As to the Y axis, if the bike changes direction, it has then rotated on that axis, albeit slowly in most cases.

 

[OldernYZer]

Second, the "coin effect". Once a rolling coin leans over, it follows a curving path that looks similar to a motorcycle turn. Each contact point is displaced slightly from the previous one, relative to the direction the coin was moving while the previous point was touching the ground. The coin follows the curving path defined by the displacement (as long as traction allows it) because there's nothing to stop it (because the previous contact point lifts off the ground while the next one touches it). This can't work for a motorcycle, because there's the whole length of the contact patch behind the leading edge of the contact patch. It's absurd to think the front few thousandths of an inch of the contact patch could generate enough force to move the entire contact patch sideways.

And yet, it does contribute, exactly as you said and for those exact reasons. Take any motorcycle or bicycle wheel and tire of any profile, pump it up 'til it's rock hard, or leave it mushy, lean it over slightly and roll it across a parking lot. It will turn in the direction of lean in a radius determined by its own radius and lean angle. I said at the outset that the whole dynamic was a complex mix of several effects. The "coin effect" is absolutely one of them. Obviously, it's not the only one or a pivoting front wheel wouldn't be required.

You are completely correct in assuming that there has to be some scrubbing and distortion of the tire tread because of the fact that the inner edge of the contact patch is taking a shorter trip than the outer edge is, but that's true even in a four-wheeled, non-banking vehicle being turned by steering the front (or rear, for that matter) tires. The steering mechanism itself in any partially sophisticated chassis has compensation for this built in to have the left and right wheels follow different arcs, but the tire itself still has a certain width of tread in contact with the ground, and the outside of that contact still goes farther than the inside.

The force that counters the lean angle and holds the bike from falling over is the lateral acceleration that causes the centripetal force outward from the turn.

 

[OldernYZer]

Following up on my post #131, I think this tread distortion mechanism for turning explains something I've always wondered about. Why does the handling change so much when the front tire is nearly worn out? After all, the change in tire shape is really pretty small if you've been riding the twisties and the sides as well and the centre are worn past the wear lines. Maybe it's because there is very little tread left to distort, so this mechanism no longer works the same because it has to distort the carcass rather than the tread.

The tire carcass can become either harder or more pliable as it wears, depending a lot on how and what it's constructed of, how much of its aging is the result of actual use rather than the passage of time, etc. The tread isn't the only thing that wears. The carcass moves against itself internally as it is repeatedly compressed and released from pressure while it rolls, creating the friction that causes the tire to heat up. Over time, this changes the tire's response to applied pressure, and that can change the rolling, or dynamic shape of the tire as viewed through the cross section in contact with the ground. That in turn subtly changes the relationship between trail and the steering axis, the tracking of the contact patches of both tires in relation to each other, etc. Even though the tire may not look a different shape, dynamically, it would surprise me if it wasn't.

Then there's the scrub issue you raised. Any change in the adhesive character of the worn tread, and any change in the size and/or shape of the contact patch will alter that, too.

 

[Eagle Six]

Eagle Six,
I only brought up the standard layout of the three axes because I referenced them in the text, and I wanted anyone reading to be on the same page.

If both tires are contacting the ground, absorbing the full weight distribution of the bike, the 'X' axis is no where near the middle of the bike, rather at ground level. That is why we refer to the rotation as 'lean' rather than 'roll'. The bike will lean left and right from that ground contact point, not at the CG of a suspended bike.

The above is a common misconception. The bike still has an X axis, and it's still above ground. Taken as a static object, leaning a bike does cause it to swing in an arc centered over the center line of the tire contact, but the CG of the bike remains in the same place. This is actually why counter steering is more effective at initiating lean. In order to lean a static bike from one side to the other, the X axis through the CG must be set in motion and swung across the center of rotation like a heavy object on a hinge. But the bike is not static when it turns. By counter steering, one in effect moves the wheels out from under the X axis, rolling the bike on that axis, rather than having to move the center of mass itself.

As to the Y axis, if the bike changes direction, it has then rotated on that axis, albeit slowly in most cases.

Yes, I understand and agree every mass has a center of gravity and therefore three axis it can be moved around by applying some type of control force when it is suspended in space. But for the purpose of turning an FJR if the weight was not suspended by the contact with earth, moving the front tire right or left, as in steering or counter steering would be a very poor method to get the bike to turn, similar to just using the rudder on an airplane or a submerged watercraft. I'm sure any engineer would disagree with my statement, that I feel an advantage by not understanding or relying on physics and geometry to figure out that the mass weight of an FJR is going to lean when counter steering regardless of the supposed CG. I will contend however, stretching my feeble mind, that the placement of weight above the ground contact will make a difference in the force required to lean the bike and/or upright it back to straight up balance.

Thank You again for the reply and the attempt to educate. I always enjoy the information even when it doesn't agree with my simplified Redneck version. My simplest answer to the thread question, "Why does a bike turn?" is, one answer, "because the operator wants to go left or right", regardless of the technique used. I know that is much to simple and far simpler than the intent of the OP. And, now that my head hurts, I need to go out and ride, and feel good I don't need to know why, but rather how. The 'how' being much more important to me than the 'why', but also understanding there are those who also need the 'why'. But for me, I think your knowledge and effort to educate may be a waste. Nevertheless I applaud your tenacity.

 

[OldernYZer]

It's never a waste. Sometimes it takes several coats to get a wall covered, and seeds sometimes take months to germinate. Besides, one thing about engaging in a technical explanation as an "authority" is that it does (or should) force one to think about the matter deeply enough to explain it. It's always good exercise.

 

[wil780]

I wasn't going to get into an argument with OldernYZer, but there's one pretty important error that many motorcycle journalists also make and I think it's worth discussing so that anyone reading this far gets a different point of view. It's in this paragraph of post #121:

Vehicles with two inline wheels steer, or turn, using a combination of dynamics that blend to varying degrees, very well or poorly, depending on the overall chassis setup. First, since it’s been brought up so many times, let’s examine counter steering. Counter steering can be used to initiate a turn by encouraging the motorcycle, bicycle, or scooter to bank, or lean into a corner. Essentially, it is a way to force the bike to roll on the X axis in spite of the angular momentum (gyro effect) of the wheels which resists its doing so. The concept is rather like that used to balance a hammer or broom on one’s finger tip with the heavy end up. Balance is maintained by shifting the low end of the mass in any direction on the X-Z plane, causing the object to roll or pitch in such a way as to remain upright and counteract the tendency to fall one way or other. When a bike running straight and upright is counter steered to the left, the contact points of the wheels are moved to the left under the CG, and the bike rolls, or falls, toward the right. Note that it is also true that one can simply shift the rider weight to the right over the CG and cause the bike to roll right in response to that without counter steering, so counter steering is not an essential element of turning, just a more effective way of initiating a turn more quickly and with less motion than shifting the rider weight.

It's easy to see that gravity is a minor reason why countersteering causes a bike to roll. Consider the following numbers, which come from simple trigonometry. (I used sine of the angle from vertical, but I didn't spend much time on the diagram and wouldn't rule out tan as the correct relation, but the numbers come out basically the same at these angles.) When the bike is upright, 100% of it's weight acts to squash the tires. At 5 degrees of lean, 8.7% of the weight acts to roll the bike, the rest is still squashing the tires. Use that 5 degree value as the standard. At 10 degrees twice as much of the weight acts to roll the bike. At 15 degrees it's almost three times as much. If gravity were the main reason a bike rolls, it should roll either much faster or much more easily as the lean angle increases. Gravity clearly does help the roll, but it's such a minor effect that it's hard to notice. What does happen is that when you want to turn right and countersteer left, the bike starts to turn left and centrifugal force leans it to the right.

Ah, what the heck, while I'm at it I'll point out the error in the next paragraph:

Gyroscopic precession does not initiate or maintain a turn at all, and in fact works specifically against any rotation of the axle in any direction from its position on the Z axis. It is a highly stabilizing influence on the chassis, owing to the mass of the rotating tire and rim assembly. The strength of the effect depends the mass involved, and on the diametrical location of the rotating mass, the farther from the center, the more effective it is. It resists leaning the bike in either direction, and that is the main reason that counter steering is used to more forcibly cause the lean than by shifting the weight of the rider.

Well, the gyroscopic force does affect leaning (rolling), but its effect is indirect. When you're countersteering to the left and leaning to the right, the gyroscopic force acts to turn the handlebars to the right. So it directly counteracts your countersteering efforts and thus increases the force you have to put into the bars. The net effect is the same, but the picture of what's happening is different. This isn't just my opinion. It's stated explicitly in the book I cited above.

Oh, and by the way you can't shift your weight to the side without having the bike move to the other side (Newton's laws and all that). So something has to resist the bike's motion to the other side if you are to move the centre of gravity by moving yourself sideways on the bike. That does happen - in effect the bike countersteers for you. So it's not really correct to say there is no countersteering when you lean the bike by shifting your weight sideways. It's just that you aren't countersteering.