Total Motorcycle Community Forums

Yes, you are completely right about the wheel having to turn about the centre of its turning circle, not the contact patch, and therefore the scrub produced would be so minimal as to be negligible (taken up by tire slip).

Now, my attempt at an explanation as to why a sportbike can turn so sharply with a huge lean and hardly any front wheel steer. When a bike is leaned over, the action of turning the front wheel doesn't just turn the contact patch, but it also moves it along the tire. The fact that it moves laterally due to leaning is obvious, but it also moves longitudinally. Take for example a hypothetical bike which has totally vertical forks (zero trail), and can lean 90 degrees (impossible, i know, but bear with me). When said bike it leaned at 90*, the entire tire sidewall can be touching the ground when the front wheel is straight. But when the wheel is ever so slightly turned in the same direction as the lean, only the front part of the tire will be touching the ground, not the bottom part, which demonstrates that the contact patch has moved forward (by 90* around the outside of the tire). Obviously this is a huge exaduration, but the same priciple can be applied to a real bike. Say a bike is leaned 45* to the left. When the front wheel is straight (along the centreline of the bike), the contact patch is dircetly below the wheel centre (below meaning along the vertical axis of the bike). When the front tire is turned a bit, the contact patch moves forward to the front part of the tire. Obviously a real bike doesn't have zero trail, and so the tire doesn't turn about the bike's vertical axis but at an angle, so this slightly reduces this effect. As an interesting point, I think if you turn the wheel so much that the contact patch moves forward of the projection of the front forks, you end up with an instantaneous trail which is negative, and the bike becomes unstable and would probably cause a high side.

Now, what difference does it make if the contact patch moves forward? When the contact patch in on the bottom of the wheel, and the wheel is rotating, the wheel will move in a forward direction only. When the contact patch is on the front of the wheel and the wheel turns, it will move in a sideways direction only. When riding a bike, it is usually somewhere in between. This is hard to explain without a diagram so I'm not sure if you understand what I am trying to say. As a demonstration, grab some sort of circular wheel (a roll of tape works well) and hold it in front of you on a table. Now lean the roll of tape to the side. Notice the contact patch may have moved laterally, but not longitudinally. Now turn the wheel about its vertical axis (which is now at an angle to the table), much like the forks would turn a cycle wheel. Notice the contact patch moves forward, and the direction it points in is not straight ahead.

Basically, the more lean you have, the less front tire turn it takes to make the bike turn, but if the front tire stays straight ahead (along the centreline of the bike), the bike will not turn no matter how much you lean. This can be demostrated with an example as well. Hop on a bicycle, and get up to speed. Then, stand up on the pedals, and lean the bike, while keeping your weight vertically above the contact patch (so only lean the bike, but not yourself). You will find that you can continue to ride straight by holding the bars completely forward. This demonstration also proves that the tire profile causes an insignificant force.

If this doesn't make sense to you, let me know. I'll try to draw a diagram and post it.

Jacob

Bike tire is curved, you can sort of recreate this by holding two beer glasses open end together, the shape isn't exactly round, but it's the same idea. Now, take one glass and roll it along the table, what happens?

Sevelturus - this is the explanation that i have just disproved. If the turning force was due to the round profile of the tire, then a thin disc would exibit no turning force as it is leaned, which is not the case. The profile does produce some turning force, but it is, I think, insignificant in the big picture due to the small difference in diameter between the outside and inside of the wheel, as well as the small lateral distance between the outside and inside of the contact patch.

Hi Jacob

OK, you've done the job - for me at least. I was aware of the longitudinal movement of the contact patch on the tyre from several articles I'd read. They had all claimed this as a factor in turning the bike, but none of them had elaborated and I'd never thought it through in enough detail to identify the mechanism (I'm fundamentally lazy). So, many thanks for making it clear.

That leaves me with just one question. I'm taking you at your word that, despite the countersteer, the wheel moves in the direction of turn relative to the bike's centreline. It makes sense, but at root I'm an empiricist (albeit a lazy one) and need to go out and check this myself.

In the meantime, one image does pop up to haunt my new found clarity, and that is something I've mentioned before. What is going on when a speedway bike corners? Of course there is a huge amount of slide taking place, but there must be some turning force at the front wheel - which is most definitely not pointing in the direction of the turn relative to the centreline, or is this an illusion (or just poor observation)? Are we talking about a whole different set of turning forces at work here? Do you have any thoughts on this?

As for my attempt to rescue the the contact patch theory from your bicycle example by reference to the radial angle, I checked out some bicycle tyres at my work today and decided to forget it!

As a sidelight, your longitudinal-displacement-of-the-contact-patch explanation does reintroduce the issue of scrub into the turning equation in, perhaps, quite a big way. If the differential rate of travel at the contact patch were the only factor in creating the turn, then the only scrub would be that caused the the finite wheelbase of the bike. (As the radius of turn would be exclusively determined by the differential distances travelled at the contact patch on a single wheel no scrub would be generated). However, if some other mechanism (logitudinal displacement) were responsible for creating a smaller radius turn than the contact patch alone, then there would be a significant amount of scrub, enough, I would guess, to generate a fair amount of heat and affect the efficiency of the bike. Under those circumstances it might even eliminate the differential at the contact patch as a useful turning factor altogether.

Just a thought.

Cheers

Richard

I see that you fellows are very interested in this subject. I heard that when a bike is in a turn, such as a 40 mph curve, that the front tire is actually turned in the opposite direction of the turn. And that the rear tire is actually running on the outside track of the front tire. Do any of you know for a FACT if this is so, or not so???

Jacob has got it right, and explained it way better than I ever could.

I'll take a wack at the counter-steering thing though....

When the bike is traveling straight and upright, the counter-steer movement tosses the bike to the outside of the direction of that bar movement(the direction you actually want to lean in to take the corner). Once the bike is heeled over to the correct side, you turn the bars into the turn slightly and the bike turns for the reasons previously stated. The counter-steering works to throw most of the bike's mass around for you, when done properly.

So why is it that when going around the turn you are actually pulling the bars away from your turn???

niterider wrote:So why is it that when going around the turn you are actually pulling the bars away from your turn???

When you countersteer, you aren't actually making the bars move away from the direction of turn (except perhaps initially and very momentarily), you are just appling a small amount of pressure on them in that direction. This is necessary to create and maintain the lean. If you don't do this, as posthumane explained, the bike will sit up and continue to move in a straight line. However, by leaning the bike in the direction of the turn you bring into play several forces and conditions which make the wheel move in the direction of the turn, also.

So it's a complicated balance of actions. To start the turn you 'countersteer' to the right to make the bike lean to he left. The bike's lean to the left then causes the front wheel to turn to the left. The front wheel's turn to the left steers the bike to the left away from its straight line motion.

While in the turn, you need to maintain pressure on the bars to the right to keep the bike leaning to the left, so the front wheel (and the bike) will continue turning to the left also.

It's no different to life, really. Things often seem to go in the opposite direction to what you intend.

Ok, I'm glad that you understood what I was talking about, as I wasn't sure if I was clear about it. I will attempt to tackle a couple other points.

First, your concern about tire slip. Yes, there is a fair amount of tire slip when a bike is cornering hard, and yes, it does generate a lot of heat. This is the primary heat source for the tires on a racing bike (accelerating and decelerating generate heat as well, but mostly in one tire at a time). The effect of tire slip in this case is exaclty the same as in a car. Basically you have two contact patches, not pointing in the same direction. This obviously is causing the bike to turn. However, due to tire slip, the contact patches will be moving partly along their longitudinal axis, and partly along their lateral axis. Tire slip at the front causes a bit of understeer and serves to increase the turning radius of the bike, and tire slip at the rear causes understeer and reduces the radius of the turn (and some instability). If both front and rear tire slip is exatly the same (which it rarely is, as it is affected by so many factors such as tire and suspension design, acceleration/deceleration, etc), then the turn radius would actually be exactly the same as without tire slip, but the centre point of the turn (the point about which the bike rotates) would move longitudinally forward along the bikes original path. Another interesting point is that acceleration on the rear tire acts only in the longitudinal direction, and therefore causes longitudinal slip at the rear only, but a fair amount of lateral slip at the rear.

Now, your other point. When you talk about "speedway bikes", I'm guessing you are referring to "dirt track" or "flat track" oval racers (ie, flat, semi-circular dirt oval, with studded tires and no brakes)? If so, the cornering mechanism is comptletely different due to the sliding action of the rear wheel. The bike is leaned over to counteract the inertial cornering force, just as on a normal bike (otherwise the driver flips to the outside of the turn, high side style), but the front tire in this type of turn is almost upright, and is being steered left/right like a car to keep the front of the bike going in the desired direction. Obviously this wouldn't work as well in a high traction situation, because it relies on the sliding action (and the forward drive) of the rear tire to maintain cornering force. While it is possible to do this same sort of powerslide on pavement, it is not the fastest way around a paved corner due to the relationship between the tire and the pavement (specifically the fact that static friction between the two is much higher than the dynamic friction when sliding).

That's my 2 cents. Hope that makes sense.

Jacob

I took one of my toy bikes and I was unable to make it turn left with the front tire pointing to the right. Also the rear tire did not track out side the front tire. The only time this will happen is on a dirt track. And as for holding the bars to the opposite side of the turn makes sense, because it is natural for the front tire to lean to the left when the bike leans left. Like when parking the bike the front tire turns to the left.