# How does it turn?

1. Dec 29, 2004

### Bartholomew

I'm not sure what forum this should be in--is it a physics puzzle, a geometry puzzle, or an engineering puzzle? It seems to be all three, so I put it here.

When you have a vehicle with two wheels and one of them is canted at an angle, and a circle can be constructed on the ground so that it has two points of tangency at the two wheels, and the lines of tangency at these points follow the directions of the two wheels, it is clear what circle the vehicle must turn in.

Now consider a bicycle. You can't construct such a circle here; the point of contact of the front wheel falls on the line of the direction of the rear wheel. In one case, when the front wheel is turned at a right angle to the rear wheel, the circle the vehicle must turn in is obvious; the front wheel traces a circle with the bike's length as the radius, and the back wheel stays put. But when, say, the front wheel is at a 45 degree angle to the back wheel, how large a circle does each wheel trace?

2. Dec 29, 2004

### brewnog

Not sure what your problem is....

Draw lines through the axles of both wheels. The intersect will be the centre of the circle.

Have I misunderstood?

3. Dec 29, 2004

### Bartholomew

Hmm, okay, but why does that work?

Edit: And which circle is that--front wheel?

Edit: Oh, I think I see--both wheels.

Last edited: Dec 29, 2004
4. Dec 29, 2004

### Gokul43201

Staff Emeritus
My preliminary thoughts :

1. If the bicycle does, in fact move entirely without slipping, the two wheels must trace out two different, though concentric circles.

2. If experimentally, one can show that it is only one circle that is being traced out, then the bicycle wheels are not rolling without slipping.

5. Dec 29, 2004

### Bartholomew

I get it entirely now.

Call A the point of contact of the back wheel, L the line of direction of the back wheel, B the point of contact of the front wheel, and M the line of direction of the front wheel. You move M back and forth (call it N once you've moved it) without changing its slope until you can construct a circle with tangent N at a point C which is on the perpendicular line to M at B, and also with tangent L at point A. It is intuitively clear that you can find such a circle by moving M "inwards." Imagine point C as another point of wheel contact with direction line N. Similarly, move L back and forth (call it O once you've moved it) without changing its slope until you can construct a circle with tangent O at a point D which is on the perpendicular line to L at A, and also with tangent M at B. In this case L will be moved "outwards." Now imagine point D as another point of wheel contact with direction line O.

So you now have four points of wheel contact: A, B, C, and D. The pair of wheels A-C clearly traces a circle with a center T at the intersection of the line drawn through their axles, and with radius AT. The pair of wheels B-D traces a circle with the same center, T, but with radius BT. You can clearly see why pairs B-D and A-C move freely on their own, and since the two pairs share the same "center of turning" T, you can see that they will move freely joined together.

So it's like a 4-wheeled vehicle. Now remove wheels C and D, and you're left with the original bicycle, except now you know how it can turn.

Edit: Or, simpler, each wheel can only turn about a point perpendicular to the line through its axle or else it skids. The intersection of these two lines is therefore the only point the two wheels can turn around.

Last edited: Dec 29, 2004
6. Dec 29, 2004

### ceptimus

It's more complicated than that.

When a bicycle banks into a turn the projections of the front and rear axles do not intersect (in three dimensions).

Because of the caster angle of the front wheel, the front axle points deeper beneath the ground than the rear axle does.

7. Dec 29, 2004

### Bartholomew

So you put it in 2 dimensions. When the bicycle is banking the points of contact and respective lines of direction can be dealt with in exactly the same way, just as if they were generated by perfectly upright wheels. Of course when the wheels are tilted then the circles which you find will be on the ground only, and will correspond to the circles traced by the points of contact, not the actual wheel centers.

8. Dec 31, 2004

### RandallB

It's more complicated than that for sure.

Even if you use a single wheel - a flat metal disk. With an axle attached to a post at some distance to hold right angles all around including a flat surface your going to have "skidding"

Now the tube shaped tire around of a real bike wheel will reduce that it is still there.
And the dynamics of the whole bike is much more complex. Note the angle used on the "steering" column for the front wheel and how the front fork is "raked" so the Axle doesn't even line up on that angle. Well beyond using A, L M, N O, D & .. .. .. to understand it.

Now a good one to get you thinking on "HOW IT WORKS" is:
Once you have your bike moving which way must you turn the front wheel to turn the bike left.

The reflex you learned as a child will respond quicker than you can think, and without telling you about it. You can sense it if you're careful - or attach strings to the end of your handlebars to override your well ingrained skills and try turning by only pulling on the left string as see if you don't fall on your right.

9. Dec 31, 2004

### Bartholomew

Skidding is when a wheel's contact point follows a path which (edit) its line of direction is not tangent to. In the situation you describe there can be no skidding.

Imagine that your flat metal disk is instead a wheel made entirely of spokes, with no rim. Each of the spokes would strike the ground and the tip would remain immobile (turning in place) until the next spoke strikes the ground and the previous spoke lifts. Never does any spoke tip skid along the ground. Then just increase the number of spokes until you have a solid disk; since there is no skidding with any arbitrary number of spokes (Edit: greater than 2), there can be no skidding as the number of spokes increases forever.

The "axle" I and Brewnog were talking about is not the literal axle; it's just shorthand for "a line segment drawn on the ground through the point of contact of a wheel, perpendicular to the direction of motion."

When the wheel is perfectly upright this corresponds to the line segment of the axle (viewed from above, and superimposed on the ground). When the wheel is not perfectly upright the literal axle does not exactly correspond to what we mean by "axle." It's just a shorthand; it's more convenient to say than "a line segment drawn on the ground through the ...."

All of the various complications you're talking about affect where the points of contact are and exactly what the line of direction of each point of contact is. But once you know the points and their directions, it doesn't matter how they were made. It's like the distance traveled by a vehicle that moves straight. If I know the speed of the vehicle, and the direction it is going, then I can tell you how far it will go after ten seconds, twenty seconds, an hour. It doesn't matter what actually went into achieving that speed; it doesn't matter whether it's a sailboat or a spaceship.

For a given set of points of contact and a given set of lines of direction, all of the turning properties of those points of contact can be determined. It doesn't matter how the wheels are angled, or how the steering is done, or what the bike is made of, or anything else like that; all that matters is the situation at the points of contact on the ground.

Last edited: Dec 31, 2004
10. Jan 1, 2005

### RandallB

OK keeping it at a "geometry puzzle" where we can keep the point of contact as a spot with an area of Zero, that is the best. I can see where using engineering or physics, or even a bike wouldn't really work in that context.
That would be a different question.

11. Jan 2, 2005

### Bartholomew

It is idealized. With rubber tires at least I think there shouldn't be much real skidding. The rubber put down first should be able to stretch slightly (so long as there is enough traction) to maintain its location until pulled up again. It would spring back when the pressure is reduced as it is pulled up, which would scuff the ground slightly but without the weight of the vehicle. That wouldn't affect the geometric prediction of the turning circle and maybe you couldn't call it "skidding." With rubber tires the contact-point representation of the vehicle's turning properties should be sufficient.

On the other side of the fence is a perfect cylinder wheel with a perfect crosswise line segment of contact. This wheel could not turn at all (on a flat surface) without skidding.

12. Jan 2, 2005

### RandallB

Now there is an interesting idea behind your thought on the cylinder wheel - there should be at least one conical solid that could extend from the the center of the circle and roll around without "skidding". Only guessing but most any size wheel tilted such that the axle pointed to the tip of the cone at the center of the circle would be a circle section of a cone that might work. And the angle of the tilt would be dependent the size of the wheel, circle, and determine the size of the cone. i.e. Many many cones would work. Not sure how to proof one way or the other.

13. Jan 3, 2005

### Kittel Knight

The real thing is a little bit more complex. You are missing the gyroscopic effect from the wheels. How does a coin turn ?

14. Jan 3, 2005

### RandallB

For the geometry problem or even the bike example (weight of passanger included) I think it'd be fair to minimize the weight in the wheel compaired to the action of the bike and its wieght enough to ignore gyro effect.

But if you have an insight on the action of a rolling coin, do share.

15. Jan 3, 2005

### NateTG

Let's say that the axles of the front and rear wheels are both parralel to the plane that the bycicle is on, and that the wheels are the same size. Then the bycicle can make a circle that keeps both wheels on the same track if the axis of the byclicle's rotation goes through the intesection of the axes of rotation of the wheels.

16. Jan 4, 2005

### Gokul43201

Staff Emeritus
I can't see this happening when the plane of the rear wheel is fixed and coplanar with the bicycle frame, while the front wheel in tilted.

Or, I'm not getting what you mean by 'axis of rotation'.

17. Jan 4, 2005

### Rogerio

...both wheels on the same track?

front
wheel
/

|-----X axesintersection
rear
wheel

It seems there will be 2 different circles, as Gokul had already said.

Last edited: Jan 4, 2005
18. Jan 4, 2005

### NateTG

You're right. I'm not thinking straight, you'd need a funky bycicle, or to tilt the bike in order to have both wheels on the same track. I forgot that the must be co-planar with the rear wheel.