Modelling bicycle trail - Tough maths?

Here's the basic model of a bicycle.

B --------------- A

A is the contact point of the front wheel on the ground and B is that of the
rear wheel. The distance (length) between A and B is fixed, let that be denoted
by R.

There are two ways in which B follows A.

1. When the front wheel is locked at an angle relative to AB whilst the bicycle
is in motion in which both trails made by A and B spiral away from the initial
position of B.

2. When the front wheel made to travel on a straght line on the ground and
eventually B moves along the trail made by A.

Given that we are to find these parametric functions (with respect to time) on
the cartesian plane (x-y plane), with initial positions B(0,0) and A(0,R), given
initially only:-

Case 1) Fixed angle relative bicycle body and speed of front wheel;
Case 2) Fixed angle relative to plane and velocity of front wheel;

How hard would the mathematics be? Also if you think this is too easy what would
the functions look like - one set for each wheel?

Assume, gound is perfectly flat and wheels perfectly circular.

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888 wrote:

Not very hard. Yes, there would be one set of functions (or a vector-
valued function) for each wheel; the wheels are not in the same places
at the same times.

Case 1 can be solved for both wheels with just some geometry and an
appeal to symmetry. Case 2 is trivial for the front wheel, but
somewhat more difficult for the rear.


- Tim

888 wrote:

I think I might have (finally) figured out question 2.

For simplicity, suppose that A follows the x-axis.
(Ax, 0) are the coordinates of A
(Bx, By) are the coordinates of B
Ths speed of A is constant at v (in the positive x direction)
r is is the distance AB
t is time

And, for symmetry, suppose that at t = 0 we have

Ax = 0
Bx = 0
By = r

A is travelling at constant v, so we know immediately that

Ax = v*t

At some arbitrary time, suppose the angle of AB with the negative
direction of the x-axis is theta. To avoid worrying about too many
cases (it'll all come right in the end), assume that 0 <= theta <=
pi/2. Instantaneously, B can only move in the direction towards A. The
component of A's velocity in this direction is v*cos(theta), so B's
speed along the line BA is also v*cos(theta). B's "horizontal" speed
(parallel to the x-axis) is therefore v*cos^2(theta). This gives us
the equation

d(Bx)/dt = v*cos^2(theta)

But we know that cos(theta) = (Ax - Bx)/r = (v*t - Bx)/r, so

d(Bx)/dt = v/r^2 * (v*t - Bx)^2

This equation gave me a whole load of grief, but, under the stated
initial conditions, I think the solution is:

Bx = v*t - r*tanh(v/r*t)

Then, using Pythagoras,

By = sqr(r^2 - (Bx - Ax)^2)
= sqr(r^2 - (Bx - v*t)^2)
= r*sech(v/r*t)

which completes the solution for the motion of B.

<matt271829-news@yahoo.co.uk> wrote in message
news:1122329628.770543.179200@g49g2000cwa.googlegr oups.com...

Hi there, I've been trying to solve this for the past few days in my spare time
and it's not easy at all - especially case 1!

I just briefly scanned through your possible solution and it looks promising.
Before I actually diving into it can you clarify what you meant by Ax, Bx etc...
I'm assuming it means Point A on the x-axis and B on the y-axis and so on. If
so, does...


....mean Ay = 0, same as the initial point I stated? Thanks!

888 wrote:

Yeah, me too!


I think that case 1 is easy to do "intuitively". It seems "obvious"
that A and B must describe concentric circles, and, as Mr Little
suggests, I think the geometry of these circles can be worked out
fairly easily.

You COULD try to solve case 1 mathematically using a method analogous
to my post, but I think it would be rather more difficult than case 2
(unless maybe you just showed that the answer found intuitively
satisfied the necessary differential equations). I think there ought to
be a far simpler yet still rigorous method that just relies on
rotational symmetry, but I have not thought about case 1 too much.


A and B can (potentially) be anywhere on the plane, so they are located
using x- and y-coordinates (I assume you understand coordinates?)

(Ax, 0) are the x- and y-coordinates of point A (i.e. the point at
which the front wheel touches the ground). I could have written this as
(Ax, Ay), but, as you say, Ay is always zero so I just wrote (Ax, 0).

(Bx, By) are the x- and y-coordinates of point B (i.e. the point at
which the back wheel touches the ground).

[Only registered users see links. ] wrote:

[...]


Actually, you've got me wondering now. I'm beginning to doubt that
every solution is concentric circles. Ho-hum... need to think some more
about this!!

[Only registered users see links. ] wrote:

False alarm... I was confusing myself unnecessarily. My opinion is back
to being that concentric circles are the only solution (apart from the
special degenerate cases of a circle and a point, or two coincident
straight lines).

<[Only registered users see links. ].uk> wrote in message
news:1122338321.877344.62420@g14g2000cwa.googlegro ups.com...

Did you meant "concentric spirals"? Maybe I'm confused. I should go and find
myself a bike and ride it round and round with wet wheels on a flat concrete
space. Then that will be very obvious...

888 wrote:

No, circles. I don't see how you can ever get spirals with the front
wheel set at a fixed angle. You may find it easier to forget the
bicycle with the potentially moving handlebars and think of a
trolley-like vehicle with one wheel welded at a wonky angle. As you
push the trolley it must just go round and round in circles, yes?

Draw two concentric circles, centred at O.
Mark any point B on the inner circle.
Draw a tangent through B, to intersect the outer circle at A
Extend the line beyond the outer circle to point E (E can be at any
distance - it's only so's we can identify angles later)
Draw line DAC tangent to the outer circle at A (C and D can be at any
distance -they're only to identify angles).

Let theta be the fixed angle of the handlebars. Thus:

EAC = theta

A is instantaneously travelling towards C, and so describes an
infinitesimal element of the outer circle. Similarly, B is
instantaneously travelling towards A and so describes an infinitesimal
element of the inner circle. (I'm not totally convinced that this
proves the paths WILL be circles ... something more seems to be
needed.)


The goal is to find OB (inner radius) and OA (outer radius) in terms of
AB and theta.

Note that OBA and OAD are right angles (OBA is a right-angled
triangle), and then continue with some simple geometry:

DAB = theta
OAB = OAD - DAB = 90 deg. - theta
BOA = 90 deg. - OAB = theta

So

OB = AB*cot(theta)
OA = AB*cosec(theta)

Or, probably to cope with theta > 90 deg. we should write

OB = |AB*cot(theta)|
OA = |AB*cosec(theta)|

[Only registered users see links. ] wrote:

[...]


Here is a more satisfactory explanation...

Draw line BA (anywhere)
Extend line BA to point D (at any distance)
Draw line AC so that angle DAC is theta
Draw a line through B perpendicular to BA, and a line through A
perpendicular to AC. These lines intersect at O (unless theta = 0,
which is a degenerate case).

Consider the instantaneous situation:

A is moving towards C (perpendicular to OA), so d(OA)/dt = 0
B is moving towards A (perpendicular to OB), so d(OB)/dt = 0

Thus OA and OB are constant, so A and B must describe circles centred
at O.