rec.autos.simulators

First of all, thanks to Gregor/Todd/Doug for the info about instant
centers a week or so ago.

I'm looking into the kinematics of suspensions. Ultimately ofcourse,
I'd like to describe it as a series of rods, and be done with it.

So let's start from 5 rods. It doesn't seem that drastic, intuitively,
to reduce the problem of the path that they describe, in 2 dimensions;
front and side views.

I've been scribbling a bit, and you could describe the rod endpoint
(at the wheel) as a line (p0,p1):
p0 = start point at the chassis
p1 = end point at the wheel
p1.x = cos(a+a0)*r
p1.y = sin(a+a0)*r

Where a0 is the static angle (when the suspension is in rest), and 'a'
is the added angle that changes when the wheel goes up or down.

Now suppose you have another rod, q, with the same type of
description. When the wheel moves, and p1 moves along to p1', I start
wondering how to find q1' (the new, moved q1).
A constraint I see is that distance(p1,q1) remains constant. Writing
this out gives a bunch of cos() and sin() arguments that seems hard to
solve.

Here's a nice Courier/fixed ASCII drawing:

    q0 o----o q1
            |
(chassis)   | (wheel side)
            |
  p0 o------o p1
     <-  r ->

Does anybody have a clue/hint as to a perhaps workable way to find
q1'? The sin/cos mess doesn't seem solvable, actually.

The above picture does look like an inverse kinematics thing. Does
anybody do this in 2x 2D (front/side) already? Or would solving in 3D
be a lot better? (although that would mean having 5 links at the same
time, which is much more costly).

Any eye-openers appreciated. ;-)

Thanks,

Ruud van Gaal
Free car sim: http://www.racesimcentral.net/
Pencil art  : http://www.racesimcentral.net/

P0 is the center of one circle, radius r0. P1 is the center of the other
circle, radius r1. P3 is/are the intersect point(s). P2 is defined as lying
on the line P0 -> P1, as the midpoint of the overlapping region of the two
circles. It can be found by simply subtracting one quarter of the overlap
value from either of the circles' radius. a is P0 to P2, and b is P1 to P2.
h is the distance from P2 to P3. If you can't visualise this then I will
draw up a picture. a, b, P0, P2, P2, d all lie along the same line (P0 ->
P1). d = a + b.

First calculate the distance d between the center of the circles. d = ||
P1 - P0 ||. If d > r0 + r1 then there are no solutions, the circles are
separate. Also, if d < || r0 - r1 || then there are no solutions because one
circle is contained within the other.

Considering the two triangles P0P2P3 and P1P2P3 we can write

a2 + h2 = r02 and b2 + h2 = r12
Using d = a + b we can solve for a,

a = (r02 - r12 + d2 ) / (2 d)

It can be readily shown that this reduces to r0 when the two circles touch
at one point, ie: d = r0 + r1

Solve for h by substituting a into the first equation, h2 = r02 - a2

So

P2 = P0 + a ( P1 - P0 ) / d
And finally, P3 = (x3,y3) is

x3 = x2 +- h ( y1 - y0 ) / d
y3 = y2 -+ h ( x1 - x0 ) / d

http://astronomy.swin.edu.au/~pbourke/geometry/2circle/

It doesn't help when copy paste converts r0 ^ 2 (r0 * r0) into r02 all over
the place...

Nick.

Ah, close enough. It's just math innit? :)

Other than that, you could also buy one of the many software programs
that do this. They are in the hunreds of dollars. Pass the plate? ;)

- Matt

I'll get back on this later, gotta run...

Ruud van Gaal
Free car sim: http://www.racer.nl/
Pencil art  : http://www.marketgraph.nl/gallery/

I only suggested using the circular method because setting up the IK is
pretty complex to use just for a 2D solution with 2 arms. It's kinda like
buying a GeForce 4 to play Solitaire :-). Oh yeah, please don't read my
first mail, check the website instead. When I was looking for that problem
(on a totally unrelated project), I found that web page and copied the text,
and it copied all the powers fine as well. But when I copy it into an email
it screws up, so I went and found the original web page instead. Shouldn't
try to take credit for somebody else's work anyway....

In 3D though (which I have no doubt you will end up using), an IK oriented
solution is the only one really useful in a realtime environment. Especially
as you will be calculating (usually) 4 times per car, times (say) 20 cars on
the racetrack every frame.

I never really thought about my solution being IK, but having looked at 3D
Studio MAX and a couple of web pages on the subject, I realise I could have
saved myself a whole bunch of work :-(

Nick.

Probably carrying coals to Newcastle here, but it's a good idea
to create a table / array of pre-calculated values on program
launch so these cpu-intensive functions don't have to be called
500 times a second ;-)

Cheers,

Uwe

--
Uwe Schuerkamp //////////////////////////// http://www.schuerkamp.de/
Herford, Germany \\\\\\\\\\\\\\\\\\\\\\\\\\ (52.0N/8.5E)
PGP Fingerprint:  2E 13 20 22 9A 3F 63 7F  67 6F E9 B1 A8 36 A4 61

Hey, hasn't anyone been listening?  Or have my posts been going
to the old bit bucket<grin>??

Olley solved this stuff with some neat approximations (and then worked
examples with a slide-rule) over 40 years ago...and now we have pulled
it all together for your analytic pleasure.

http://www.millikenresearch.com/olley.html

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The thing seems to be that you can do suspension geometry in a million
ways; many ways of which are simplified but very useful still. The
idea is to get the definition right (for the users/creators of the
suspension geometry) and always be able to switch to a different
implementation when you feel like it's raining too much outside. ;-)

Got it. Too bad shipping costs are a bit high from the US (tried
Amazon in the UK but they didn't have it yet).

Ruud van Gaal
Free car sim: http://www.racer.nl/
Pencil art  : http://www.marketgraph.nl/gallery/

But still, indeed, 500 times/sec times 4 suspension times 20 cars...
Hmm.

Ruud van Gaal
Free car sim: http://www.racer.nl/
Pencil art  : http://www.marketgraph.nl/gallery/

Chassis Design is also being sold by the IMechE in England -- they might
have cheaper shipping to Europe (but the basic price might be higher...)