rec.autos.simulators

first, the notations

P = position in world coordinate
G = position of the center of gravity of the solid (CG)
Fp = force at P
Fs = force of the solid on the centre of gravity (CG)
T  = torque of the solid
m = mass of the solid
Omega = angular velocity of the solid
Alpha = angular accleration of the solid
Vs = linear velocity of the solid
As = linear acceleration of the solid

L = angular momentum of the solid
I = Intertia tensor of the solid

Vp = Velocity at P
Ap = acceleration at P

x = vector product
. = dot product

i know that

R = P - G

Vp = Vs + Omega x R

Ap = As + Alpha x R + Omega x ( Omega x R )

what the value of  Fp ?????
Do i have to use the Angular momentum, because like Omega is derivation
Alpha, L is the derivation of the Torque.

Fp = Fs + Torque x R + L x ( L x R ) ???

i know how to compute Force and Torque from a Force on P

Fs = Fp
Torque = R x Force.

It could help me to calculate real load even on a solid block, and
compute contact forces and constraint forces.

best regards,
--
Sebastien TIXIER - Game Developer
Dynamics and Car Physics
http://www.racesimcentral.net/
GPLRank Normal:-44.24 Monster:-124.44

m*Ap (using F=ma)

I think you could possibly use the angular acceleration at P to get the
force, but it would just give you the same force (assuming that the angular
acceleration and linear acceleration are consistent).

Is that what you're asking?

You know the total acceleration at p due to both angular and linear
components, and you want to know what force would have produced that, right?

Do you know As, the linear acceleration of the solid? If so then it's just

Fp = m*As.

If you don't know As (only Ap), can you explain more precisely what this is
for? I never had a need do this in my rigid body simulator, so if you give
an explanation of what you're trying to achieve, maybe there's a way around
it. (Think it should be possible to solve, but I'm just curious as to the
problem before attempting it)

Mike.

well, i have the Force , torque, angular momentum, inertia and stuff of my
solid.

Now if i want to constraint 2 solid by one point.

S1 connect to S2 on point a on S1 and b on S2 ( a= b in the world )

I want to know Fa and Fb because Fa is the force of S1 on S2 due to the
constraint and Fb is the force of S on S1.

Also i could compute the load on a point.

Imagine a block laied on a table. If consider taht there is 4 contact with the
table at each block corner. at equilibrium if i know the force at each corner i
know that the contact reponses will void this forces along the normal of the
table. So i could compute reaction forces then friction forces, i could also
compute the load at the contacts.

so what i know is all the values of the solid ( force,torque, linear and angular
acceleration and velocity ) ,
and i want to know what is the force at a particular position of the solid.

I don't think Fp = M As is good. because there is also torque and angular
velocity in the solid.
The real value would be Fp = Mp Ap... but i don't know Mp.

.... i'm going back home ... make a drawing on paint shop and post the link, it
will be easier to explain ;o)

regards,

--
Sebastien TIXIER - Game Developer
Dynamics and Car Physics
http://www.eden-studios.fr
GPLRank Normal:-44.24 Monster:-124.44

But if you know As, you don't need to take torque into account. Applying the
force Fp at any point on the body would result in the same As. To explain
more clearly, imagine applying a force at some point P on your body. You get
the linear acceleration and torque of the body by:

As = Fp/Ms
Ts = Fp x (P-C) (angular acceleration can be derived from this)

So if you're given As and Ts, you can work backwords using either of these 2
equations and get the same force.

the thing i don't understand is

if a solid have a torque and No linear acceleration you get Fp = As * Ms = 0.

If i take an exemple more complicated

imagine a block wich translate up slowly and at the same time have a torque. You
can find a point in the solid that have no force : the rolling center, at the
rolling center you have V = A = F = 0.. However the solid have force and torque
! so Fp = 0 , but  M As != 0 and Ts x ( C - P ) != 0 ...

your solution seems to simple for me or maybe i miss something else :o(

--
Sbastien TIXIER - Game Developer
Dynamics and Car Physics
Overall GPLRank -44.24
Monster GPLRank -114.44

Just curious as to why you guys are so into the science of this stuff.

Simracer

I am into physics too, not the heavy stuff you guys are into, but it does
interest me. I hope you got what you were looking for, can't wait to hear of
your upcoming project.

Ah, I see what you're getting at now. The answer to this is that there is no
_single_ force that could result in this situation.

I mentioned in my first reply that the angular and linear accelerations need
to consistent for this equation to work. By this I meant that they had to be
produced by the same single force.

Any force applied to a body will result in both linear and angular
acceleration. However, it is possible to apply 2 forces to a body and
produce torque but no linear accleration, eg Apply F1 at some (non central)
point P - produces angular and linear acceleration. Apply F2 at COM of body,
such that the linear acceleration produced cancels the linear acceleration
from F1. Since this 2nd force is applied at the COM, it will not give any
torque, leaving the angular acceleration unchanged. However, after applying
both forces, you will have no linear acceleration.

So what you really need is a way of getting all forces applied to reach a
given angular/linear acceleration combination.

Hmm, I'll have to think about this one. My initial instinct is that it will
be impossible to resolve (more unknowns than equations), but that is only an
instinct at this point. I'll give it some thought and if I can help one way
or the other I'll let you know.

Mike.

yeah i totaly agree with that. For exemple a wheel is attach to is center so the
link
produces the force F2 in your exemple and if someoby pull the wheel  the wheel
will turn but not move.
Then i'd like to know the force/torque on any other point on the wheel.

yep, i also have the position of all unkown forces ( contact and constraint
forces , say n forces).
So maybe the solution is to solve this =>

Sum( Fi ) = Fs
Sum( Ri x Fi ) = Ts

A theroem say also that the reaction forces between 2 solid are always equal and
opposite.
Glups, it seems that i'll need a big gaussien matrix to resolv those equations
!!! ;o)

No there is less unkowns than equation because we know that between 2 contact or
2 constraint the reaction are equal and opposite. So we need a equation system
for each solid that have common forces.

exemple : with have 3 solids S1 S2 S3 that are connected like this S1 and S2 on
J , S2 and S3 on K

Fi/Ti force/torque of solid i
G = gravity
we can write

F1  = FJ + M1.G
M1 = FJ x RJ1

F2 = -FJ - FK + M2.G
M2 = -FJ x RJ2 - FK x RK2

F3 = FK + M3.G
M3 = FK x RK3

RJ1 = position of J in S1 regarding to the center of gravity

so we have 6 equations and 2 unkowns ( FJ, FK ), am i right ?

--
Sbastien TIXIER - Game Developer
Dynamics and Car Physics
Overall GPLRank -44.24
Monster GPLRank -114.44

Now note the name of the ng... ;-)

Jonny

anyway we starting to really need a comp.auto.simulators or autos.simulators.coding , thing like that ;o)

Never been a star in solid body maths, sorry, can't help you much
here. Restraint forces (torques rather) are something I will get into
when I try to couple 3 differentials in a row.

Good luck. :)

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