Formula/Algorithm to apply force to an arbitrary point on a polygon


by MTK
Tags: apply, arbitrary, force, formula or algorithm, point, polygon
MTK
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#1
Oct26-09, 08:32 AM
P: 13
I am trying to code a 2D rigid body physics engine in Java, and I am having some trouble figuring out how an object will move and rotate if the force is off-center.

Basically, what I can't seem to figure out is a way to find out the x, y, and rotational acceleration when a force is applied at an arbitrary point on an arbitrary polygon. I tried searching, but there were only something like 2-3 relavent results, and I couldn't really understand them well.
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A.T.
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#2
Oct26-09, 11:05 AM
P: 3,551
Quote Quote by MTK View Post
I am trying to code a 2D rigid body physics engine in Java, and I am having some trouble figuring out how an object will move and rotate if the force is off-center.
Linear:
http://en.wikipedia.org/wiki/Newton%...27s_second_law

Angular:
http://en.wikipedia.org/wiki/Angular...ions_of_motion

Both accelerations happen.
MTK
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#3
Oct26-09, 11:32 AM
P: 13
I found this:

http://en.wikipedia.org/wiki/File:To..._and_force.svg

So from this I understand that if the force is all parallel, then there is no rotation and all acceleration is linear, and if it is all perpendicular, then the linear/angular ratio depends somehow on the mass, mass distribution, and how far from the center of mass the force is applied.

MTK
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#4
Oct26-09, 05:05 PM
P: 13

Formula/Algorithm to apply force to an arbitrary point on a polygon


So I made it that only the component of the force vector that is parallel with the coener of mass causes linear acceleration.

But I still don't know how to tell the ratio of linear to angular acceleration caused by the perpendicular component, and how to know how much torque is produced.
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#5
Oct26-09, 05:36 PM
P: 3,551
Quote Quote by MTK View Post
and if it is all perpendicular, then the linear/angular ratio depends somehow on the mass, mass distribution, and how far from the center of mass the force is applied.
There is no linear/angular ratio. The instantaneous linear/angular accelerations are independent.
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#6
Oct26-09, 08:01 PM
P: 13
Then how do you explain this?
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A.T.
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#7
Oct27-09, 06:58 AM
P: 3,551
Quote Quote by MTK View Post
Then how do you explain this?
Both forces produce the same instantaneous linear acceleration, but different instantaneous angular accelerations.
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#8
Oct27-09, 07:32 AM
P: 13
So, for example, if I have a circular object, and I apply a force right on the edge (like a tangent on the egde of the circle), will that apply just as much linear acceleration as pushing the circle right in the middle?
Kricket
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#9
Oct27-09, 07:49 AM
P: 14
When you apply a force to an object, be it a polygon or otherwise, you do two calculations:

1. Apply the force to the center-of-gravity (COG)
2. Take the cross product of the radius to the point of application, with the force, to obtain the torque

Why, you might ask? Think of the object as a system of separate particles. When you apply a force to a particle that is offset from the COG (of the system!), immediately that particle wants to tug away at F / (mass of the particle). However, since it's linked to all the other particles in your object, you get this system of internal forces - the particle pulls on the system, and the system pulls back. Your input force is therefore distributed to all the particles in the system, through their constraints.

The rotation is a totally different game. It's the effect that your force has on the orientation of the system - but torque-induced rotation is caused entirely by INTERNAL forces. Thus, if you start dividing your torque into perpendicular/parallel components, you're actually throwing away some of the input force.
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#10
Oct27-09, 07:55 AM
P: 3,551
Quote Quote by MTK View Post
So, for example, if I have a circular object, and I apply a force right on the edge (like a tangent on the egde of the circle), will that apply just as much linear acceleration as pushing the circle right in the middle?
Yes.
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#11
Oct27-09, 08:05 AM
P: 13
But my Vector class is 2D, but it appears that the cross product must be 3D. But when to think of it, only the Z axis could be perpendicular to the x, y plane, so can I just calculate the Z axis and ignore X and Y?

And how exactly do you convert a 3D vector into a scalar rotation value?
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#12
Oct27-09, 11:13 AM
P: 3,551
Quote Quote by MTK View Post
But my Vector class is 2D, but it appears that the cross product must be 3D. But when to think of it, only the Z axis could be perpendicular to the x, y plane, so can I just calculate the Z axis and ignore X and Y?
XY of the cross product will be zero anyway if Z=0 in both input vectors
Quote Quote by MTK View Post
And how exactly do you convert a 3D vector into a scalar rotation value?
Generally you take the norm, but here it is just Z.
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#13
Oct27-09, 11:45 AM
P: 13
I am still wondering if this is all correct, because if I apply force to the edge of an object in my simulation, it accelerates sideways, but hardly spins at all. This is nothing like what I observe in real life or this applet:

http://www.myphysicslab.com/collision.html

And these are my formulas:

velocity = velocity.add(force.scale(1/mass))
dt += offset.crossProduct(force) / momentOfInertia

(dt is the angular velocity, in radians per second)
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#14
Oct27-09, 11:50 AM
P: 3,551
Quote Quote by MTK View Post
I am still wondering if this is all correct, because if I apply force to the edge of an object in my simulation, it accelerates sideways, but hardlt spins at all.
Reduce the moment of Inertia if you want it to spin more. How is your force determined? Does the force direction rotate with the object, or does it always point in the same direction?
MTK
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#15
Oct27-09, 12:06 PM
P: 13
I figured out the problem, I forgot to convert meters to pixels and seconds to frames in a few places.

Now the simulation feels much more correct.
MTK
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#16
Oct27-09, 03:43 PM
P: 13
Now I need to implement bouncing off the walls...


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