Game Development Reference
In-Depth Information
ApplyRopeImpulse( float P)
bodies[0].ApplyImpulse(d[0] P);
for ( int i=1; i < N; i++)
bodies[i].ApplyImpulse((d[i] d[i 1]) P);
bodies[N].ApplyImpulse(d[N 1] ∗− P);
// same procedure is used for applying test impulse,
// except body impulses are accumulated in a special set of P,L
float MeasureVelocityResponse()
// compute v test and w test for each body as velocity changes
// due to test impulses
// compute v[i] = bodies[i].v test + (bodies[i].w test ˆ
// pt[i] bodies[i].center)
float dv = v[N] d[N] v[0] d[0];
for ( int i=1; i < N; i++)
dv += v[i] (d[i 1] d[i]);
return dv; //now 1/dv can be precomputed
The final push will be to add friction in the pull direction (see Figure 8.5 ) .
The first thing to do is separate contact vertices with static and dynamic fric-
tion. It would make sense to assume that all contacts with tan( α ) (the friction
coefficient) are static and the rest are dynamic, applying friction impulse in the
direction opposite to the current slide velocity, in accordance with the Coulomb
law (in reality some of the latter might actually become static during the solv-
ing process, but not addressing that immediately in the same frame sounds like a
reasonable approximation).
Static friction contacts can be handled by simply splitting the rope constraint
into several parts so that each one has no static contacts in the middle (series of
consecutive static contacts with the same object can be merged).
For the dynamic sliding contacts, the velocity Equation (8.3) thankfully re-
mains the same. To find the impulses, we'll use the same idea that impulses
applied to each contact vertex sum up to zero. Assuming that P i and P i are the
(scalar) impulses that pull a vertex towards its left and right neighbors, respec-
tively, and P i is the normal impulse from the object, we can write the following:
(P i
+P i )cos α =P i ,
(P i
+P i )sin α = μ P i .
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