Real-Time Physics Simulation Forum

Re: Help building Jacobian using partial derivative

2014-02-02T01:42:41+00:00

Yes, I'm familiar with that approach. I was specifically interested in getting comfortable with the mathematics, though. I'm going to be exploring building the hessian matrices of a reduced coordinate system, which is a much harder problem! So I wanted to exercise my math chops and make sure I can do problems I know the answer to.

Statistics: Posted by Numsgil — Sun Feb 02, 2014 1:42 am

Re: Help building Jacobian using partial derivative

2014-02-02T01:12:47+00:00

Yes, the common way is to build the time derivative and then find the Jacobian by inspection. You can find some details e.g. in the excellent book 'Computational Dynamics' by Shabana.

Basically you have C( x(t), t ) = 0
Now you can build the time derivative which has a special form: dC/dt = del_C / del_x * dx / dt + del_C / del_t
Where del_* is the partial derivative.

The constraint functions we usually use in games don't dependent explicitly on time so the last term is usually zero. The del_C / del_x is the famous Jacobian matrix. Each row of the Jacobian is the transposed gradient vector of each scalar constraint function c_i in the constraint vector C = ( c1( x, t ), c2( x, t ), cn( x, t ) )^T

For reference: http://en.wikipedia.org/wiki/Total_derivative

In your particular example above for a revolute joint in 2D you write down the constraint function for a revolute joint:
C = x2 + R2 * r2' - x1 - R1 * r1' = x2 + r2 - x1 - r1 // The primes should denote body fixed coordinates
dC/dt = = J * v = v2 + cross( omega2, r2 ) - v1 - cross( omega1, r1 )

From here we identify the Jacobian by inspection as:
J = ( -I skew( r1 ) I -skew( r2 )

Where I is the 2x2 identity matrix and skew( r ) is the skew symmetric cross product matrix.

You can find a bunch of examples in the Box2D joint implementations. Erin added a bunch of derivations at the top of the corresponding *.cpp files.

HTH,
-Dirk

Statistics: Posted by Dirk Gregorius — Sun Feb 02, 2014 1:12 am

Re: Help building Jacobian using partial derivative

2014-02-01T21:38:31+00:00

Ah, hmm, I think I figured it out.

Code:

d cos θ(t)    -sin θ(t) * dθ(t)---------- = ------------------- = -sin θ(t)  dθ(t)              dθ(t)

That is, I'm not finding the derivative with respect to time, but with respect to θ(t). So the derivatives on the top and bottom cancel, and w drops out of my Jacobian.

Statistics: Posted by Numsgil — Sat Feb 01, 2014 9:38 pm

Help building Jacobian using partial derivative

2014-01-30T21:03:32+00:00

I'm trying to get my mathematical ducks in a row and finally tackle building Jacobians for a constraint using partial derivatives. In the past I've sort of messed around with what I want the constraint to do in velocity space and built up the Jacobian that way. But this time I'd like to tackle doing it from first principles from the position-space constraint equation.

As a medium-difficulty problem, I'm tackling a revolute joint pinning a point on a rigid body to a fixed point in space in 2D. Can someone take a look at what I have and help me correct any misunderstandings? I'm not being super formal below with the notation, as many of the matrices below are actually block matrices, but hopefully it makes sense. Also, apologies in advance for this massive wall of text

...

Let [θ] be the 2D rotation matrix for an angle θ. That is:

Code:

| cos θ  -sin θ | = [θ]| sin θ   cos θ |

Let [θ, p] be the 2D affine transformation matrix for orientation θ and translation p. That is:

Code:

| cos θ  -sin θ  p.x | = | [θ] p | = [θ, p]| sin θ   cos θ  p.y |   | 0 0 1 || 0       0        1 |

Then for my revolute joint I want:

[θ,p] * r = { c.x; c.y; 1 }, where r is the position of the pivot in the body's local space, and [θ,p] is the local-to-world transformation matrix for the body, and c is the position we're pinning the body to in world space.

Let X = { θ; p.x; p.y } be the 3x1 column vector of the degrees of freedom of the system. ie: X is the position and orientation of the body.

We can write the above as f(X) = { c.x ; c.y ; 1 }. Then take the derivative. Using the chain rule, this becomes:

f'(X) * X' = {0;0;0}

f'(X) is the Jacobian, and X' is the change in the degrees of freedom over time, ie: the body's linear and angular velocity.

So now we want to take the Jacobian of the affine transformation matrix.

First:

Code:

cos' θ = -θ' * sin θ = -w * sin θsin' θ =  θ' * cos θ =  w * cos θw = θ' = angular velocity

So then:

Code:

[θ]' = | cos' θ -sin' θ | = w * | -sin θ -cos θ | = w [θ] [#]       | cos' θ  sin' θ |       | -sin θ  cos θ |

where [#] is a 2x2 matrix which takes a vector and maps it to its perpindicular. ie:

Code:

[#] = | 0 -1 |      | 1  0 |

So now we take [θ,p] and find the partial derivatives of it with respect to x, y, and θ. First, we subdivide the matrix in terms of sub functions:

Code:

F1 = { cos θ, -sin θ } dot r + p.xF2 = { sin θ,  cos θ } dot r + p.yF3 = 1

And then we take the partial derivative:

Code:

J = | I,    w [θ] [#] r |    | 0, 0, 0           |

And of course, being all 0s, the last row can just be dropped in practice. That gives a 2x3 matrix, which makes sense: two degrees of freedom have been removed from the system.

...

But now my question: I have a w term (the angular velocity) in my Jacobian, which I don't think is right! I feel like the above is almost what I expect, but I can't figure out how to get rid of that w term.

Statistics: Posted by Numsgil — Thu Jan 30, 2014 9:03 pm