(programmer*) minimalist ≅ mathematical experimentalist
Published
Edited
Apr 16, 2018
6 stars
// Create a Clifford Algebra with 2,0,1 metric.
Algebra(2,0,1,()=>{
// In the inverse kinematics problem we need to calculate joint angles of
// a kinematic chain so its base remains fixed and its tip reaches a given
// target. This is a highly non-linear problem with many solutions. We
// implement a solver that tries to minimize the differences on all
// remaining degrees of freedom.
// This algorithm readily translates to 3D, is efficient and very well
// received by artists in a cg animation context.
// Translate dist along line.
var translator = (line,dist)=>1+0.5*dist*(line.Normalized*1e012);
// Create chain of points. (l = number of segments in chain)
var l=6, d=3/l, points=Array.apply([],{length:l+1}).map((x,i)=>1e12+(1.5-i*d)*1e02);
// solver. Last point in chain is the target. Set tip to target, then cycle to base and back
// restoring original lengths.
var IK=(c)=>{
// Run four relaxation steps
for (var i,j=0;j<4; j++) {
// Set the tip to the target. (this will change the length of the last segment.)
c[l-1].set(c[l]);
// Run backwards to the base and restore the lengths along the chain.
for (i=l-2; i>0; i--) c[i].set(translator(c[i]&c[i+1],-d)>>>(c[i+1]));
// Loop the other way from base to tip again restoring all lengths.
for (i=1; i<l; i++) c[i].set(translator(c[i-1]&c[i],d)>>>(c[i-1]));
}};
// Evaluate IK, render points and lines, target last (so its on top)
return this.graph([].concat(
()=>IK(points), points[0], "Base", // run IK and render base
0x0000ff,points.slice(1,l), // render joint points. [A,B,C,..]
points.map((x,i,a)=>[a[i-1],x]).slice(1,l), // render line segments. [[A,B],[B,C],..]
0x000000,points[l],"Target" // render target in black
),{grid:true});
})
Algebra = require('ganja.js')
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