90 lines
2.7 KiB
JavaScript
Executable file
90 lines
2.7 KiB
JavaScript
Executable file
/**
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* Based on:
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* https://stackoverflow.com/questions/7348009/y-coordinate-for-a-given-x-cubic-bezier
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* https://math.stackexchange.com/questions/26846/is-there-an-explicit-form-for-cubic-b%C3%A9zier-curves
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* TODO: Reduce rounding error
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*/
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/**
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* EXPERIMENTAL
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* Given a cubic-bezier curve, get the x value (time) given
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* the y value (progression).
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* Ex: cubic-bezier(0.32, 0.72, 0, 1);
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* P0: (0, 0)
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* P1: (0.32, 0.72)
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* P2: (0, 1)
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* P3: (1, 1)
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*
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* If you give a cubic bezier curve that never reaches the
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* provided progression, this function will return an empty array.
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*/
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const getTimeGivenProgression = (p0, p1, p2, p3, progression) => {
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return solveCubicBezier(p0[1], p1[1], p2[1], p3[1], progression).map(tValue => {
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return solveCubicParametricEquation(p0[0], p1[0], p2[0], p3[0], tValue);
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});
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};
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/**
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* Solve a cubic equation in one dimension (time)
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*/
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const solveCubicParametricEquation = (p0, p1, p2, p3, t) => {
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const partA = (3 * p1) * Math.pow(t - 1, 2);
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const partB = (-3 * p2 * t) + (3 * p2) + (p3 * t);
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const partC = p0 * Math.pow(t - 1, 3);
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return t * (partA + (t * partB)) - partC;
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};
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/**
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* Find the `t` value for a cubic bezier using Cardano's formula
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*/
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const solveCubicBezier = (p0, p1, p2, p3, refPoint) => {
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p0 -= refPoint;
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p1 -= refPoint;
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p2 -= refPoint;
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p3 -= refPoint;
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const roots = solveCubicEquation(p3 - 3 * p2 + 3 * p1 - p0, 3 * p2 - 6 * p1 + 3 * p0, 3 * p1 - 3 * p0, p0);
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return roots.filter(root => root >= 0 && root <= 1);
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};
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const solveQuadraticEquation = (a, b, c) => {
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const discriminant = b * b - 4 * a * c;
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if (discriminant < 0) {
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return [];
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}
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else {
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return [
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(-b + Math.sqrt(discriminant)) / (2 * a),
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(-b - Math.sqrt(discriminant)) / (2 * a)
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];
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}
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};
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const solveCubicEquation = (a, b, c, d) => {
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if (a === 0) {
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return solveQuadraticEquation(b, c, d);
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}
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b /= a;
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c /= a;
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d /= a;
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const p = (3 * c - b * b) / 3;
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const q = (2 * b * b * b - 9 * b * c + 27 * d) / 27;
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if (p === 0) {
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return [Math.pow(-q, 1 / 3)];
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}
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else if (q === 0) {
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return [Math.sqrt(-p), -Math.sqrt(-p)];
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}
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const discriminant = Math.pow(q / 2, 2) + Math.pow(p / 3, 3);
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if (discriminant === 0) {
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return [Math.pow(q / 2, 1 / 2) - b / 3];
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}
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else if (discriminant > 0) {
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return [Math.pow(-(q / 2) + Math.sqrt(discriminant), 1 / 3) - Math.pow((q / 2) + Math.sqrt(discriminant), 1 / 3) - b / 3];
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}
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const r = Math.sqrt(Math.pow(-(p / 3), 3));
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const phi = Math.acos(-(q / (2 * Math.sqrt(Math.pow(-(p / 3), 3)))));
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const s = 2 * Math.pow(r, 1 / 3);
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return [
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s * Math.cos(phi / 3) - b / 3,
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s * Math.cos((phi + 2 * Math.PI) / 3) - b / 3,
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s * Math.cos((phi + 4 * Math.PI) / 3) - b / 3
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];
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};
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export { getTimeGivenProgression as g };
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