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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,135 @@ | ||
| module.exports = (ca, cb, r, righthand) => { | ||
| // @affineplane.sphere2.tangentCircle(ca, cb, r, righthand) | ||
| // | ||
| // Find a circle C that is left-hand tangent to the circles A and B. | ||
| // Use the fourth parameter to switch to right-hand tangent. | ||
| // Note the coordinate system: x-axis points right and y-axis points down. | ||
| // | ||
| // If the gap between A and B is too large for C to be tangent to both | ||
| // then the resulting circle is tangent with A and as close to B as possible. | ||
| // Under the righthand flag, the preference is reversed. | ||
| // | ||
| // Parameters: | ||
| // ca | ||
| // a sphere2 {x,y,r}, the circle A. | ||
| // cb | ||
| // a sphere2 {x,y,r}, the circle B. | ||
| // r | ||
| // a number, a dist2, the radius of the circle C. | ||
| // righthand | ||
| // a boolean, default false. Set true to find right-hand tangent circle. | ||
| // | ||
| // Returns: | ||
| // a sphere2 {x,y,r} | ||
| // | ||
| // See also sphere2.tangentCircles for efficient computation of both hands with just one function call. | ||
| // | ||
|
|
||
| // Sort circles largest first. This avoids special treatment of cases | ||
| // where the point H (the point C projected on the AB line) is outside AB. | ||
| // However, we need to maintain handedness with respect to the original argument order. | ||
| let swapped = false | ||
| let hand = (righthand ? 1 : -1) | ||
| if (ca.r < cb.r) { | ||
| // Swap | ||
| const ct = cb | ||
| cb = ca | ||
| ca = ct | ||
| swapped = true | ||
| hand = -hand | ||
| } | ||
|
|
||
| // Vector from A to B, of length c. | ||
| const vab = { | ||
| x: cb.x - ca.x, | ||
| y: cb.y - ca.y | ||
| } | ||
| // Triangle sides | ||
| const a = cb.r + r // length of BC | ||
| const b = ca.r + r // length of AC | ||
| const c = Math.sqrt(vab.x * vab.x + vab.y * vab.y) // ca.r + cb.r for tangent A and B | ||
| // Check special cases. | ||
| const epsilon = 1000 * Number.EPSILON | ||
| if (-epsilon < c && c < epsilon) { | ||
| // The circles A and B are concentric i.e. the distance between their centers is zero. | ||
| // The circle C can be tangent to both only if A and B have equal radius. | ||
| // The best compromise is to let C be tangent to the one with the largest radius (A). | ||
| // The tangent position for the circle C is found at an arbitrary angle (choose 0 deg). | ||
| return { | ||
| x: ca.x + b, // because b = ca.r + r and ca.r is always >= cb.r. | ||
| y: ca.y, | ||
| r | ||
| } | ||
| } | ||
| if (c + cb.r <= ca.r) { | ||
| // The circle A fully covers the circle B. | ||
| // In other words, the circle B is completely inside the circle A. | ||
| // The best compromise is to let the circle C be tangent to the circle A | ||
| // at a position closest to the circle B. | ||
| const vac = { | ||
| x: vab.x * b / c, | ||
| y: vab.y * b / c | ||
| } | ||
| // The center point of the circle C: vc = va + vac | ||
| return { | ||
| x: ca.x + vac.x, | ||
| y: ca.y + vac.y, | ||
| r | ||
| } | ||
| } | ||
| if (a + b < c) { | ||
| // The circles A and B are so far away that the circle C cannot connect them. | ||
| // Find a circle that is tangent with A and has center point along AB. | ||
| // To maintain the original circle order regardless of their radii, | ||
| // we treat both orders separately. | ||
| if (!swapped !== !righthand) { // Coerce to bool, then XOR | ||
| // Swapped xor right-handed order. | ||
| // Prefer the circle B. | ||
| const vbc = { | ||
| x: -vab.x * a / c, | ||
| y: -vab.y * a / c | ||
| } | ||
| // The center point of the circle C: vc = vb + vbc | ||
| return { | ||
| x: cb.x + vbc.x, | ||
| y: cb.y + vbc.y, | ||
| r | ||
| } | ||
| } // else | ||
| // Original order. Prefer the circle A. | ||
| const vac = { | ||
| x: vab.x * b / c, | ||
| y: vab.y * b / c | ||
| } | ||
| // The center point of the circle C: vc = va + vac | ||
| return { | ||
| x: ca.x + vac.x, | ||
| y: ca.y + vac.y, | ||
| r | ||
| } | ||
| } | ||
| // After excluding the special cases above, | ||
| // it is now possible to find a circle that is | ||
| // tangent to both circles A and B. | ||
| const p = (a + b + c) / 2 | ||
| const A = Math.sqrt(p * (p - a) * (p - b) * (p - c)) | ||
| const h = 2 * A / c | ||
| const w = Math.sqrt(b * b - h * h) | ||
| // Vector from A to H, of length w. | ||
| const vw = { | ||
| x: vab.x * w / c, | ||
| y: vab.y * w / c | ||
| } | ||
| // Vector from H to C, of length h. | ||
| const vh = { | ||
| x: hand * -vw.y * h / w, | ||
| y: hand * vw.x * h / w | ||
| } | ||
| // Find the center point of C. | ||
| // A + vw + vh | ||
| return { | ||
| x: ca.x + vw.x + vh.x, | ||
| y: ca.y + vw.y + vh.y, | ||
| r | ||
| } | ||
| } |
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