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@@ -44,6 +44,25 @@ JPH_INLINE float ACos(float inX)
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return Vec4::sReplicate(inX).ACos().GetX();
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}
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+/// An approximation of ACos, max error is 4.2e-3 over the entire range [-1, 1], is approximately 2.5x faster than ACos
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+JPH_INLINE float ACosApproximate(float inX)
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+{
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+ // See: https://www.johndcook.com/blog/2022/09/06/inverse-cosine-near-1/
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+ // Taylor of cos(x) = 1 - x^2 / 2 + ...
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+ // Substitute x = sqrt(2 y) we get: cos(sqrt(2 y)) = 1 - y
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+ // Substitute z = 1 - y we get: cos(sqrt(2 (1 - z))) = z <=> acos(z) = sqrt(2 (1 - z))
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+ // To avoid the discontinuity at 1, instead of using the Taylor expansion of acos(x) we use acos(x) / sqrt(2 (1 - x)) = 1 + (1 - x) / 12 + ...
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+ // Since the approximation was made at 1, it has quite a large error at 0 meaning that if we want to extend to the
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+ // range [-1, 1] by mirroring the range [0, 1], the value at 0+ is not the same as 0-.
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+ // So we observe that the form of the Taylor expansion is f(x) = sqrt(1 - x) * (a + b x) and we fit the function so that f(0) = pi / 2
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+ // this gives us a = pi / 2. f(1) = 0 regardless of b. We search for a constant b that minimizes the error in the range [0, 1].
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+ float abs_x = min(abs(inX), 1.0f); // Ensure that we don't get a value larger than 1
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+ float val = sqrt(1.0f - abs_x) * (JPH_PI / 2 - 0.175394f * abs_x);
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+
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+ // Our approximation is valid in the range [0, 1], extend it to the range [-1, 1]
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+ return inX < 0? JPH_PI - val : val;
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+}
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+
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/// Arc tangent of x (returns value in the range [-PI / 2, PI / 2])
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JPH_INLINE float ATan(float inX)
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{
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