# SplineCurves

样条曲线：多段三阶贝塞尔曲线的联合。三阶插值（取自 Unreal）

FVector FMath::CubicInterp(const FVector& P0, const FVector& T0, const FVector& P1, const FVector& T1, const float& A)
{
	static_assert(PLATFORM_ENABLE_VECTORINTRINSICS == 1, "Requires SSE intrinsics.");
	FVector res;

	const float A2 = A  * A;
	const float A3 = A2 * A;

	float s0 = (2 * A3) - (3 * A2) + 1;
	float s1 = A3 - (2 * A2) + A;
	float s2 = (A3 - A2);
	float s3 = (-2 * A3) + (3 * A2);

	VectorRegister v0 = VectorMultiply(VectorLoadFloat1(&s0), VectorLoadFloat3(&P0));
	VectorRegister v1 = VectorMultiply(VectorLoadFloat1(&s1), VectorLoadFloat3(&T0));
	VectorRegister v2 = VectorMultiply(VectorLoadFloat1(&s2), VectorLoadFloat3(&T1));
	VectorRegister v3 = VectorMultiply(VectorLoadFloat1(&s3), VectorLoadFloat3(&P1));

	VectorStoreFloat3(VectorAdd(VectorAdd(v0, v1), VectorAdd(v2, v3)), &res);

	return res;
} 

• 其中 P0、P1、T0、T1 分别是起点、终点、起点、起点切线、终点切线。

样条曲线插值推导：

• 三阶贝塞尔曲线公式，其中 t 取值范围为 [0~1]，标识插值系数：

$$H(t) = at^3 + bt^2 + ct + d$$

• 假定 P0、P1、T0、T1 对曲线的影响系数随插值系数的关系如下：

  $$H(t) = P0s_0(t) + T0s_1(t) + P1s_2(t) + T1s_3(t)$$

• 该多项式要满足这四个条件：

  • t = 0 时 H (t) = P0；s0 (0) = 1，s1 (0) = 0，s2 (0) = 0，s3 (0) = 0
  • t = 1 时 H (t) = P1；s0 (1) = 0，s1 (1) = 0，s2 (1) = 1，s3 (1) = 0
  • t = 0 时 H (t)' = T0；s0'(0) = 0，s1'(0) = 1，s2'(0) = 0，s3'(0) = 0
  • t = 1 时 H (t)' = T1；s0'(1) = 0，s1'(1) = 0，s2'(1) = 0，s3'(1) = 1

• 分别带入求解 s0、s1、s2、s3，以 s0 为例：

\begin{flalign*} & t = 0,\; H(0) =at^3 + bt^2 + ct + d = 1, \; d = 1 \\ & t = 1,\; H(1) =at^3 + bt^2 + ct + d = 0, \; a + b + c + d = 0 \\ & t = 1,\; H(0)' =3at^2 + 2bt + c = 0, \; c = 0 \\ & t = 1,\; H(1)' =3at^2 + 2bt + c = 0, \; 3a + 2b + c = 0 \\ & a = 2,\; b = -3,\; c = 0,\; d = 1,\; s_0(t) = 2t^3 - 3t^2 + 1 \end{flalign*}

• 依次求解 s0、s1、s2、s3：

\begin{flalign*} & s_0(t) = 2t^3 - 3t^2 + 1 \\ & s_1(t) = t^3 - 2t^2 + t \\ & s_2(t) = -2t^3 + 3t^2 \\ & s_3(t) = t^3 - t^2 \end{flalign*}

最终得到以上 Unreal 的三阶插值函数。