Question

这可能更像是一个线性代数问题，但我说我有一个SCNVector，我想要一个新的SCNVector，它与y轴（或任何轴）的原始角度成一定角度。理想情况下：

extension SCNVector3 {  
    // assume dot, cross, length, +, - functions are available.
    enum Axis {
        case x, y, z
    }
    func rotatedVector(aroundAxis: Axis, angle: Float) -> SCNVector3 {
        // code from smart person goes here
    }
}

e.g。 (0,0,-1).rotatedVector(aroundAxis: y, angle: pi/2) = (1,0,0)

谢谢！

Answer 1

一般情况下使用Rodrigues' rotation formula。

Rodrigues的＆＃39;旋转公式是一种有效的旋转算法给定轴和旋转角度的空间矢量

对于初始矢量 v ，旋转轴单位矢量 k 和角度θ结果

v _rot = v * cos（theta）+（ k x v ）sin（theta）+ k *（ k .dot。 v ）*（1 - cos（theta））

Answer 2

尝试使用四元数 https://developer.apple.com/documentation/accelerate/working_with_quaternions

extension SCNVector3 {
    enum Axis {
        case x, y, z
        
        func getAxisVector() -> simd_float3 {
            switch self {
            case .x:
                return simd_float3(1,0,0)
            case .y:
                return simd_float3(0,1,0)
            case .z:
                return simd_float3(0,0,1)
            }
        }
    }

    func rotatedVector(aroundAxis: Axis, angle: Float) -> SCNVector3 {
        /// create quaternion with angle in radians and your axis
        let q = simd_quatf(angle: angle, axis: aroundAxis.getAxisVector())
        
        /// use ACT method of quaternion
        let simdVector = q.act(simd_float3(self))
        
        return SCNVector3(simdVector)
    }
}

使用：

let a = SCNVector3(5,0,0)

let b = a.rotatedVector(aroundAxis: SCNVector3.Axis.y, angle: -.pi/2)

// SCNVector3(x: 0.0, y: 0.0, z: 5.0)

您还可以围绕任何向量旋转：

let simdVector = q.act(simd_normalize(simd_float3(x: 0.75, y: 0.75, z: -0.2)))

Answer 3

Well thanks to a comment on gamedev.net I think I have what I need. It doesn't work for arbitrary angles but as it turns out I only need 90° so this works for me. Here's the solution:

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围绕轴旋转SCNVector3

3 个答案: