Question

我在SCNScene中有一个SCNBox。在场景动画后，SCNBox更改的方向可以通过检查其presentationNode.orientation来查看。此值在SCNVector4中返回。如何从SCNVector4中返回的值确定SCNBox的哪一侧是朝上的一面？

我试图规范化数据并提出以下数据

Side 1 = (-x, 0, 0, +x)
Side 2 = (0, -x, 0 +y)
Side 3 = (-x, +x, -y, y)
Side 4 = (x, x, y, y)
Side 5 = (-x, 0, +y, 0)
Side 6 = (-x, -y, +y, -x)

不幸的是，这并不总是正确的，对它进行检查有时会偶尔返回无效的边。

是否有一种可靠的方法可以从几何体的方向属性中确定SCNBox的朝上一面？

编辑：根据Toyos的回答，我提出了以下代码，但这些代码并不起作用。希望这有助于更接近最终目标。我还使用在Extracting vertices from scenekit找到的代码来获取我的SCNBoxNode的顶点。

- (NSNumber *)valueForRotation:(SCNVector4)rotation andGeometry:(SCNGeometry*)geometry {
    SCNVector4 inverse = SCNVector4Make(rotation.x, rotation.y, rotation.z, -rotation.w);

    CATransform3D transform = CATransform3DMakeRotation(inverse.w, inverse.x, inverse.y, inverse.z);

    GLKMatrix4 matrix = GLKMatrix4Make(transform.m11, transform.m12, transform.m13, transform.m14, transform.m21, transform.m22, transform.m23, transform.m24, transform.m31, transform.m32, transform.m33, transform.m34, transform.m41, transform.m42, transform.m43, transform.m44);

    GLKVector4 vector = GLKVector4Make(rotation.x, rotation.y, rotation.z, rotation.w);

    GLKVector4 finalVector = GLKMatrix4MultiplyVector4(matrix, vector);

    NSArray *vertexSources = [geometry geometrySourcesForSemantic:SCNGeometrySourceSemanticVertex];

    SCNGeometrySource *vertexSource = vertexSources[0]; // TODO: Parse all the sources

    NSInteger stride = vertexSource.dataStride; // in bytes
    NSInteger offset = vertexSource.dataOffset; // in bytes

    NSInteger componentsPerVector = vertexSource.componentsPerVector;
    NSInteger bytesPerVector = componentsPerVector * vertexSource.bytesPerComponent;
    NSInteger vectorCount = vertexSource.vectorCount;

    SCNVector3 vertices[vectorCount]; // A new array for vertices

    // for each vector, read the bytes
    NSLog(@"vetor count %i",vectorCount);
    float highestProduct = 0;
    int highestVector = -1;
    NSMutableArray *highVectors;
    for (NSInteger i=0; i<vectorCount; i++) {

        // Assuming that bytes per component is 4 (a float)
        // If it was 8 then it would be a double (aka CGFloat)
        float vectorData[componentsPerVector];

        // The range of bytes for this vector
        NSRange byteRange = NSMakeRange(i*stride + offset, // Start at current stride + offset
                                        bytesPerVector);   // and read the lenght of one vector

        // Read into the vector data buffer
        [vertexSource.data getBytes:&vectorData range:byteRange];

        // At this point you can read the data from the float array
        float x = vectorData[0];
        float y = vectorData[1];
        float z = vectorData[2];

        // ... Maybe even save it as an SCNVector3 for later use ...
        vertices[i] = SCNVector3Make(x, y, z);

        // ... or just log it 
        NSLog(@"x:%f, y:%f, z:%f", x, y, z);
        float product = (x * finalVector.x) + (y * finalVector.y) +  (z * finalVector.z);
        if (product > highestProduct) {
            highestProduct = product;
            highestVector = i;
        }

    }

    NSLog(@"highestProduct = %f",highestProduct);
    NSLog(@"highestVector = %i",highestVector);
    NSLog(@"top verticy = %f, %f, %f",vertices[highestVector].x,vertices[highestVector].y,vertices[highestVector].z);

    return [NSNumber numberWithInt:highestVector];
}

Answer 1

这是一个返回面朝上的面部索引的方法。它假定“boxNode”是由6个面组成的框，具有以下（任意）顺序：前/右/后/左/上/下。它返回面朝上的面部索引。别忘了导入。对于任意网格，你必须使用面法线而不是“boxNormals”（这是不明显的计算，因为SceneKit网格每个顶点有一个法线，而不是每个面一个法线，所以你必须计算每个面的法线自己）。

- (NSUInteger) boxUpIndex:(SCNNode *)boxNode
{
    SCNVector4 rotation = boxNode.rotation;
    SCNVector4 invRotation = rotation; invRotation.w = -invRotation.w;

    SCNVector3 up = SCNVector3Make(0,1,0);

    //rotate up by invRotation
    SCNMatrix4 transform = SCNMatrix4MakeRotation(invRotation.w, invRotation.x, invRotation.y, invRotation.z);
    GLKMatrix4 glkTransform = SCNMatrix4ToGLKMatrix4(transform);
    GLKVector3 glkUp = SCNVector3ToGLKVector3(up);
    GLKVector3 rotatedUp = GLKMatrix4MultiplyVector3(glkTransform, glkUp);

    //build box normals (arbitrary order here)
    GLKVector3 boxNormals[6] = {{{0,0,1}},
        {{1,0,0}},
        {{0,0,-1}},
        {{-1,0,0}},
        {{0,1,0}},
        {{0,-1,0}},
    };

    int bestIndex = 0;
    float maxDot = -1;

    for(int i=0; i<6; i++){
        float dot = GLKVector3DotProduct(boxNormals[i], rotatedUp);
        if(dot > maxDot){
            maxDot = dot;
            bestIndex = i;
        }
    }

    return bestIndex;
}

Answer 2

node.orientation会给你一个四元数。您也可以使用node.eulerAngles或node.rotation（轴角度），这样可以简化数学运算。

在所有情况下，我认为您需要将反向旋转应用于向量[0,1,0]（向上向量）。结果向量（V）将为您提供正面的方向。然后找到正面，比较框面和V的点积。最高点积是面朝上的面。

（另一种方法是旋转盒子的面，用[0,1,0]做点积。）

得到＆＃34; up＆＃34;来自Orientation的SCNNode的一面

2 个答案: