我试图做(除其他之外)与矩阵计算。这就是我快要完成了。当矩阵中存在一个常量(不是矩阵)时,如何进行加法,矩阵乘法?
deltaX = -419.375
deltaY = -99.352
deltaZ = -591.349
deltaXYZ = np.array([deltaX,deltaY,deltaZ]).reshape(3,1)
omegaX = 0.850458
omegaY = 1.817245
omegaZ= -7.862245
omegaXR=radians(omegaX/3600)
omegaYR=radians(omegaY/3600)
omegaZR=radians(omegaZ/3600)
delta = (0.99496/(10**6))
x = 3240036.3696
y = 990578.5272
z = 5385763.1648
swerefkoordinates = np.array([x,y,z]).reshape(3,1)
rZ=np.array([cos(omegaZR),sin(omegaZR),0,-sin(omegaZR),cos(omegaZR),0,0,0,1]).reshape(3,3)
rY=np.array([cos(omegaYR),0,-sin(omegaYR),0,1,0,sin(omegaYR),0,cos(omegaYR)]).reshape(3,3)
rX=np.array([1,0,0,0,cos(omegaXR),sin(omegaXR),0,-sin(omegaXR),cos(omegaXR)]).reshape(3,3)
R=np.dot(rZ,rY,rX)
到目前为止好
接下来的这个计算的线虽然给我一些麻烦,我的意思是错误的答案。
RR92 = deltaXYZ+(delta+1)*np.dot(R,swerefkoordinates)
我想要做的是以下公式(RR92)。
[XYZ] = [DELTAX,移动deltaY,deltaZ] +(1 +增量)R [X,Y,Z]
我有一个很难可视化所述式(和正在不允许添加图片)。无论如何,无论在[]中是什么格式的矩阵:3行一列。
什么是您的公司直接回答? – swatchai
这将是: [3240036.3696 990578.5272 5385763.1648】 (当然在矩阵) – Nedim