How do you get 3D range of object (highlighted in red below) given its lowest (PL) and highest (PH) (x,y,z) coordinates and orientation of object? Image below is a top view of a box.
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I've found a way already
Given:
1)$$H_1 = L\sin(\theta) + W\cos(\theta)$$
2)$$H_2 = L\cos(\theta) + W\sin(\theta)$$
Then...
3)$$L = \frac{H_2 - W\sin\theta} {\cos(\theta)}$$
Substitute... $$H_1 = \frac{H_2-W\sin\theta} {\cos\theta}sin\theta + W\cos\theta$$ $$H_1 = \frac{H_2\sin\theta} {\cos\theta} - \frac{W\sin^2\theta}{\cos\theta} + Wcos\theta$$ $$H_1 = \frac{H_2\sin\theta} {\cos\theta} - \frac{W(1-\cos^2\theta)}{\cos\theta} + W\cos\theta$$ $$H_1 = \frac{H_2\sin\theta} {\cos\theta} - \frac{W}{\cos\theta} + W\cos\theta + W\cos\theta$$ $$H_1 = \frac{H_2\sin\theta} {\cos\theta} - W\left(\frac{1}{\cos\theta} + \cos\theta + \cos\theta\right)$$ $$W\left(\frac{1}{\cos\theta} + \cos\theta + \cos\theta\right) = \frac{H_2\sin\theta} {\cos\theta} - H_1$$ $$W = \frac{(H_2\tan\theta - H_1){\frac{1-2\cos^2\theta}{\cos\theta}}$$ 4)$$W = \frac{(H_2\tan\theta - H_1)\cos\theta}{1-2\cos^2\theta}$$
Since you have W, substitute W of formula 3 to get L.
Once you have L, you can get the points easily.
P.S. This does not apply if angle is 45 degrees.
tjvg1991
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