How to visualize angle of projection in Oblique Parallel Projection?

Question

We know that in Oblique Parallel Projection Point $(x,y,z)$ is projected to position $(x_p,y_p)$ on the view plane.Projector (oblique) from $(x,y,z)$ to $(x_p,y_p)$ makes an angle $\alpha$ with the line (L) on the projection plane that joins $(x_p,y_p)$ and $(x,y).$ Line $L$ is at an angle $\phi$ with the horizontal direction in the projection plane.See this image1:

And in Oblique Parallel Projection Angles, distances, and parallel lines in the plane are projected accurately.For example see below image2:

My question is where is the angle $\alpha$ in image2, I mean I see the angle $\phi$ on the image , so where is $\alpha$ in that image to understand better?

N. B:1 -- I am following Hearn and Baker book which screenshot like this.

N. B. -- I want to understand just intuition in easy way rather than details.

Answer 1

• The green arrow is the out-of-plane vector
• the light blue arrow is the plane vector

Calculation of $a$ angle

If you got the coordinates $x,y,z$ and $x_p,y_p,0$ then the most generic way to use is through the dot product and the inverse cosine.

I.e.

• the vector in the projection plane will have direction:

$$v_{plane} = \begin{bmatrix} x-x_p\\ y-y_p\\ 0\end{bmatrix}$$

and the unit direction vector would be:

$$e_{plane} = \frac{1}{||v_{plane}||}\begin{bmatrix} x-x_p\\ y-y_p\\0\end{bmatrix}=\frac{1}{\sqrt{(x-x_p)^2+ (y-y_p)^2}}\begin{bmatrix} x-x_p\\ y-y_p\\ 0\end{bmatrix}$$

Similarly, the out of plane vector is:

$$v_{out} = \begin{bmatrix} x-x_p\\ y-y_p\\ z\end{bmatrix}$$

and the corresponding unit direction vector would be:

$$e_{out} = \frac{1}{||v_{out}||}\begin{bmatrix} x-x_p\\ y-y_p\\z\end{bmatrix}=\frac{1}{\sqrt{(x-x_p)^2+ (y-y_p)^2 + z^2}}\begin{bmatrix} x-x_p\\ y-y_p\\ z\end{bmatrix}$$

The dot product of the unit vectors would be:

$$e_{out} \cdot e_{plane} = \frac{1}{\sqrt{(x-x_p)^2+ (y-y_p)^2 + z^2}}\begin{bmatrix} x-x_p\\ y-y_p\\ z\end{bmatrix}\cdot \frac{1}{\sqrt{(x-x_p)^2+ (y-y_p)^2}}\begin{bmatrix} x-x_p\\ y-y_p\\ 0\end{bmatrix}$$ $$e_{out} \cdot e_{plane} = \frac{( (x-x_p)^2 + (y-y_p)^2 )}{\sqrt{((x-x_p)^2+ (y-y_p)^2 + z^2) ((x-x_p)^2+ (y-y_p)^2)}} $$

$$e_{out} \cdot e_{plane} = \frac{\sqrt{ (x-x_p)^2 + (y-y_p)^2 }}{\sqrt{(x-x_p)^2+ (y-y_p)^2 + z^2 }} $$

At the same time:

$$e_{out} \cdot e_{plane} = ||e_{out}|| \cdot ||e_{plane}|| \cdot \cos(a)= 1\cdot 1\cdot \cos(a) = \cos(a) $$

therefore:

$$\cos(a) = \frac{\sqrt{ (x-x_p)^2 + (y-y_p)^2 }}{\sqrt{(x-x_p)^2+ (y-y_p)^2 + z^2 }} $$

$$a = \arccos\left( \frac{\sqrt{ (x-x_p)^2 + (y-y_p)^2 }}{\sqrt{(x-x_p)^2+ (y-y_p)^2 + z^2 }}\right) $$

(PS: please check the derivation for any mistakes/errors, because the final equation seems a bit too elegant)

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Calculation of $\phi$

Calculation for $\phi$ angle can be expressed by the same equation but you can select a point on the x axis e.g. $(x, y, z) = (1,0, 0).$

In that case the equation will have the following form:

$$e_{out} \cdot e_{plane} = \frac{1}{\sqrt{(x-x_p)^2+ (y-y_p)^2 + 0^2}}\begin{bmatrix} x-x_p\\ y-y_p\\ 0\end{bmatrix}\cdot \begin{bmatrix} 1\\ 0\\ 0\end{bmatrix}$$

$$\phi = \arccos\left( \frac{x-x_p}{\sqrt{(x-x_p)^2+ (y-y_p)^2 }}\right) $$

Answer 2

The angle $\alpha \ $is the angle of projection, meaning it has no shadow on the plane to be shown or measured.

However, if we know the length, L1, as shown on the figure and measure its projection from the figure, l1, then:

$$\pi/2 -\alpha=arctan(l_1/L_1), \quad \alpha=\pi/2-arctan(l_1/L_1)$$

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i apologize for poor sketch, but it conveys the message.