How to visualize angle of projection in Oblique Parallel Projection?

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.

• Which page have you got to? Dec 4 '21 at 21:32
• @Solar it has 315 in Hearn and Baker book. Dec 4 '21 at 21:35
• I think it is the angle between the -Zv axis and +Yv axis, which is 90 degrees in this case.
– r13
Dec 4 '21 at 22:31
• @r13 could you draw the image and level the $\alpha$ where is exist in the image.. Dec 4 '21 at 22:37
• It is the angle in between the horizontal baseline and vertical line on the Yv, Zv plane.
– r13
Dec 4 '21 at 22:52 • 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)

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:

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

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

• could you show the image with projection and after projection with Projector and angle $\alpha$ and $\phi$, by which I can understand easily. Dec 5 '21 at 8:57
• apologies, instead of $\phi$ I should have written $a$. I will correct it later, since I am out on a family excursion. Dec 5 '21 at 8:59
• OK thank you.. Please do it as soon as possible. Dec 5 '21 at 9:01
• The "out of plane" refers to the vector which does not exists within the plane you see in the image. It is the vector connecting $(x,y,z)$ to $(x_p, y_p)$. The plane vector is the one connecting $(x,y)$ to $(x_p, y_p)$. Dec 5 '21 at 14:12
• It's a similar process to the above. I'm finding the able between the in plane vector and the unit vector of the X axis. Dec 5 '21 at 14:53

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:

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

. i apologize for poor sketch, but it conveys the message. • it's the right-side view of cube or front view? You treat $a$ as $\alpha$? Dec 5 '21 at 5:51
• it's the right side view. you need to click on the figure and enlarge it. yes i treat a, as alpha cause my sketch up is limited font. Dec 5 '21 at 6:31
• I have one request could you show the whole image where original image projects to projected image with angle levelling? I will be helpful to got felt better as for beginner. Dec 5 '21 at 7:07
• yes. its late here. I do it tomorrow. but basically its going to be too busy. the concept it L1 and its image l1 are the sides of triangle. but i do it tomorrow. Dec 5 '21 at 7:11
• Thank you kamran. Hats off to you. And kindly you mention $\alpha$ and $\phi$ Dec 5 '21 at 7:24