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Milosvukovic

Geometry: Axonometric (slant) projection of a square

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Posted (edited)

Hi there, 

This 'problem' torments me since a while:

Let us draw one projection of a square, and the projection is seen as a rhomboid - so, draw some arbitrary rhomboid. (Imagine a square tile that you see from a certain angle - it will become slanted and therefore distorted and seen as a series of romboids as you change the angle of view. Have in mind that distance at/from which you see it does not matter - only the angle.)

Q: How to find the 2nd projection of this square, and how to draw this square (full size, seen 'en face', as if seen directly under 90°)?  See the image below. 

Milos

Screenshot_20191204_163212_com.google.android.gm.jpg

P. S. You can rotate the image by 90 deg or change the shape of the rhomboid as well, the case is meant to be general and thereby doesn't change the essence.

Edited by Milosvukovic
To bring closer the point

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Moderator Note

Moved to Mathematics (seems most appropriate).

 

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Posted (edited)

To get projected coordinate, 3D world original coordinate must be multiplied by projection matrix.

https://en.wikipedia.org/wiki/3D_projection

In orthogonal projection matrix, one 3D axis is simply cleared (multiplied by 0).

What you need to do is construction of inverse of projection matrix.

https://www.google.com/search?q=inverse+projection+matrix

(in some cases it is not possible to reverse projection, to get original coordinate, like in orthogonal projection were you lost certain axis data)

 

Axonometric projection that you gave is similar to isometric projection.

https://en.wikipedia.org/wiki/Isometric_projection

In the above article there are showed matrices which can be used to get such projection.

 

In your case it's simple as you know that [math]\alpha[/math] used to be originally 90 degrees, [math]\beta =\frac{360-2*\alpha}{2}[/math] was also 90 degrees, and length of each Rhombus edge used to be (prior projection) the same value (projection of square, not rectangle).

 

Edited by Sensei

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Thank you kindly for such a quick response, I will study it and reply ASAP.  Milos

I forgot to mention that the problem should be tried to be solved by means and rules of descriptive geometry, if possible.

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