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I'm looking for the mathematical expression converting a 3D coordinate (x0,y0,z0) to a 2D (x1,y1) coordinate in a curvilinear perspective of radius R where the values of x1 and y1 are the angles of views {-90° .. +90°} of the original point.

image
(source: ntua.gr)

(image via http://www.ntua.gr/arch/geometry/mbk/histor.htm )

Thanks !

Glorfindel
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Pierre
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2 Answers2

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About one year later , the solution was really simple. For a point having the coordinates:

(x1,y1,z1)

Then, to transform this point in a curvilinear drawing of radius R:

dist=sqrt(x1^2 + y1^2 + z1^2)

x= R*(1+x/dist)
y= R*(1+y/dist)

I can now generate my own drawings (image via wikipedia) :-)

Curvilinear drawing.

Community
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Pierre
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  • Impressive :) Also swapping distance out for Z (or whatever way round you have the coords) gives you a more conventional perspective. It's so simple that the R and "1/" parts may not be needed if you have (0, 0) in the centre of the screen. – alan2here Feb 26 '20 at 01:44
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You may first need to use a transformation matrix to project the 3D object on a 2D plane. http://en.wikipedia.org/wiki/Graphical_projection, choose the one that best fits your needs.

As a second step, you will then want to use the general conversions to bring the coordinates into the Euclidian space. http://en.wikipedia.org/wiki/Curvilinear_coordinates

Alexandre Bell
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  • thanks but this doesn't help much. The wikipedia page is full of mathematical functions. Which one should I choose ? – Pierre Aug 03 '09 at 19:23
  • Well, that is why to my knowledge there is no single formula to convert those coordinates because it will depend on the type of 3d projection you want to do into a Cartesian 2d plane. You need to decide that first. Chose a projection that best fits you and apply its formula to translate your coordinates to the cartesian system. Now you will be ready to apply the simple formulas in the second webpage to translate your cartesian coordinates into an Euclidian coordinate system. – Alexandre Bell Aug 03 '09 at 21:05
  • Problem diagnosis: Good question, too good wikipedia article. Happens to me too from time to time. Solution: Use google to search for supplementary information or ask on a math forum. Also, maybe Wolfram Research offers better organized information. – Stefan Steiger Mar 05 '10 at 11:25