Perspective in blender

Support Basics & Interface
#1

Hi all!
For a project I’m doing I need to recreate a 3d scene from images taken from blender. I use the rendered sequence with perspective for the textures and I use the z-buffer to get the depth information. In this case I use the orthographic images. Now my problem is that from the depth information I create the 3d space, but to correctly apply the texture I have to do the perspective correction of the 3d data. Since blender does this too (when I select or deselect the orthographic projection option), I wonder if anyone knows the exact formulas that blender uses to do this calculation. Now i’m using these formulas. It seems that the values I use do not allow for a perfect fit of the images. I’ve tried searching in the source code too, but without luck. Does anyone knows which formulas I should use to correctly convert orthographic images to perspective ones?

thanks

#2

So no one can help me? I just need to know how to modify the orthographic images in order to superimpose them to the perspective images created with blender.
Thanks again!

#3

If you have the distance from camera (z-depth), the screen coordinates (x, y) and the viewing angle (lens), then getting the 3d coordinates in camera-local space should be simple trigonometry…

#4

Actually what I need is to tranform the orthographic coordinates into perspective coordinates. The basic step is to divide the x,y coordinates by their z-depth, then some other multiplications should be done. Those are described in the link posted above. I tried various values for the focal length, but some are good when the objects are far from the camera, some are good when the objects are near, so I think that Blender does some calculations. I’m looking exactly for this, I would like to know how Blender converts the orthographic coordinates into perspective ones.
Thanks

#5

Ah, I see, try
(px, py) = (xw, yh)/(2zsin v)
where x, y and z are you known coordinates and (w, h) are the dimensions of your canvas and v is the viewing angle.

#6

So x,y and Z are the coordinates of the 3d space (I get those from a blender render using z buffer right?), w and h are the dimension of the image and v is the field of view as specified in blender?

PS: to be more correct, I get the actual x and y by using the scale value from blender

#7

Yes, you correctly understood my intentions :slight_smile:
Does it work?

#8

I cannot test it today because I’m on a different computer, I will try that tomorrow or at worst the next week. Thanks a lot for your help for the moment, I’ll keep you up to date!

#9

Ok, cool, hope it turns out to be useful :slight_smile:

#10

Just one question, should x and y be centered? I mean, if x goes from 10 to 30 BU, should I subtract 20 (x coordinate of the image center) to the coordinates?

#11

I would think not, since your original coordinate is an image coordinate, but you might have to. If so you should add the centered x-value to half the image width in the end, I suppose…

#12

Hi, I tried using the formula you proposed, using an image of 768 * 575 pixels. It does not seem to work well, if I remove the 2 it’s much better, but still it’s not perfectly superposable to the blender generated perspective file. The x,y and z I used are equal to the blender units in the image (i.e. in orthographic I have a scale of 50, so x goes from -25 to 25, z limits are 0 to 100 so z goes from 0 to 100). also, I have a lens of 35, which is 49.13 in degrees, and that the value i used in the sin.

#13

I’ve also found thisdiscussion, wich deals with a similar problem. In the blender source code I found a couple of matrix, perspective_m4 and orthographic_m4, which are similar to the ones on the link, but add a few things, they are like this:

perspective

nearClip * 2.0f/Xdelta 0 0 0
0 nearClip * 2.0f/Ydelta 0 0
(right + left)/Xdelta (top + bottom)/Ydelta -(farClip + nearClip)/Zdelta -1
0 0 (-2.0f * nearClip * farClip)/Zdelta 0

orto

2.0f/Xdelta 0 0 0
0 2.0f/Ydelta 0 0
0 0 -2.0f/Zdelta 0
-(right + left)/Xdelta; -(top + bottom)/Ydelta -(farClip + nearClip)/Zdelta 0

Those are a bit confusing, do you have any idea about this?
Thanks