Recent content by Thomas Kundera

Hi, Thanks everyone. I finally found my error. Now my result is compatible with Povray rendering of a similar scene and looks comparable to some ISS picts showing Earth shape from orbit. Thanks again. PS: Error is stupid: y=-h in the circle equation, not y=+h...

Hi, Thanks again for your answer. Got some time again, so I'm back on the subject. Because I felt I answered already, seems I'm not clear enough. My camera position is fully described by the single parameter "h" that gives its altitude to the surface, as I'm only considering a camera hold...

I used that equation for the same thing, so I ask:\begin{equation} \left\{ \begin{array}{l} x^2 + z^2 = l^2\\ y = h' \end{array} \right. \end{equation} With $$h' = \frac{rh}{r + h}$$ and $$ l = r \frac{\sqrt{h (2 r + h)}}{r + h} $$ Yes, I defined this centra projection in the...

Would you suggest me to use this forme instead of the one I used above ? It's more general and flexible, I agree.

Yes, thanks.

I'm not sure about your general form above (shouldn't there be some squared terms somewhere?). Anyway, once you have this set (that I computed as being a circle as said above), you want to find it's image by a central projection defined by the point P and the plane as seen of figure above. And...

I believe that's pretty standard, should be the same definition as the one you'll find on Wikipedia (https://en.wikipedia.org/wiki/Horizon) : horizon at point P is the union of points of intersections of the tangent to the Earth passing by P with the Earth (the Earth being assimilated to a...

Hi, I'm trying to simulates how the horizon looks at different altitude, first using a camera hold "'straight" (plane of projection is vertical). So I first computed the horizon shape from a given altitude (it's a circle), and then trying to project that circle on my camera focal plane. Seems...

Hi, Thanks, for your remark. Imagine you standing by the sea, holding a camera body straight, and taking a picture. You won't see an ellipse as horizon, but something that looks like a straight line, that is in fact a very large hyperbola. The higher you get, the more you'll see it curved...

Thanks for your answer. Numerical application used $$\epsilon=50mm$$ and viewport $$24 \times 36mm$$ which are pretty standard camera setting (default for any so called "full-frame" DSLD). That's what I want to simulate. NB: both axis should be in mm, would be nicer, but I didnt yet fixed...

AS: If this is not the right place to ask this question, please let me know where I could try. Thanks. AS: I had to validate the fact I used the template to post my assignment, however I didn't (but it wouldn't post otherwise - btw it's not an "assignment"). Hello everyone. I need help with a...