Projection: Line of Sight on a Sphere
- Author:
- Ryan Hirst
In the flat image with the center O, aligned with Boxy's eye, the head of the pillar falls at Q'.
Notes:
Perspective: With the vanishing point on the horizon, the surface is infinite: Boxy and the pillar live on one branch of a hyperbola. The horizon lines are the asymptotes of the hyperbola.
On any real surface, there will be a finite distance to the horizon. In the previous worksheet
http://www.geogebratube.org/material/show/id/32720
the cube moves smoothly past the horizon. However, no such space exists, so there is a curious optical illusion: the points of reference appear to change. So. The correct place for the discontinuity is the horizon. A smooth mapping to infinity comes at the cost of fact. To know the distance of an object above the horizon, in real life, we need additional points of reference.
I prefer reality.
First slew or problems solved. Don't imply infinite surfaces. Know your mapping.
2) Discard the mystery:
Place an object in the world.
Look at it.
Have a second person look at it from another location.
Let us define perspective as the relationships among the image you see, the image I see, and the neutral dimensions of the object. Introduce additional complications one at a time: lens properties, occlusion, etc.
Whether projection "takes" points to lines and vice versa is a silly question. Faced with an argument to the contrary, focus your eyes. Can you do it anyway? Good. Then this is not a question for logic; we make it true by deciding. The question then remains, how to project, to achieve a set of desired properties.
3) The distortion of the "inifinte sheet" approach is very small for an enormous surface like the earth, and a perspective position a few feet above it, but for flexible,3D (machine) perspective I don't think it will not do. I will not take straight lines to points in the viewport, or maybe at all.
4) Go through Desargues' Theorem anyway. How to carry lines and points around? I hope I run into Gauss again soon.