I need to find the 2 points of the visual horizon, of a curved face.
I have:
- XYZ of the 4 corner points
- XYZ of the 2 curved edge bezier points
And I need to calculate either:
- XY of the 2 horizon points
- XYZ of the 2 horizon points
Note: I got a solution the last time I asked this question, but it only found the extrema of the curves, not the horizon points, which changes based on the position and rotation of both curves in respect to each other.
You don't say how your surface is defined, only that it is bounded by two quadratic Bézier curves. There are lots of ways to build such a surface, and each way of building it would have a different horizon. So this answer is going to be guesswork.
The horizon consists of those points on the surface where the vector from the camera to the point is tangent to the surface, as shown here:
A quadratic Bézier curve has parametric equation
B(t) = (1 − t)2 P0 + 2(1 − t)t P1 + t2 P2
differentiating that with respect to t gives us the tangent to the curve:
B′(t) = 2(t − 1) P0 + 2(1 − 2t) P1 + 2t P2
and this is parallel with the vector from the camera (at the origin) to the curve if
B(t) × B′(t) = 0
Solve this for t and you'll have the point on the curve at the horizon. How you can extend this to the horizon for the whole surface depends on how your surface is constructed. (Maybe you can just find the horizon points for the curves at each end of the surface and join them with a straight line?)
Does it work to first rotate the curves so that the connection between the corner points is horizontal and then calculating the extrema?
To test it visually, you can rotate your example image by about 150 degree:
Rotated image http://www.freeimagehosting.net/uploads/ad502509e9.png
Note the extrema of this curve isn't exactly where you want it, but this could be caused by several factors, for example the way you marked the horizon points doesn't seem to be that exact.
What you're looking for is actually called a silhouette, not a horizon.
The most simple method of doing this is finding the boundary between the surface parts in which the normal is directed towards the camera (dot product is negative) and the surface parts in which the normal is directed away from the camera (dot product is positive).
With a triangle mesh you can do this directly by using the normals. with NURBS you can probably find a closed formula which does this.
来源：https://stackoverflow.com/questions/574732/calculate-the-horizon-of-a-curved-face-not-extrema