Geek Answers Handbook

Calculate the horizon of a curved face? - Not extremes

I need to find 2 points of the visual horizon, a curved face.

I have:

  • XYZ from 4 corner points
  • XYZ of two curved bezier points

And I need to calculate:

  • XY from two points of the horizon
  • XYZ from two points of the horizon

Note: I got the solution last time . I asked this question, but he found only the extrema of the curves, and not the horizontal points, which vary depending on the position and rotation of both curves relative to each other.

+5
source share
3 answers

, , , . ​​, . .

, , :

Tangent to bezier curve

B (t) = (1 - t) 2 P 0 + 2 (1 - t) t P 1 + t 2 P 2

t :

B '(t) = 2 (t - 1) P 0 + 2 (1 - 2 t) P 1 + 2 < > P < > 2 >

( ) ,

B (t) × B '(t) = 0

t, . , , . ( , ?)

+2

, , ?

, 150 :

http://www.freeimagehosting.net/uploads/ad502509e9.png

, , , , , , , , .

+2

What you are looking for is actually called a strong silhouette. , not the horizon.
The easiest way to do this is to find the boundary between the surface parts in which the normal is directed towards the camera (the point product is negative), and the parts of the surface in which the normal is directed from the camera (the point product is positive),

With a triangular grid, you can do this directly using normals. with NURBS you can find a closed formula that does this.

+1
source

More articles:


All Articles