Re: Altitude of a point in a triangle

Erik Max Francis wrote:

> > Huh, I don't see how it accomplishes that.  It seems to me that you're
> > solving for a point that is along the line defined by p + sn where s
is
> > a scalar scale value, and appears to completely disregard the
"vertical"
> > vector which is what you're actually trying to compute.
> 
> No, the solution I specified solves for the intersection between a line 
> parallel to the y axis which contains the point and the plane of the 
> triangle.  Which is exactly the interpolated point on the triangle.

Upon pondering and the bit-flip (between posting my last post and it
showing up on the server, yay) of realizing 'doh "dot 0" means on the
plane' I see what you're getting at.  I'm left wondering if the
confusion of one who does this for a living might be amplified for
someone who doesn't.

So yeah, choose your Y so the point lies on the plane.  Simple enough
and absolutely not what came to mind upon reading the "how-to"
initially.  Strangely enough the solutions we provided are algebraically
equivalent; they were simply provided with different means of looking at
the problem.

MissElaine: the algebraic solution then is:

P is some point on the plane, one of the verts of your triangle
V is your starting point above the plane
N is the normal of your plane

height = Py - (( ((Vx-Px)/Nx) + ((Vz-Pz)/Nz) ) / Ny)

....I think.  Y'might want to work through the problem yourself to
validate.  ;)

-tom!

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