Re: Projecting segment with normals

cbull <cbull@xxxxxxxxxxxxxx> wrote:

> "Hans-Bernhard Broeker" <broeker@xxxxxxxxxxxxxxxxxxxxx> wrote in message
> news:3lbkqcF11piefU3@xxxxxxxxxxxxxxxxx

> > Those are two separate problems combined into one. You had better
> > look at each on its own. First find out what's the behaviour of
> > "point along the segment offset by an interpolated vector" in 3D.

> I don't understand what do you mean by this. I don't want to project points
> that are offseted from the original segment. I want to project points of the
> segment along vectors produced by the interpolation.

That was quite unclearly worded in the OP. You're using the word
"project" in a somewhat non-standard meaning there. So let's write
this down in completely clear mathematics:

P(t) = interpol(t, A, B) + s(t) * interpol(t, V1, V2)

where t goes from 0 to 1, s(t) (>= 0?) is chosen such that P(t) is a
point in the given plane

P(t) . N = d

for given N and d, and interpol(t, x, y) is linear interpolation
between x and y. For a given t that yields 4 equations for 4 unknowns
(the 3 coordinates of P(t), plus s(t)). Substituting P(t) into
the plane normal equation gives

t * A.N + (1-t) * B.N + s(t) * (t * (V1.N) + (1-t) * (V2.N)) = d

that's a simple linear equation for s(t):

s(t) = (d - t*A.N - (1-t)*B.N) / (t*V1.N + (1-t)*V2.N)

Substitute that into P(t), and you have your curve on the plane. The
resulting curve would be ratio of polynomials in t.

--
Hans-Bernhard Broeker (broeker@xxxxxxxxxxxxxxxxxxxxx)
Even if all the snow were burnt, ashes would remain.
.

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