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#### interpolation of barycentric coordinates

```Hello all,
I have points P1, P2 lying in a triangle A, B, C. I'm able to compute
barycentric coordinates of P1 and P2. Is it possible to interpolate
somehow the barycentric coordinates to obtain points which lie on the
line segment P1P2? Of course, I could interpolate linearly between P1
and P2, but I would like to interpolate bar. coordinates instead. Is
that possible?

Thanks.

Jindra
```
 0
jparus (92)
4/12/2008 1:39:18 PM
comp.graphics.algorithms 6674 articles. 0 followers.

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```On Apr 12, 3:39 pm, jindra <jpa...@gmail.com> wrote:
> Hello all,
> I have points P1, P2 lying in a triangle A, B, C. I'm able to compute
> barycentric coordinates of P1 and P2. Is it possible to interpolate
> somehow the barycentric coordinates to obtain points which lie on the
> line segment P1P2? Of course, I could interpolate linearly between P1
> and P2, but I would like to interpolate bar. coordinates instead. Is
> that possible?
>

It's exactly the same thing. The barycentric coordinates computed from
any triangle or generic cartesian coordinates are related to one
another by an affine transformation.
```
 0
gatti1 (74)
4/12/2008 4:36:04 PM
```On Apr 12, 6:36 pm, Lorenzo Gatti <ga...@dsdata.it> wrote:
> On Apr 12, 3:39 pm, jindra <jpa...@gmail.com> wrote:
>
> > Hello all,
> > I have points P1, P2 lying in a triangle A, B, C. I'm able to compute
> > barycentric coordinates of P1 and P2. Is it possible to interpolate
> > somehow the barycentric coordinates to obtain points which lie on the
> > line segment P1P2? Of course, I could interpolate linearly between P1
> > and P2, but I would like to interpolate bar. coordinates instead. Is
> > that possible?
>
> It's exactly the same thing. The barycentric coordinates computed from
> any triangle or generic cartesian coordinates are related to one
> another by an affine transformation.

I'm sorry, but I don't follow now. You suggest to linearly interpolate
the barycentric coordinates? Eg., let's have (u1,v1,w1) as
barycentric coordinates of P1, and (u2, v2, w2) as barycentric
coordinates of P2 (with respect to some triangle A, B, C). I'm looking
for a function f(t) which for a given t returns barycentric
coordinates (u_t, v_t, w_t) so that A*u_t + B*v_t + C*w_t = Q, where Q
is a point on the line P1, P2.

Jindra
```
 0
jparus (92)
4/13/2008 7:17:24 AM
```> I'm sorry, but I don't follow now. You suggest to linearly interpolate
> the barycentric coordinates? Eg., let's have (u1,v1,w1) as
> barycentric coordinates of P1, and (u2, v2, w2) as barycentric
> coordinates of P2 (with respect to some triangle A, B, C). I'm looking
> for a function f(t) which for a given t returns barycentric
> coordinates (u_t, v_t, w_t) so that A*u_t + B*v_t + C*w_t = Q, where Q
> is a point on the line P1, P2.

It's correct. Consider

P1 = u1 * A + v1 * B + w1 * C
u1 + v1 + w1 = 1
P2 = u2 * A + v2 * B + w2 * C
u2 + v2 + w2 = 1

Q = P(t) = (1 - t)P1 + tP2 =
((1 - t)u1 + tu2) * A +
((1 - t)v1 + tv2) * B +
((1 - t)w1 + tw2) * C

Checking:

(1 - t)u1 + tu2 +
(1 - t)v1 + tv2 +
(1 - t)w1 + tw2
=
(1 - t)(u1 + v1 + w1) +
t(u2 + v2 + w2)
= (1 - t) + t = 1

--
kaba.hilvi.org
```
 0
none87 (1162)
4/13/2008 9:55:44 AM
```On Apr 13, 11:55 am, Kaba <n...@here.com> wrote:
> > I'm sorry, but I don't follow now. You suggest to linearly interpolate
> > the barycentric coordinates? Eg., let's have (u1,v1,w1) as
> > barycentric coordinates of P1, and (u2, v2, w2) as barycentric
> > coordinates of P2 (with respect to some triangle A, B, C). I'm looking
> > for a function f(t) which for a given t returns barycentric
> > coordinates (u_t, v_t, w_t) so that A*u_t + B*v_t + C*w_t = Q, where Q
> > is a point on the line P1, P2.
>
> It's correct. Consider
>
> P1 = u1 * A + v1 * B + w1 * C
> u1 + v1 + w1 = 1
> P2 = u2 * A + v2 * B + w2 * C
> u2 + v2 + w2 = 1
>
> Q = P(t) = (1 - t)P1 + tP2 =
> ((1 - t)u1 + tu2) * A +
> ((1 - t)v1 + tv2) * B +
> ((1 - t)w1 + tw2) * C
>
> Checking:
>
> (1 - t)u1 + tu2 +
> (1 - t)v1 + tv2 +
> (1 - t)w1 + tw2
> =
> (1 - t)(u1 + v1 + w1) +
> t(u2 + v2 + w2)
> = (1 - t) + t = 1
>
> --
> kaba.hilvi.org

Very nice, thank you very much.

Jindra
```
 0
jparus (92)
4/13/2008 1:00:19 PM

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