On 2020-10-31 13:58, nop head wrote:

The problem was presented with this illustration
http://forum.openscad.org/file/t2988/span-area.jpg

As far as I can tell you can do exactly what the figure indicates, add
the vector d to each point to arrive at the parallelogram. But maybe I
am missing something?

Carsten Arnholm

Yes it would be easy to calculate the forth vector to complete the polygon
but then how do you draw it with text, which are 2D operations operating in
the XY plane?

On Sat, 31 Oct 2020 at 13:05, arnholm@arnholm.org wrote:

On 2020-10-31 14:10, nop head wrote:

I have not studied the details, but obviously you can compute the plane
equation from 3 points.  Or even simpler, create a vector from P1 to P2
and compute a couple of cross products (the first with vector d) to
arrive at the 3 first columns of the transformation matrix. The fourth
is simply choosing an origin. So I thing you can transform the text with
the resulting matrix.

Carsten Arnholm

That is pretty much exactly what I did.

On Sat, 31 Oct 2020 at 13:22, arnholm@arnholm.org wrote:

Here is a simpler version. Instead of computing the shear and applying it I
simply set up the X and Y axes along the edges of the parallelogram and
make Z orthogonal to both with a cross product.

module fill(p1,p2,d)
{
// this is the vector
v=p2-p1;

// length of vector
l=norm(v);

// length of d
h=norm(d);

x = unit(v);
y = unit(d);
z = unit(cross(v, d));
m = [[x.x, y.x, z.x, p1.x],
     [x.y, y.y, z.y, p1.y],
     [x.z, y.z, z.z, p1.z],
     [0,   0,   0,   1   ]];

multmatrix(m) area(l,h,"p1","p2","p1+d","p2+d");

}

On Sat, 31 Oct 2020 at 13:37, nop head nop.head@gmail.com wrote:

On 2020-11-01 10:15, nop head wrote:

That's pretty much it, except with this you are assuming it is a
rectangle, not a parallelogram. In the above, d is typically not
orthogonal to v, so the matrix m is then incorrect. That's why I said "a
couple of cross products" and not just one as in your code. If you cross
z with x you can arrive at a new y and thus arrive at an orthogonal m.

Carsten Arholm

My previous iteration had two cross products so that the new axes were
mutually orthogonal because I thought that was a requirement. But it seems
not as aligning the axes with the parallelogram works to skew the rectangle
to map onto it. Presumably the matrix values are identical to the version
with orthogonal axes post multiplied by the skew matrix. The end result
looks the same.

On Sun, 1 Nov 2020 at 09:46, arnholm@arnholm.org wrote: