Re: Implicit equations for surfaces as determinants of basis matrices

On May 11, 3:18 am, OwlHoot <ravensd...@xxxxxxxxxxxxxx> wrote:
On May 11, 2:04 am, Dave the Funkatron <dave.rud...@xxxxxxxx> wrote:



[...]

That all seems good, though I'm not sure why the "bases" at the top of
this post are called bases. They don't seem to be bases in the linear
algebra sense.

Ah but they are linear - in the coefficients you're trying
to find.

For example, if the points you are given are on a circle
in the Oxy plane then this must have the general form:

  (x - u)^2 + (y - v)^2 = r^2

where (u, v) is the centre and r is the radius, neither
of which you know.

Now that can be expanded as:

 (x^2 + y^2) - 2.u.x - 2.v.y + (u^2 + v^2 - r^2) = 0

where you don't know 2u, 2v, and (u^2 + v^2 - r^2).

But for each numerical point (x, y) you're given, you
_do_ know all three of x^2 + y^2, x, and y !

So if point (x1, y1) lies on the circle you have a _linear_
equation in the unknowns u, v, and w := (u^2 + v^2 - r^2) :

  1 + a.u  + b.v  +  c.w  =  0

where a, b, c are numeric values defined by:

  a  =  - 2.x1 / (x1^2 + y2^2)

  b  =  - 2.y1 / (x1^2 + y2^2)

  c  =       1 / (x1^2 + y2^2)



Just checking, but y2 should be y1, correct?

Anyway, I think what you are saying makes sense, but something is
still lost on me. I think you are saying to solve the linear system

[ 1, a1, b1, c1 ] [ 1 ] [ 0 ]
[ 1, a2, b2, c2 ] [ u ] = [ 0 ]
[ 1, a3, b3, c3 ] [ v ] [ 0 ]
[ 1, a4, b4, c4 ] [ w ] [ 0 ]

But, I am apparently missing something, because I would only have
three points for a circle (and really only three variables to solve
for). So I have to do something about that last row in the matrix, do
I not?

By the way, your method of fitting the curve seems different than the
one in Pratt's paper. At this point, I'll take any solution that
works, but I'm also curious to understand what he was doing. His
matrices were different, and he uses its determinant to derive the
coefficients, though I'm still unclear as to exactly how it is done.


Well I hope that hasn't confused you even more!

Cheers

John Ramsden

Thanks. Actually, that does give me some insight.
.

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