PanoTools mailing list archive
|Date/Time:||Fri, 05 Jan 2001 19:42:58 -0800|
|Subject:||Re: Number of Control Points, was Re:|
Helmut Dersch wrote:
> John Blommers wrote:
> > Haw! Your are right. Each control point represents two variables to be
> > optimized.
> No. Each control point represents one equation.
It seems to me a single control point could represent two equations.
If the error function to be optimized is the Cartesian distance between the
two mapped locations of the control point, then yes, the cost is
by a single equation.
However, if you are minimizing that distance with respect to two variables
get two partial derivatives that we want to find the zeros of. This should
if you hold r and v constant and optimize y and p -- the transformations
More generally, we would want to optimize r and v as well as y and p. Now
equations become non-linear because of the rotation, but if you consider a
region the curves should be close enough to linear within that region. So
initial estimates of the variables are wildly wrong, I think adding a
point is sufficient to give a unique solution.
Certainly if you add the lens correction factors into the mix, the
gets to be more of a factor. If we want to optimize a, b, and c, I suspect
points really are needed because one or two points doesn't give a unique
That is, if we set b to 0 and optimize a for a single point, then change b
again optimize a, we should find a different value of a that gives the same
This would be like looking for the lowest point in a trough that has a flat
Add a second point, and we can optimize a and b with c held constant (now
looking for the intersection of two troughs, and hope they're not
parallel). A third
point would allow us to find c.
Incidentally, I use panotools for single- and double-row panos of less than
360-degree coverage, with no fisheye lens (yet), so I find a rough fixed
of the correction factors a, b, and c to be adequate. Therefore, I do not
P.S. -- This kind of thinking takes me way back. It's been a long time
since college for me.