| Re: Finding all angles within a range of directions; an algorithm question [message #30278 is a reply to message #30266] |
Thu, 18 April 2002 02:27   |
Struan Gray
Messages: 178
Registered: December 1995
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Senior Member |
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Craig Markwardt, craigmnet@cow.physics.wisc.edu writes:
>
> tbowers0@yahoo.com writes:
>>
>> Wow Struan! Your explanation was a bit beyond me. So, instead of just
>> replying "Uhh.. Huh?" I did some surfing on rotation matrices and
>> stuff and found the Matrix and Quaternion FAQ at
>> http://skal.planet-d.net/demo/matrixfaq.htm. Educated myself a bit,
>> but I'm still unclear. If I understand:
>
> You definitely want to use the dot-product approach. It's fast, and
> there's no room for screwing up the signs like there is in rotation
> matrices. And, as you say, you can make a different cut, other than
> 90 degrees.
I am more than happy to bow to Craig's superior knowledge, but in case you
want to do it the hard way here's a more coherent (I hope) explanation of what
I meant.
Rotations are coordinate transformations, and if you have an array of
coordinates you can transform them all at once by doing a matrix
multiplication between the array of coordinates and a rotation matrix. In
deciding which method to use you have to weigh the possible speed penalty of
doing a whole series of dot products against the difficulties of constructing
the correct rotation matrix.
The rotation you want is the one which will rotate the current theta=0
direction into the surface normal of your plate. Find that, and you then just
rotate all the other coordinates and find the ones who have a new theta
greater than zero. The rotation is about the vector formed by the cross
product of the plate normal and the original zenith, with an angle found by
vector geometry.
As Craig said, the real problem is keeping track of all your sign
conventions for rotations and directions. This is a common problem in
astronomy (it is roughly the same as asking which stars will be above the
'horizon') so some of the astro libraries (or Craig's) may have already done
the hard work in your particular case.
Another place the hard work has been done is the IDL object graphics
libraries. IDLgrModel objects have a method called GetCTM which returns the
correct transformation matrix to map coordinates in the Model's internal
coordinate system to those of any of the models higher up the tree, and
finally the View object itself. In this case all coordinates are cartesian,
so you would still have to do a sperical-polar to cartesian conversion, but
IDL would handle the signs properly.
Once you have the transformation matrix you apply it to an array which
lists the theta and phi values for each data entry in your matrix. You then
have to find those that match a criterion for visibility, which varies with
the coordinate system you've ended up in: i.e. theta<pi/2 or z>0. Finding
elements of an array which match a condition is an old chestnut here in the
newsgroup, and there are essentially three ways of doing it.
First, and most obvious, is WHERE. Here the only difficulty is taking the
one dimensional indices WHERE returns and reforming them to index your
original two or more dimensional array.
The second technique uses HISTOGRAM's useful REVERSE_INDICES keyword to
identify all elments with a particular value or which lie in a range of
values. This is usually faster then WHERE, but you have to parse the
REVERSE_INDICES array correctly.
The third technique uses a direct compare on the whole array at once, i.e.
if z is an array of tranformed z-values, just write:
visible = z>0
and visible will be an array of the same size as z with 1's where z is
greater than zero and 0's where it's not. This is faster yet, but whether it
is useful depends on what you want to do. It makes it very easy to increment
or add up those values:
new_z = z + visible*increment
conditional_total = total(z*visible)
So there you go, more than you ever wanted to know. Happy hunting and let
us know what you end up with.
Struan
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