Path: utzoo!attcan!uunet!lll-winken!lll-tis!ames!elroy!mahendo!wlbr!etn-rad!shaw
From: shaw@etn-rad.UUCP (Howard Shaw)
Newsgroups: comp.graphics
Subject: Re: Projective transformations
Message-ID: <560@etn-rad.UUCP>
Date: 27 Jul 88 18:03:39 GMT
References: <559@etn-rad.UUCP> <1161@eos.UUCP>
Reply-To: shaw@etn-rad.UUCP (Howard Shaw)
Organization: Eaton Inc. IMSD, Westlake Village, CA
Lines: 35

In article <1161@eos.UUCP> jbm@eos.UUCP (Jeffrey Mulligan) writes:
>From article <559@etn-rad.UUCP>, by shaw@etn-rad.UUCP (Howard Shaw):
>
>< I believe the mapping from the original image space to the new image 
>< space is a bilinear fractional transformation:
>< 
><         a*x + b*y + c                     f*x + g*y + h
><     X = -------------                 Y = --------------
><         d*x + e*y + 1                     d*x + e*y + 1
>< 
>< Any known easy way to get the coefficients a,b,...?  Even if I could
>< only get four (x,y) <-> (X,Y) pairs, I could get the coefficients by
>< curve-fitting.  Any help would be greatly appreciated.   -HS
>
>If you have four points, the transformation is determined;  8 equations,
>8 unknowns.  Set up an 8 by 8 matrix which operates on the vector
>of coefficients, then invert this matrix.  I have written program
>which does this, but it doesn't handle special cases very nicely
>right now.
>
>-- 
>
>	Jeff Mulligan (jbm@aurora.arc.nasa.gov)
>	NASA/Ames Research Ctr., Mail Stop 239-3, Moffet Field CA, 94035
>	(415) 694-6290

Thanks for the help, but let me quote from the original posting:

" I am given the viewing point (Lat, Long, and Alt, relative to the 
assumed to be known Lat/Long for some points on the original image), 
and the viewing angles (degrees clockwise from North, and degrees 
down from the local horizontal)."

That is, I know how to solve the problem, given four (x,y) <-> (X,Y) 
pairs.  However, what do I do, if given, instead, the viewing point 
and viewing angles?  Sorry if my original posting was unclear. -HS