March 2015
Beginner to intermediate
840 pages
25h 12m
English
Content preview from GPS Satellite Surveying, 4th Edition
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Substituting (7.8.16) back into (7.8.15) we present it in the form
where
Then, the solution to the problem (7.8.17) is given by the following expressions:
7.8.19 
which are obtained the same way that expression (7.8.11) was obtained above.
We note that the matrix 
now has a form similar to (7.8.13) for 
. Since the matrix 
has a form similar to (7.8.14) for all 
, then columns 
for all , which follows from expression (3.1.15). It means that the algorithm described in ...
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