Trigonometry, Logs and Map ProjectionsDate: 10/01/97 at 10:35:54 From: Gavin Lotter Subject: Trigonometry, logs and map projections Dear Doc, I have a headache from trying to figure out the inverse function used by the Mercator map projection, a very simple map projection. To project a point of longitude and latitude onto a piece of paper, the Mercator projection uses X as the longitude and Y is determined by the formula Y = log_e( sec(A) + tan(A) ) where A is the latitude. My problem is that I need to determine the latitude from the projected point, that is Y is known. The formula then becomes sec(A) + tan(A) = e^Y where A is the latitude which I need to determine and Y is the projected latitude. It appears as though I have two unknowns ( sec(A) and tan(A) ) with one equation. How can I possibly solve for A, the original unprojected latitude? Thanks for a great service and for one of the best sites on the Web. Gavin Lotter Date: 10/01/97 at 12:30:16 From: Doctor Rob Subject: Re: Trigonometry, logs and map projections Flattery will get you everywhere! :-) Thanks for the kind comment. You need one more fact to realize that you don't really have two unknowns here. That is, for every angle A, [sec(A)]^2 = 1 + [tan(A)]^2. You could use this to rewrite your equation as Sqrt[1 + tan(A)^2] + tan(A) = e^Y. If you transpose tan(A) to the other side, then square, you will see that 1 + tan(A)^2 = e^(2*y) - 2*tan(A)*e^Y + tan(A)^2. Now subtract tan(A)^2 from both sides, and solve for tan(A): tan(A) = [e^(2*Y)-1]/2*e^Y, = [e^Y - e^(-Y)]/2, = sinh(Y), (if you are familiar with hyperbolic functions). Thus A = Arctan([e^Y - e^(-Y)]/2), = Arctan(sinh[Y]). That ought to take care of this problem. -Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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