- #1

WellHydrated

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## Homework Statement

I need to find the inverse of the relative mercator equation. That is, given an origin latitude/longitude, find the current latitude/longitude from an x,y point. Because longitude is easy to calculate, we will ignore that, but if anyone is interested in the formula I will post it up. I'll also post up what each variable means if anyone is interested, but I don't want to convolute this post as the context isn't important - this is just trigonimic algebra. I have tried to solve in Maple but I guess it lacks the trignomic indentity needed?

## Homework Equations

Need to rearrange the below equations for L.

x = w * log((c + s) / (c - s)) / (2 * PI)

*(1)*

c = cos(R * (o + L)

*(2)*

s = sin(R * (o - L)

*(3)*

## The Attempt at a Solution

Rearranging equation 1 to find u, the "log term"

u = (c + s) / (c - s)

*(4)*

x = w * log(u) / (2 * PI)

u = e^((2 * y * PI) / w)

Rearranging equation 2 to find L:

c = cos(R * (o + L)

L = (2 * arccos(c) / R) - o

*(5)*

Rearranging equation 3 to find L:

s = sin(R * (o - L)

L = o - (2 * arcsin(s) / R)

*(6)*

Rearranging equation 4 to find c:

c = ((u + 1) * s) / (u - 1)

*(7)*

Equating 5 and 6:

(2 * arccos(c) / R) - o = o - (2 * arcsin(s) / R)

*(8)*

Substituting equation 7 into equation 8:

(2 * arccos((u + 1) * s) / (u - 1)/ R) - o = o - (2 * arcsin(s) / R)

*(9)*

This is where I get stuck. I need to rearrange equation 9 for s, then it will just be simple substitution in order to get the equation I need for L. I have managed to reduce it to a more readable state (you may or may not want to go down this path).

o * R = arccos((S * (U + 1)) / (U - 1)) + arcsin(S)

Now we have S stuck in an arcsin and an arccos with the latter S being involved in a more complex arc.

Any help would be very much appreciated!

Thanks,

Mickey