Trigonomic algebra to find the inverse of a relative mercator projection

In summary, the poster is seeking help to rearrange an equation in order to find the inverse of the relative mercator equation. They have provided equations and attempted solutions, but are currently stuck and seeking assistance. They clarify that they need to rearrange the equation for S.
  • #1
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Hi guys, my first post here. This isn't homework or anything, just a personal project.

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
 
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  • #2
So to clarify, I need to rearrange the following equation for S

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

Related to Trigonomic algebra to find the inverse of a relative mercator projection

1. What is trigonometric algebra and how is it used in finding the inverse of a relative Mercator projection?

Trigonometric algebra is a branch of mathematics that deals with the relationships between angles and sides of triangles. In the context of finding the inverse of a relative Mercator projection, it involves using trigonometric functions such as sine, cosine, and tangent to calculate the coordinates of points on the projection.

2. Why is finding the inverse of a relative Mercator projection important?

Finding the inverse of a relative Mercator projection is important because it allows us to accurately convert between the two-dimensional coordinates on the projection and the corresponding real-world locations on the Earth's surface. This is crucial for various mapping and navigation purposes.

3. What is a relative Mercator projection and how is it different from a regular Mercator projection?

A relative Mercator projection is a type of map projection that aims to preserve the relative sizes and shapes of landmasses, as opposed to the regular Mercator projection which distorts the size and shape of landmasses as they get closer to the poles. This is achieved by using a mathematical formula that takes into account the latitude of a location.

4. Can trigonometric algebra be used to find the inverse of other map projections?

Yes, trigonometric algebra can be used to find the inverse of other map projections that involve angular distortions. However, the specific equations and methods may vary depending on the type of projection.

5. Are there any limitations or challenges in using trigonometric algebra to find the inverse of a relative Mercator projection?

One limitation of using trigonometric algebra to find the inverse of a relative Mercator projection is that it can only accurately calculate the coordinates of points within a certain range of latitudes. This is because the projection itself becomes increasingly distorted at higher latitudes, making the mathematical equations less accurate.

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