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View Full Version : Any Math Majors in the House? Help Wanted.

ender_wiggin
08-16-2011, 07:25 PM
I want to draw latitude and longitude lines on a map I'm making for my homebrew, but it's a pretty unique map. I suspect I will need the help of a real mathematician.

The sole landmass in my setting is an archipelego that begins at the north pole and runs to the southpole. It does so however, not using the shortest path but one that winds around the globe such that it completes exactly one rotation by the time it reaches the south pole (a loxodrome aka rhumb line). I want to draw a rectangular map such that the horizontal midline of the map corresponds to the axis of the archipelego, which means I'm projecting a curved structure onto a straight 2D plane. In other words on the map the archipelego is rendered as a straight line, with the poles at either end of the map, but the archipelego actually follows a rhumb line and thus is curved. Due to this excellent site, I don't have any trouble getting the art right.

Now, the hard part about this is figuring out where the longitude and latitude lines go. I have an idea in my head but I want to be precise.

I was an engineer in undergrad, so I have the background required to figure this out, but it's been a long time since I've done any of that. I need someone who knows what they're doing to point me in the right direction. It took me about 4 hours just to figure out the length of the archipelego in terms of the planet's radius, so you can see how out of shape I am.

As an aside, I think the solution lies in calculating an equation for x and y of cartesian coordinates as a function of psi & theta of the spherical coordinates of the globe. Then I can use a graphing program to hold either of the angles constant and draw the results. That's pretty much as far as I can surmise.

cantab
08-16-2011, 07:37 PM
Your archipelago will be a series of parallel lines on a Mercator map, which you can then easily rearrange. However, such a map won't preserve areas over the archipelago, islands near the poles will be enlarged. Since you are mapping a narrow strip, it seems to me that it should be possible to preserve areas well.

ender_wiggin
08-16-2011, 07:49 PM
I do not want to use a Mercator projection because that means the ends of the archipelego will be dilated, like you pointed out. Instead I want the long axis of the archipelego to be preserved in terms of scale, with the dilation occuring on the lateral sides of the island chain. Since it is only 5k miles wide (compared to 30k long) the map should be able to be rendered with minimal dilation entirely. The problem arises because the long/lat lines are no longer straight like on a Mercator.

Hai-Etlik
08-16-2011, 10:56 PM
Well, I majored in Comp Sci, and have studied non-euclidean geometry and GIS. So not a math major but I know may way around a projection.

First off, a Rhumb line that is not a meridian or parallel never reaches the poles but just keeps getting closer and closer as it spirals around. You will need some other spiral to make one turn about the earth while traversing from pole to pole.

Now, you could take any full globe map in any normal cylindrical projection that has a finite distance between the poles, and draw a straight diagonal line on it from one corner to the opposite corner and you would have a spiral that makes exactly one turn about the planet in the way you want. Depending on the projection you would get different spirals of course. You could then rotate the map to make the straight line run vertical or horizontal as you see fit, and trim off the excess.

Doing this would produce some nasty distortion, this is unavoidable for any projection that could make this work. You will also have a map that is a non-rectangular parallelogram after cropping off the excess.

An alternative is to take the spiral defined as above, make distance along it one of your axes (Which will require a spiral for which it is practical to compute distance along it), and distance along a great circle perpendicular to the spiral the other axis. This would make for the nicest and most usable map, but the math isn't going to be fun.

There seems to be some material online about Spherical Helices for use in antennae. These are curves about a sphere that maintain a constant distance from themselves. I'd expect that distance along such curves has been covered somewhere if you are willing to dig into it.

ender_wiggin
08-16-2011, 11:44 PM
An alternative is to take the spiral defined as above, make distance along it one of your axes (Which will require a spiral for which it is practical to compute distance along it), and distance along a great circle perpendicular to the spiral the other axis. This would make for the nicest and most usable map, but the math isn't going to be fun.

This way, for sure. I've worked on and played in this setting for 10+ years. I really love doing this, so no compromises. And I have time on my side.

Now, if we compute distance along the vertical axis by traveling along the great circle perpendicular to the 'backbone' spiral, won't those great circles intersect at some point away from the spiral? I assume to fix this you have to distort the image as you travel farther away from the spiral to avoid this. And I'm ok with that, but how does that affect the math?

cantab
08-16-2011, 11:49 PM
First off, a Rhumb line that is not a meridian or parallel never reaches the poles but just keeps getting closer and closer as it spirals around.I thought of this, but considered that if there are islands at both poles, those islands can "hold" the infinity of turns, and leave only one turn between them. Essentially, the rhumb line is being clipped at a latitude less than 90N/S.

ender_wiggin
08-17-2011, 12:52 AM
Wait a minute, I think I answered my own question. If we force the spiral into a straight line, then all the great circles are parallel to each other, and at whatever point they intersect is the maximum width of the map.

Hai-Etlik
08-17-2011, 01:02 AM
Wait a minute, I think I answered my own question. If we force the spiral into a straight line, then all the great circles are parallel to each other, and at whatever point they intersect is the maximum width of the map.

They aren't parallel, two distinct great circles are never parallel. That's pretty much the defining trait of elliptic geometry.

The specific points of intersection will vary, but for any two that are fixed by points close together on the spiral, the intersection points will be quite a long ways off from the spiral, approximately 90° of central angle.

ender_wiggin
08-17-2011, 01:51 AM
They aren't parallel, two distinct great circles are never parallel. That's pretty much the defining trait of elliptic geometry.

The specific points of intersection will vary, but for any two that are fixed by points close together on the spiral, the intersection points will be quite a long ways off from the spiral, approximately 90° of central angle.

In a real sphere, yes the great circles can't be parallel, but in 2D projections they can look parallel on paper (ie in Mercator all the longitude lines are parallel). Since great circles never have any curve except for the geodesicurve, and we force the spiral into a straight line, it should stand that all the great circles run parallel to each other off the page. Of course, in order to account for the fact that they aren't actually parallel on the sphere, we have to distort the map. The point of intersection is the maximum point of distortion, which should be the maximum width of the map (like the poles in Mercator).

If you happen to find any links that explain the equations to actually implement this, please do share.

Hai-Etlik
08-17-2011, 02:03 AM
In a real sphere, yes the great circles can't be parallel, but in 2D projections they can look parallel on paper (ie in Mercator all the longitude lines are parallel). Since great circles never have any curve except for the geodesicurve, and we force the spiral into a straight line, it should stand that all the great circles run parallel to each other off the page. Of course, in order to account for the fact that they aren't actually parallel on the sphere, we have to distort the map. The point of intersection is the maximum point of distortion, which should be the maximum width of the map (like the poles in Mercator).

Yes, their images under the projection would be parallel, but that is not the same thing as them being parallel themselves. The distortion would actually be more like that in the Equidistant Cylindrical projection than the Mercator projection.

waldronate
08-17-2011, 02:56 AM
Perhaps http://en.wikipedia.org/wiki/Space-oblique_Mercator_projection with the orbit angle of inclination and rotation periods selected to travel along the centerline of your archipelago?

ender_wiggin
08-17-2011, 08:09 AM
I thought of this, but considered that if there are islands at both poles, those islands can "hold" the infinity of turns, and leave only one turn between them. Essentially, the rhumb line is being clipped at a latitude less than 90N/S.

That's a good point. I can start the rhumb line arbitrarily close to the actual pole. Although upon looking at the curve closely, I don't think a loxodrome is what I had in mind. I'm looking for a curve that hits every latitude and longitude exactly once. To do this I think the bearing changes.

Perhaps http://en.wikipedia.org/wiki/Space-oblique_Mercator_projection with the orbit angle of inclination and rotation periods selected to travel along the centerline of your archipelago?

That's an interesting find, thanks. Although the scanner satellite in that photo is a straight line, not a curve.

waldronate
08-17-2011, 03:16 PM
the scanner satellite in that photo is a straight line, not a curve.

The earth turns under the satellite, making the the ground track a spiral. If the orbital inclination is polar (satellite scans pole to pole) and the rotation of the globe is faster than the orbital period of the satellite (say, exactly twice as fast), then the ground path should be a spiral that traces from one pole to another in the time that the globe rotates one full revolution. But math was never my strong suit, so take that estimate with a grain of salt.

I've attached the example image for this projection from Snyder's An Album of Map Projections (USGS Professional paper 1453).
37864

ender_wiggin
08-17-2011, 10:04 PM
Wow, you're right. This looks very promising, thanks! Do you have a link to documentation explaining the projection and how to use his equations? The wikipedia article is a good intro but is a bit lacking for the inexperienced. I will look into this and post the results here. Thanks again.

Btw if anyone else also has insight into this problem, I'm all ears.

gilgamec
08-20-2011, 02:05 PM
Wow, you're right. This looks very promising, thanks! Do you have a link to documentation explaining the projection and how to use his equations? The wikipedia article is a good intro but is a bit lacking for the inexperienced. I will look into this and post the results here. Thanks again.As waldronate hasn't posted this, I'll mention that Snyder's "Album" can be downloaded from the USGS at http://pubs.usgs.gov/pp/1453/report.pdf. In addition, Snyder's "Map Projections - A Working Manual" goes into much greater detail about the equations, how they are derived, and how to use them; it's online at http://pubs.usgs.gov/pp/1395/report.pdf.

ender_wiggin
08-21-2011, 10:50 AM
Ah, yes. Now there is the only issue of implementing this monster. Currently trying to brush up on my MATLAB..