Graphic: a global grid but not emanating from poles

AI Thread Summary
The discussion focuses on creating a fictional map of Earth with a grid system that has its poles at Tierra del Fuego and its antipodes, rather than the traditional poles. The goal is to establish a crude grid that indicates distances and directions to a specific point, the Tunguska Explosion site, using Great Circles. Participants suggest various tools and methods, including gnomonic projections and azimuthal map generators, to simplify the mapping process. There is a particular interest in using 3D CAD software, though some express concerns about the learning curve associated with it. Ultimately, the aim is to create a functional reference for distance and direction without requiring high accuracy.
DaveC426913
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This is for a fictional project, but the tools I need will more likely be found here in Earth forum, I think. Move to fiction if necessary.

I want to render a map of Earth with the "poles" of the grid in a different place. Say, the South "pole" of the grid is at Tierra del Fuego (ish) in S. America. The North "pole" of the grid naturally will be at its antipodes. A Great Circle will define all points equidistant from these poles - a tilted equator.

What I'm really after is not about an actual axis of rotation. I want to make a crude grid (just 30 or even degrees per segment or so) that will show - for any chosen point on the Earth - how far it is, and in what direction it is (a Great Circle) to Juan de Fuca and its antipodes.

I've started by using Google maps to "measure distance" from point to point, which gives me partial Great Circles. These make crude "longitudinal" lines, but now I want crude "latitude" lines (which are essentially all points equidistant from the poles).

Google maps is Mercator Projection-like so the lines form sine-like curves - very difficult to manually draw in latitudes.

Can anyone suggest how I might do this with only a modicum of effort? Any online tools that might allow me to draw poles and lines on a map of Earth?
 
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3D cad models of the Earth are available . You need suitable 3D cad software but if available you can manipulate the model in any way you like .

Certainly you could overlay any grid system and set any axis as principle axis .

Also possible are :

Calcs of distance and surface areas , zoom and orientation control , cut planes , local area selection , flat plane projection ...and more .
 
What are you trying to do? The latitude you cite is suspiciously near
54.73561° = arccos[ sqr(1/3) ]
which is related to the tetrahedron. That is, if you set down a tetrahedron at this latitude, it would function like a sundial. That is, one of the rising edges would run parallel to the axis of the Earth.
Knowing this, the task is easy. Recalling that the dual of the tetrahedron is also a tetrahedron, we can make a gnomonic projection map, actually 4 gnomonic projections onto 4 triangles.
This means that we put the Earth into a tetrahedron and make a gnomonic projection. The 4 points of this tetrahedron are:
1) south pole
2, 3) latitude 35.26439° north, longitude 60°, -60°
4) antipode: latitude 35.26439° north, longitude 180°

You can have this tetrahedron map unfold into a big triangle map of the entire Earth with your ground zero at the center. The north pole would be at the center in one of the sub-triangles. Sundial equations apply to the gnomonic projection. All great circles are straight lines. Your "tilted equator" is the horizon of ground zero. Direction is azimuth. Distance is altitude. Conic sections, circles, ellipses, hyperbolas, and a parabola would map out distances. These can be drawn on MS Paint and it's not really difficult.
 
Nidum said:
3D cad models of the Earth are available . You need suitable 3D cad software but if available you can manipulate the model in any way you like .

Certainly you could overlay any grid system and set any axis as principle axis .

Also possible are :

Calcs of distance and surface areas , zoom and orientation control , cut planes , local area selection , flat plane projection ...and more .
Thanks but I've found the entry barriers and steep initial learning curves for 3D CAD software to be little too much for me.

Maybe I'm thinking of 3DStudio Max etc. Maybe there are super-basic ones.
 
Helios said:
What are you trying to do? The latitude you cite is suspiciously near
54.73561° = arccos[ sqr(1/3) ]
which is related to the tetrahedron.

No, I simply want to have a quick dirty reference from any point on Earth, as to the distance and direction to my "special point".

OK, fine, my special point is The Tunguska Explosion site, in Siberia.

Don't judge me! I told you this was fiction. :cool:

It is close enough to 60N 100E to use those coords.
Helios said:
You can have this tetrahedron map unfold into a big triangle map of the entire Earth with your ground zero at the center. The north pole would be at the center in one of the sub-triangles.
This would be perfect.
It doesn't have to be accurate, just 'good enough'.

How would I do this?
 
Your platonic solids reference gave me an idea that mgiht simplify things.

If an actual graphic (containing lines) ends up being too much effort, I think I might settle for just a twenty, or merely a dozen, points, distributed around the Earth.

So, if I were to calculate my coords at the vertices of a icosahedron or even merely a dodecahedron that would be sufficient.

dodGn-s30.png

This would be perfect, though I'd like to find one with Russia nearer the cetnre. (or rearrange this one).

I'd still need to figure out the angles (through the solid) from vertex to vertex.
 
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At the latitude 60.916666° N, I don't think we can exploit the tetrahedron geometry.
You can still make a gnomonic map with concentric circles for distance. Pick some angle interval (theta) and draw circles with radius = cot(theta). Each circle represents a distance R cot(theta), where R = 3959 mi = 6371 km (Earth radius).
A helpful equation is
cos( D ) = sin( Φ1 )sin( Φ2 ) + cos( Φ1 )cos( Φ2 )cos(δλ)
gives the spherical distance between any two points given latitude Φ and longitude λ.
One degree = 2 π R/ 360°
 
  • #10
https://en.wikipedia.org/wiki/Azimuthal_equidistant_projection
"The azimuthal equidistant projection is an azimuthal map projection. It has the useful properties that all points on the map are at proportionately correct distances from the center point, and that all points on the map are at the correct azimuth (direction) from the center point.

http://ns6t.net/azimuth/azimuth.html

Request an Azimuthal Map
I'm not recommending this particular generator, but it's an example of generators that are out there. You can pick Tunguska Impact site as the center of the projection and go from there. As with all projections, you lose something to get something.
 
  • #11
CapnGranite said:
http://ns6t.net/azimuth/azimuth.html
Request an Azimuthal Map
I'm not recommending this particular generator, but it's an example of generators that are out there. You can pick Tunguska Impact site as the center of the projection and go from there. As with all projections, you lose something to get something.
Wow. That is absulotely perfect!

Thank you!

I have only one thing left to do now, which is determine the angle of declination at the latitudes.
Angle X:
picture-angle-tangent-chord.jpg

http://www.mathwarehouse.com/geometry/circle/angle-tangent-and-chord.php

There are only 3 azimuthal latitudes the mapmaker generates: 45, equator and -45, so this should be easy.
So...
A⊕C = 45°: X = ?
A⊕C = 90°: X= ?
A⊕C = 135°: X= ?
(where ⊕ is the centre)

Looking over the exercises, I think I can flip tje calculation.

Is it as simple as halving the arc angle?
A⊕C = 45°: X = 22.5
A⊕C = 90°: X= 45
A⊕C = 135°: X= 67.5
?
Well, A⊕C = 90°: X= 45 works out correctly, since it would form a right triangle, so that's good.
 
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  • #12
Perfect!
I still have to estimate both declination angle and direction but it's pretty straightforward now.
Oh, and I guess I should put some instructions for doing so...
Bigger-AzimuthalMap.png
 
  • #14
LaurieMD said:
http://wdc.kugi.kyoto-u.ac.jp/index.html

I like the map above!
Found a link to add some data...if you want your compasses to be exact! :)
Thanks but I've actually corrupted the compass roses. The roses - and all the red lines are not magnetic; they're actually geographic/true. (i,e, they point to the true geo poles)

So, I can reference my new fabricated grid (the straight, radial grey lines) and say "at location X on Earth, a straight line running to "Tunguska North" is x degrees off from True geo North (the red curves)".

Halifax is the only point referenced on the map. From Halifax, an arrow that points directly to "Tunguska North (the straight radial grey line) will actually point 11.25 degrees East of True North (the curvy red lines).

Here is a closeup of Halifax:
Closeup-AzimuthalMap.png
 
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  • #15
You do know, of course, that the Canadian Security Intelligence Service had tracked the surviving alien family from the crash at Tunguska to Halifax and that you are not their offspring?
 
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  • #16
CapnGranite said:
You do know, of course, that the Canadian Security Intelligence Service had tracked the surviving alien family from the crash at Tunguska to Halifax and that you are not their offspring?
That's just what they're expecting you to think!
 
  • #17
Very interesting. I'll try 3D cad software.
 

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