## Blending equal-area map projections

For the discussion of math. Duh.

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Qaanol
The Cheshirest Catamount
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### Blending equal-area map projections

I want to create an equal-area map projection that is a blend of the Hammer and Mollweide projections, but I’m not sure how to do it or if it’s even possible.

Near a certain location (0°N, 150°E) I want the map to be as the Mollweide projection. Near the antipodal point (0°N, 30°W) I want it to be as the Hammer projection. And I want them to smoothly meld throughout the rest of the map, while maintaining the equal-area property.

My first thought is, for each point (θ,φ) on the globe, find the angular distance from (0°N, 150°E), call it α, and take the weighted average of the projections of that point under each mapping, as (α/π)Hammer(θ,φ)+(1-α/π)Mollweide(θ,φ). However, I haven’t been able to show whether or not this preserves areas.

Any ideas?
wee free kings

PM 2Ring
Posts: 3659
Joined: Mon Jan 26, 2009 3:19 pm UTC
Location: Mid north coast, NSW, Australia

### Re: Blending equal-area map projections

I have no idea if your weighted mean will preserve areas, but it sounds reasonable. I guess you could brute-force test it by writing a program to project small circles & spherical triangles of known area.

It's a pity that Daniel "daan" Strebe doesn't post here any more. You could contact him via the mapthematics site. However, he might not be interested in doing such consultation for free.

Nicias
Posts: 163
Joined: Tue Aug 13, 2013 4:22 pm UTC

### Re: Blending equal-area map projections

You can just math this.

Call you new map N(\theta, \phi). You just need to find the determinant of the Jacobian of your map. If it isn't cos(theta), you projection isn't equal area.

After messing around, the math appears to be a PITA. I can't seem to find a formula for | J(N) | that doesn't require an explicit formula for the Mollweide Projection, which does not exist.

Nicias
Posts: 163
Joined: Tue Aug 13, 2013 4:22 pm UTC

### Re: Blending equal-area map projections

Ok, doesn't work. Here is the Mathematica code I used:

Code: Select all

ClearAll[Evaluate[Context[]<>"*"]]\$Assumptions=  -Pi<=p<=Pi && -Pi/2<=l<=Pi/2 ;sol=NDSolve[{2T'[p]+2Cos[2T[p]]T'[p]==Pi Cos[p] ,T[0]==0},T,{p,-1.5,1.5}];Theta=sol[[1]][[1]][[2]];p0=0;l0=0;M[l_,p_]:=Sqrt[2]*{2 l/Pi*Cos[T[p]],Sin[T[p]]}H[l_,p_]:= Sqrt[2]*{2 Cos[p]Sin[l/2],Sin[p]}/Sqrt[1+Cos[p]Cos[l/2]]Alpha[l_,p_]:=ArcCos[Sin[p]Sin[p0]+Cos[p]Cos[p0]Cos[l-l0]]PM[l_,p_]:=(Alpha[l,p]* H[l,p]+(Pi-Alpha[l,p])*M[l,p])/Pi Jac[l_,p_]:=D[PM[x,y],{{x,y}}]/. {T->Theta, x->l, y->p}Error[l_, p_]:=(Det[Jac[l,p]]-Cos[p])Sec[p];Plot3D[Error[l,p], {l,-3,3},{p,-1.5,1.5}]

The Error ranges from -0.1 to 0.05. So the area is off by up to -10% in some areas and +5% in other.
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Strebe
Posts: 6
Joined: Sat Nov 19, 2011 9:48 pm UTC

### Re: Blending equal-area map projections

In general, weighted averages do not preserve areas. Here is a link to my paper that describes how to do this for •any• two equal-area projections:
Regards
— daan Strebe

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