I want to create an equalarea 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 equalarea 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)+(1α/π)Mollweide(θ,φ). However, I haven’t been able to show whether or not this preserves areas.
Any ideas?
Blending equalarea map projections
Moderators: gmalivuk, Moderators General, Prelates
Blending equalarea map projections
wee free kings
Re: Blending equalarea map projections
I have no idea if your weighted mean will preserve areas, but it sounds reasonable. I guess you could bruteforce 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.
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.
Re: Blending equalarea 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.
[edit]
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.
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.
[edit]
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.
Re: Blending equalarea map projections
Ok, doesn't work. Here is the Mathematica code I used:
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.
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[ll0]]
PM[l_,p_]:=(Alpha[l,p]* H[l,p]+(PiAlpha[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.
 Attachments

 foo.png (32.81 KiB) Viewed 2902 times
Re: Blending equalarea 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 equalarea projections:
Regards
— daan Strebe
Regards
— daan Strebe
Who is online
Users browsing this forum: No registered users and 3 guests