# csdms-contrib/slepian_alpha

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 function [Ao4p,varargout]=spharea(c11,cmn) % [Ao4p,XY]=SPHAREA(c11,cmn) % [Ao4p,XY]=SPHAREA(region) % [Ao4p,XY]=SPHAREA(XY) % [Ao4p,XY]=SPHAREA(TH,sord) % % Computes FRACTIONAL surface areas on the unit sphere. % % INPUT: % % c11 [x,y]/[lon,lat]-coordinates of the upper left corner % cmn [x,y]/[lon,lat]-coordinates of the bottom right corner % All coordinates are given in degrees. % Both inputs may be matrices of size (Mx2); OR % region A string name with an approved region such as 'africa', OR % a cell array containing the region name and buffer such as {'greenland' 0.5} % XY its coordinates (such as supplied from 'africa' etc), OR % TH A colatitudinal radius [in degrees] % sord 0 Allowable option for region string % 1 Single cap of radius TH [no default - must specify] % 2 Double cap left by SUBTRACTING belt of width 2TH % 3 Equatorial belt of width 2TH % % OUTPUT: % % Ao4p Area as a fraction of the area of the unit sphere. % Multiply by 4*pi*radius^2 for a known sphere. % XY The outlines of the region used, if available % % EXAMPLE I: the fractional surface area of the Earth is % % spharea([0 90],[360 -90]) % and the continental coverage % spharea('africa')+spharea('eurasia')+spharea('namerica')+... % spharea('samerica')+spharea('greenland')+spharea('australia') % % EXAMPLE 2: area between one degree of longitude for all latitudes - % this goes as the cosine of the latitude of course % % latsup=90:-1:-89; latsdwn=89:-1:-90; % lonslft=zeros(size(latsup)); lonsrgt=ones(size(latsup)); % c11=[lonslft' latsup']; cmn=[lonsrgt' latsdwn']; subplot(121) % imagesc([0.5 0.5],[90 -90],spharea(c11,cmn)); colormap gray; axis xy % subplot(122); % imagesc([0.5 0.5],[90 -90],cos(linspace(-pi/2,pi/2,length(latsup)))'); % axis xy % % EXAMPLE 3: % % [a,xy]=spharea('australia'); a-areaint(xy(:,2),xy(:,1)) % % SEE ALSO: % % RCENTER, AREAINT % % Last modified by charig-at-princeton.edu, 05/14/2015 % Last modified by fjsimons-at-alum.mit.edu, 07/21/2014 defval('XY',NaN) if nargin==2 && ~all(cmn(:)==0) if all(size(c11)==size(cmn)) && length(cmn)~=1 % Conversion to radians c11=c11*pi/180; cmn=cmn*pi/180; lon1=c11(:,1); lat1=c11(:,2); lon2=cmn(:,1); lat2=cmn(:,2); Ao4p=abs(lon1-lon2).*abs(sin(lat1)-sin(lat2))/4/pi; elseif length(cmn)==1 % This means that the variable SORD is specified TH=c11; sord=cmn; switch sord case 1 % Area of the single cap of width TH Ao4p=(1-cos(TH*pi/180))/2; case 2 % Area of the double cap each of width 90-TH Ao4p=(1-sin(TH*pi/180)); case 3 % Area of the belt of width 2TH Ao4p=sin(TH*pi/180); end end else defval('ngl',200) if isstr(c11) % Specify the region by name defval('N',10); region=c11; eval(sprintf('XY=%s(%i);',region,N)); elseif iscell(c11) % Specify the region by name defval('N',10); region=c11{1}; eval(sprintf('XY=%s(%i,%f);',region,N,c11{2})); % Do this so dphregion below is correct for these coordinates region=XY; N=NaN; else % Specify the region by the coordinates of the bounding curve XY=c11; region=XY; N=NaN; end % Calculate northernmost and southernmost colatitude thN=90-max(XY(:,2)); thN=thN*pi/180; thS=90-min(XY(:,2)); thS=thS*pi/180; % Calculate Gauss-Legendre points intv=cos([thS thN]); % Make this number high nGL=min(ngl,size(XY,1)/2); [w,x,Ngg]=gausslegendrecof(nGL,[],intv); % Now get the longitudinal intersections [phint,thp,php]=dphregion(acos(x)*180/pi,N,region); % plot(thp,php) phint=phint*pi/180; % This is not ridiculous as I may be giving it multiple intervals I=coscos(acos(x),0,0,phint); Ao4p=w(:)'*I/4/pi; end % Provide desired output varns={XY}; varargout=varns(1:nargout-1);