How do I change the colour in each sector depending on the angle?

I created 6 sectors via linspace and patch function. However I can't figure out how to fill in each sector depending on the heading. e.g from sample code, when heading 90 degree that sector should be '5' or yellow, the other sectors towards the left/right should be getting more and more 'blue' or towards the value '1'. It will be great if I can define the starting value for each sector or reducing as it move towards the left/right of the heading.

Many thanks for the help in advance.

x0=0
y0=0
r=5
angleDEG=90
angleRAD=angleDEG/180*pi

HSpread=45
HSpreadRAD=HSpread/180*pi

t = linspace(angleRAD-HSpreadRAD,angleRAD+HSpreadRAD,6); %(begining angle, end angle, resolution)
x = [x0 (x0+r*cos(t)) x0];
y = [y0 (y0+r*sin(t)) y0];
c = linspace(1,5,length(y));

figure; 
a=patch(x,y,c)
a.FaceAlpha = 0.3;
colorbar
axis equal

Currently this is what I got (pic insert):

Answer 1

[DGM]

I don't know about how to wrangle patch() to do it right. It probably can, but I don't have that much hair to spare. This is a version using surf() instead. Anyone is free to trump this with a solution using patch().

% parameters
x0 = 0;
y0 = 0;
rmax = 5;
th0 = 45; % start angle
thInc = 90; % included angle

rsteps = 10;
thsteps = 6;

ang = [th0 th0+thInc]/180*pi;
r = linspace(0,rmax,rsteps);
t = linspace(ang(1),ang(2),thsteps).';
x = x0 + r.*cos(t);
y = y0 + r.*sin(t);
z = ones(thsteps,rsteps).*0; % just plot everything in x-y plane
c = repmat(8*(0.5-abs((t-min(ang))/range(ang)-0.5)) + 1,[1 rsteps]);

a = surf(x,y,z,c);
a.FaceAlpha = 0.5;

% do the border
xl = [x0 x0+rmax*cos(t.') x0];
yl = [y0 y0+rmax*sin(t.') y0];
line(xl,yl,'color','k')

shading interp
colorbar
caxis([1 5])
axis equal
view(2)

[Julian Tan]

Thanks for the help! Appreciate it.

[Julian Tan]

Tried with 'Patch', somehow got it but can't get the colour (e.g yellow) to start from (0,0).

'Surt' works well just that for my application it is too heavy.

This is what I got, I think by 'logic' as it filled up the 'polygon' it shoud generate the above by DGM, somehow it din't, but did the work 'half correct'.

x0=0
y0=0
rmax=8
rsteps=1
thsteps=15

angleDEG=90
angleRAD=angleDEG/180*pi
HSpread=45
HSpreadRAD=HSpread/180*pi

ang = [angleDEG-HSpread angleDEG+HSpread];

% r = linspace(0,rmax,rsteps);
t = linspace(angleDEG-HSpread,angleDEG+HSpread,thsteps); %angel in degree

x = [x0 x0+rmax*cosd(t) x0]; %in degree
y = [y0 y0+rmax*sind(t) y0];

caa = repmat(rmax*(0.5-abs((t-min(ang))/range(ang)-0.5))+1,[1 rsteps]); 
c=[0 caa 0]
a=patch(x,y,c)
a.FaceAlpha = 0.8;

colorbar
axis equal

The colour, instead of start from the origin (0,0), it appears to start from x=0 axis.

Anyone have any idea why it is like that or possible to to use any '2D' code to achieve the same task?

Many Thanks

[Julian Tan]

I tried doing this, but I think I missing the '3rd coordinate' hence ends up with thin line. Still trying to figure out how.

x0=0
y0=0
rmax=8
rsteps=1
thsteps=5

angleDEG=90
angleRAD=angleDEG/180*pi
HSpread=45
HSpreadRAD=HSpread/180*pi

ang = [angleDEG-HSpread angleDEG+HSpread];

% r = linspace(0,rmax,rsteps);
t = linspace(angleDEG-HSpread,angleDEG+HSpread,thsteps); %angel in degree

x = [x0 x0+rmax*cosd(t) x0]; %in degree
y = [y0 y0+rmax*sind(t) y0];

xc=rmax*cosd(t);
xx=[0; 1; 0];
xxt=xx*xc+(zeros(3,thsteps)+x0);

yc=rmax*sind(t);
yy=[0; 1; 0];
yyt=yy*yc+(zeros(3,thsteps)+y0);

caa = repmat(rmax*(0.5-abs((t-min(ang))/range(ang)-0.5))+1,[3 rsteps]); 
c=caa
a=patch(xxt,yyt,c)
a.FaceAlpha = 1;

colorbar
axis equal

[Julian Tan]

Finally Manage to solved it. In a very inefficeint manner I persume since I am still new with with Matlab.

Hope anyone of you can shorten it, with codes that are much more efficient. Looking forward to learn from...

x0=0
y0=0
rmax=8
rsteps=1
thsteps=50

angleDEG=90
angleRAD=angleDEG/180*pi
HSpread=45
HSpreadRAD=HSpread/180*pi

ang = [angleDEG-HSpread angleDEG+HSpread];

% r = linspace(0,rmax,rsteps);
t = linspace(angleDEG-HSpread,angleDEG+HSpread,thsteps); %angel in degree
tt= [1:1:thsteps]
ratio=abs((max(ang)-min(ang))/(thsteps-1));
tnext=t+ratio;
% x = [x0 x0+rmax*cosd(t) x0]; %in degree
% y = [y0 y0+rmax*sind(t) y0];

xc=rmax*cosd(t);
xx=[0; 1; 0];
xnext=rmax*cosd(tnext);
xxnext=[0; 0; 1];
xxt=xx*xc+xxnext*xnext+(zeros(3,thsteps)+x0)

yc=rmax*sind(t);
yy=[0; 1; 0];
ynext=rmax*sind(tnext);
yynext=[0; 0; 1];
yyt=yy*yc+yynext*ynext+(zeros(3,thsteps)+y0);

caa = repmat(rmax*(0.5-abs((t-min(ang))/range(ang)-0.5))+1,[3 rsteps]); 
c=caa
a=patch(xxt,yyt,c,'LineStyle','none')
a.FaceAlpha = 1;



colorbar
axis equal

[DGM]

I don't know. That looks pretty good to me. If it works well for your needs, I'd roll with it.

[Julian Tan]

Thanks for the inital valuable insight. Learn great deal from your codes.

All is good excpet the angle need to be trimmed (equal angle)/2, due to the way the polygons were drawn and filled. The above can see 'tilted' slgihtly (equal angle)/2 to the rgiht. Once that is done, it is perfect.

AngleDEG_trim=(HSpreadDEG)/(thsteps-1) 
NewAngleDEG_trim=angleDEG-AngleDEG_trim