Aligning axes labels in 3D plots

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broken_arrow
broken_arrow on 1 Apr 2022
Commented: Dave B on 10 Sep 2023 at 21:28
I've been playing around for several hours trying to get the x and y labels in a 3D plot to align properly. The following code creates an isometric view, for which the axis angles should be 30° (which is correctly computed).
figure;
axh = axes;
Z = peaks(20);
surf(Z)
xlabel('x-axis');
ylabel('y-axis');
azimuth = -45;
elevation = 35.264;
% Isometric view, c. f. https://en.wikipedia.org/wiki/Isometric_projection
view(axh,azimuth,elevation);
camproj % returns 'orthographic'
unitx = [1;0;0];
unity = [0;1;0];
unitz = [0;0;1];
projectedunitx = rotx(elevation) * rotz(-azimuth) * unitx;
projectedunity = rotx(elevation) * rotz(-azimuth) * unity;
xlabelangle = atan2d(projectedunitx(3),projectedunitx(1)) %#ok
ylabelangle = -(180 - atan2d(projectedunity(3),projectedunity(1))) %#ok
xlabelhandle = axh.XLabel;
ylabelhandle = axh.YLabel;
xlabelhandle.Rotation = xlabelangle;
ylabelhandle.Rotation = ylabelangle;
xlimits = xlim(axh);
ylimits = ylim(axh);
zlimits = zlim(axh);
xmean = mean(xlimits);
ymean = mean(ylimits);
xbottom = xlimits(1);
ybottom = ylimits(1);
zbottom = zlimits(1);
xlabelhandle.Position = [xmean ybottom zbottom];
ylabelhandle.Position = [xbottom ymean zbottom];
Yet in the plot the labels don't align exactly parallel to the axes:
The error is relatively small, but I'd like to have an exact solution. It appears Matlab doesn't exactly adhere to the rules of orthographic projection because in a truly isometric view (which is orthographic), the axes angles are 30°. Is there some way to get the alignment (mathematically) right? (I'm aware of this FEX function https://mathworks.com/matlabcentral/fileexchange/49542-phymhan-matlab-axis-label-alignment but it doesn't seem to work in R2020b.)

Accepted Answer

Dave B
Dave B on 1 Apr 2022
The differrence between your labels and the axes is because MATLAB stretches an axes to fill the space of its container - if you made your figure wider the angles would become flatter. To get the exact angle, axis equal should do the trick:
figure;
axh = axes;
Z = peaks(20);
surf(Z)
xlabel('x-axis');
ylabel('y-axis');
azimuth = -45;
elevation = 35.264;
% Isometric view, c. f. https://en.wikipedia.org/wiki/Isometric_projection
view(axh,azimuth,elevation);
camproj % returns 'orthographic'
ans = 'orthographic'
unitx = [1;0;0];
unity = [0;1;0];
unitz = [0;0;1];
projectedunitx = rotx(elevation) * rotz(-azimuth) * unitx;
projectedunity = rotx(elevation) * rotz(-azimuth) * unity;
xlabelangle = atan2d(projectedunitx(3),projectedunitx(1)) %#ok
xlabelangle = 29.9998
ylabelangle = -(180 - atan2d(projectedunity(3),projectedunity(1))) %#ok
ylabelangle = -29.9998
xlabelhandle = axh.XLabel;
ylabelhandle = axh.YLabel;
xlabelhandle.Rotation = xlabelangle;
ylabelhandle.Rotation = ylabelangle;
xlimits = xlim(axh);
ylimits = ylim(axh);
zlimits = zlim(axh);
xmean = mean(xlimits);
ymean = mean(ylimits);
xbottom = xlimits(1);
ybottom = ylimits(1);
zbottom = zlimits(1);
xlabelhandle.Position = [xmean ybottom zbottom];
ylabelhandle.Position = [xbottom ymean zbottom];
axis equal
  6 Comments
Dave B
Dave B on 10 Sep 2023 at 21:28
@Giovanni de amici - I used rotx and rotz because they were included in the code snippet in the question. But these are quite simple one liners...if you don't have access to the toolbox mentioned above you'll find they're trivial to implement (and the documentation page for rotx has a clear description of what the matrices look like, although you can find these elsewhere of course).
I think, for instance, you could implement your own version rotx as:
function T = myrotx(ang)
T = [1 0 0; ...
0 cosd(ang) -sind(ang); ...
0 sind(ang) cosd(ang)];
end

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