Plotting within 0 to 360

I need to have a plot between 0 and 360 using my data, and as the values should be the same I did the following:

i = 0:1:360;

for i = 1:1:length(i)

if beta_1 < 0

beta_1 = beta_1 + 360;

end

if beta < 0

beta = beta + 360;

end

plot(beta_1, alpha_1,':r')

title ('Elevation and Azimath Angles of Day 1 and Day 182')

xlabel ('Azimuth Angle (degrees Celsius)')

ylabel ('Elevation Angle (degrees Celsius)')

hold on

plot(beta, alpha, ':k')

legend ('Day 1', 'Day 182')

ylim([0 70])

hold off

end

but it looks like the values are not going being added by 360

[Image Analyst]

What is beta, beta_1, alpha, and alpha_1?

[Sarah Kneer]

They are some angle values I collected

[Image Analyst]

Please attach them so we can use them.

save('answers.mat',  'beta', 'beta_1', 'alpha', 'alpha_1');

then use the paper clip icon.

When you do this:

	if beta_1 < 0
		beta_1 = beta_1 + 360;
	end
	if beta < 0
		beta = beta + 360;
	end

how does any of that depend on the iteration? Nothing changes so it's the same with every iteration. Are they a scalar, or an vector?

Did yoiu perhaps mean this:

	if beta_1(i) < 0
		beta_1(i) = beta_1(i) + 360;
	end

so that the beta_1 value changes during each iteration?

[Sarah Kneer]

Okay, I think I understand what you mean, I just did the last bit that you suggested:

if beta_1(i) < 0

beta_1(i) = beta_1(i) + 360;

end

(for beta as well) and the plot just came out the right way, thank you very much.

Answer 1

[Steven Lord]

You're plotting individual points as separate lines, so they're not connected and the default marker is 'none'.

h = plot(1, 1);

h.Marker

ans = 'none'

If you want to do this, either create vectors of data and plot those vectors after the loop or create an animatedline and addpoints to that animatedline in the loop.

Vectors of data:

x = zeros(1, 10);
y = zeros(1, 10);
for k = 1:10
    x(k) = k;
    y(k) = k.^2;
end
plot(x, y, ':k')

Animated line:

figure
h = animatedline('Color', 'r', 'LineStyle', '--');
for k = 1:10
    addpoints(h, k, k.^2);
end

[Sarah Kneer]

Okay thank you very much, I appreciate that