The second argument to unwrap is not the period of the input data, but the tolerance. The function always unwraps data assuming a 2π interval. That is, it wants to see that x(i)-x(i+1) is larger than the tolerance before unwrapping, and smaller after unwrapping. In the case of a tolerance of pi/2, if, for example, x(i)=0 and x(i+1)=3, the jump is larger than the tolerance, but adding or subtracting 2*pi to x(i+1) does not improve things.

One work-around would be to multiply the input by 2, and divide by 2 after unwrapping:

unwrap(theta_atan * 2) / 2

However, it is always best to use atan2 to obtain an angle.

Answer from Cris Luengo on Stack Overflow
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MathWorks
mathworks.com › matlabcentral › answers › 362558-stumped-again-converting-python-to-matlab
Stumped again... Converting Python to Matlab - MATLAB Answers - MATLAB Central
October 22, 2017 - https://www.mathworks.com/matl... frame · theta = zeros(numPanels, 1); for i = 1 : numPanels · theta(i) = atan2( yp(i+1)-yp(i), xp(i+1)-xp(i) ); end ·...
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stackoverflow.com › questions › 61615692 › converting-matlab-into-python
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May 5, 2020 - import math from matplotlib import pyplot as plt import numpy as np dx=42; dy=-17; ab=10.5; bc=42; cd=24; be=65; ef=5; ml = np.arange(0, 2*math.pi, math.pi/180) alpha = -90*math.pi/180; P=2*cd*(dx-ab+math.cos(theta)); Q=2*cd*(dy-ab+math.ain(theta)); R=(dx**2)+(dy**2)+(ab**2)+(cd**2)-(bc**2)-(2*ab+(math.cos(theta))+dy+(math.sin(theta))); math.cos(phi)=((-P*R)-(Q*math.sqrt((P**2)+(Q**2)-(R**2))))/((P**2)+(Q**2)); math.sin(phi)=((-Q*R)+(P*math.sqrt((P**2)+(Q**2)-(R**2))))/((P**2)+(Q**2)); phi=math.atan2(sinphi,cosphi); math.cospsi=((dx+cd+(math.cos(phi)))-(ab+math.cos(theta)))/bc; math.sinpsi=((d
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The second argument to unwrap is not the period of the input data, but the tolerance. The function always unwraps data assuming a 2π interval. That is, it wants to see that x(i)-x(i+1) is larger than the tolerance before unwrapping, and smaller after unwrapping. In the case of a tolerance of pi/2, if, for example, x(i)=0 and x(i+1)=3, the jump is larger than the tolerance, but adding or subtracting 2*pi to x(i+1) does not improve things.

One work-around would be to multiply the input by 2, and divide by 2 after unwrapping:

unwrap(theta_atan * 2) / 2

However, it is always best to use atan2 to obtain an angle.

Answer from Cris Luengo on Stack Overflow
🌐
W3Schools
w3schools.com › python › ref_math_atan2.asp
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help.scilab.org › m2sci_atan2
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February 15, 2023 - However, the range of theta exceeds 360 degrees, so the value of atan2 output is not properly. ... https://www.mathworks.com/matlabcentral/answers/1912880-changing-the-atan2-function-so-that-it-ranges-from-0-to-4-pi#comment_2618305
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Johns Hopkins University
math.jhu.edu › ~shiffman › 370 › help › techdoc › ref › atan2.html
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Reddit
reddit.com › r/matlab › help needed to understand the error in measuring angles between three coordinate point using atan2
r/matlab on Reddit: Help needed to understand the error in measuring angles between three coordinate point using atan2
September 3, 2022 -

Hi everyone,

I was using the following formula to find the angle between three coordinate values (i.e., P1, P0, P2; finding angle at P0)

ang = (180/pi)*(atan2(abs(det([P2-P0;P1-P0])),dot(P2-P0,P1-P0)));

formula taken from: https://in.mathworks.com/matlabcentral/answers/57736-how-to-calculate-degree-between-3-points-in-matlab#answer_69886

So, it worked fine when I took the coordinate values as ;

P0 = [3,1];
P1 = [1,3];
P2 = [4,4];
figure;
plot(P0(1),P0(2),"*"); hold on;
plot(P1(1),P1(2),"*"); hold on;
plot(P2(1),P2(2),"*"); hold on;
ylim([0 7])
xlim([0 5])
legend("P0","P1","P2")
% P0 is the center where the angle would be
ang = (180/pi)*(atan2(abs(det([P2-P0;P1-P0])),dot(P2-P0,P1-P0))); % formula to get angle
% ang = 63.439 deg

But when I took three coordinate values as the following then it gave me a strange angle value, which is probably the supplmentary angle at P0.

P1 = [-16.49,-17.69];
P0 = [-25.83,-21.73];
P2 = [-40.77,-18.10]
figure;
plot(P0(1),P0(2),"*"); hold on;
plot(P1(1),P1(2),"*"); hold on;
plot(P2(1),P2(2),"*"); hold on;
legend("P0","P1","P2");
ang = (180/pi)*(atan2(abs(det([P2-P0;P1-P0])),dot(P2-P0,P1-P0)));
% ang = 142.9526 deg ; seems like the supplementary angle at P0.

Why is it happening? Any clues?

I need to apply this formula to calculate angle over 100s of coordinates and but if this kind of non-uniformity would happen then I wouldn't be able to use it.

Any help would be appreciated!

Thank you.

I have also asked in the Matlab community at Mathworks:

https://in.mathworks.com/matlabcentral/answers/1793970-error-in-finding-the-angle-between-three-points-using-atan2