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Same equation, different answers from Pylab and Octave

I am porting code created in octave into pylab. One of the ported equations gives dramatically different results in python than it does in octave.

The best way to explain is to show plots generated by octave and pylab from the same equation.

Here is a simplified snippet of the original equation in octave. In this small test script, the result of function with phi held at zero is plotted from ~ (-pi,pi):

clear
clc
close all

L1 = 4.25; % left servo arm length
L2 = 5.75; % left linkage length
L3 = 5.75; % right linkage length
L4 = 4.25; % right servo arm length
L5 = 11/2; % distance from origin to left servo
L6 = 11/2; % distance from origin to right servo

theta_array = [-pi+0.1:0.01:pi-0.1];
phi = 0/180*pi;

for i = 1 : length(theta_array)

theta = theta_array(i);

A(i) = -L3*(-((2*cos(theta)*L1*(sin(phi)*L4-sin(theta)*L1)-2*sin(theta)*L1*(L6+L5-cos(phi)*L4-cos(theta)*L1))/(2*L3*sqrt((L6+L5-cos(phi)*L4-cos(theta)*L1)^2+(sin(phi)*L4-sin(theta)*L1)^2))-((2*sin(theta)*L1*(L6+L5-cos(phi)*L4-cos(theta)*L1)-2*cos(theta)*L1*(sin(phi)*L4-sin(theta)*L1))*(-(L6+L5-cos(phi)*L4-cos(theta)*L1)^2-(sin(phi)*L4-sin(theta)*L1)^2-L3^2+L2^2))/(4*L3*((L6+L5-cos(phi)*L4-cos(theta)*L1)^2+(sin(phi)*L4-sin(theta)*L1)^2)^(3/2)))/sqrt(1-(-(L6+L5-cos(phi)*L4-cos(theta)*L1)^2-(sin(phi)*L4-sin(theta)*L1)^2-L3^2+L2^2)^2/(4*L3^2*((L6+L5-cos(phi)*L4-cos(theta)*L1)^2+(sin(phi)*L4-sin(theta)*L1)^2)))-((cos(theta)*L1)/sqrt((L6+L5-cos(phi)*L4-cos(theta)*L1)^2+(sin(phi)*L4-sin(theta)*L1)^2)-((sin(theta)*L1-sin(phi)*L4)*(2*sin(theta)*L1*(L6+L5-cos(phi)*L4-cos(theta)*L1)-2*cos(theta)*L1*(sin(phi)*L4-sin(theta)*L1)))/(2*((L6+L5-cos(phi)*L4-cos(theta)*L1)^2+(sin(phi)*L4-sin(theta)*L1)^2)^(3/2)))/sqrt(1-(sin(theta)*L1-sin(phi)*L4)^2/((L6+L5-cos(phi)*L4-cos(theta)*L1)^2+(sin(phi)*L4-sin(theta)*L1)^2)))*sin(acos((-(L6+L5-cos(phi)*L4-cos(theta)*L1)^2-(sin(phi)*L4-sin(theta)*L1)^2-L3^2+L2^2)/(2*L3*sqrt((L6+L5-cos(phi)*L4-cos(theta)*L1)^2+(sin(phi)*L4-sin(theta)*L1)^2)))-asin((sin(theta)*L1-sin(phi)*L4)/sqrt((L6+L5-cos(phi)*L4-cos(theta)*L1)^2+(sin(phi)*L4-sin(theta)*L1)^2)));

end

plot(theta_array,A)

The resulting octave plot looks like this:

Octave result

The same equation was copied and pasted from octave into python with '^' replaced with '**', 'acos' replaced with 'arccos', and 'asin' replaced with 'arcsin'. The same range of theta was plotted with phi held at zero:

from pylab import *

# physical setup
L1 = 4.25; # left servo arm length
L2 = 5.75; # left linkage length
L3 = 5.75; # right linkage length
L4 = 4.25; # right servo arm length
L5 = 11.0/2.0; # distance from origin to left servo
L6 = 11.0/2.0; # distance from origin to right servo

theta = arange(-pi+0.1,pi-0.1,0.01);
phi = 0/180.0*pi

def func(theta,phi):

A = -L3*(-((2*cos(theta)*L1*(sin(phi)*L4-sin(theta)*L1)-2*sin(theta)*L1*(L6+L5-cos(phi)*L4-cos(theta)*L1))/(2*L3*sqrt((L6+L5-cos(phi)*L4-cos(theta)*L1)**2+(sin(phi)*L4-sin(theta)*L1)**2))-((2*sin(theta)*L1*(L6+L5-cos(phi)*L4-cos(theta)*L1)-2*cos(theta)*L1*(sin(phi)*L4-sin(theta)*L1))*(-(L6+L5-cos(phi)*L4-cos(theta)*L1)**2-(sin(phi)*L4-sin(theta)*L1)**2-L3**2+L2**2))/(4*L3*((L6+L5-cos(phi)*L4-cos(theta)*L1)**2+(sin(phi)*L4-sin(theta)*L1)**2)**(3/2)))/sqrt(1-(-(L6+L5-cos(phi)*L4-cos(theta)*L1)**2-(sin(phi)*L4-sin(theta)*L1)**2-L3**2+L2**2)**2/(4*L3**2*((L6+L5-cos(phi)*L4-cos(theta)*L1)**2+(sin(phi)*L4-sin(theta)*L1)**2)))-((cos(theta)*L1)/sqrt((L6+L5-cos(phi)*L4-cos(theta)*L1)**2+(sin((phi)*L4-sin(theta)*L1)**2)-((sin(theta)*L1-sin(phi)*L4)*(2*sin(theta)*L1*(L6+L5-cos(phi)*L4-cos(theta)*L1)-2*cos(theta)*L1*(sin(phi)*L4-sin(theta)*L1)))/(2*((L6+L5-cos(phi)*L4-cos(theta)*L1)**2+(sin(phi)*L4-sin(theta)*L1)**2)**(3/2)))/sqrt(1-(sin(theta)*L1-sin(phi)*L4)**2/((L6+L5-cos(phi)*L4-cos(theta)*L1)**2+(sin(phi)*L4-sin(theta)*L1)**2)))*sin(arccos((-(L6+L5-cos(phi)*L4-cos(theta)*L1)**2-(sin(phi)*L4-sin(theta)*L1)**2-L3**2+L2**2)/(2*L3*sqrt((L6+L5-cos(phi)*L4-cos(theta)*L1)**2+(sin(phi)*L4-sin(theta)*L1)**2)))-arcsin((sin(theta)*L1-sin(phi)*L4)/sqrt((L6+L5-cos(phi)*L4-cos(theta)*L1)**2+(sin(phi)*L4-sin(theta)*L1)**2)))

return A

f = figure();
a = f.add_subplot(111);

a.plot(theta,func(theta,phi))

ginput(1, timeout=-1); # wait for user to click so we dont lose the plot

Python's result looks like this: Python result

I cant determine what is causing the differences, Any ideas?

like image 813
Inverse_Jacobian Avatar asked Nov 07 '11 01:11

Inverse_Jacobian


1 Answers

Try from __future__ import division to eliminate errors arising from floor division.

like image 58
Raymond Hettinger Avatar answered Oct 18 '22 09:10

Raymond Hettinger