How to shorten these values?

enter image description here

They are the result of a matrix like this. enter image description here

When I individually run sin(theta1) and such functions, it would give me the correct value as a zero or a one (the angles I am working with are 0 or 90 degrees)

In some cases they may go on to be values like 1.574.

I know there is a round off function in MATLAB, but then I would have to add that function to every element individually. Is there any easier way to achieve this?

PS.: ST stands for sin(Theta) SA for Sin(Alpha) CA for cos(Alpha) and so on..

PPS. : I tried eval function, not working at all.

Edit 1: The code I am using is as follows:


syms q1 q2 q3 q4 q5;

%DH Parameters for the robot:
robot.DH.theta= '[pi/2 pi/2 0 q(4) q(5)]';
robot.DH.d='[q(1) q(2) q(3) 0.1 0.020]';
robot.DH.a='[0 0 0 0 0]';
robot.DH.alpha= '[pi/2 pi/2 0 pi/2 0]';

% We input the joint parameters:
q = [q1 q2 q3 q4 q5];

%Storing the evaulated values of 'q'

A01=dh(Theta(1), d(1), a(1), alpha(1));
A12=dh(Theta(2), d(2), a(2), alpha(2));
A23=dh(Theta(3), d(3), a(3), alpha(3));
A34=dh(Theta(4), d(4), a(4), alpha(4));
A45=dh(Theta(5), d(5), a(5), alpha(5));
A05 = A01*A12*A23*A34*A45;

Where, dh is a function that comes from a predefined library. It basically substitutes the four values into a generalized form of the matrix I posted as the second image.


2 Answers 2


As you are using the symbolic toolbox, you should declare your angles as symbolic so that the cosine of ${\pi/2}$ is zero instead of numerically estimated to something close to zero. Also so you can convert your expression to a function via the matlabFunction() native function from matlab.

Alternatively on matlab you can use round on a matrix and it's rounding every single entry individually.

  • $\begingroup$ Hey Staub thank you for your suggestion, I got the answer when I by passed all the eval functions and everything and I directly substituted theta = [pi/2 pi/2 0 q4 q5] as you had suggested. $\endgroup$
    – Ln_r1
    Commented Nov 25, 2017 at 18:12
  • $\begingroup$ however I don't understand why it calculates numerically when I do it the way I have written at present, since q(1), q (2)...q(5) points to symbolic values q1 q2 .. q5 respectively. Am I missing some key concept here? $\endgroup$
    – Ln_r1
    Commented Nov 25, 2017 at 18:16
  • $\begingroup$ Basically, the issue was in your cos/sin of 0 ad $\pi/2$ if you don't give them as symbolic somehow matlab starts to do a numerical approximation as it does for any number. Hence despite having your parameters q1,... symbolic matlab was puzzled by the rest of your expression $\endgroup$
    – N. Staub
    Commented Nov 27, 2017 at 7:36

The best solution that comes to my mind in vpa(). It stands for variable precision arithmetic. If you have a symbolic expression X (I believe a matrix with symbolic entries should work as well), then vpa(X,d) will return all the constants within X as a number with at most d significant digits.

  • $\begingroup$ Hi Thank you for your suggestion, however I am getting 6.123e-16 in place of the huge fraction. So this still has not solved my problem. $\endgroup$
    – Ln_r1
    Commented Oct 4, 2017 at 11:55
  • $\begingroup$ @Ln_R How did you insert the values for q4 and q5 into the expression. You might done something there such that MATLAB is converting it to a double after which further calculations can have errors of that magnitude. $\endgroup$
    – fibonatic
    Commented Oct 4, 2017 at 16:39
  • $\begingroup$ q4 and q5 were put in as symbolic terms (using syms). I'm sorry I wasn't clear but q4 and q5 are not supposed to be substituted with values. I have edited my question with the entire code to give you a better idea. $\endgroup$
    – Ln_r1
    Commented Oct 6, 2017 at 14:05

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