How to solve a complicated algebraic equation with symbolic toolbox?

Hi guys,

In the following image there are some expression of some parameters.

I need to solve the equality deriving from expression (17), i.e. in order to get .

I tried to solve the problem by hand but it involves many substitutions and it is full of computations.

Is there a way to perform this task by using the symbolic toolbox with as little effort as possible?

Answer 1

[Torsten]

Either

delta_w = k*pi (k integer)

or

delta_w = acos(((p1*e2)^2 + (p2*e1)^2 - (p1-p2)^2)/(2*p1*p2*e1*e2))

[Giuseppe]

Can you show me the code that allows you to get the second expression of your answer?

[Torsten]

Here is the "code":

k2^2-k1*k3 = 0
-> (a*b)^2 - (b^2+c^2)*(a^2-c^2) = 0
-> c^2*(a^2-b^2-c^2) = 0
-> c= 0 or a^2-b^2-c^2 = 0 
c = 0 -> dw = k*pi (k integer)
a^2-b^2-c^2 = 0
-> (p1-p2)^2 - (p1*e2*cos(dw)-p2*e1)^2 - (p1*e2*sin(dw))^2 = 0
-> (p1-p2)^2 -p1^2*e2^2*(cos^2(dw)+sin^2(dw)) + 2*p1*p2*e2*e2*cos(dw) - p2^2*e1^2 = 0
-> (p1-p2)^2 - p1^2*e2^2 * 1 + 2*p1*p2*e1*e2*cos(dw) - p2^2*e1^2 = 0

I think you can take it from here.

Here is MATLAB code:

syms p1 p2 e1 e2 dw
a = p1-p2;
b = p1*e2*cos(dw)-p2*e1;
c = -p1*e2*sin(dw);
k1 = b^2+c^2;
k2 = a*b;
k3 = a^2-c^2;
det = k2^2-k1*k3;
F = solve(det==0,dw)

[Giuseppe]

Yes, sorry. I meant the MATLAB code! Thank you.

[Torsten]

Since I don't have MATLAB available at the moment:

Do you get the same formula from MATLAB as the one from my hand calculation or

         /      _____________________________________________________________\
         |     /  /  2   2                     2   2     2               2\  |
         |    /  -\e1 *p2  - 2*e1*e2*p1*p2 + e2 *p1  - p1  + 2*p1*p2 - p2 /  |
+/-2*atan|   /   ----------------------------------------------------------- |
         |  /        2   2                     2   2     2               2   |
         \\/       e1 *p2  + 2*e1*e2*p1*p2 + e2 *p1  - p1  + 2*p1*p2 - p2    /