How to solve the equation to get the result in atan2 form?

Question

Shiv Karpoor 2022-3-21

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1676519-how-to-solve-the-equation-to-get-the-result-in-atan2-form

评论： Shiv Karpoor 2022-3-23

Hello MATLAB Community,

I am trying to solve an equation in MATLAB, but I don't know which function or how to solve it.

syms phi theta psi
ROT = [cos(theta)*cos(phi)+sin(psi)*sin(theta)*sin(phi), -cos(theta)*sin(phi)+sin(psi)*sin(theta)*cos(phi), cos(psi)*sin(theta); ...
    cos(psi)*sin(phi), cos(psi)*cos(phi), -sin(psi); ...
    -sin(theta)*cos(phi)+sin(psi)*cos(theta)*sin(phi), sin(theta)*sin(phi)+sin(psi)*cos(theta)*cos(phi), cos(psi)*cos(theta)];
eqn = ROT(1,2) == ROT(2,1)

I tried to use functions like solve, simplify, etc. but I am not getting it in the desired form as shown below:

psi = atan2(sin(phi)*sin(theta),cos(phi)+cos(theta))

Can anyone please help me with any suggestions.

Thank you in advance!!

I really appreciate your help.

Kind regards,

Shiv

5 个评论
显示 3更早的评论隐藏 3更早的评论

Shiv Karpoor 2022-3-22

Matlab Solution

Mohammed Hamaidi 2022-3-23

在 MATLAB Online 中打开

Be careful. The equation has solution under some strict assumptions.

Example:

sin(x)+cos(x)==4

, has no solution,

Matlab gives:

>> solve(cos(x)+sin(x)==4,x)
 
ans =
 
 -log(- 28i^(1/2)/2 + (2 + 2i))*1i
   -log(28i^(1/2)/2 + (2 + 2i))*1i

So, you have to give the assumptions/conditions

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Torsten 2022-3-21

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1676519-how-to-solve-the-equation-to-get-the-result-in-atan2-form#answer_923214

编辑：Torsten 2022-3-21

在 MATLAB Online 中打开

syms phi theta psi
ROT = [cos(theta)*cos(phi)+sin(psi)*sin(theta)*sin(phi), -cos(theta)*sin(phi)+sin(psi)*sin(theta)*cos(phi), cos(psi)*sin(theta); ...
       cos(psi)*sin(phi), cos(psi)*cos(phi), -sin(psi); ...
       -sin(theta)*cos(phi)+sin(psi)*cos(theta)*sin(phi), sin(theta)*sin(phi)+sin(psi)*cos(theta)*cos(phi), cos(psi)*cos(theta)];
eqn = ROT(1,2) == ROT(2,1);
sol = solve(eqn,psi)

11 个评论
显示 9更早的评论隐藏 9更早的评论

Shiv Karpoor 2022-3-23

在 MATLAB Online 中打开

How did you get the above solution,

I am getting the solution like this, after doing what you said:

sol =
 
 -log(((2*exp(phi*2i) + exp(phi*4i) + 2*exp(theta*2i) + exp(theta*4i) - 12*exp(phi*2i)*exp(theta*2i) + 2*exp(phi*2i)*exp(theta*4i) + 2*exp(phi*4i)*exp(theta*2i) + exp(phi*4i)*exp(theta*4i) + 16*exp(phi*2i)*exp(theta*2i)*cos(theta)^2*sin(phi)^2 + 1)^(1/2) + 4*exp(phi*1i)*exp(theta*1i)*cos(theta)*sin(phi))/(exp(phi*2i) - exp(theta*2i) - exp(phi*2i)*exp(theta*2i) + 1 - exp(theta*1i)*2i + exp(phi*2i)*exp(theta*1i)*2i))*1i
-log(-((2*exp(phi*2i) + exp(phi*4i) + 2*exp(theta*2i) + exp(theta*4i) - 12*exp(phi*2i)*exp(theta*2i) + 2*exp(phi*2i)*exp(theta*4i) + 2*exp(phi*4i)*exp(theta*2i) + exp(phi*4i)*exp(theta*4i) + 16*exp(phi*2i)*exp(theta*2i)*cos(theta)^2*sin(phi)^2 + 1)^(1/2) - 4*exp(phi*1i)*exp(theta*1i)*cos(theta)*sin(phi))/(exp(phi*2i) - exp(theta*2i) - exp(phi*2i)*exp(theta*2i) + 1 - exp(theta*1i)*2i + exp(phi*2i)*exp(theta*1i)*2i))*1i
 
>> pretty(sol)
/      / #1 + 4 exp(phi 1i) exp(theta 1i) cos(theta) sin(phi) \     \
|  -log| ---------------------------------------------------- | 1i  |
|      \                          #2                          /     |
|                                                                   |
|     /   #1 - 4 exp(phi 1i) exp(theta 1i) cos(theta) sin(phi) \    |
| -log| - ---------------------------------------------------- | 1i |
\     \                            #2                          /    /
where
   #1 == sqrt(2 #3 + exp(phi 4i) + 2 #4 + exp(theta 4i) - 12 #3 #4 + 2 #3 exp(theta 4i) + 2 exp(phi 4i) #4
                                                       2         2
      + exp(phi 4i) exp(theta 4i) + 16 #3 #4 cos(theta)  sin(phi)  + 1)
   #2 == #3 - #4 - #3 #4 + 1 - exp(theta 1i) 2i + #3 exp(theta 1i) 2i
   #3 == exp(phi 2i)
   #4 == exp(theta 2i)

Shiv Karpoor 2022-3-23

在 MATLAB Online 中打开

I got this:

soln =
 
angle((- exp(phi*2i)*1i + exp(theta*2i)*1i - exp(phi*2i + theta*2i)*1i + 2*(exp(phi*2i) - 2*exp(phi*2i + theta*2i) + exp(phi*2i + theta*4i))^(1/2) + 1i)/(exp(phi*2i) - exp(theta*2i) - exp(phi*2i + theta*2i) + 1 - exp(theta*1i)*2i + exp(phi*2i + theta*1i)*2i))
 angle(-(exp(phi*2i)*1i - exp(theta*2i)*1i + exp(phi*2i + theta*2i)*1i + 2*(exp(phi*2i) - 2*exp(phi*2i + theta*2i) + exp(phi*2i + theta*4i))^(1/2) - 1i)/(exp(phi*2i) - exp(theta*2i) - exp(phi*2i + theta*2i) + 1 - exp(theta*1i)*2i + exp(phi*2i + theta*1i)*2i))
 
>> pretty(soln)
/      / - exp(phi 2i) 1i + exp(theta 2i) 1i - #3 1i + #2 + 1i \ \
| angle| ----------------------------------------------------- | |
|      \                           #1                          / |
|                                                                |
|       /   exp(phi 2i) 1i - exp(theta 2i) 1i + #3 1i + #2 - i \ |
|  angle| - -------------------------------------------------- | |
\       \                           #1                         / /
where
   #1 == exp(phi 2i) - exp(theta 2i) - #3 + 1 - exp(theta 1i) 2i + exp(phi 2i + theta 1i) 2i
   #2 == sqrt(exp(phi 2i) - 2 #3 + exp(phi 2i + theta 4i)) 2
   #3 == exp(phi 2i + theta 2i)

I think, I might have to stick with written mathematical solution for now and use matlab to get the value of psi after substituting the values of theta & phi into the eqn.

Thank you so much, for suggestions

I really appreciate it

Shiv Karpoor 2022-3-23

编辑：Shiv Karpoor 2022-3-23

在 MATLAB Online 中打开

Hi David,

I found the ROT equation from a journal paper,

I was using the above ROT equation, because the desired form was given by the journal paper.

The actual ROT that I am using is :

eROT = [(cos(phi)*cos(psi))-(cos(theta)*sin(phi)*sin(psi)) -(cos(phi)*sin(psi))-(cos(theta)*cos(psi)*sin(phi)) (sin(phi)*sin(theta));...
    (cos(psi)*sin(phi))+(cos(phi)*cos(theta)*sin(psi)) (cos(phi)*cos(theta)*cos(psi)-sin(phi)*sin(psi)) -(cos(phi)*sin(theta));...
    (sin(theta)*sin(psi)) (cos(psi)*sin(theta)) cos(theta)];

For my eROT, I am unable to find the desired form. But I guarantee you that if you take any euler angle transformations, you will get the desired form with probably different variables.

You can solve the below and see that I am able to get the psi result:

syms psi
d2r = pi/180;
r2d = 180/pi;
phi = 10*d2r;
theta = 15*d2r;
eROT = [(cos(phi)*cos(psi))-(cos(theta)*sin(phi)*sin(psi)) -(cos(phi)*sin(psi))-(cos(theta)*cos(psi)*sin(phi)) (sin(phi)*sin(theta));...
    (cos(psi)*sin(phi))+(cos(phi)*cos(theta)*sin(psi)) (cos(phi)*cos(theta)*cos(psi)-sin(phi)*sin(psi)) -(cos(phi)*sin(theta));...
    (sin(theta)*sin(psi)) (cos(psi)*sin(theta)) cos(theta)];
eqn = eROT(1,2) == eROT(2,1)
[Psi] = double(vpasolve(eqn))

For this eROT, Psi will always be opposite of phi and vice versa.

But What I need is to solve the eqn symbolically, and get the desired form without any numerical value. (like a formula).

If its possible, then Great !! or else I'll solve it by hand. (anyways I am getting the value of psi through matlab, I'll figure out the formula by hand)

Torsten 2022-3-23

编辑：Torsten 2022-3-23

在 MATLAB Online 中打开

You didn't get David's point.

Your formula to calculate psi as

psi = atan2(sin(phi)*sin(theta),cos(phi)+cos(theta))

seems to be wrong.

But now you know the solution to write psi in terms of phi and theta in symbolic form (the expression from octave). Why don't you simply copy it and use it in your code or whereever you need it ?

Shiv Karpoor 2022-3-23