How can I get analytical solution of trigonometric equations?
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the constants are:
k11 = (16*V1*V1)/(n^3*(pi)^2*(2*pi*f)*L)
k22 = (16*V2*V2)/(n^3*(pi)^2*(2*pi*f)*L)
k33 = (16*V3*V3)/(n^3*(pi)^2*(2*pi*f)*L)
k12 = (8*V1*V2)/(n^3*(pi)^2*(2*pi*f)*L)
k13 = (8*V1*V3)/(n^3*(pi)^2*(2*pi*f)*L)
k23 = (8*V2*V3)/(n^3*(pi)^2*(2*pi*f)*L)
The equations are:
P1 = (k12.*cos(x(1)*pi/360).*cos(x(2)*pi/360).*sin(x(4)*pi/180))+(k13.*cos(x(1)*pi/360).*cos(x(3)*pi/360).*sin(x(5)*pi/180))
P2 = -(k12.*cos(x(1)*pi/360).*cos(x(2)*pi/360).*sin(x(4)*pi/180))+(k23.*cos(x(2)*pi/360).*cos(x(3)*pi/360).*sin((x(5)-x(4))*pi/180))
P3 = -(k13.*cos(x(1)*pi/360).*cos(x(3)*pi/360).*sin(x(5)*pi/180))+(k23.*cos(x(2)*pi/360).*cos(x(3)*pi/360).*sin((x(4)-x(5))*pi/180))
Q1 = (k11.*cos(x(1)*pi/360).*cos(x(1)*pi/360))-(k12.*cos(x(1)*pi/360).*cos(x(2)*pi/360).*cos(x(4)*pi/180))-(k13.*cos(x(1)*pi/360).*cos(x(3)*pi/360).*cos(x(5)*pi/180))
Q2 = -(k12.*cos(x(1)*pi/360).*cos(x(2)*pi/360).*cos(x(4)*pi/180))+(k22.*cos(x(2)*pi/360).*cos(x(2)*pi/360))-(k23.*cos(x(2)*pi/360).*cos(x(3)*pi/360).*cos((x(5)-x(4))*pi/180))
Q3 = -(k13.*cos(x(1)*pi/360).*cos(x(3)*pi/360).*cos(x(5)*pi/180))-(k23.*cos(x(2)*pi/360).*cos(x(3)*pi/360).*cos((x(5)-x(4))*pi/180))+(k33.*cos(x(3)*pi/360).*cos(x(3)*pi/360))
How can I solve for the angles x(1), x(2), x(3), x(4) and x(5)? Can anyone please help me to solve these equations?
8 Comments
Mukul
on 22 Jun 2018
John D'Errico
on 22 Jun 2018
Did you try using solve? What does it do? If you tried nothing, then why not?
Note: there is a very good chance there is no analytical solution.
Mukul
on 22 Jun 2018
Walter Roberson
on 22 Jun 2018
Wait -- P1, P2, P3, Q1, Q2, Q3 are intended to be definite values, not names of the equations that are assumed to have value equal to zero ?
Mukul
on 26 Jun 2018
Walter Roberson
on 26 Jun 2018
There is not necessarily any error in your code. The system is just difficult to solve.
My work so far shows that for each x1 there are two x2, and that for each x2 there are four x3. Computation is slow, so I have not gotten further than that quite yet.
Mukul
on 26 Jun 2018
Accepted Answer
Walter Roberson
on 22 Jun 2018
Analytic solution:
x(1) = 180 + 360*Z1
x(2) = 180 + 360*Z2;
x(3) = 180 + 360*Z3;
x(4) and x(5) arbitrary (that is, the above 3 together solve all 5 equations)
Here, Z1, Z2, and Z3 represent arbitrary integers
6 Comments
Mukul
on 22 Jun 2018
Walter Roberson
on 22 Jun 2018
As I wrote,
Here, Z1, Z2, and Z3 represent arbitrary integers
For example you could use Z1 = -34834756 and Z2 = 0 and Z3 = 12345
That is, your functions are periodic, and the solutions repeat every 360 infinitely in both directions.
"Can you please tell me how you get this"
I used Maple and I asked it to solve eqn_P1 for x1 asking for all solutions. That gave me the 180 + (integer of your choice)*360 for x1. I substituted that x1 into eqn_P2 through eqn_Q3 and asked to solve that for x2, getting out 180 + (integer of your choice)*360 for x2. I substituted the x1 and x2 values into eqn_P3 through eqn_Q3 and saw that eqn_P3, eqn_Q1, and eqn_Q2 became identical to 0, indicating that they were linear dependent on the first two equations. eqn_Q3 still had a value that involved x3 so I solved for that, getting out 180 + (integer of your choice)*360 for x3. Those set of 3 values together are solutions no matter what the x4 and x5 values are.
Mukul
on 23 Jun 2018
Edited: Walter Roberson
on 26 Jun 2018
Walter Roberson
on 26 Jun 2018
You can solve the first equation set, eqn1, for the first variable x1. Substitute that into the remaining equations, getting eqn2, one shorter than before. Solve the first equation of eqn2 for the second variable x2, getting two solutions. Substitute the first solution of x2 into the remaining variables for eqn2, getting eqn3 that is one shorter still.
Now, solving the first equation of eqn3 (which was the third equation of the original set) with respect to x3, x4, or x5 pretty much fails, so we do not get anywhere with that.
Solve the second equation of eqn3 (which was the fourth equation of the original set) with respect to x3. Substitute that into eqn3([1,3, 4]), getting eqn4 with three equations.
At this point, eqn4(1) is the same equation that we failed to solve for x3, x4, or x5, and we have made it even more complicated by the substitution of x3, so we will not be able to get anywhere easily by solving it.
We can now try to solve eqn4(2) [which used to be the fifth equation] for x4. But it will take a long time. I do not know yet if it can work; my system has been computing it for hours so far.
If eventually we do get a solution, take the first value, substitute it into eqn4(3) [which used to be the sixth equation] and try to solve for x5. This will leave eqn4(1) to substitute x5 into, and as all variables would now have been substituted in, you would check to see if the equation is consistent.
If the equation is not consistent, then back up to the last place you had a choice about which of multiple solutions you substituted, and try the next possibility and move forward from there.
If all goes well, eventually you get a consistent solution.
It is not likely that you will get a consistent solution for six trig equations in five variables.
If you are trying to solve for analytical equations of motion of a multi-jointed arm, then be advised that most such systems people try to solve are analytically inconsistent or else have multiple fairly different solutions.
Mukul
on 28 Jun 2018
Walter Roberson
on 28 Jun 2018
Solving for x(4) and x(5) both failed at the place I was indicating was taking a long time. I did not go back to try substituting in the other choices.
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