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System of nonlinear equations

  1. Aug 11, 2005 #1

    I have a system of trigonometric equations from which I should find theta1,..., theta5. Is it possible you can give me a hint on how to proceed. Thanks.

    theta, phi, psi, Px, Py, Pz, l1, l2, l3, l4, l5, d1, d2, d3, d4, d5 are all constants.

    Cos[t1+t2] Cos[t3+t4] Cos[t5]+Sin[t1+t2] Sin[t5]=Cos[phi] Cos[theta]

    Cos[t5] Sin[t1+t2]-Cos[t1+t2] Cos[t3+t4] Sin[t5]=Cos[theta] Sin[phi] Sin[psi]-Cos[psi] Sin[theta]

    Cos[t1+t2] Sin[t3+t4]=Cos[psi] Cos[theta] Sin[phi]+Sin[psi] Sin[theta]

    l1Cos[t1]+Cos[t1+t2] (l2+l3 Cos[t3]+Cos[t3+t4] (l4+l5 Cos[t5])+d5 Sin[t3+t4])+Sin[t1+t2] (d3+d4+l5 Sin[t5])=Px

    Cos[t3+t4] Cos[t5] Sin[t1+t2]-Cos[t1+t2] Sin[t5]=Cos[phi] Sin[theta]

    -Cos[t1+t2] Cos[t5]-Cos[t3+t4] Sin[t1+t2] Sin[t5]=Cos[psi] Cos[theta]+Sin[phi] Sin[psi] Sin[theta]

    l1Sin[t1]+Sin[t1+t2] (l2+l3 Cos[t3]+Cos[t3+t4] (l4+l5 Cos[t5])+d5 Sin[t3+t4])-Cos[t1+t2] (d3+d4+l5 Sin[t5])=Py

    Sin[t1+t2] Sin[t3+t4]=-Cos[theta] Sin[psi]+Cos[psi] Sin[phi] Sin[theta] Cos[t5] Sin[t3+t4]-Sin[phi]-Sin[t3+t4] Sin[t5]=Cos[phi] Sin[psi]


    d1+d2-d5 Cos[t3+t4]+l3 Sin[t3]+(l4+l5 Cos[t5]) Sin[t3+t4]=Pz
  2. jcsd
  3. Aug 11, 2005 #2
    Yikes! A little more readability can't hurt:
    (https://www.physicsforums.com/showthread.php?t=8997) << read this!

    Last edited by a moderator: Aug 11, 2005
  4. Aug 12, 2005 #3


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    Feel like posting the values for all these constants?

    Then me anyway, in some desperate attempt at approaching it, I would then convert each to:

    [tex]t1=f(t1,t2,t3,t4,t5; constants)[/tex]


    and so on and then use iteration of some sort to analyze if it converges to a solution. There is a sufficiency condition for this sort of iteration to converge to a solution and involves the partials of each function above.

    Oh yea, I'd rely heavily on Mathematica too. :smile:

    Edit: I just noticed you have 10 equation and one in particular:

    [tex]-\cos \left( t_3+t_4 \right) =\cos \phi \cos \psi [/tex]

    You can immediately start cleaning them up by substituting this one and it's Sin equivalent.
    Last edited: Aug 12, 2005
  5. Aug 12, 2005 #4


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    I change my mind. I can do that. There're 9 equations in 9 unknowns. For example:


    What are the rest?
  6. Aug 12, 2005 #5
    Thanks a lot. I managed to get the reduction you mentioned.
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