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Transpose problem

  1. Aug 17, 2009 #1
    Hey guys,

    So I've actually learnt a fair bit about trig identities the last few weeks and beginning to understand how they actually work thanks to Irrational, Mute and some prompting from Hurkyl.

    I'm still having trouble with transposing which I think should be fairly simple. The equation is solving steering geometry with two points of rotation that locates two points in space. The equation which works out the length between two points is:

    D = | P1 - P2 | with | | denoting the length of the resulting 3D vector.

    I know D (the length) and I also know P2 (the x,y,z position of point 2). P1 is what I'm having trouble with. The equation for P1 is:

    P1 = Pf + (i1f * Rf * cos(Vf)) + (i2f * Rf * sin(Vf))

    I know Pf (the x,y,z position of the axis of rotation), i1f (the x projection vector of P1's rotation axis), Rf(the length between P1 and the axis of rotation) and i2f(the y projection vector of P1's rotation axis).

    From those I know all but Vf, which is an angle from P1 to the x axis and P1, which is solvable by substituting it into the 1st equation D = | P1 - P2 |.

    Vf is where I'm having the problem. How can I transpose it to make it the subject. I've tried using the identity R * cos(x - alpha) but still have the same problem of having 2 instances of Vf in the equation as in P1.x, P1.y and P1.z will all use different values for R and alpha. I hope this makes sense. If not I'll post up the images from the white paper I'm using. Any help is much appreciated.
     
  2. jcsd
  3. Aug 17, 2009 #2
    I've attached the excerpt anyway:

    steering.jpg
     
  4. Aug 27, 2009 #3
    Can anyone help here or point me in the right direction? I'm really stuck with this and it doesn't matter which theorem I apply, I still end up with 3 instances of x on the same side of the equation.
     
  5. Aug 27, 2009 #4
    After simplification, you should be left with an equation of the form k = Acos(t) + Bsin(t). Using the angle addition identities, you should be able to get k = Csin(t + p) where [itex]C = \sqrt{A^2 + B^2}[/itex] and p = arctan(A/B).
     
  6. Aug 27, 2009 #5
    Thanks for responding slider. When I expand it out, I have something like this:

    k = [tex]\sqrt{}[/tex] (A.xCos(t) + B.xSin(t))[tex]^{2}[/tex] + (A.yCos(t) + B.ySin(t))[tex]^{2}[/tex] + (A.zCos(t) + B.zSin(t))[tex]^{2}[/tex]

    Is there a way to simplify it because A and B are 3D vectors? I can use the Rcos(x-alpha) like you suggested but I still end up with 3 instances of t on the right side for each component of A and B.
     
  7. Aug 27, 2009 #6
    My path was to first get the equation:
    [tex]\frac{P_1 - P_f}{R_f} = i_{1f}\cos V_f + i_{2f}\sin V_f[/tex]
    Then we have
    [tex]
    C = \sqrt{i_{1f}^2 + i_{2f}^2} [/tex]
    [tex]\theta = \arctan \frac{i_{1f}}{i_{2f}}
    [/tex]
    Expanding and simplifying, we get
    [tex]
    \frac{P_1 - P_f}{R_f} = \sqrt{i_{1f}^2 + i_{2f}^2}\sin (\arctan \frac{i_{1f}}{i_{2f}} + V_f)
    [/tex]
    or
    [tex]
    V_f = \arcsin\left(\frac{P_1 - P_f}{R_f\sqrt{i_{1f}^2 + i_{2f}^2}}\right) - \arctan\frac{i_{1f}}{i_{2f}}
    [/tex]

    Edit: Oh, I see, the i's are vectors. you will have to use one of the 3 component equations.
     
  8. Aug 30, 2009 #7
    Ok thanks. So assuming I use the x component for the equation I still have two unknowns being Vf and P1.x. Do I need another equation to work out what P1 is?
     
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