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Help with extracting alpha from: -alpha-asin(sin(alpha)(r/l)

  1. Jan 4, 2012 #1
    I am using the Law of Cosines to extract and angle I need from a shaft that is turning on an air engine by piston oscillation and I am having a brain fart on if there are any identities I am missing that can help me pull '[itex]\alpha[/itex]' from the right side of my equation:

    acos(([itex](h-pl)^{2}[/itex]-[itex]r^{2}[/itex]-[itex]l^{2}[/itex])/(-2*r*l))-[itex]\pi[/itex]=-[itex]\alpha[/itex]-asin(sin([itex]\alpha[/itex])(r/l))

    I would like to solve the entire equation for [itex]\alpha[/itex] and the other variables can be treated like constants.

    Thanks for the help or attempts in advance.
     
    Last edited: Jan 4, 2012
  2. jcsd
  3. Jan 4, 2012 #2

    Simon Bridge

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    well - you didn't close the brackets in the arcsine, so it's hard to tell - but asin(sin(x))=x
     
  4. Jan 4, 2012 #3
    I fixed the bracket issue, I apologize about that.

    I am concerned with the (r/l) term inside of the equation. I understand that asin(sin(x))=x .. but I'm certain that asin(sin(x)(r/l)) --DNE-- (r/l)x
     
  5. Jan 4, 2012 #4

    Simon Bridge

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    Oh I think I see, it wasn't clear to me if the r/l was multiplied with the alpha or the sine-alpha, or with the arcsine.
    This what you mean?

    [tex]\sin^{-1}( \frac{r}{l}\sin\alpha )[/tex]

    ... afaik you can't extract alpha from this sort of equation - you need to be cleverer in the setup or use an approximation (or use a numerical method).

    Equation is of form: x+sin-1(A.sin(x))=b ... solve for x. argh. (assuming the RHS is all constants)

    Note: the arcsin will be undefined for some alpha, depending on the value of rl - you want |sin(α)| < l/r (something...)

    Where did you start from?
     
    Last edited: Jan 4, 2012
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