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Property of a generalised helix

  1. Sep 27, 2014 #1

    CAF123

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    1. The problem statement, all variables and given/known data
    A generalized helix is a space curve whose unit tangent makes ##T## makes a constant angle ##\theta## with the a fixed unit vector ##A## in Euclidean space, I.e ##T \cdot A = \cos \theta = \text{const}##. Prove that if the torsion ##\tau \neq 0## everywhere then the space curve is a generalized helix iff ## \kappa/\tau = \text{const}##

    2. Relevant equations

    Frenet-Serret Formulae


    3. The attempt at a solution

    ##(\Rightarrow)##WLOG, assume curve is unit speed parametrised. Then $$\frac{d}{ds} (T \cdot A) = T' \cdot A = \kappa N \cdot A \Rightarrow N \cdot A = 0$$ Then $$ \frac{d}{ds} (N \cdot A) = (- \kappa T + \tau B) \cdot A = 0$$ using the Frenet serret formulae. This is equal to ##-\kappa \cos \theta + \tau B \cdot A = 0##. I would nearly be there if I can show that ##B \cdot A = \text{const}##. I managed to see that ##A## lives in the plane spanned by T and B. So ##A = aT + bB## for some a and b. Then ##A \cdot A = 1 = a \cos \theta + b B \cdot A \Rightarrow B \cdot A = (1 - a \cos \theta)/b## which certainly is constant provided a and b are constant. But since T and B revolve around the curve, I think a and b are parameter dependent. So I don't think my argument works. Thanks for any help here. Any hints on how to start the reverse implication would be great too, thanks.
     
    Last edited: Sep 27, 2014
  2. jcsd
  3. Sep 27, 2014 #2

    CAF123

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    Has anyone any pointers? thanks.
     
  4. Sep 28, 2014 #3
    Consider ##{d \over ds} \left( B \cdot A \right) ##.
     
  5. Sep 28, 2014 #4

    CAF123

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    Hi voko! Thanks, I got it from that. Do you have any hints on the reverse implication please?
     
  6. Sep 28, 2014 #5
    Since you have ##N \cdot A = 0##, ##A## must be in the T-B plane, as you concluded earlier. Your coefficients ##a## and ##b## are simply ##\cos \theta## and ##\sin \theta##. What condition do you get for ##\kappa, \tau## and ##\theta## from that? Can you use it to complete the proof?
     
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