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Puzzled with the loxodrome ( spherical spiral ) equation

  1. Oct 6, 2014 #1
    Hi All!

    the mathworld web site http://mathworld.wolfram.com/SphericalSpiral.html claims that the loxodrome is given by the parametric equations
    ##x=cos(t) cos(c)##
    ##y=sin(t) cos(c)##
    ##z=-sin(c)##

    Why so?
    Now, as far as I can see, since the spherical coordinates are
    ##x=sin\phi cos\theta##
    ##y=sin\phi sin\theta##
    ##z=cos\phi##

    Then the loxodrome equations look like the derivative of the radius vector ##{\mathbf r}## with respect to the zenith angle ##\phi## in spherical coordinates, namely

    ##\frac{dx}{d\phi} = cos \phi cos \theta##
    ##\frac{dy}{d\phi} = cos \phi sin \theta##
    ##\frac{dz}{d\phi} = -sin \phi##

    where ##\theta, \phi## are the usual spherical angles, but in the loxodrome equations they are just replaced with ##t## and ##c## respectively.

    Would anyone know why is the loxodrome defined in exactly the way shown above?
    Thanks!
     
  2. jcsd
  3. Oct 11, 2014 #2
    Thanks for the post! Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
  4. Oct 12, 2014 #3

    lavinia

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    Maybe this will help.

    A loxodrome is defined to intersect each meridian on the sphere at a constant angle. In polar coordinates,it projects to a logarithmic spiral.
     
    Last edited: Oct 13, 2014
  5. Oct 13, 2014 #4
    I have been collecting articles and textbook chapters that may shed some light on the derivation. For instance, this link https://answers.yahoo.com/question/index?qid=20110605025959AAzWdiY
    works the problem backwards - assuming a loxodrome, they try to arrive at the constant angle condition.
    Still working on this ....

    I am trying to approach it by using tangent vectors and the necessary derivatives and then the dot-product for the angle.
     
  6. Oct 13, 2014 #5

    lavinia

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    Just calculate the derivatives.what is the problem?
     
  7. Oct 13, 2014 #6
    The vector tangent to the meridian ##\theta = ##const is
    ##\frac{d{\mathbf r}}{d\phi} = (cos\phi cos\theta, cos\phi sin\theta, -sin\phi)##

    If the curve is ##{\mathbf r}(t) = r(sin\phi(t) cos\theta(t), sin\phi(t) sin\theta(t), cos\phi(t))## then its tangent vector is

    ##{\mathbf r}^{\prime}(t)=r(\phi^{\prime}cos\phi cos\theta-\theta^{\prime} sin\phi sin \theta, \phi^{\prime} cos \phi sin \theta + \theta^{\prime} sin \phi soc \theta, -\phi^{\prime} sin \phi)##

    Then the angle between ##\frac{d{\mathbf r}}{d\phi}## and ##{\mathbf r}^{\prime}(t)## is ##\alpha##,

    ##cos\alpha = \frac{\frac{d{\mathbf r}}{d\phi} \cdot {\mathbf r}^{\prime}(t)}{|\frac{d{\mathbf r}}{d\phi}| |{\mathbf r}^{\prime}(t)|}##

    Using the above derivatives, plug in into the expression for ##cos \alpha## and get

    ##cos\alpha = \frac{\phi^{\prime}}{\sqrt{\phi^{\prime 2}+\theta^{\prime 2} sin^{2} \phi}}##

    Now, assume ##\alpha##=const for all t.
    Then we get the ODE
    ##C d\theta = \frac{d\phi}{sin\phi}## which yields

    ##\theta(\phi) = C_1 \ln \tan \frac{\phi}{2} + C_2##

    This is the loxodrome equation. Now, somehow I need to find out how do we get to the parametric equations (see #1) from here...
     
  8. Oct 20, 2014 #7
    here is the answer... It turns out, the "loxodrome parametric equations" are actually a special case of the oblate spheroid coordinates
    ##x=a cosh(\mu) cos(\nu)cos(\phi)##
    ##y=a cosh(\mu) cos(\nu)sin(\phi)##
    ##z=a sinh(\mu) sin(\nu))##

    when ##abs(\mu)## is large enough. Then ##a cosh(\mu) = a sinh (\mu) = R=\mbox{const}## and the above becomes the parametrization of a sphere
    ##x=R cos(\nu)cos(\phi)##
    ##y=R cos(\nu)sin(\phi)##
    ##z=-R sin(\nu))##

    even though in oblate spheroid coordinates ##\mu## is not supposed to be negative, the above should work fine to parametrize a sphere (correct me if I'm wrong).
     
  9. Oct 20, 2014 #8
    Now the only thing I need to sort out is why does Mathworld stipulate ##c=tan^{-1}(at)##. Some say (see here https://answers.yahoo.com/question/index?qid=20110605025959AAzWdiY) that because this way
    ##cos(c) =\frac{1}{\sqrt{1+a^2t^2}}## and
    ##sin(c)=\frac{at}{\sqrt{1+a^2t^2}}##

    But then we can equally well may set them to be

    ##cos(c)=t## and
    ##sin(c)=\sqrt{1-t^2}##...

    Perhaps this needs another thread on its own,...
     
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