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Spacecraft path with polar coordinates

  1. Apr 4, 2013 #1
    There is a circular gate rotating at a constant angular speed of ω. The circular gate has a tunnel across its diameter. The mission is to pass through the gate. (That is, come in one side of the gate, travel the whole diameter, and exit at the other side.)

    Also, craft is neutrally buoyant, the craft has four thrusters (in the i, -i, j, and -j directions), and sudden movements cause "slosh" in the fuel tank that destroys the craft.

    My thought was to describe the position of the craft in the complex plane [tex] z=r \cdot e^{i\theta}[/tex] where r and θ depend on time t. Let's say at time 0 the craft is in front of the tunnel opening.

    Differentiating once, I get [tex] v = (\dot r +i r \dot \theta)e^{i\theta} [/tex]
    which describes a velocity of [tex]\dot r[/tex] in the radial direction and a velocity of [tex] r \dot \theta[/tex] in the perpendicular direction.

    Differentiating again, I get [tex] a = (\ddot r - r {\dot \theta}^2 )+i(2\dot r \dot \theta + r \ddot \theta)] e^{i\theta}[/tex] which has the familiar expressions for acceleration in the radial and perpendicular directions.

    Now, [tex] \dot \theta = \omega [/tex] which is constant. We also want to move smoothly, so we want the radial speed to be constant as well. Thus, our acceleration is [tex]a = [(-r \omega^2) + i(2\dot r \dot \theta + r \ddot \theta)]e^{i\theta}.[/tex]

    Now, all that matters is that the craft gets through the tunnel, so I set [tex]\dot r \equiv -1[/tex] until the craft gets to the center and then [tex] \dot r \equiv 1 [/tex] after we get to the center.

    Thus, the acceleration before getting to the center is [tex] a = [(-r {\dot \theta}^2)+i(-2\omega)]e^{i\theta}.[/tex]

    Since the thrusters are in the i and j directions, I multiply the expression out, substitute cosθ = x/r, and sinθ = y/r, and r = √x^2+y^2 to obtain [tex]\left(-\omega^2 \cdot x+2\omega \cdot \frac{y}{\sqrt{x^2+y^2}}\right)+i \left(-\omega^2 \cdot y - 2\omega \frac {x}{\sqrt{x^2+y^2}} \right)[/tex].

    Of course, the real part represents the horizontal acceleration and the imaginary part gives the vertical.

    The thrust would then be given by the mass times the acceleration.

    This is apparently incorrect, since I have run it on a simulation. Does anyone see where I went wrong?
     
    Last edited: Apr 4, 2013
  2. jcsd
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