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I have a spaceship moving in a 2D plane and wish to intercept another body which is also moving in the same plane. To do this I need to find a series of accelerations over a series of times that will result in the relative velocity = zero and also the relative distance = 0. Maximum acceleration is fixed but there is no upper limit on speed.

To simplify the problem, I first calculate the relative velocity in polar coordinates, then switch the local axes so that in local coordinates, the spaceship is positioned at the origin with a known velocity and I want it to arrive at a static target (at angle = 0) then stop.

Calculating this in 1D is quite simple and can be solved by equating the various kinematic equations. This can be solved in both parallel and perpendicular directions separately, but even though we know that $a,max^2 = a,ll^2 + a,T^2$, I cannot figure out how to combine the two or calculate a suitable ratio of accelerations even approximately.

Just to summarise, we know distance, angle, relative velocity and maximum acceleration, and are looking to find a series of 2 (or more) accelerations over an equal number of time periods. The end state is the two bodies sharing the same coordinates with no relative velocity.

This has had me stumped for a week so I would be very grateful for a solution!

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You could first do the manouver in the $x$-direction to make sure that the $x$-coordinate is constant and equal to the $x$-coordinate of the target (assuming you are in a reference frame where the target is stationary). This is the same problem as the one-dimensional problem. Once that you have done that you do the same in the $y$-direction. Then you are done.

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  • $\begingroup$ If I understand you correctly, you are proposing doing two separate calculations in x and y which have known accelerations in each direction. The problem is that while I know total acceleration, I do not know how much should be in each direction! $\endgroup$ – Harry Crofton Apr 9 '16 at 21:35
  • $\begingroup$ I don't think you understand my solution. First you do the manouver in the x-direction. With this i mean you accellerate in the x-direction only. After you finish this manouver, say at time $t_0$, you start accellerating in the y-direction until you reach your destination. $\endgroup$ – Ward Beullens Apr 9 '16 at 21:45
  • $\begingroup$ Thank you for your clarification. Your response does strictly answer the question and is admirably simple, but a spaceship would want to take a more direct route to the target. Should I ask a new question, or edit this one to get a new answer? $\endgroup$ – Harry Crofton Apr 10 '16 at 9:15
  • $\begingroup$ I think you should ask a new question that refers to this one. Make sure to ask a precise question. Don't ask for a "more direct" route, but maybe for a route that minimizes the total travelled time or total travelled distance. $\endgroup$ – Ward Beullens Apr 10 '16 at 9:23
  • $\begingroup$ But I can give a more direct route myself: You can do the 2 manouvers simultaneously by just adding the acceleration in the 2 directions. The problem with this is that the maximal acceleration might be exceeded. But if you solve the 2 1-dimensional problems with respect to a maximal acceleration of $\frac{a_{max}}{\sqrt{2}}$, then the sum of the 2 acceleration will stay below $a_{max}$ at all times. $\endgroup$ – Ward Beullens Apr 10 '16 at 9:26

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