1
\$\begingroup\$

We know for an object:

  • Initial point O(x,y)
  • Initial speed (s_x,s_y)
  • Constant vector module aceleration (A)
  • A point P(i,j) where we have to pass through in the future.

We need:

  • A point W(x, y) so the angle alpha (or only the angle alpha), formed between O and W, can be used to calculate acceleration in both axis:

    Ax = A * cos(alpha)

    Ay = A * sin(alpha)

so that the object passes through W. enter image description here

How do I define an elliptic motion system knowing two points of the orbit (and Vo)?

I'll appreciate alternatives to design a guided rocket too.

My attempt for solving this:

  1. Calculate module of initial point O, initial speed So, destination point D. (Aceleration module A is given)

  2. Use these terms to calculate the time we need to arrive to D with the equation:

    D = O + So*t + 1/2 A * t^2

  3. Find out the value of alpha from this equation, knowing time:

    Dx = Ox + Sox * t + 1/2 A * cos(alpha) * t^2

I tried to implement this, but for some reason it doesn't work. Should this theorically work?

\$\endgroup\$
  • \$\begingroup\$ Do you mean you want the theta to produce the dotted path here? I'm not sure you can achieve it with an arbitrary constant force. Is the 'pass through here' point fixed, or just to help visualize your question? \$\endgroup\$ – Chris Mills-Price Oct 9 '14 at 17:13
  • \$\begingroup\$ Aceleration module is constant, but not their components (I'm looking to change angle of aceleration only, to have it passing through the point). "Pass through here" is point fixed \$\endgroup\$ – freesoul Oct 10 '14 at 0:05
  • \$\begingroup\$ Ah, I took "aim point" to mean "point the rocket is ultimately aimed at/intended to reach", rather than directly related to the acceleration vector, hence my wrong 'dotted path' image. I'll think on this some more and get back to you if you or Pieter haven't solved it in the meantime. \$\endgroup\$ – Chris Mills-Price Oct 10 '14 at 0:31
  • \$\begingroup\$ @freesoul: No, your equations will not apply here because they are only for the case where both the magnitude and direction of acceleration is constant, whereas your description of the problem is that only the magnitude of the acceleration is constant, with its direction being directed towards a fixed aim pont. \$\endgroup\$ – Pieter Geerkens Oct 10 '14 at 12:15
2
\$\begingroup\$

If I understand your problem correctly, you wish to determine, for a given initial location, initial velocity, desired target, and constant magnitude acceleration, what constant point the acceleration must be aimed at in order to pass through the desired target point at some future time.

This is the definition of circular motion around the aiming point, for a circle of radius r passing through both the initial location and the target, where by the well-known kinetic equations of circular motion the radius can be derived as:

r = v^2 / a

where v and a are respectively the magnitude of the starting velocity and acceleration.

Now it is sufficient to obtain, using some basic Euclidean geometry, the location of the point that is distance r from both the starting point and the target location, on the desired side so that the intersection occurs in the future rather than the past.

\$\endgroup\$
  • \$\begingroup\$ Hello Pieter, thanks for your answer. However while the acceleration module of the object is constant, their x and y component are not. I'm looking to get an angle of this vector to have the object passing through a point. (Thats why I look for a "new point to aim", that could give me the angle). Maybe this is related to elliptic motion? \$\endgroup\$ – freesoul Oct 10 '14 at 0:03
  • \$\begingroup\$ @freesoul: Exactly! for circular motion the acceleration is constantly directed towards a fixed point, the center of the circular orbit. The angle of the acceleration will rotate with the rocket's motion, maintaining a fixed angle relative to the orientation of the rocket but rotating relative to a fixed (actually any inertial) observer. \$\endgroup\$ – Pieter Geerkens Oct 10 '14 at 3:44
  • 1
    \$\begingroup\$ @PieterGeerkens: What a simple and elegant solution! I was about to try to solve differential equations and I've totally forgot about this simple special case of circular orbit. \$\endgroup\$ – Podgorskiy Oct 10 '14 at 7:08
  • 1
    \$\begingroup\$ Generally, I think that there would be infinite set of aim points. One of them would produce a circular orbit. And one point at an infinite distance would produce a parabola. This two cases is a special cases of ecliptic motion.However, I'm not definitely sure, because the central force is constant, not inversely proportional to the square of the distance as in classical equation. \$\endgroup\$ – Podgorskiy Oct 10 '14 at 7:15
  • 1
    \$\begingroup\$ There is an example I've wrote for the case when the aim point located at infinite distance, so the module and direction of acceleration is constant. Rocket moves along a parabolic trajectory. \$\endgroup\$ – Podgorskiy Oct 10 '14 at 7:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.