0
$\begingroup$

I want to intercept two objects, making both their time and velocity the same. One has acceleration to achieve this goal, while the other is moving at a constant rate. Furthermore, I want them to intersect at a specific intersection point.

The easiest permutation of this is a 1 dimensional line with the faster, unpowered object reaching the intersect point first. The powered object has to accelerate at a specific time to finish acceleration at the intersect point matching velocity at the intercept. I am pretty sure it has to be a system of equations, but I'm having a lot of trouble trying to figure out how to set it up.

I know that obj1: x2 = x1 +v1*t +.5at^2 and obj2: x2 = vt + x1

The x2 values should be equal so: x1 + v1t + .5at^2 = vt + x1

but I'm not quite sure what to do next because the vs and xs are specific to their objects.

Edit: I thought about this some more and a simpler version would be an accelerating object reaching a specified position with a specified velocity at a specified time, since the target will always reach the point at the same time. Still working this out, but maybe I'm a bit closer.

Thanks

$\endgroup$
  • $\begingroup$ If they have to arrive at the destination point with the same velocity, the accelerating one can never catch up. You have given it a maximum velocity equal to the speed of the one with constant speed. The accelerating one will always be slower except at the instant of interception. $\endgroup$ – C. Towne Springer May 7 '18 at 0:22
  • $\begingroup$ No, not always. I've given this some thought and there are two constraints. The first is that the acceleration must end with the velocities the same at the intersection point. The instant before that, there is a difference in velocity. The second constraint is that if the non-accelerating interceptee would be faster and yet reach the intercept point after the interceptor, there is no acceleration to make this possible. If one is faster and they reach the interception point at the same time, it is impossible without infinite acceleration. $\endgroup$ – Oblivion May 7 '18 at 1:01
  • $\begingroup$ A positive way to phrase the second constraint is that if the relative speed of the target is positive, the acceleration required also has to be positive and more importantly the relative time to intercept of the interceptor must be positive as well. $\endgroup$ – Oblivion May 7 '18 at 1:07
  • $\begingroup$ I thought of a simpler example. If the targetv = 0, this is easy because the time constraint is removed. In fact, now that I think about it, the interceptor just has to reach the right velocity in the right time which simplifies the target equation to the it will take to get there. So a simpler example would be changing velocity to arrive at a position and velocity at a time, not complicated by two vehicles. $\endgroup$ – Oblivion May 7 '18 at 1:31
  • $\begingroup$ @C.TowneSpringer I apologize for rambling on, this is my intuitive reasoning based on my limited physics knowledge. Does this constitute a physics impossibility with the given constraints? $\endgroup$ – Oblivion May 7 '18 at 1:49
0
$\begingroup$

I've been working on this for a few days and was finally able to figure this out. Upon realizing that it was a problem involving only one powered object with a few constraints, I was able to solve v2^2=v1^2+2aDx for the acceleration required and then use v2=v1+at to get the time required to accelerate.

Constraints:

  1. the acceleration finishes at the intercept point.
  2. The target will reach the intercept first if faster (and last if slower) if no action is taken, allowing time for the interceptor to match velocity

I knew I was frustrated for a reason. That was easy. The actual problem is a 3d rendezvous of a spacecraft with another, which I guess just goes to show that simplifying things to the concepts you're having trouble with can help a great deal. I took it from a 3d, 2 object scenario to 1d and one object, representing the other as a constraint for time to intercept. Now on to learning vector math which I suspect won't give me as many problems as this physics I learned in high school :-/

| cite | improve this answer | |
$\endgroup$

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.