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In a two-dimensional environment:

Given:

  • a starting point (x1,y1)
  • an initial velocity (speed and 2D heading/direction)
  • a constant rate of turning (yaw, since there is no roll or pitch in only two dimensions)
  • a constant acceleration of the object from itself (not like gravity where the direction of the acceleration is relatively constant - "down" - but like a rocket that is propelling itself)
  • a time change t1 to t2

What is the formula for the final point (x2,y2)? What kind of physics problem do I have on my hands? Is this just a variation on kinematics with variable acceleration (varying direction of acceleration instead of amount), or is it qualitatively different?

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  • $\begingroup$ Intuitively I think that this should result in a path which might be visualized as a quarter of an ellipse and I was trying to work off of physicsclassroom.com/class/1DKin/Lesson-6/Kinematic-Equations and numericana.com/answer/ellipse.htm but I'm not sure this is the correct direction/method. $\endgroup$ – Stephen T. Robbins Jan 8 '16 at 0:21
  • $\begingroup$ Is this vehicle like a car with wheels, so it cannot slide sideways, or is it like a rocket, that can go sideways? $\endgroup$ – Mike Dunlavey Jan 8 '16 at 1:52
  • $\begingroup$ Like a car, no sliding. (turning without accelerating) If it is headed north/up from (0,0) and turns 90 degrees to the right at a speed of pi/10 per second, then after 10 seconds it will be at (2,2). $\endgroup$ – Stephen T. Robbins Jan 8 '16 at 2:23
  • $\begingroup$ OK, this is like a car. Now the constant rate of turning. Acceleration comes into play. Do you mean the steering wheel is held at a constant angle, so the front wheels are at a constant angle, so there is a fixed center about which the car is turning, in which case the curve is a circular arc? Or do you mean that as the speed increases the steering wheel is turned back toward 0 so as to keep the yaw rate constant? Just trying to clarify the question. $\endgroup$ – Mike Dunlavey Jan 8 '16 at 13:33
  • $\begingroup$ That is a great question that I had not thought to clarify. In this case, I definitely want the yaw rate to be constant. $\endgroup$ – Stephen T. Robbins Jan 8 '16 at 18:58
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OK, If acceleration is constant over the interval, that means velocity $v$ changes at a constant rate over the interval. If steering wheel position (and thus path curvature) is $\phi$, and yaw rate is proportional to $v\phi$, that means $\phi$ must vary as $1/v$. To put it another way, turning radius $r$ must vary proportionally to $v$.

This describes some kind of spiral, which might be well-known, but I don't know what it is (where both $v$ and $r$ vary linearly with time and $v/r$ is constant).

Without knowing that (or even if you do know it) I would use a suitable ODE solver, such as Runge-Kutta, or even Euler if I didn't mind the accuracy vs. time tradeoff.

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  • $\begingroup$ Ah, yeah, you're right. It is going to be a spiral if it keeps going. I confused myself by only considering one short section. $\endgroup$ – Stephen T. Robbins Jan 8 '16 at 19:59
  • $\begingroup$ @StephenT.Robbins: I'm tempted to say Achimedean spiral where radius is linear in $\theta$, but I'm not sure. $\endgroup$ – Mike Dunlavey Jan 8 '16 at 20:10
  • $\begingroup$ I found the Achimedean spiral pretty quickly and it does look at least very close though I'm still trying to nail down the correct equations to be using. Still, it helps a lot to at least be looking at the correct shape set! I am hoping it turns out to be the Archimedes spiral because that would make things simpler/easier. $\endgroup$ – Stephen T. Robbins Jan 8 '16 at 21:31
  • $\begingroup$ @StephenT.Robbins: I just realized, it $d\theta/dt$ is constant, that means $dr/dt$ is constant, so $dv/dt$ is constant, and that's your constant acceleration. So yes, the spiral is Archimedean. $\endgroup$ – Mike Dunlavey Jan 9 '16 at 15:37

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