0
$\begingroup$

I would like to calculate the $R(t)$ of an object that accelerates at a constant magnitude, but at a changing direction (making the acceleration always tangent to the trajectory. Something like a rocket engine, or aerodynamic drag(with neglect for complex factors that make acceleration non-constant)).

I would pick a predetermined $R(t)$, say $R(t)=ti-\frac{9.81t^2}{2}j$. Then, I would like to apply my acceleration, which probably cannot be done with $W(R(t)) = R(t)-\frac{dR}{ds}$, because the derivative changes too.

I thought of making an equation that looks like this: $$ R(t) = R(t) + C\frac{dR}{ds} $$ where $C$ denotes the magnitute of the constant acceleration. I am not sure whether this would give me what I am looking for though.

I also heard that I might need differential equations for this problem, but I don't know how to construct one.

  • 0
    What is $R(t)$? ${}{}$ – 2017-02-03
  • 0
    @copper.hat The position vector – 2017-02-03
  • 0
    I don't understand what you are trying to do. Are you give some $\ddot{R}(t)$ that satisfies $\| \ddot{R}(t) \| = 1$, for example, and you want to find an equation for $R$? – 2017-02-03
  • 0
    @copper.hat He means $R''(t) = \frac{\alpha R'(t)}{||R'(t)||}$, I believe. – 2017-02-03
  • 0
    @Guangliang: Surely $R''(t) = \alpha {R(t) \over \|R(t)\|}$? – 2017-02-03
  • 0
    If we take your words literally, you have set an impossible task. Acceleration cannot be tangent to a curved trajectory. Is it possible that what you really want is that the _component_ of acceleration tangent to the trajectory is constant? The total acceleration would be composed of that component plus some component perpendicular to the trajectory. – 2017-02-03

1 Answers 1

0

Your $R(t)$ is really a point with three components. You have a set of coupled differential equations. You should also define $V(t)=\frac d{dt}R(t)$ as the velocity and $A(t)=\frac d{dt}V(t)$ as the acceleration, each with three components. You compute $A(t)$ by dividing the force on the object by its mass.

The naive way to approach this is to take a small time step $\Delta t$. At the start you have the conditions $V(t), R(t)$ and you update the conditions by $$V(t+\Delta t)=V(t)+A(t)\Delta t\\R(t+\Delta t)=R(t)+V(t)\Delta t$$ This is called the Euler method of integrating differential equations. It is conceptually simple-the change in position is just the velocity times the time step. The problem is also easy to see-the velocity is changing during the time step and we are ignoring that fact in updating the position. This leads to accumulating error in the solution. Reducing the time step can help with this, but you run into numerical accuracy and maybe compute time issues if you make it too small. There are more complex algorithms that are more accurate. You can consult any numerical analysis textbook for a discussion. I like chapter 16 of Numerical Recipes as a starting point. It is free online. Others will have their favorites.