Calculate the position of rocket acted upon simultaneously by multiple thrusters?

Question

I'm looking for an equation that will let me predict the position of a rocket after a period of time given that it is acted upon by multiple forces.

By multiple forces I mean the main thruster force and a second attitude thruster, one that affects the direction of the rocket.

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>|<  attitude thrusters
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^^^ 
main thrusters

I can easily find equations to predict the position of the rocket if only the main thruster is firing, but when we introduce the attitude thruster because it rotates the ship it consequently modifies the direction that the main thruster is acting in. I've looked into this a fair bit trying to combine equations of torque with linear motion equations but I'm still confused?

fyi. My maths level is not particularly high I achieved an A at alevel in mechanics

Thanks

Piers

Answer 1

The following is the simplest treatment I can think of..

The rocket has length $L$ meters and mass $M$ kg.

Main thrusters provide thrust of $T_m$ newtons directly along the line connecting the bottom of the rocket to the centre of mass

Attitude thrusters provide a thrust of $\pm T_a$ newtons perpendicular to the line connecting the bottom of the rocket to the centre of mass. This thrust is applied a distance $l \le L/2$ from the centre of mass.

The state of the rocket is specified by the co-ordinates of the centre of mass $\vec x (t)$ and the angle of orientation $ \theta (t)$of the rocket to positive x-axis.

for the simplest treatment assume that $\theta (t)$ is constant while the main thrusters are being applied. Then the effect of the main thrusters can be calculated from Newton's second law.

$$\vec T_m = M \frac {d^2 \vec x }{dt^2} = T_m(\cos(\theta), \sin(\theta) )$$

which is assumed to be constant so that the solution is

$$ \vec x (t) = \vec x_0 + \vec v_0 t + \frac { \vec T_m }{2 M }t^2 $$

when the attitude thrusters are applied the orientation angle $\theta (t) $ will change according to the rotational equivalent of Newton's second law $\tau = I \alpha$

The moment of inertia of a rod of length $L$ and mass $M$ about its centre of mass works out to be $I= \frac {1}{12}M L^2$

$$ \tau = I \alpha \Rightarrow T_a l = \frac {1}{12}M L^2 \frac {d^2 \theta}{d t^2} $$

so that

$$ \frac {d^2 \theta}{d t^2} = \frac{12T_a l }{M L^2 } $$

whose solution is

$$ \theta(t) = \theta_0 + \omega _0 t + \frac{6 T_a l }{M L^2 } t^2 $$

the attitude thrusters will also accelerate the centre the centre of mass directly, but we can ignore this ( at first approximation ) on the grounds that $T_m >> T_a$

Answer 2

If the two thrusters fire continuously and we assume the main thruster fires exactly along the axis, so it makes no torque, the attitude is given by $\theta(t) = \theta_0 + \omega _0 t + \frac{6 T_a l }{M L^2 } t^2$ as shown by WW1. If we define $\theta=0$ to be the direction the rocket is pointed at $t=0$ and assume there was no rotation at the start, this becomes $\theta(t) = \frac{6 T_a l }{M L^2 } t^2$. Then the acceleration becomes $\vec {\ddot x}=\frac{T_m}M (\cos \theta \hat x + \sin \theta \hat y)$ In principle you can integrate this to get the position as a function of time. As $\cos (at^2)$ has no elementary integral, you are reduced to numeric integration. Numeric integration handles multiple forces and torques quite well. At each time, you compute the total force and torque acting on the vehicle and feed that into your integrator. This will be discussed in any numerical analysis text.

Answer 3

If both thrusters operate simultaneously , the $\theta (t)$ equation would be unchanged since the main thrusters provide zero torque. Coupling the equations for the centre of mass co-ordinates gives ..

$$ \frac{d^2}{dt^2}\left ( x_1(t)\right) = \frac{T_m}{M} \cos(\theta_0 + \omega _0 t + k_a t^2) $$

$$ \frac{d^2}{dt^2}(x_2(t)) = \frac{T_m}{M} \sin(\theta_0 + \omega _0 t + k_a t^2) $$

where $k_a = \frac{6 T_a l }{M L^2 }$

At this point, I think we have to give up on finding an analytic solution. This system would need to be solved numerically by taking tiny time increments and keeping track of $x_1(t)\; v_1(t)\;x_2(t)\; v_2(t)\; \theta (t)\; \text{ and } \omega(t).$

We may as well include the effect of the attitude thrusters on the centre of mass co-ordinates,so that the DEs to be solved numerically are ...

$$ \frac{d^2}{dt^2}\left ( x_1(t)\right) = \frac{T_m}{M} \cos(\theta_0 + \omega _0 t + k_a t^2) - \frac{T_a}{M} \sin(\theta_0 + \omega _0 t + k_a t^2) $$

$$ \frac{d^2}{dt^2}(x_2(t)) = \frac{T_m}{M} \sin(\theta_0 + \omega _0 t + k_a t^2) + \frac{T_a}{M} \cos(\theta_0 + \omega _0 t + k_a t^2)$$

You might want to consider an scheme in which the attitude thrusters are only active for a short period of time $\Delta t$ during which they fire in one direction for half the time and the opposite direction for the other half of the time so that $\omega (t) = 0 $ unless the attitude thrusters are firing, ( surely you don't want your rocket to be constantly spinning, it would be very wasteful to fire the main thrusters while the rocket spins. )