2
$\begingroup$

I am trying to simulate a rocket launch.

Let's say all the forces on the rocket are applied at its center of mass.
Let's say the center of mass of the rocket is currently at point named P1.
Let's say the goal point the rocket will have to reach is named P2.

I can simulate some forces applied to the rocket (such as weight, thrust and drag).
I can then compute the acceleration and then the speed and finally update the position.

I use 2 angles, one for the direction of the path the rocket is actually following and the other one need to be the angle of attack.

Given the forces, how can I compute the angle of attack which the rocket shall have so as to follow a straight line from P1 to P2?

Consider the following image: Rocket with angle of attack

The red line is the path the rocket shall follow from P1 to P2.
The green vector is the thrust.
The blue line is the line that the rocket shall point so as to have an angle of attach named 'a' (orange color) which will let the rocket follow the red line.

For example I can extract the following equation:

$$ \frac{P2_x - P1_x}{P2_y - P1_y} = \frac{engine_x - drag_x}{engine_y - weight - drag_y} $$

The left fraction contains the coordinates of the 2 points.
The right fraction contains all the forces that act upon the rocket.

For example $engine_x$ is the component of the engine's power in the horizontal axis. And so on.

I need one more equation in order to solve them together and find $engine_x$ and $engine_y$.

If this equation is wrong then correct me. I just need those equations.

For example maybe the second equation can be like:

$$ engine ^ 2 = engine_x ^ 2 + engine_y ^ 2 $$

Because I know the power of the engine, then I can compute $engine_x$ and $engine_y$ and the angle of attack might then be their arctangent.

NOTE: Let's say I also know $drag_x$, $drag_y$, the $weight$ and of course the points' location.

$\endgroup$
  • $\begingroup$ It's a rocket at low speed. $\endgroup$ – thanopi57 Aug 1 '18 at 21:42
  • $\begingroup$ you also need the angle of the gravity vector $\endgroup$ – JCRM Aug 1 '18 at 21:51
  • $\begingroup$ Yes, the image is only demonstrating the thrust, not the other forces indeed. But I can simulate those also. $\endgroup$ – thanopi57 Aug 1 '18 at 21:58
  • $\begingroup$ $Engine_x = cos(i)Engine$ and $Engine_y = sin(i)Engine$ where i is the angle from the horizontal and you are correct, $engine=\sqrt{engine_x^2+engine_y^2}$ $\endgroup$ – JCRM Aug 2 '18 at 10:20
2
$\begingroup$

Consider a spherical rocket (so that drag does not depend on angle of attack). Then, what you want is that the force on the rocket perpendicular to the red line is zero. So you can compute the component of gravity in that direction (drag will always act along the red line) and choose a to make the component of thrust perpendicular to the red line equal and opposite to the component of gravity.

For a realistically shaped rocket travelling at any significant speed, the drag will depend heavily on the angle of attack, in both strength and direction, so you will need to take that into account.

So let $\theta$ be the angle between your red line and the horizontal. The gravitational force perpendicular to the line is then $Mg\cos(\theta)$ ($M$ the mass of the rocket). The trust in the other direction perpendicular to that line is $T\sin(a)$ ($T$ the thrust of the engine) so your equation is $$a = \sin^{-1}\left(Mg\cos(\theta)\over T\right)$$

$\endgroup$
  • $\begingroup$ Can you provide me with equations? (I updated my question). If so, I will accept your answer. For now, I upvoted it. $\endgroup$ – thanopi57 Aug 2 '18 at 6:53
  • $\begingroup$ this doesn't account for lift, $\endgroup$ – JCRM Aug 2 '18 at 7:26
  • $\begingroup$ It doesn't account for lift, but I got the point! Maybe then the component of the lift is simply added (or substracted) to the numerator of the inner fraction. $\endgroup$ – thanopi57 Aug 2 '18 at 7:32
  • $\begingroup$ Lift will surely depend on the angle of attack, though, so allowing for it is not really possible in this approbation $\endgroup$ – Steve Linton Aug 2 '18 at 7:51
  • $\begingroup$ Im guessing I forgot lift in my calculations. I will take a look at it. What is generating the lift though? The rocket does not have pitch-fins (supposedly). Can I integrate the lift to the engine's power somehow? $\endgroup$ – thanopi57 Aug 2 '18 at 8:00
2
$\begingroup$

lift on the rocket is dependant on angle a and velocity and the vehicle design. calculating it is non-trivial across the range of subsonic, transonic, supersonic and hypersonic regimes.

$g$ force of gravity
$M$ mass of rocket
$T$ thrust of rocket
$l$ calculated lift
$w$ component of the force due to gravity perpendicular to $\tiny{\overrightarrow{P1 P2}}$

The force due to gravity is $Mg$, but you only need the component of that acting perpendicular to the direction of travel.

$w = \sqrt{\frac{Mg}{(P2_y-P1_y)^2\left((P2_x-P1_x)^2+(P2_y-P1_y)^2\right)}}$

Subtract this from your calculated value for lift (lift works upwards, gravity downwards) to get the net force causing you to deviate from $\tiny{\overrightarrow{P1 P2}}$. This gives you the force your engine needs to exert to remain on course.

$a = arcsin\left(\frac{l-w}{T}\right)$

The problem here, unfortunately, is as $a$ changes, so the calculated for $l$ changes, so it's very difficult to work it out. If $\tiny\left(\frac{l-w}{T}\right)$ is ever greater than 1, this means the force you need to keep the rocket on course is greater than its engine can provide, so you aren't going to reach your destination.

$\endgroup$

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.