# Rocket Calculations, wind resistance, Rectangular hyperbola

1. Oct 10, 2009

### 3physqs

The Basis of this question is that:
* Rockets launched at an angle follow the path of a rectangular hyperbola when thrust greater than their mass is produced.
* That rockets fall in the path of a parabola when thrust is no longer produced, this only applies when the rocket has both x and y inertia, however wind resistance skews this fall

given a rocket that produced a constant thrust of T, was launched at a angle of theta, and burnt for a time period of M, and any other variables, is it possible to calculate where the rocket will land?

2. Oct 12, 2009

### 3physqs

The rocket flight is illustrated using MathGV. Come one Can anyone answer ?

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3. Oct 12, 2009

### Mentallic

The solution to this question is probably way out of my scope (and yes, I'm nearly certain you can express the range of the rocket in terms of the variables given), but I would like to give it a shot.

Before I begin, I'm unsure about one thing so I need clarification on that: When the rocket is launched at an angle theta, the thrust is obviously being produced at that angle too. However, as the rocket moves through its path of projection and is affected by gravity, does the thrust angle change? Maybe the angle the rocket is pointing at is always changing to be tangential to its instantaneous velocity vector?
I hope I'm being clear.

4. Oct 12, 2009

### 3physqs

Hi Metallic,
Due to the rockets center of gravity not being where the thrust is being produced, gravity will affect the direction that the rocket is travelling, however as the rocket becomes faster and faster, the change in direction per distance is lessening, so therefore i deduced that the flight of the rocket must follow a graph like a east west hyperbola, or x to the power of a fractional index.
my working is that:
picture 1
converting the launch angle to a gradient
tanθ = y1/1
therefore tanθ = y1
rise over run tanθ/1 = tanθ
finding the power of X
facts we know:
to the power of a fractional index
as a guess (1 - gravitational constant/acceleration of rocket). because the power will never go below 0, if the gravitational constant of the rocket is greater than the acceleration of the rocket it will not launch

Guessed equation

f(x)=tanθ(x)$$_{\frac{gravitational constant}{acceleration of rocket}}$$

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5. Oct 12, 2009

### 3physqs

if the equation is right:
f(x)=tanθ(x) ^ gravitational constant/ acceleration of rocket. this equation does not factor wind resistance. how could this be added ?

Given this equation how can one work out the parabola that if formed from the decent when the rocket is no longer firing, if i have one point on the parabola which is the vertex, and x and y forces at that point, taking into account wind resistance?

thanks to whoever replies

6. Oct 12, 2009

### 3physqs

sorry guys
f(x)=tanθ(x) ^ 1- gravitational constant/ acceleration of rocket

7. Oct 12, 2009

### 3physqs

i have done some research and figured that the equation of a parabola can be calculated with 3 points on the parabola, however i only have one point and the x and y forces.
anyway i have made an excel which with three points you can calculate the equation of a parabola, it is attached
this is for the stage where the rocket is no longer generating thrust
the theory is that if there were three points infinitesimally close, calculated using the vertex and the changing x force and the changing y force, the reason they should be close is because of the effect of air resistance on the calculation,
for anyone who can figure the first section of the problem
when the second root of the parabola is generated this should be where the rocket hits the ground, apart from the effect of air resistance

so the question should be modified to state,
how can air resistance be taken into account into the first and second part the problem given that air resistance is a squared relationship,
i am not familiar with vector mathematics or complex numbers :( but i am sure that is the key to solving this problem. any engineers?

once this problem is solved i plan to test the accuracy of the results with model rockets, given they produce constant thrust

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• ###### Parabola Calculator.xls
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8. Oct 12, 2009

### 3physqs

using all the ideas i had and using your own knowledge could anyone solve the following question showing how they did it?

in a 2 dimensional model (x,y)
a rocket is launched on a flat plain, at a angle of 45 degrees with the ground, there is no change in air pressure as it rises the starting air pressure being 101.33 kpa, the rocket fires for 10 seconds producing a constant thrust of 50 newtons, the rocket has a mass of one kg and this does not change as the combustion of the rocket proceeds, taking into account air resistance, what is the x value of the landing point? how many km does the rocket travel?, what is the highest point the rocket reaches?
thank you to whoever replies

9. Oct 12, 2009

### Mentallic

What's really hindering any progress for me to find a solution (although I can only offer an approximate solution at best given all the variables) is the difficulty posed by the changing angle of thrust. I was going to assume the centre of gravity is in the centre of the rocket, but this doesn't help tell me how quickly it "flattens out" with respect to the horizontal.
It would depend on many factors and frankly until we figure this out, giving a good approximate solution will be quite daunting..

10. Oct 12, 2009

### 3physqs

Hi Metallic,
Would it be Possible to find the angle of thrust using the derivative of my proposed equation ?

it is possible to determine the original equation by substituting enough infinitesimally close points?

11. Oct 12, 2009

### Mentallic

Yes it would be possible to find the angle of thrust by taking the derivative of your equation. But is your equation anywhere close to the real deal?
If it were true, then we would have:
let the gravity constant be on Earth = g
let the acceleration of the rocket = a

So $$y=tan\theta.x^{\frac{1}{k}}$$

Actually, the derivative would be quite complicated. Only because $\theta$ is a variable and not a constant.

Just an idea:
If we remove all other variables including gravity for a second and watch the path of the rocket in its flight while providing thrust, assuming it's nose still dips towards the horizontal as its centre of gravity (even though gravity isn't present) balances out the rocket, then the fractional index power would be a good guess.

So its height above the ground at any time h, and displacement along the horizontal x is given by $$h=c.x^\frac{1}{k}$$ where c and k are constants

Then $$\frac{dh}{dx}=\frac{c}{k}.x^{\frac{1}{k}-1}$$

But $$\frac{dh}{dx}$$ is the gradient and the direction the rocket is pointed at which is $$tan\theta$$

So $$tan\theta=c.x^{\frac{1}{k}-1}$$

Now another problem. Once we bring gravity back into this, the fractional index guess won't work because gravity will force the rocket to be pushed further down as the rocket straightens out. This wouldn't follow such a simple path.

Real life problems become big problems :tongue2: maybe the physics section will help you out more.

12. Oct 12, 2009

thanks,