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Speaker
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Calculating the trajectory of rockets
« on: Oct 2nd, 2003, 12:15am » |
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I need to find a quick way to calculate the trajectory of a missile. I should say quick and easy. Here is the situation.
We have built water rockets. Water rockets are made from 2 liter pet bottles. The bottle is turned upside down and a nozzle is screwed onto its mouth. Then, fins are attached for guidance and a nose cone to increase aerodynamics as well as to provide a place to put the parachute.
The rocket is then filled with about 2/3 liter of water. Then it is placed on a stand that holds it vertical to the ground, or almost vertical. The base of the stand mates with the nozzle and provides a way to pump in air, which is done with a standard bicycle pump. After about 30 pumps a triggering mechanism similar to the cable for a handbrake on a bicycle is used to release the nozzle. When this happens the pressurized air and water are quickly ejected and the rocket shoots up into the air.
We can measure the distance they travel by pacing it off, or comparing it to the size of the nearby football field. However, the height is only conjecture.
How can I measure how high the rocket goes?
There are two scenarios:
1. The parachute fails to open and the rocket travels in a smooth arc until it impacts.
2. The parachute opens as the rocket passes the heighest point in its path, and the rocket falls almost straight down.
When we do this we are all outstanding in the field... of rocketry 
So the only tools we would normally have are paper and pencil, maybe a calculator, maybe a protractor to measure the angle of flight...
So, how can we measure the height of the rockets?
Thanks for your help.
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Sir Col
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Re: Calculating the trajectory of rockets
« Reply #1 on: Oct 2nd, 2003, 1:35am » |
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This is a very complex problem, Speaker.
I'm sure Aero_Guy will be more of an authority, but I can help with part of this...
It is dependent on a number of variables that change throughout the ascent. For simplicity we shall assume 'perfect' conditions. Initially, the weight of the rocket (including water) and pressure would need to be known. As the valve is released, the rocket will accelerate due to the sudden release of pressure. As the rocket ascends, the pressure will reduce at a rapid rate, so too will the weight of the rocket (as the water evacuates). Once the pressure and contents are spent, the rocket will continue upwards with the momentum gained, but as gravity acts against it, it will decelerate to a speed of zero (maximum height), before descending to earth.
The whole ascent needs, therefore, to be modelled on two parts: (i) propulsion phase, (ii) continued (unassisted ascent) phase.
Phase (ii) is fairly easy to calculate using standard Newtonian mechanics. Using v2=u2+2as, where v is final velocity (which will be zero at maximum height), u is initial velocity (after propulsion phase ends), a is acceleration (in this case, acceleration due to gravity: -g), and s is distance travelled. So we get 0=u2–2gs; that is s=u2/2g.
I don't know how to calculate the final velocity nor the height at the end of phase (i).
I did find a page containing some links to 'water rocket enthusiasts' websites, though:
http://www.sumrallworks.com/rockets/main.php3?area=links&link_type=r ank
I'm interested to know how phase (i) is calculated, and look forward to a lesson from some of our resident professors.
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Speaker
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Re: Calculating the trajectory of rockets
« Reply #2 on: Oct 2nd, 2003, 1:42am » |
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Thanks Sir Col.
I will try to apply what you have given me, and will check out the site.
After trying to find a way to figure this out, I have determined that if I ever do figure it out, I want to make a table that lists some of the expected variables and gives expected distances and heights.
If it is possible.
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Sir Col
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Re: Calculating the trajectory of rockets
« Reply #3 on: Oct 2nd, 2003, 3:13am » |
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I had another thought. We could apply the principle of conservation of energy to the problem. When the system is primed, there is an amount of energy in the system; let's call this, X. At the maximum height, all of this will have been converted to potential energy: mgh. When the bottle returns to the ground it will all have been converted to kinetic energy: (1/2)mv2. If we don't use a parachute, which would effectively cause an indeterminate amount energy to be lost from the system through drag, we could measure the velocity of impact. The mass of the bottle at the maximum height and when it impacts is equal, so mgh=(1/2)mv2, giving h=(1/2)v2/g.
Obviously the parachute is normally used, but you could do some experiments to link the initial variables: pressure and mass of water against height reached.
By combining the distance travelled, s, during phase (ii), the known potential energy at the end of phase (ii), we can probably work out the velocity at the start of phase (ii) – as the system will be made up of potential and kinetic energy (all the stored energy will be spent).
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towr
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Re: Calculating the trajectory of rockets
« Reply #4 on: Oct 2nd, 2003, 3:59am » |
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you can get the height (approximately) from the amount of time the thing is in the air (without a parachute).
x = 1/2 g (t/2) 2 (t/2 because the way up takes as long as the way down, and at the top the vertical velocity = 0)
where g = 9.81
You could also just wait to see how long the thing goes up, and use that time (in case there's a working parachute). It shouldn't be too hard to see when it approximately stops climing. (use a stopwatch)
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| « Last Edit: Oct 2nd, 2003, 4:04am by towr » |
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Sir Col
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Re: Calculating the trajectory of rockets
« Reply #5 on: Oct 2nd, 2003, 4:56am » |
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I've just realised a problem with our method: terminal velocity. A normal projectile, projected upwards, will follow a symmetrical path. The length of time on the way up is equal to the length of time on the way down. However, the initial propulsion phase ruins the symmetry. After the water/pressure has expired, the rocket will continue to travel up as a normal projectile with initial velocity, u, for a distance, s, in time, t, until the velocity is reduced to zero due to gravity. Then it will travel down a distance, s, in time, t, reaching a final velocity, v = u. At this moment, the rocket will be at the same height as when the propulsion system ended. After this, the rocket will continue to gather speed as it falls towards the earth. When it strikes the ground, the speed will be greater than the initial velocity when the energy propulsion expired.
What I'm not sure about is the effect of terminal velocity. If it reaches maximum velocity a significant distance above the ground, we cannot use the principle that kinetic energy upon impact = potential energy at maximum height. The drag, causing a maximum downward velocity, expends some of the energy in the system.
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otter
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Re: Calculating the trajectory of rockets
« Reply #6 on: Oct 2nd, 2003, 8:04am » |
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Try this web site for Estes model rockets: http://www.estesrockets.com and in particular a pdf file entitled "Elementary Mathematics of Model Rocket Flight" at http://www.estesrockets.com/Elem_Math.pdf, which includes a section entitled "How High Did It Go?"
I hope this helps.
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| « Last Edit: Oct 2nd, 2003, 8:18am by otter » |
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Speaker
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Re: Calculating the trajectory of rockets
« Reply #7 on: Oct 5th, 2003, 7:55pm » |
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Thanks otter. I have downloaded the pdf file and am currently studying. It is harder than I expected. However, I am fortunate in that the other people launching rockets with us are better at math than I am. So, working together we should be able to figure it out.
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Sir Col
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Re: Calculating the trajectory of rockets
« Reply #8 on: Oct 6th, 2003, 3:36am » |
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I had a read of the document, Speaker. Unfortunately, it doesn't provide any methods to model the trajectory of the rocket, rather it describes how to use a surveying technique.
It suggest you use right angle trigonometry...
Stand a suitable distance from the launch site (sufficiently far away to see the rocket reach its maximum height before it begins to fall towards earth), and measure the angle of elevation (from the horizontal to the maximum height). As you know how far from the launch site you are, and assuming the rocket travels straight upwards, the height can be estimated with, h=d tanA (make sure your calculator is in degrees, if you're measuring angle in degrees); where h is the height, d is the horizontal distance from the launch site, and A is the angle of elevation.
The document suggests, towards the end, about using a formula to model the path mathematically, however, it would require information that you do not possess; e.g. propulsion impulse in weight/time units.
The surveying technique they recommend is perhaps the best, albeit a little crude. Regardless of accurary, if applied as a measure for everyone in a contest, it would at least be consistent.
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otter
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Re: Calculating the trajectory of rockets
« Reply #9 on: Oct 6th, 2003, 6:51am » |
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on Oct 5th, 2003, 7:55pm, Speaker wrote:| Thanks otter. I have downloaded the pdf file and am currently studying. It is harder than I expected. However, I am fortunate in that the other people launching rockets with us are better at math than I am. So, working together we should be able to figure it out. |
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Speaker, quite a few years ago, I was doing model rocketry as a hobby along with my brother. We used the approach recommended by the PDF and made a rudimentary surveyor's transit using a camera tripod upon which was mounted, via a wooden structure, a protractor to measure the angle and an old .22 calibre rifle scope with crosshairs. The whole apparatus costs us about $25 and we were able to get the parts at second hand stores. The camera tripod had telescoping legs which made it easy to level and by using the panning handle and tilt head lock, we could track the apogee of the flight and lock the readings into place. We were really quite pleased with the outcome and wound up giving it to the club when we left. I wish I could find an old photo or sketch so I could send it to you.
With the advent of laser transits, the old optical ones can sometimes be had, although they are not exactly cheap. Maybe you could find one on eBay or something.
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| « Last Edit: Oct 6th, 2003, 8:40am by otter » |
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Speaker
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Re: Calculating the trajectory of rockets
« Reply #10 on: Oct 6th, 2003, 6:13pm » |
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That sounds like the best idea, the surveyors transit to measure the angle of elevation at apogee. I will bring this up at the next meeting and see about getting some people to work on it. The oldest of these people is about 18, but they are all exceedingly intelligent. So, if I can provide them with some of the equipment and the theory, they should be able to get it to work, if I just stand back and let them do the hard bits.
I also found a table with the tangents. I tried to do this by multiplying the distance from the launch point to the point of impact by the tangent. But, my calculations gave me an altitude of almost 1000 meters. Of course the distances were all from memory.
Thanks again.
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