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PaulS1950

Altitude calculations for kids...

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Is there an easy way to calculat the approximate altitude (maximum possible) if you know the weight of a model rocket and the engines total impulse that I can teach to 1st through 5th graders?

Thanks,

Paul - thinking of starting a model rocketry club for beginners through a local school.

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That's like trying to calculate how fast a car can go by the size of its fuel tank and the car's weight. I's not the size of the fuel tank, but how quickly the engine burns the fuel, which produces an amount of power, etc, and aerodynamics plays a key role, etc.

 

You'll need average thrust, burn duration and weight. Get the thrust and duration from the engine specs. It may be better to calculate the performance of the powered phase and then the unpowered phase, because that's the way it happens in real life. IT may help to use an Excel spreadsheet that calculates acceleration, velocity and distance versus time. If you can get ahold of a good model rocket book, all the better.

 

Here's a NASA page on rocket motors.

 

Here's an Estes webpage on model rocket motors.

 

I suggest working the kids through the science and math before throwing performance data at them. With the prospects of launching a real rocket, kids will tolerate lots of studying (and remember to stress safety). Kids are not stupid if given a chance to understand the material and with good examples. Use lots of real-life analogies to explain some of the more esoteric facts. This study can touch on integral calculus, area under the curve, center of mass, aerodynamics, etc. It's good for the kids who can grasp it, and it's of no consequence for those that can't.

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That's like trying to calculate how fast a car can go by the size of its fuel tank and the car's weight. I's not the size of the fuel tank, but how quickly the engine burns the fuel, which produces an amount of power, etc, and aerodynamics plays a key role, etc.

 

That's why he has the engine's total impulse, rather than the thrust. Most model rocket motors are labeled with their impulse in Newtons.

 

Now, I can't think of a direct way to do this without loads of approximation. You'd have to assume the impulse is instantaneous to make a quick solution out of kinematics, but of course that's false. So I'm not sure how you'd go about it.

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That's why he has the engine's total impulse, rather than the thrust. Most model rocket motors are labeled with their impulse in Newtons.

 

Now, I can't think of a direct way to do this without loads of approximation. You'd have to assume the impulse is instantaneous to make a quick solution out of kinematics, but of course that's false. So I'm not sure how you'd go about it.

 

actually, most of the engines come with a burn rate chart showing when Thrust reaches its peak (approx. .2 seconds) and when the thrust reaches its constant rate (approx. .4 seconds) and the aversage thrust. The total impulse is also listed. Model rockets are weighed before launch (with engine) so I have the weight in grams.

 

As for the math on the NASA page it is way too complex to teach the kids - most of them have no more than general math background. The maximum altitude figure I am looking for would be the theoretical maximum (even excluding air resistance) height that the rocket would never exceed. (or even achieve in the real world)

I have searched both the NASA site and the Estes site (at least the parts I can access) and not found what I am looking for, which is why I am asking here.

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Okay, here we go. I'm familiar with Estes motors, and they have designations such as "C6-3", where the "C" represents total impulse (such as 10 N-sec), and the "6" represents the average thrust in Newtons (6 N).

 

From there (or the actual specs from the maker), compute (or obtain) the burn duration (such as 10 N-sec / 6 N = 1.67 sec). Then compute the velocity and altitude attained (v=aT and s=½aT², using a=F/m and T = time of powered flight) during the powered phase. Compute the force F to equal the average thrust minus the average weight of the rocket (use an average motor weight = initial motor weight + ½ of propellant weight) in order to obtain the resultant force that accelerates the rocket upward. (That is, if a 4 oz rocket has a thrust of only 4 oz, it simply hovers (ie, 4 oz - 4 oz = 0))

 

Then use the height from which to drop an object (in a vacuum) to attain this velocity, which represents the ballistic (unpowered) phase, where t = time of ballistic phase. This thinking is actually the ballistic phase in reverse: If a rocket attains 100 ft/sec upward at burnout, it will then travel a distance upward retarded by gravity which, if dropped downward through that same distance, would give it 100 ft/sec downward. I think I'm correct in this assumption.

 

s=½gt² and v=gt

 

then t=v/g and so t²=v²/g²

 

so s=½g(v²/g²) = ½v²/g = ½(aT)²/g = ½a²T²/g

 

and add this ballistic height to the powered height.

 

s = ½aT² + ½a²T²/g = ½aT²(1+a/g)

 

As you said, this will be a rough calculation. (I hope my work is correct.)

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Okay, here we go. I'm familiar with Estes motors, and they have designations such as "C6-3", where the "C" represents total impulse (such as 10 N-sec), and the "6" represents the average thrust in Newtons (6 N).

 

From there (or the actual specs from the maker), compute (or obtain) the burn duration (such as 10 N-sec / 6 N = 1.67 sec). Then compute the velocity and altitude attained (v=aT and s=½aT², using a=F/m and T = time of powered flight) during the powered phase. Compute the force F to equal the average thrust minus the average weight of the rocket (use an average motor weight = initial motor weight + ½ of propellant weight) in order to obtain the resultant force that accelerates the rocket upward. (That is, if a 4 oz rocket has a thrust of only 4 oz, it simply hovers (ie, 4 oz - 4 oz = 0))

 

Then use the height from which to drop an object (in a vacuum) to attain this velocity, which represents the ballistic (unpowered) phase, where t = time of ballistic phase. This thinking is actually the ballistic phase in reverse: If a rocket attains 100 ft/sec upward at burnout, it will then travel a distance upward retarded by gravity which, if dropped downward through that same distance, would give it 100 ft/sec downward. I think I'm correct in this assumption.

 

s=½gt² and v=gt

 

then t=v/g and so t²=v²/g²

 

so s=½g(v²/g²) = ½v²/g = ½(aT)²/g = ½a²T²/g

 

and add this ballistic height to the powered height.

 

s = ½aT² + ½a²T²/g = ½aT²(1+a/g)

 

As you said, this will be a rough calculation. (I hope my work is correct.)

 

You could teach kids rocket thrust by telling them that fuel gets used up to create force. This force is also called thrust. The fuel makes the materials create an explosion that continues, usually a wood fire would make smoke, yes? If there is [combustion] you would have pressure forming and then carbon, physical or atomic [displacement] and you would have [propulsion].

 

Riddle me this then, why can't you say that the [displacement] where the rocket moves as much as it's [combustion] rate equalling a progressive [propulsion] rate?

 

If there is ten [combustion], and that makes five [propulsion], then should be a rate of two [displacement] for that fuel, if you are working with round numbers, and I can't see people working with little numbers for a sexy time...

 

So, I hope that clears that mess up! And that must be as good as it gets. Now repeat with me chilren; x [combustion] makes y [propulsion] makes z [displacement]...

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Okay, here we go. I'm familiar with Estes motors, and they have designations such as "C6-3", where the "C" represents total impulse (such as 10 N-sec), and the "6" represents the average thrust in Newtons (6 N).

 

From there (or the actual specs from the maker), compute (or obtain) the burn duration (such as 10 N-sec / 6 N = 1.67 sec). Then compute the velocity and altitude attained (v=aT and s=½aT², using a=F/m and T = time of powered flight) during the powered phase. Compute the force F to equal the average thrust minus the average weight of the rocket (use an average motor weight = initial motor weight + ½ of propellant weight) in order to obtain the resultant force that accelerates the rocket upward. (That is, if a 4 oz rocket has a thrust of only 4 oz, it simply hovers (ie, 4 oz - 4 oz = 0))

 

Then use the height from which to drop an object (in a vacuum) to attain this velocity, which represents the ballistic (unpowered) phase, where t = time of ballistic phase. This thinking is actually the ballistic phase in reverse: If a rocket attains 100 ft/sec upward at burnout, it will then travel a distance upward retarded by gravity which, if dropped downward through that same distance, would give it 100 ft/sec downward. I think I'm correct in this assumption.

 

s=½gt² and v=gt

 

then t=v/g and so t²=v²/g²

 

so s=½g(v²/g²) = ½v²/g = ½(aT)²/g = ½a²T²/g

 

and add this ballistic height to the powered height.

 

s = ½aT² + ½a²T²/g = ½aT²(1+a/g)

 

As you said, this will be a rough calculation. (I hope my work is correct.)

 

Ewmon,

Thank you very much for your help. I think it will be helpful. I may have to resolve myself to giving them the math and let those that can deal with it help teach those who don't. Sometimes kids can accomplish what we adults can't.

You have simplified this a lot more than I could on my own so you are my hero !

Thanks,

Paul

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