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  #1  
Old 10-16-2008, 01:59 PM
tomhale tomhale is offline
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Default how much rocket power would be required to lift 10,000 lbs 3-4 feet

Hi, does anybody have an educated guess as to the amount of rocket power required to lift 10,000 lbs three to four feet off the ground. I am picturing four different rockets, one at each corner of a frame supporting this weight. Is this feasible with existing technology? Would the rockets be too large to allow for the platform to perform any other functions? Any conjectures welcome.
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Old 10-16-2008, 03:06 PM
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Quote:
Originally Posted by tomhale
Hi, does anybody have an educated guess as to the amount of rocket power required to lift 10,000 lbs three to four feet off the ground. I am picturing four different rockets, one at each corner of a frame supporting this weight. Is this feasible with existing technology? Would the rockets be too large to allow for the platform to perform any other functions? Any conjectures welcome.

Hmmmmmm.....sounds like a physics problem.........where is my physics book? I'm sure there's a formula for figuring this out........something to do with force = something......a mime is a terrible thing to waste......
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Old 10-16-2008, 03:27 PM
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About 40,000 lb/ft or 10 ton/ft

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Old 10-16-2008, 03:47 PM
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Depends. Do you want to just flip the mass up and let it drop, or do you want it to ascend in a stable manner to a height of 4 feet and then hover? If you want it to hover, then for how long? Do you want it to then descend back down to the ground in a stable, controlled manner (soft landing)? And are you, by any chance, thinking of eventually having this platform perform this maneuver someplace other than on Earth?

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Old 10-16-2008, 03:54 PM
tomhale tomhale is offline
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it would be to quickly flip the mass up and then drop and it would be on planet earth.
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Old 10-16-2008, 04:09 PM
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Doug Sams Doug Sams is offline
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Quote:
Originally Posted by tomhale
Hi, does anybody have an educated guess as to the amount of rocket power required to lift 10,000 lbs three to four feet off the ground. I am picturing four different rockets, one at each corner of a frame supporting this weight. Is this feasible with existing technology? Would the rockets be too large to allow for the platform to perform any other functions? Any conjectures welcome.
Assuming you want it to hover, four feet off the ground, you need four rocket motors, each with 2500 pounds of thrust (10,000 pounds total). The total rocket power - ie, total impulse - is determined by how long you want it to hover.

The amount of work (energy) to get the 10K pounds off the ground to 4' is Force times Distance, 10,000pounds times 4 feet = 40,000pound-feet of work. The additional energy required to hover is much more difficult to determine. The thrust (force) is easy enough (10,000 pounds) but since you're hovering and not moving, the distance is 0, and thus no additional work is being done. So a different formula would be required to caclulate the energy.

It'd probably be easier to just measure the fuel flow rate and find a corresponding work equivalent and use that to approximate the energy consumed during hover.

BTW, hovering is a great example of why rocket-powered backpacks are not common. It takes a lot of impulse to hover and thus lots of fuel, rendering such devices useful for only very short hops, on the order of 10's of seconds (~90-100 seconds, IIRC).

And that's why most rocket flights are more like drag races than duration events (oval track races). The rocket rapidly accelerates to a very high speed in a very short time, and then coasts the rest of the way, for the majority of the flight duration. This is typical whether you're flying Alphas on A motors or Saturns to the moon. Most of the time, you're coasting, and thus not burning any fuel. Hovering ain't a strong point for rockets.

Maybe a helicopter would work better for you

Doug

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Old 10-16-2008, 05:00 PM
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it would not be for hovering but would be for a quick burst to raise the weight 4 feet and then drop it.
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Old 10-16-2008, 10:03 PM
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Quote:
Originally Posted by tomhale
it would not be for hovering but would be for a quick burst to raise the weight 4 feet and then drop it.
At four feet, the potential energy (PE) would be m*g*h=10,000pounds*32ft/sec²*4'. (=4536kg*9.8m/s²*1.2192m)

If the rocket motor had an infinitessimally short motor burn and imparted a velocity to the object at ground level, it would need to have a kinetic energy (KE) equal to the potential energy at apogee. Thus, ½mv²=mgh. Rearranging, we get v²=2gh, or v=sqrt(2gh).

Impulse equals momentum, so Ft=mv. Thus, the impulse required to raise it to four feet will be Ft=10,000pounds*sqrt(2gh) = 4536kg*sqrt(2*9.8m/s²*1.2192m)=22,173Ns, which is just a little over a full N motor (20,480Ns).

But, this is the net impulse - ie, the net force after subtracting gravity. That means that the motor's thrust must be much greater than the weight of the object. IOW, you want a very fast burning motor. Something on the order of 0.1 seconds strikes me as about right.

During the 0.1 second motor burn, the force of gravity acting on the object will be subtracted from the motor's thrust. Therefore, the motor's impulse must be bolstered to compensate. The force of gravity acting on the object is 10,000 pounds or 4448 newtons. Times 0.1 seconds, this amounts to needing an additional 444.8 newton-seconds of impulse added to the motor.

Thus, the total motor setup would be 222178 newtons (221,733 + 444.8) or 49948 pounds for a duration of 0.1 seconds.

Since the motor burnout occurs at 0.1 seconds - slighty later than the 0 length time used in the initial setup - the object will already be above the ground when the max velocity is achieved, so the apogee may be a bit higher than four feet. Thus, further tweaking is required to get exactly 4 feet; some of impulse added to compensate for gravity would need to come back out. As you can see, multiple iterations may be required to get more exact numbers, but for a good first pass, the numbers above are pretty close.

HTH.

Doug

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  #9  
Old 10-16-2008, 11:17 PM
A Fish Named Wallyum A Fish Named Wallyum is offline
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Quote:
Originally Posted by Doug Sams
At four feet, the potential energy (PE) would be m*g*h=10,000pounds*32ft/sec²*4'. (=4536kg*9.8m/s²*1.2192m)

If the rocket motor had an infinitessimally short motor burn and imparted a velocity to the object at ground level, it would need to have a kinetic energy (KE) equal to the potential energy at apogee. Thus, ½mv²=mgh. Rearranging, we get v²=2gh, or v=sqrt(2gh).

Impulse equals momentum, so Ft=mv. Thus, the impulse required to raise it to four feet will be Ft=10,000pounds*sqrt(2gh) = 4536kg*sqrt(2*9.8m/s²*1.2192m)=22,173Ns, which is just a little over a full N motor (20,480Ns).

But, this is the net impulse - ie, the net force after subtracting gravity. That means that the motor's thrust must be much greater than the weight of the object. IOW, you want a very fast burning motor. Something on the order of 0.1 seconds strikes me as about right.

During the 0.1 second motor burn, the force of gravity acting on the object will be subtracted from the motor's thrust. Therefore, the motor's impulse must be bolstered to compensate. The force of gravity acting on the object is 10,000 pounds or 4448 newtons. Times 0.1 seconds, this amounts to needing an additional 444.8 newton-seconds of impulse added to the motor.

Thus, the total motor setup would be 222178 newtons (221,733 + 444.8) or 49948 pounds for a duration of 0.1 seconds.

Since the motor burnout occurs at 0.1 seconds - slighty later than the 0 length time used in the initial setup - the object will already be above the ground when the max velocity is achieved, so the apogee may be a bit higher than four feet. Thus, further tweaking is required to get exactly 4 feet; some of impulse added to compensate for gravity would need to come back out. As you can see, multiple iterations may be required to get more exact numbers, but for a good first pass, the numbers above are pretty close.

HTH.

Doug

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  #10  
Old 10-16-2008, 11:20 PM
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Hey....did you remember to do the calcs with the weight of the motors too?
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