| Topic: |
Science > Physics |
| User: |
"mikecoon" |
| Date: |
10 Aug 2003 06:44:51 AM |
| Object: |
hovering horsepower |
Lets say a rocket produces just enough thrust to hover above the
ground at 50 feet. How do you calculate the amount to power needed for
ther rocket of mass M to hover at 50 feet?
now i know that no work is being done, yet, the rocket is obviously
expending energy to hover. any thoughts?
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| User: "Gregory L. Hansen" |
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| Title: Re: hovering horsepower |
10 Aug 2003 08:06:54 AM |
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In article <b38a4794.0308100344.26b281fa@posting.google.com>,
mikecoon <mikecoon@prodigy.net> wrote:
Lets say a rocket produces just enough thrust to hover above the
ground at 50 feet. How do you calculate the amount to power needed for
ther rocket of mass M to hover at 50 feet?
now i know that no work is being done, yet, the rocket is obviously
expending energy to hover. any thoughts?
Force
F = dp/dt = (dm/dt) v_e
dm/dt is rate that mass is ejected (e.g. kilograms per second), v_e is
exhaust velocity.
Kinetic energy
K = 1/2 m v^2
Power is energy per unit time,
P = K/t = 1/2 dm/dt v_e^2
Use force equation to eliminate dm/dt,
P = 1/2 F v_e
Or to eliminate v_e,
P = 1/2 F^2 / (dm/dt)
This assumes zero temperature and a parallel flow, but with the
inefficiencies you'll get about the same curve scaled up. The more mass
you throw down, the less energy it takes to stay up there. That's why
airliners have big turbofan engines. Combat planes have turbojet engines
without the fans because they need kick-***** performance, but they suck up
the fuel as if they didn't have to pay for it.
--
"A good plan executed right now is far better than a perfect plan
executed next week."
-Gen. George S. Patton
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| User: "Ian Stirling" |
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| Title: Re: hovering horsepower |
10 Aug 2003 11:06:30 AM |
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Gregory L. Hansen <glhansen@steel.ucs.indiana.edu> wrote:
<snip>
This assumes zero temperature and a parallel flow, but with the
inefficiencies you'll get about the same curve scaled up. The more mass
you throw down, the less energy it takes to stay up there. That's why
airliners have big turbofan engines. Combat planes have turbojet engines
That's right, but not as far as it goes.
The wings are even more important than the engine.
Considering an airliner tooling along at M0.8 or so, the engines may be
pushing out air at around M1.?.
But, the greatest mass of air is moved by the wings.
At a L/D of around 20, this will be around 15m/s, and is a hell of a lot
more air moved than just the engines.
without the fans because they need kick-***** performance, but they suck up
the fuel as if they didn't have to pay for it.
As most current combat aircraft have a need for supersonic flight, then
they really need the exhaust to come out as fast as possible, in order
to keep them up there.
--
http://inquisitor.i.am/ | mailto:inquisitor@i.am | Ian Stirling.
---------------------------+-------------------------+--------------------------
If God hadn't intended us to eat animals,
He wouldn't have made them out of MEAT! - John Cleese
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| User: "Old Man" |
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| Title: Re: hovering horsepower |
11 Aug 2003 06:49:57 PM |
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mikecoon <mikecoon@prodigy.net> wrote in message
news:b38a4794.0308100344.26b281fa@posting.google.com...
Lets say a rocket produces just enough thrust to hover above the
ground at 50 feet. How do you calculate the amount to power needed for
ther rocket of mass M to hover at 50 feet?
now i know that no work is being done, yet, the rocket is obviously
expending energy to hover. any thoughts?
The same power that is required at 1 meter or 10 meters. For exhaust
velocity, v_e, rocket mass, m, and acceleration of gravity, g:
Energy E = (1/2) m*v_e^2
Power P = dE / dt = (1/2) (dm / dt) v_e^2
Force F = mg = dp / dt = d(mv) / dt = ( dm / dt ) v_e
Power = m*g*v_e / 2 independent of altitude.
[Old Man]
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| User: "Old Man" |
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| Title: Re: hovering horsepower |
11 Aug 2003 06:49:57 PM |
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mikecoon <mikecoon@prodigy.net> wrote in message
news:b38a4794.0308100344.26b281fa@posting.google.com...
Lets say a rocket produces just enough thrust to hover above the
ground at 50 feet. How do you calculate the amount to power needed for
ther rocket of mass M to hover at 50 feet?
now i know that no work is being done, yet, the rocket is obviously
expending energy to hover. any thoughts?
The same power that is required at 1 meter or 10 meters. For exhaust
velocity, v_e, rocket mass, m, and acceleration of gravity, g:
Energy E = (1/2) m*v_e^2
Power P = dE / dt = (1/2) (dm / dt) v_e^2
Force F = mg = dp / dt = d(mv) / dt = ( dm / dt ) v_e
Power = m*g*v_e / 2 independent of altitude.
[Old Man]
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| User: "Ian Stirling" |
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| Title: Re: hovering horsepower |
10 Aug 2003 07:33:09 AM |
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mikecoon <mikecoon@prodigy.net> wrote:
Lets say a rocket produces just enough thrust to hover above the
ground at 50 feet. How do you calculate the amount to power needed for
ther rocket of mass M to hover at 50 feet?
now i know that no work is being done, yet, the rocket is obviously
expending energy to hover. any thoughts?
It depends on how much mass you are moving.
For each kilo, you need to supply ~10N in order to make it hover.
If you are creating this force by throwing mass downwards, at typical
rocket exhaust speeds (3Km/s), you need 3g/second of propellant, and
0.5*3g*3000m/s^2 ^2 = 13500W.
Decrease the speed a hundredfold, you increase the amount of propellant
by a hundred to 300g/s, and the power goes down to 0.5*300g*30m/s^2 ^2 =
135W.
This is fairly close to what helicopters do.
Decrease the speed again to 3m/s, and you get into the range of gliders,
at 13W/Kg/s.
--
http://inquisitor.i.am/ | mailto:inquisitor@i.am | Ian Stirling.
---------------------------+-------------------------+--------------------------
If it can't be expressed in figures, it is not science, it is opinion.
-- Robert A Heinlein.
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| User: "Sam Wormley" |
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| Title: Re: hovering horsepower |
10 Aug 2003 12:15:16 PM |
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mikecoon wrote:
Lets say a rocket produces just enough thrust to hover above the
ground at 50 feet. How do you calculate the amount to power needed for
ther rocket of mass M to hover at 50 feet?
now i know that no work is being done, yet, the rocket is obviously
expending energy to hover. any thoughts?
Once at 50 ft, the net force is zero, i.e., the opposing forces must
be equal and opposite--you can figure it out from there.
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| User: "mikecoon" |
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| Title: Re: hovering horsepower |
11 Aug 2003 02:47:20 PM |
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thanks to everyone.
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| User: "Bill Vajk" |
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| Title: Re: hovering horsepower |
10 Aug 2003 08:44:42 AM |
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mikecoon wrote:
Lets say a rocket produces just enough thrust to hover above the
ground at 50 feet. How do you calculate the amount to power needed for
ther rocket of mass M to hover at 50 feet?
now i know that no work is being done, yet, the rocket is obviously
expending energy to hover. any thoughts?
Hovering requires that thrust = weight (idealized
for no wind conditions, natch.)
The moment you think about "how much does it weigh"
you immediately realize that it will not continue to
weight the same as you burn off fuel. So the third
thought has to do with a control system, either
shaped propellant or a thrust control system.
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