Pounds thrust to horsepower
Do a Hover - it avoids G
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Back to the beginning
How do you put across to the general public a feel for the ‘power’ (NB the quotes) that the jet pilot has under the control of his (or her) left hand at brake release?
Not easy without using a comparison. A possible comparison is Michael Schumacher (they may not know about Alonso yet) uses 900 horsepower to get his X ton car to 100 mph in Y seconds. This pilot has a T ton aeroplane that gets to 100 mph in S seconds – if M Schumacher’s car weighed that much it would need H Horsepower to do the same thing.
Replace the letters with your own relevant numbers on the basis that the same HP per ton will give the same acceleration – ignoring drag differences.
How do you put across to the general public a feel for the ‘power’ (NB the quotes) that the jet pilot has under the control of his (or her) left hand at brake release?
Not easy without using a comparison. A possible comparison is Michael Schumacher (they may not know about Alonso yet) uses 900 horsepower to get his X ton car to 100 mph in Y seconds. This pilot has a T ton aeroplane that gets to 100 mph in S seconds – if M Schumacher’s car weighed that much it would need H Horsepower to do the same thing.
Replace the letters with your own relevant numbers on the basis that the same HP per ton will give the same acceleration – ignoring drag differences.
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"a stopped aircraft is producing no power"
Old Smokey could perhaps come up with a formula as to the effect upon mother earth (change of rotational velocity) of say a jumbo at T/O power on the brakes. Effect is of course dependent on orientation (heading) and latitude I imagine.
Blue Skies,
Brian
Old Smokey could perhaps come up with a formula as to the effect upon mother earth (change of rotational velocity) of say a jumbo at T/O power on the brakes. Effect is of course dependent on orientation (heading) and latitude I imagine.
Blue Skies,
Brian
Do a Hover - it avoids G
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Mirv
I have done a bit of Googling and it apears that 1530 BHP per ton is needed to give an F1 car an initial 1 g accel before drag rears its head.
Always assuming you are happy to accept that there is some merit in considering BHP per ton as one way of visualising what makes things accelerate these F1 numbers enable us to consider your mate on the brakes.
If your man has 33000 LB of thrust and a T/W of say 0.8, you could reasonably say he is about to charge down the runway with the equivalent power of 25 F1 motor cars.
The 'about to' is for all the reasons that have been made clear earlier.
If his thrust weight ratio is 1 then the 25 becomes 31.
This is based on an F1 car having 900 BHP.
Should you want to use some other car then divide 900 by the BHP of your other car and multiply the 25 or 31 by this number.
Any help?
JF
I have done a bit of Googling and it apears that 1530 BHP per ton is needed to give an F1 car an initial 1 g accel before drag rears its head.
Always assuming you are happy to accept that there is some merit in considering BHP per ton as one way of visualising what makes things accelerate these F1 numbers enable us to consider your mate on the brakes.
If your man has 33000 LB of thrust and a T/W of say 0.8, you could reasonably say he is about to charge down the runway with the equivalent power of 25 F1 motor cars.
The 'about to' is for all the reasons that have been made clear earlier.
If his thrust weight ratio is 1 then the 25 becomes 31.
This is based on an F1 car having 900 BHP.
Should you want to use some other car then divide 900 by the BHP of your other car and multiply the 25 or 31 by this number.
Any help?
JF
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But hovering, while it's a very useful characteristic, should not be confused with doing work in physical terms. While it may carry an impressive payload, a hovering machine has zero velocity and is thus doing zero work - i.e. zero HP generated to the payload.
Consider a turbine in a power plant, Since it isn't physically moving, or translating, by your definition, it's doing zero work
In the case of a stationary jet engine...
The basic equation is..
Work done = Force x Distance
Power = work/time
so
Power = Force x Distance/Time
Distance/Time = Velocity so
Power = Force x Velocity
I think the force can be calculated from the pressure after the turbine x the area of the turbine. Velocity is the velocity of the exhaust.
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If your man has 33000 LB of thrust and a T/W of say 0.8, you could reasonably say he is about to charge down the runway with the equivalent power of 25 F1 motor cars.
will have lost around 15% of its thrust by
the time you get to Vr...
Cheers,
M
Lots of confusion here!
Stationary on the runway, brakes on and max thrust, is zero power because no work is being done on the aircraft. You could get the same effect by attaching a rope to the nose of the aircraft going out horizontally, then to a pulley, and to a 20 tonne weight suspended just above the ground. Aircraft still feels the same thrust - no power.
I am currently converting from 737 to A320. Instead of saying "thrust set", I now have to say "power set". Of course, I asked my instructor how much power we had when stationary, and even suggested that when the N1s were stable I should say "power increasing with speed". I blame the French.
To go right back to the original question, I think a good answer would be the bhp of an engine that would drive a good ducted prop to produce the thrust in question.
Stationary on the runway, brakes on and max thrust, is zero power because no work is being done on the aircraft. You could get the same effect by attaching a rope to the nose of the aircraft going out horizontally, then to a pulley, and to a 20 tonne weight suspended just above the ground. Aircraft still feels the same thrust - no power.
I am currently converting from 737 to A320. Instead of saying "thrust set", I now have to say "power set". Of course, I asked my instructor how much power we had when stationary, and even suggested that when the N1s were stable I should say "power increasing with speed". I blame the French.
To go right back to the original question, I think a good answer would be the bhp of an engine that would drive a good ducted prop to produce the thrust in question.
Perhaps it would make the semantics easier to deal with if the work done were thought of as USEFUL work done. Certainly the work done by the helicopter engine is converted into useful work by it's rotor if it accelerates enough air to allow it to hover and prevent an acceleration toward the earth's core at a rate of 32' per second per second. Likewise for any work done by a machine that can be directed or converted to useful purpose such as in the electrical power generation (energy conversion) plant.
In the case of thrust horsepower (a term of somewhat limited usefulness itself) calculation for a jet airplane, thrust required to sustain a given flight speed is simply multiplied by the flight speed (converted to the appropriate units of measurement) to arrive at a thrust horsepower as per the basic equation: Power = force X distance/time. Let's say two identical airplanes producing equal engine thrust of 20,000 lbs are compared using the same formula. One airplane is configured for landing while the other is clean. Both airplanes will stabilize at the speed where total drag is equal to thrust applied. One airplane might be doing 150 kts while the other does 300 kts. With the same thrust applied in both cases, the thrust horsepower figure is about double for the "clean" higher speed airplane. You can see that thrust horsepower is a rather poor form of of comparison for the purposes the thread originator has in mind. Perhaps it would be far more descriptive to solve for equivalent shaft horsepower in presenting relative comparisons between jet engines and the more familiar auto engines to the non-aviation public. Unfortunately, I cannot presently locate my powerplant theory books to find the applicable formula at the moment. IIRC, the idea is to find the additional shaft HP needed to duplicate the value of the jet pipe thrust with additional shaft ouput. This amount is then added to the known shaft horsepower output of the engine as measured on the shaft and referred to as ESHP. Torque can be measured directly on some TP and TS engines but require factory test data for TF engines because torque is not read directly since it would be of little or no operational value. As Smokey points out, for jets, it's all about the thrust!
On the subject of a conversion factor for HP of an engine to lbs of thrust produced by a propeller or rotor driven by that engine, that too seems to be somewhat more complex a problem than it might appear at first glance. The relationship of power input to propeller or rotor thrust output is dependant on many factors that extend beyond the depth of my studies, but a few things seem apparent when considering the following:
By making some rough estimations, typical GA trainer engine/prop combos appear to produce no more than about 3 lbs static thrust/HP. At Reno last year, the Texan II military turboprop trainer had a placard claiming 3,500 lbs ST on about 1,000 HP. Carter Copter claim 1,140 lbs. ST on 240 HP for it's propeller. The UH-60A would hover at a TOW of 20,000 lbs at about 90% torque from it's two 1,543 SHP engines. With some of that torque being used to drive the tail rotor, It seems safe to say that It's rotor had to produce at least 8 lbs ST/SHP. Larger and slower turning props and rotors appear to convert a greater proportion of input power into thrust output. Perhaps there are reduced losses from slippage, spanwise flow and compressability effects when the delta V across the disc is reduced and the area of the disc increased. I would very much like to read some comments from someone well-schooled in the black art of propeller or rotor design!
Best regards,
Westhawk
In the case of thrust horsepower (a term of somewhat limited usefulness itself) calculation for a jet airplane, thrust required to sustain a given flight speed is simply multiplied by the flight speed (converted to the appropriate units of measurement) to arrive at a thrust horsepower as per the basic equation: Power = force X distance/time. Let's say two identical airplanes producing equal engine thrust of 20,000 lbs are compared using the same formula. One airplane is configured for landing while the other is clean. Both airplanes will stabilize at the speed where total drag is equal to thrust applied. One airplane might be doing 150 kts while the other does 300 kts. With the same thrust applied in both cases, the thrust horsepower figure is about double for the "clean" higher speed airplane. You can see that thrust horsepower is a rather poor form of of comparison for the purposes the thread originator has in mind. Perhaps it would be far more descriptive to solve for equivalent shaft horsepower in presenting relative comparisons between jet engines and the more familiar auto engines to the non-aviation public. Unfortunately, I cannot presently locate my powerplant theory books to find the applicable formula at the moment. IIRC, the idea is to find the additional shaft HP needed to duplicate the value of the jet pipe thrust with additional shaft ouput. This amount is then added to the known shaft horsepower output of the engine as measured on the shaft and referred to as ESHP. Torque can be measured directly on some TP and TS engines but require factory test data for TF engines because torque is not read directly since it would be of little or no operational value. As Smokey points out, for jets, it's all about the thrust!
On the subject of a conversion factor for HP of an engine to lbs of thrust produced by a propeller or rotor driven by that engine, that too seems to be somewhat more complex a problem than it might appear at first glance. The relationship of power input to propeller or rotor thrust output is dependant on many factors that extend beyond the depth of my studies, but a few things seem apparent when considering the following:
By making some rough estimations, typical GA trainer engine/prop combos appear to produce no more than about 3 lbs static thrust/HP. At Reno last year, the Texan II military turboprop trainer had a placard claiming 3,500 lbs ST on about 1,000 HP. Carter Copter claim 1,140 lbs. ST on 240 HP for it's propeller. The UH-60A would hover at a TOW of 20,000 lbs at about 90% torque from it's two 1,543 SHP engines. With some of that torque being used to drive the tail rotor, It seems safe to say that It's rotor had to produce at least 8 lbs ST/SHP. Larger and slower turning props and rotors appear to convert a greater proportion of input power into thrust output. Perhaps there are reduced losses from slippage, spanwise flow and compressability effects when the delta V across the disc is reduced and the area of the disc increased. I would very much like to read some comments from someone well-schooled in the black art of propeller or rotor design!
Best regards,
Westhawk
