View Full Version : Pounds thrust to horsepower

13th Oct 2005, 06:56
Hello all,

can any of the gurus help...I am writing a commentary for a jet handling display, but need to convert lbs thrust into horsepower so the ice cream lickers can relate to it. Something along the lines of..."FLTLT Farnarkle's aircraft has 35,000lbs of thrust, this is XXXXhp, or the equivalent of about 147 Ford Lasers" etc etc

Thanks in advance for any help received.


Old Smokey
13th Oct 2005, 09:11

The conversion from Thrust to Horsepower is not a straight conversion of units, such as Kilometres per Hour to Miles per Hour, it depends upon the Thrust and the speed of the aircraft, as derived from the basic relationship where -

Power = Force X Velocity.

As a straight conversion to eliminate the constants of the various units used, the following formula applies -

Pa = Ta V / 325, where -

Pa = Propulsive Power available in Horse Power, Ta = Thrust available in Pounds, and V = Velocity in Knots.

In the example that you quoted, FLTLT Farnarkle's aircraft, whilst producing 35,000lbs of thrust, is producing absolutely ZERO Power whilst holding on the brakes, but 53846 HP when moving at 500 Knots (TAS).

Your Ford Laser equivalent made me remember one of my son's "young science" books where it quoted the Concorde as producing the equivalent power of 55000 Mini Minors. Has the Ford Laser replaced the Mini Minor as the layman's unit of equivalent power? :O

If you get the Thrust of that, it will add more Power to your commentary. :ok:


Old Smokey (who has owned a Mini Minor and a Ford Laser)

13th Oct 2005, 10:11
Old Smokey,

thanks very much for taking the time. I appreciated that the relationship was not linear; I was after someone who could explain a complex problem in simple terms, which is exactly what you have done. Thankyou very much!

In my world the Ford Laser has replaced EVERYTHING as the yardstick of equivalent power! (But only if it is off-blue, has two hubcaps and is rusty).



Sven Sixtoo
13th Oct 2005, 17:45
But I'd love to see the gearing needed to get anything moving at 500 kts while attached to 53846 horses.


13th Oct 2005, 20:51
Smokey, you wrote

"In the example that you quoted, FLTLT Farnarkle's aircraft, whilst producing 35,000lbs of thrust, is producing absolutely ZERO Power whilst holding on the brakes,...."

I've always had trouble with similar statements. Would I be correct in suggesting that you mean that while certainly producing thrust, no power is being used ** to accelerate ** the aircraft. After all, there is added kinetic energy (of the air accelerated backwards), and heat energy (pardon the poor term) of the air and perhaps tires/brakes, hence the presence of power produced by the aircraft.

13th Oct 2005, 22:53
Somewhere in the dim dark ages i recall something along the lines of 2 to 3 lb/thrust per gg..with a average of 2.5...

14th Oct 2005, 13:21
Not very scientific I know, but the "Mythbusters" programme on Discovery tried the myth about the car with the JATO rocket flying into the hillside, and they worked on the basis of 1lb thrust = 2hp, although I'd imagine there will be differences between US and UK calculations.

Mark 1
14th Oct 2005, 13:27
As has been said above, direct comparisons are not simple.

An RB211 for instance, may be quoted with a static thrust of 50,000 lbs; this is the force going through the pylon, but it isn't doing any work.

In flight, the thrust reduces maybe down to 20-30,000lbs, but it is now working and generating power.

A similar engine used in a generating plant or a marine engine would produce around 30,000 SHP. This is similar to the amount of power absorbed by the fan to produce take-off thrust.

These numbers are all from vague memory, but it might give you a rough idea

Old Smokey
15th Oct 2005, 05:09
hawk37, This is one where we can really get down to semantics. The Power Vs Thrust arguments on these forums go on endlessly, when someone can show me a diagram of the forces affecting flight (Thrust, Drag, Lift, and Weight) that includes Power, I'll surrender. For the jet aircraft (which this topic is all about), Power is only looked at as an abstract in examining Rate of Climb, at all other times direct reference to Thrust applies.

You did mention "added kinetic energy (of the air accelerated backwards), and heat energy (pardon the poor term) of the air and perhaps tires/brakes, hence the presence of power produced by the aircraft", well the air accelerated backwards is what produced the Thrust, which the aircraft requires. With respect to Noise, heat generated etc., note again the formula that I posted refers to Ta, the "a" implies AVAILABLE after all other losses.

Mark 1, as Power is the rate of doing Work, could your statement "but it is now working and generating power" be re-written, by substitution, as "but it is now working and generating a rate of doing work".

We have been raised in the mechanical era where we have an early association with Power in motor vehicles. Motor vehicles need Power to generate Torque to turn the wheels, Propeller aeroplanes need Power to produce Thrust via their propellers, so do ships. Jet aircraft skip the transformation process by directly producing thrust.

And why do ships and aeroplanes convert their Power to Thrust? - because they need Thrust, not Power, to function.

If these words can't convince you, then allow me to finish with a direct unmodified quote from the Performance Engineering text book of the world's second oldest airline (QANTAS), maybe KLM (the oldest) sees it differently -

"The output of the jet engine (including fan jet) is thrust. There is no place to measure or observe horsepower. Power is transmitted by shafts from turbines to the compressors (and to fan where employed); incidentally the energy absorbed by the compressors far exceeds the useful output of the engine. The useful output is thrust (and that's what we want), so it is useless from any technical viewpoint to go backwards to figure out horsepower".


Old Smokey

15th Oct 2005, 12:41
Smokey, there is a place to measure shaft HP in a high-bypass fanjet that makes some sort of sense, but you'd have to be privy to some internal figures to work the sums.

Recall that one HP is 33000 ft-lb of work per minute. In rotational terms, think of a winch with a one-foot radius. One revolution of the winch pulls the rope 6.28 feet, and if the load is 33000 lb., the torque is 33000 lb-feet.

So (example 1) - if the shaft turns 1.00 rpm, with a torque of 33000 lb-ft, the amout of work done in a minute is 33000 x 6.28 ft-lbs = 6.28 HP.

Scaling back this particular problem to example 2, If the shaft turns 1.00 rpm, with a load of 33000/6.28 lb on the rope, the torque is 5252 lb-ft, and the work rate is 1.00 hp.

ERGO: Shaft HP = RPM x Torque (lb-ft) / 5252 (stick that in your personal QRH!)

SO: if you know the torque on the fan shaft, and you know the fan rpm (derived from the % gage), you CAN calculate the horsepower driving the fan. And at static conditions, it's ROUGHLY the same number as the static thrust (lbs.) of the engine.

15th Oct 2005, 16:23

Some thoughts...

It would seem that a basic assumption has been made by most posters and this ignores the fact that work is being carried out by the engine, even when the aircraft is stationary.

Let’s break down the stages: -

1) Aircraft stationary, engine stopped.... no work being carried out, therefore no power. (Everyone agree?)

2) Aircraft stationary, engine started (idle)... work being carried out by engine accelerating cold stationary air in front of engine, to warmer, flowing air at rear of engine (i.e mass flow through engine) The power (horsepower /kilowatts etc) produced by the turbine is absorbed in a number of ways.

To simplify let us ignore the thermal effects and identify the main power source and two main power "sinks" for the turbine: -

Source: - Combustion produces hot gases which expand towards turbine section. Therefore, for a given mass flow rate, there is a given source of energy which we can call S1, which can be converted from mass flow to mechanical energy)

a) Power is used to drive the turbine/compressor - this is not useful work as far as the aircraft is concerned when it comes to thrust and producing acceleration, but nevertheless cannot be ignored. (Call it T1)

b) Power is used to drive the fan/turbine - if the fan was 100% efficient (assume) then the work per second (power) absorbed by the fan will be completely converted to thrust. (Ignoring flow losses etc) Call this T2.

Therefore remaining energy in gas stream exiting engine, S2 = S1 - T1 - T2.

There are two sources of thrust from the engine: -

1) The fan (T2)
2) Jet efflux from the engine core (S2)

Ignoring frictional losses within the engine (both frictional losses and transfer losses) the energy contained by these two gas streams S2 and T2 would be equal to S1-T1 i.e. power from combustion less power required to drive compressor. Also assume S1-T1 is a constant, C.

Therefore, maximum energy available for thrust is C = S2 + T2

Now hopefully, the crux…..

1) Maximum energy would be obtained when the engine is running at 100% (i.e max fuel flow rate, producing lots of hot expanding gas to drive turbine), giving C max.
2) And, maximum power (work/second) is when the gas flow must be accelerated from zero to max in engine and max exhaust gas flow is exhausting to a stationary body of gas.

These two conditions will be met when take-off power is set and the aircraft is stationary.

Point two above also explains why thrust reduces as altitude / temp increase.

Mind you, all this doesn’t help convert lbs thrust to shp!!!



I my attempt to explain this I may have muddied the waters some. Unfortunately, I know what I mean, but on re-reading my post I feel I have not expressed myself particularly well.

15th Oct 2005, 17:51

There is no way to convert thrust to horsepower.
It might be some approx tumb rules that apply in some cases, but thats all.


Is meassured in Newton.
It is defined by Newtons 2nd law -

F = m * a
F = kg * m/s^2 = N = Newton

1N = 0,102 kp = 0,225 lbf


Is meassured in Watt -
energy pr second

P = E/t
P = J/s (joules/sec) = W = Watt

1 hk = 0,7355 kW = 75 kpm/h = 632 kcal/h
1 hp = 0,7457 kW

As you can see, these have nothing in common.
One can offcourse meassure some brake power
on a jet engine shaft, but this is sort of pointless
since the shaft isn't pushing the acf forward, the
MASS accelerating through the engine is... .



16th Oct 2005, 17:58

My only contention was with the statement that a stopped aircraft is producing no power. If this were so, then gas turbines at powerplants would be producing no power either. Shuttlebus has brought this up.

16th Oct 2005, 18:10
I have a model plane that weighs 2Kg and will climb vertically forever on about 2 bhp. So (ignoring drag) this works out about 2.2 Lbs of thrust per HP... but I think that's a very conservative estimate.

How much power does a helicopter need to hover a certain weight?

17th Oct 2005, 01:32
How much power does a helicopter need to hover a certain weight?

That's a question a good helo perf guy/gal can readily answer, I'm sure, and is one measure of the vehicle efficiency.

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.

So it's a case of zero efficiency in physical terms. Whatever horsepower is used to hover the machine, there's zero horsepower output.

17th Oct 2005, 02:35

Hope this helps,

Convert Thrust to Horsepower (http://www.aerospaceweb.org/question/propulsion/q0195.shtml)


CY&B :ugh:

17th Oct 2005, 03:41
Barit, something about your logic seems flawed.

Consider a turbine in a power plant. Since it isn't physically moving, or translating, by your definition, it's doing zero work. However, we know there is all that electrical energy being generated. Take the power output of the plant, multiply by the time, and you'll get the work done in producing that power. Actual value will be a bit more, due to inefficiencies.

Old Smokey
17th Oct 2005, 04:09
My only contention was with the statement that a stopped aircraft is producing no power.
In any analysis of Power, you have to be very careful to consider whether you are looking at the engine alone, or the engine/airframe combination. From the point of view of aircraft performance, we are only interested in the engine/airframe combination, and Ta (Thrust AVAILABLE after all other losses).

If we examine the engine standing alone, for a stationary aircraft with Takeoff Engine Power / Thrust applied, the total Power produced within the engine is enormous. This includes the internally required power for engine operation (far in excess of that which is useful for a turbine engine), noise and heat generation, operation of accessories such as generators, hydraulic pumps etc. Thus, the engine is internally producing significant power, and a lot of Thrust, whether you are speaking of the Jet, Turbo-Prop, or Piston-Prop. Maximum Thrust is also being produced, but, as the aircraft is not moving, the engine/airframe combination is producing no Power, in other words, performance is Zero. The same can be said for the hovering helicopter.

This is all alluded to in the quote that I made from the QANTAS P/E manual, and from the excellent link provided by CaptYanknBank.

Aircraft performance is all about Thrust AVAILABLE (Ta), not the power produced within the engine/s viewed in isolation to the engine/airframe combination.


Old Smokey

17th Oct 2005, 12:34
Hawk, I understand your dilemma. In the case of power generation, there IS output power from the turbine, and it's rotational work - i.e. rpm with torque.

The rpm will be a constant based on the power grid frequency of the country; that's typically 3000 rpm in a 50 hz. country.

The torque of the generator drive shaft will be directly proportional to the load current leaving the generator.

And shaft horsepower is the product of rpm and torque. Look at my response to Old Smokey a few days ago for the math.

OH YES - and the useful power from the generator output is the product of voltage x current!

Old Smokey
17th Oct 2005, 13:28
Sorry barit1, intermittent responses from me on this thread between flights, which kind of developed into a conversation with hawk37, Sorry that I did not see your response, good reasoning, I think we agree in most respects :O

Regards, and sorry for the apparent rudeness.

Old Smokey

17th Oct 2005, 14:46
Smokey, e-mail me a San Miguel and we'll be all square... ;)

Old Smokey
17th Oct 2005, 17:33
Working on it barit1, just gotta work out how to squeeze that San Mig bottle into the scanner and it's on the way :ok:


Old Smokey

John Farley
17th Oct 2005, 19:32
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.

Brian Abraham
18th Oct 2005, 07:58
"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,

John Farley
18th Oct 2005, 14:16

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?


18th Oct 2005, 14:35
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.

Yup kow that... but the OP asked about thrust and horsepower. I mentioned a heli because the thrust will be equal to the weight in the hover - making the numbers easier to obtain.

Consider a turbine in a power plant, Since it isn't physically moving, or translating, by your definition, it's doing zero work

No that's incorrect. It doesn't have to move to do work - just moving something else is enough.

In the case of a stationary jet engine...

The basic equation is..

Work done = Force x Distance

Power = work/time


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.

18th Oct 2005, 15:32
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.

Maybe - also, don't forget that a turbofan engine
will have lost around 15% of its thrust by
the time you get to Vr... :p



18th Oct 2005, 19:36
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.

19th Oct 2005, 00:36
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,


19th Oct 2005, 07:02
I find GOOGLE works just fine.

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