Thrust & Horsepower

[Niall Hanley 0]

The technical definition is better explained using the SI units, so 52,000lbs thrust = 23,600 Kg (Newtons) of thrust, this means that if the engines were attached to a plane that weighed 23,600 Kg, those engines at max power would give an acceleration of 1ms^-2 [if no external forces act upon the system], however, for simplicities sake, lets just say the 737 weighs 47,200 kg, that means that the engines at max rated thrust would give the plane an acceleration of 0.5ms^-2. The we get a bit more technical, engines only give off their max rated thrust in ISA conditions, ie. Sea level, 15C, 1013.25 hpa ect... Then the bits about external factors, that would be wheel drag, an uphill slope, ect. That help?

Thanks, that explains it all, you sound like my physics teacher with your "SI Units" lol. Any way, I was always curious about pounds of thrust so thanks for that.

[Thomas_Kenobi 0]

Back to the OP question:I've found using a measure of mass to indicate thrust irritating since it first came up in high school physics, exactly because it leads to confusion such as here. I'll try too do my small part to clarify this: Warning: Math ():

• Lbs. (or kgs in the sane parts of worlds ) (symbol m) is a measure of mass.
• Weight (symbol W) on the other hand is the force exerted by the Earth (or any other large body) to another body and according to Newton's 2nd law is directly proportional to mass:
  • W=m*g

  [*]Weight can also be calculated by Newton's law of universal gravitation:

  • W = G*M*m/r², where G is the gravitational constant, M is the mass of one object (in this case Earth) and r the distance between the centres of the two

Equating the two we have g=G*M/r² Now since we consider Earth to be roughly uniform in shape and composition and assuming we are at water level, acceleration (symbol g) is approximately 9.81 metres/second². This is not entirely accurate as neither assumptions are correct but the error is small enough for the purposes of a thought experiment. So on the surface of the Earth, if we accept a small error, a specific mass will always correspond to a specific gravitational force. e.g. 1 kg of mass is pulled by the Earth with 9.81 Newtons of force Practical:For the more practical explanation now. Since 1 kg of mass is pulled by the Earth with a force of 9.81 Newtons, if you apply 9.81 Newtons of force in an opposite vector (i.e. upwards) the mass will hover (not climb! You need more force to start accelerating upwards). Those 9.81 Newtons of force are called 1 kg of thrust. Same thing for lbs, just convert lbs to kg and use the same formulas. Thus 4.45 Newtons of force is called 1 lbs of thrust. So if you strapped one of those 26k lbs rated engines to an object weighing 26k (incl. the weight of the engine itself) it would be enough to make that object hover. The plane weighs a lot more of course but it doesn't matter because the force that counteracts weight is lift and that is produced by the wings not the engines. The engines provide forward motion which is needed to create lift. As to how that happens, that's an issue of aerodynamics and another matter altogether. Hope that has helped :) @Ronan:

The technical definition is better explained using the SI units, so 52,000lbs thrust = 23,600 Kg (Newtons) of thrust, this means that if the engines were attached to a plane that weighed 23,600 Kg, those engines at max power would give an acceleration of 1ms^-2 [if no external forces act upon the system], however, for simplicities sake, lets just say the 737 weighs 47,200 kg, that means that the engines at max rated thrust would give the plane an acceleration of 0.5ms^-2.

Sorry but what you say is wrong. A 23600kg thrust would not cause a 1ms^-2 acceleration of a 23600kg object. It would in fact provide 0ms^-2. It's like having two people pulling a rope at opposite ends with an equal amount of force. The rope wouldn't go anywhere.That if said thrust were pointing up. If it were pointing horizontally as in the case of the plane then it doesn't interact with weight at all as it would be at a 90 degree angle to it (while on the ground or at 0 degrees angle of attack).

[Airbus Commander 563]

@Ronan: Sorry but what you say is wrong. A 23600kg thrust would not cause a 1ms^-2 acceleration of a 23600kg object. It would in fact provide 0ms^-2. It's like having two people pulling a rope at opposite ends with an equal amount of force. The rope wouldn't go anywhere.That if said thrust were pointing up. If it were pointing horizontally as in the case of the plane then it doesn't interact with weight at all as it would be at a 90 degree angle to it (while on the ground or at 0 degrees angle of attack).

Listen here now, admitted, it wouldn't give 1ms^-2, but that was accounted for with the "External Forces". It'd give 0.99ms^-2. It's not like "two people on a rope". Going from the equation

• F=m*a, thence a=F/m
• F=23,600N
• m=23,600kg
• 23,600/23600 = 1
• And since SI unit for a [acceleration] the acceleration is equal to 1ms^-2.

[Thomas_Kenobi 0]

Listen here now, admitted, it wouldn't give 1ms^-2, but that was accounted for with the "External Forces". It'd give 0.99ms^-2. It's not like "two people on a rope". Going from the equation

• F=m*a, thence a=F/m
• F=23,600N
• m=23,600kg
• 23,600/23600 = 1
• And since SI unit for a [acceleration] the acceleration is equal to 1ms^-2.

Ah right, now I see what you mean. We are talking about things 90 degrees apart (literally ).I was talking about thrust applied on an opposite vector to weight (i.e. vertical), in which case weight and thrust counteract each other (thus a=0ms^-2) whereas you were talking about horizontal thrust in an ideal setting with no resistance present. Confusion fixed

[Airbus Commander 563]

Ah right, now I see what you mean. We are talking about things 90 degrees apart (literally ).I was talking about thrust applied on an opposite vector to weight (i.e. vertical), in which case weight and thrust counteract each other (thus a=0ms^-2) whereas you were talking about horizontal thrust in an ideal setting with no resistance present. Confusion fixed

Now, no offence intended, but have you ever seen a 737's pointing in the vertical plane?

[Thomas_Kenobi 0]

Now, no offence intended, but have you ever seen a 737's pointing in the vertical plane?

Hmm, now there's an interesting home experiment Okay seriously now. I used the vertical plane to better dispel any confusion stemming from using a unit of mass to describe weight. From experience this is often present with this issue so I thought it best to put the two on the same axis (like a rocket) to simplify the thought experiment and make the difference (and the connection between the two) perfectly clear.

[MrKen 3]

Hmm, now there's an interesting home experiment Okay seriously now. I used the vertical plane to better dispel any confusion stemming from using a unit of mass to describe weight. From experience this is often present with this issue so I thought it best to put the two on the same axis (like a rocket) to simplify the thought experiment and make the difference (and the connection between the two) perfectly clear.

The measure of thrust is better explained on a vertical plane. Basically, 1 lb of thrust (force) is enough to maintain 0g acceleration of a 1 lb object if applied vertically.  It would hover.  2 lb of thrust would cause the 1lb object to accelerate upward at a rate of 1g or 32ft/s^2 (9.8m/s^2) That's why the F-16 (26,500lb gross weight) can accelerate going straight up.  Its GE F110-GE-132 turbofan produces 32,500 lbs of thrust.  Not all F-16's are equipped with that engine, but it's the most powerful that I've read about. On a horizontal axis, ignoring drag, rolling friction, etc... If 1lb thrust (force) is applied to a 1lb object it would accelerate at 32ft/s^2 (9.8m/s^2). For the OP, there really isn't a practical conversion from thrust (force) to horsepower (work).  HP is a measure of torque vs rpm.

[badderjet 181]

Practical:For the more practical explanation now. Since 1 kg of mass is pulled by the Earth with a force of 9.81 Newtons, if you apply 9.81 Newtons of force in an opposite vector (i.e. upwards) the mass will hover (not climb! You need more force to start accelerating upwards).

Nitpicking here, but the bold part isn't precise. It very well can climb. It however can't transition to a climb (as you said, it can't accelerate upwards). During a steady climb all forces are balanced (same as in level flight!), so there is absolutely zero surplus force in either direction, they are the same - and we're still able to climb, not only hover! Weight opposes the upward force precisely. A greater force pointing upwards will only change the vertical speed, which obvoulsy is nothing else than a vertical acceleration. Same applies to descent obviously. I think that's all Newton #1 IIRC. No force, no acceleration. That however never means you can't already be in a climb or descent. Sorry for the rant.

[aaronrash 0]

The technical definition is better explained using the SI units, so 52,000lbs thrust = 23,600 Kg (Newtons) of thrust, this means that if the engines were attached to a plane that weighed 23,600 Kg, those engines at max power would give an acceleration of 1ms^-2 [if no external forces act upon the system], however, for simplicities sake, lets just say the 737 weighs 47,200 kg, that means that the engines at max rated thrust would give the plane an acceleration of 0.5ms^-2. The we get a bit more technical, engines only give off their max rated thrust in ISA conditions, ie. Sea level, 15C, 1013.25 hpa ect... Then the bits about external factors, that would be wheel drag, an uphill slope, ect. That help?

Cleared it right up. Thanks Ronan

[Chock 17,769]

Explanations aside, the really important thing here for actually flying a jet is that ratings for engines are determined in an International Standard Atmosphere, that's the part that really matters when working out take offs and such, because that is the baseline that all your various reduced thrust take offs are going to be working from, so if you like, the lbs of thrust a jet engine generates is the 'brochure figure' for engine thrust on a fairly nice day, it could, and probably will be different from that in most everyday situations. More about that: http://en.wikipedia....dard_Atmosphere And more about how you can make use of it on a 737 if you like all that 'saving wear and tear and fuel' malarkey, not that you really have to be arsed about that in a sim: http://www.b737mrg.n...akeoffperfo.pdf Al

[PaulKent 0]

this is an interesting topic, especially for a physicist (by degree) like myself. Those explanations seem pretty good to me. Its important to stress, that an unbalanced force will cause an acceleration. Even in a steady climb there is no acceleration, so W=L and T=D. Its like circular motion at constant speed.....despite speed remaining constant there is a change of direction and hence a change in velocity, hence there must be an acceleration. if an engine produces 20,000lb of thrust then it produces 20000/2.2*9.8N of force 89,000N. Now newtons second law states that the Force is equal to the rate of change of momentum (mv) which assuming the mass is constant (reasonable) reduces to F=ma. The F of course is the unbalanced force. But of course there is friction initially between the tyres and tarmac. This is the static friction. Wikipedia states that the coefficient of static friction is 1 for dry tarmac on rubber, so the static friction on an aircraft weighing 9.1T would be F_static=c*mg=1*9100*9.8. This gives a static friction of 89000N. In this case the unbalanced force is (thrust-friction) is zero and there will be no acceleration. In space, in the absence of friction the acceleration would be Thrust/mass=89000/9100=9.8m/s2, or 1g Since an aircraft would have two engines, the thrust would be 178,000N. With the friction of 89000N the acceleration would be 89000/9100=9.8m/s2 or 1g So we can think of thrust as the force required to accelerate a mass at 1g in the absence of friction. So a 20K lb engine can move a 20k lb aircraft at an acceleration of 1g with no friction hope that helps paul

[timers15 11]

Oh what a fun topic. :) One definite fact is that if you have a 120000 lb airplane in level flight 120000 lbs worth of air is being shoved down at 1 G. The extra energy to hold the plane in the sky comes from the air collapsing in the wake of the plane. Some bizarre vortex swirls around the wings where the wings are the axis. The radius of this is the same as the wing span.This is what causes ground effect - the vortex kinda of ducts with the ground. Most importantly Newton is happy.It took me years to figure this out. I still haven't met a physics instructor who understands this. Their usual line is "Bernoulli Principle".After I invent time travel and killer cyborgs Mr. Bernoulli's getting a surprise - for no other reason than to simplify this discussion in the future. Tim Dobrowolsky.

[G550flyer 1,052]

The we get a bit more technical, engines only give off their max rated thrust in ISA conditions, ie. Sea level, 15C, 1013.25 hpa ect...

Now days engines will continue to make rated power at the higher temps. For example the CF-6-50C will adjust to keep a rated thrust of 52,500 up to 30c at sea level.The BR710 on the Gulfstream will keep a rated thrust of 14,750 up to ISA + 20(35c) at sea level.

[opherben 0]

This varies according to aircraft manufacturer requirements and has been designed from the mid sixties, e.g. CH-53D. The AH-64 and UH-60 were designed to retain SL performance to 4000 feet at 35 deg C.

[BIGSKY 317]

For a turboprop you can use:Fn = SHPx375xProp Efficiency / Speed in MPHFn = Total Net Thrust in pounds.For an estimate, roughly 80% on average for prop. efficiency. Prop Efficiency is also put into the equation in hundredths; 100% efficiency = 1.0; 80% efficiency = .80.Example with a turboprop rated at 1200 SHP with 80% efficiency at 350 MPH:FN = 1200x375x.80 / 350=1028.57 lb. of thrust