StrongEagle
21st Sep 2009, 08:03
Hello Folks,
While doing a Google search of ‘thrust versus horsepower’ I came upon this PPRuNe article dating back to 2005: http://www.pprune.org/tech-log/194018-pounds-thrust-horsepower.html. This article, along with others I read provided valuable information into the entire subject but missed the point of why this question is asked in the first place (although the original poster did ask for information for the ‘ice cream lickers’).
Ice cream lickers (I am one) want to know how much POWER it will take to push that big machine up to the point it will take off because, when expressed in this manner, the ‘power’ of an airplane can be compared to the power of a top fuel dragster in the quarter mile or the quarter mile/top speed times of the latest Ferrari, or maybe the power required to drive the largest container ships in the world. I don't think many are concerned about the power required to keep the plane at 33,000 ft.
So, I think this question is not answered by trying to relate thrust directly (which as the article above and several others imply, is essentially a fruitless endeavor) but rather to approach it from a simple perspective. If it were the wheels that drove the airplane down the runway, how much power would be required to accelerate to the point of Vr?
To answer this question, I went to the Boeing website and pulled up some info on the 747-400. I also did a bunch of web searches to develop this back of the envelope approach to answering the power question.
I came up with the following.
MTOW – 875,000 lbs (396,890 kg) for the non ER version and for the purposes of this calculation, I’ll take the max weight.
Time from V0 to Vr – Varies from reference to reference but around 45 seconds, as little as 20 seconds for an empty airplane into a good headwind.
Estimated acceleration – Not much out there but I found 7.3 ft/sec^2 (2.23 m/sec^2).
Take off roll distance (not F.A.R. runway length requirements) – Estimates of 6000 ft (1829 m) to 7000 ft (2134m), admittedly all SWAG’s, and hence part of the reason for this post.
Speed at Vr – 178 mph to 184 mph = 261 ft/sec to 270 ft/sec.
Engine thrust – 60,000 lbs to 63,000 lbs, irrelevant for this computation.
Assumption – Acceleration will be uniform from V0 to Vr – Probably not exactly correct since parasitic drag increases with speed, somewhat offset by more thrust once the engine is moving faster than V0, but probably OK for this back of the envelope exercise. Also, there should not be any induced drag until Vr when the weight of the airplane transfers to the wings.
Computation
The first step is to see how much variation there is in the estimates above since they are all related and to see if they properly relate to one thing known with fair certainty: Vr.
Acceleration and time, solving for velocity: V = a * t
Vr = 7.3 * 45 = 328.5 ft/sec = 224 mph (hmmm – What! It aint right on Google???)
But, if I use the 7.3 number and compute for time I get
t = Vr / a = 270 / 7.3 = 37 secs
I’ve got a range of Vr speeds and 2 take off roll distances so I see what this means in terms of acceleration.
Velocity and distance and solving for acceleration: V^2 = 2a * X
a = V^2 / 2X = (4 possible combinations as below)
a = 261^ 2/ 2 * 6000 = 5.7 ft/sec^2
a = 270^ 2/ 2 * 6000 = 6.1 ft/sec^2
a = 261^ 2/ 2 * 7000 = 4.9 ft/sec^2
a = 270^ 2/ 2 * 7000 = 5.2 ft/sec^2
Let’s see what Vr looks like with the highpoint of these acceleration numbers and the 45 seconds estimate found on the web
Vr = a * t = 6.1 * 45 = 274.5 ft/sec = 187 mph – a bit too fast but after playing around I find that if I make the take off distance I get
a = 270^ 2/ 2 * 6100 = 6.0 ft/sec^2
and
Vr = a * t = 6.0 * 45 = 270 ft/sec = 184 mph – on the money!
For fun, I took the 37 seconds and 7.3 ft/sec^2 computed above and used it to compute what the rollout distance must be to achieve a Vr of 184 mph (270 ft/sec).
X = Vr^2 / 2*a = 270^2 / 2*7.3 = 4993 ft take off length
There are lots of possible combinations of acceleration, time, and distance that provide the correct Vr – for simplicity’s sake, I’ll use just these two.
Computing Horsepower.
F = M * a
But an 875,000 lb airplane has an English units mass of
875,000 lbs / (32 ft / sec^2) = 27,344 slugs (why am I doing this in English units?)
For both examples:
F = 6.0 * 27,344 = 164064 lbs
F = 7.3 * 27,344 = 199 611 lbs
Power = (F * X) / t = (164, 064 * 6100) / 45 = 22, 239,787 ft-lbs / sec
1 HP = 550 ft-lbs / sec so this is
40, 436 HP or
Power = (F * X) / t = (199, 611 * 4993) / 37 = 26,936,695 ft-lbs / sec
1 HP = 550 ft-lbs / sec so this is
48,976 HP – a pretty large variation!
Conclusion
I realize this computation is completely unrelated to thrust mechanics, engine shaft speed and torque, etc, but it does provide a ‘feel’, relative to other ‘understandable’ engine sizes, of what it takes to get a large airplane into the air.
I’d appreciate any insight into the variables used to make the computations because there was always something missing. For example, the 45 second roll out time as for a 747-400 but made no reference to weight, altitude, headwinds, temperature, etc. Similarly, I have no idea if a 6100 (or 4993) foot rollout distance is reasonable for a standard day at max weight. For that matter would 184 mph (160 kts) really be the Vr for a fully loaded aircraft?
Many thanks.
While doing a Google search of ‘thrust versus horsepower’ I came upon this PPRuNe article dating back to 2005: http://www.pprune.org/tech-log/194018-pounds-thrust-horsepower.html. This article, along with others I read provided valuable information into the entire subject but missed the point of why this question is asked in the first place (although the original poster did ask for information for the ‘ice cream lickers’).
Ice cream lickers (I am one) want to know how much POWER it will take to push that big machine up to the point it will take off because, when expressed in this manner, the ‘power’ of an airplane can be compared to the power of a top fuel dragster in the quarter mile or the quarter mile/top speed times of the latest Ferrari, or maybe the power required to drive the largest container ships in the world. I don't think many are concerned about the power required to keep the plane at 33,000 ft.
So, I think this question is not answered by trying to relate thrust directly (which as the article above and several others imply, is essentially a fruitless endeavor) but rather to approach it from a simple perspective. If it were the wheels that drove the airplane down the runway, how much power would be required to accelerate to the point of Vr?
To answer this question, I went to the Boeing website and pulled up some info on the 747-400. I also did a bunch of web searches to develop this back of the envelope approach to answering the power question.
I came up with the following.
MTOW – 875,000 lbs (396,890 kg) for the non ER version and for the purposes of this calculation, I’ll take the max weight.
Time from V0 to Vr – Varies from reference to reference but around 45 seconds, as little as 20 seconds for an empty airplane into a good headwind.
Estimated acceleration – Not much out there but I found 7.3 ft/sec^2 (2.23 m/sec^2).
Take off roll distance (not F.A.R. runway length requirements) – Estimates of 6000 ft (1829 m) to 7000 ft (2134m), admittedly all SWAG’s, and hence part of the reason for this post.
Speed at Vr – 178 mph to 184 mph = 261 ft/sec to 270 ft/sec.
Engine thrust – 60,000 lbs to 63,000 lbs, irrelevant for this computation.
Assumption – Acceleration will be uniform from V0 to Vr – Probably not exactly correct since parasitic drag increases with speed, somewhat offset by more thrust once the engine is moving faster than V0, but probably OK for this back of the envelope exercise. Also, there should not be any induced drag until Vr when the weight of the airplane transfers to the wings.
Computation
The first step is to see how much variation there is in the estimates above since they are all related and to see if they properly relate to one thing known with fair certainty: Vr.
Acceleration and time, solving for velocity: V = a * t
Vr = 7.3 * 45 = 328.5 ft/sec = 224 mph (hmmm – What! It aint right on Google???)
But, if I use the 7.3 number and compute for time I get
t = Vr / a = 270 / 7.3 = 37 secs
I’ve got a range of Vr speeds and 2 take off roll distances so I see what this means in terms of acceleration.
Velocity and distance and solving for acceleration: V^2 = 2a * X
a = V^2 / 2X = (4 possible combinations as below)
a = 261^ 2/ 2 * 6000 = 5.7 ft/sec^2
a = 270^ 2/ 2 * 6000 = 6.1 ft/sec^2
a = 261^ 2/ 2 * 7000 = 4.9 ft/sec^2
a = 270^ 2/ 2 * 7000 = 5.2 ft/sec^2
Let’s see what Vr looks like with the highpoint of these acceleration numbers and the 45 seconds estimate found on the web
Vr = a * t = 6.1 * 45 = 274.5 ft/sec = 187 mph – a bit too fast but after playing around I find that if I make the take off distance I get
a = 270^ 2/ 2 * 6100 = 6.0 ft/sec^2
and
Vr = a * t = 6.0 * 45 = 270 ft/sec = 184 mph – on the money!
For fun, I took the 37 seconds and 7.3 ft/sec^2 computed above and used it to compute what the rollout distance must be to achieve a Vr of 184 mph (270 ft/sec).
X = Vr^2 / 2*a = 270^2 / 2*7.3 = 4993 ft take off length
There are lots of possible combinations of acceleration, time, and distance that provide the correct Vr – for simplicity’s sake, I’ll use just these two.
Computing Horsepower.
F = M * a
But an 875,000 lb airplane has an English units mass of
875,000 lbs / (32 ft / sec^2) = 27,344 slugs (why am I doing this in English units?)
For both examples:
F = 6.0 * 27,344 = 164064 lbs
F = 7.3 * 27,344 = 199 611 lbs
Power = (F * X) / t = (164, 064 * 6100) / 45 = 22, 239,787 ft-lbs / sec
1 HP = 550 ft-lbs / sec so this is
40, 436 HP or
Power = (F * X) / t = (199, 611 * 4993) / 37 = 26,936,695 ft-lbs / sec
1 HP = 550 ft-lbs / sec so this is
48,976 HP – a pretty large variation!
Conclusion
I realize this computation is completely unrelated to thrust mechanics, engine shaft speed and torque, etc, but it does provide a ‘feel’, relative to other ‘understandable’ engine sizes, of what it takes to get a large airplane into the air.
I’d appreciate any insight into the variables used to make the computations because there was always something missing. For example, the 45 second roll out time as for a 747-400 but made no reference to weight, altitude, headwinds, temperature, etc. Similarly, I have no idea if a 6100 (or 4993) foot rollout distance is reasonable for a standard day at max weight. For that matter would 184 mph (160 kts) really be the Vr for a fully loaded aircraft?
Many thanks.