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#1
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prop rpm question
At 1000 rpm or so, my airplane will taxi and get up to, what, 15-20
kts? But at double the rpm it will fly at 80-90 kts, though it would take a long time to take off. Surely double the rpm produces more than double the propellor thrust...or does it? Anyway, it seems very nonlinear, that is, double the rpm and I get much more than double the performance. Why is that? |
#2
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prop rpm question
"Bob Fry" wrote in message
... At 1000 rpm or so, my airplane will taxi and get up to, what, 15-20 kts? But at double the rpm it will fly at 80-90 kts, though it would take a long time to take off. Surely double the rpm produces more than double the propellor thrust...or does it? Anyway, it seems very nonlinear, that is, double the rpm and I get much more than double the performance. Why is that? The laws of physics (certainly those relating to mechanics - velocity, acceleration, thrust, drag and all that) are rarely linear. On the ground your aircraft is probably not in an optimum attitude for drag reduction, so it's probably not fair to compare it with an aircraft in the sky. And will your aircraft fly straight and level at 1000rpm? If so, how fast? The main thing dictating how your aircraft performs is drag. "Normal" drag increases with the square of the speed you fly at - so if you double the speed, you roughly quadruple the drag (hence everything has a terminal velocity when falling to earth - as you get faster, the drag increases faster than your speed increases, and you stop accelerating once drag equals the acceleration caused by gravity). Remember also that at low speeds you have induced drag, which is high at low speeds but vanishes as you get faster. The moral of the story, though, is that if you think something ought to behave in a linear manner, you're probably mistaken. Cheers, D. |
#3
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prop rpm question
Uh, thanks for trying...I guess:
DC "Bob Fry" wrote in message DC ... At 1000 rpm or so, my airplane will taxi and get up to, what, 15-20 kts? But at double the rpm it will fly at 80-90 kts, though it would take a long time to take off. Surely double the rpm produces more than double the propellor thrust...or does it? Anyway, it seems very nonlinear, that is, double the rpm and I get much more than double the performance. Why is that? DC The laws of physics (certainly those relating to mechanics - DC velocity, acceleration, thrust, drag and all that) are rarely DC linear. Eh? F=ma and many others are. DC On the ground your aircraft is probably not in an DC optimum attitude for drag reduction, so it's probably not fair DC to compare it with an aircraft in the sky. And will your DC aircraft fly straight and level at 1000rpm? If so, how fast? Kee-rist. Drag, and attitude (angle of attack, really) have little to do with the explanation I was looking for. No, of course it won't fly straight and level at 1000 rpm. That's nearly full idle landing rpm. DC The main thing dictating how your aircraft performs is DC drag. "Normal" drag increases with the square of the speed you DC fly at I think you mean parasitic drag. DC - so if you double the speed, you roughly quadruple the DC drag (hence everything has a terminal velocity when falling to DC earth - as you get faster, the drag increases faster than your DC speed increases, and you stop accelerating once drag equals DC the acceleration caused by gravity). Remember also that at low DC speeds you have induced drag, which is high at low speeds but DC vanishes as you get faster. Induced drag--drag caused by the wing producing lift at a vector not perpendicular to flight--never vanishes unless lift vanishes. Anyway I'll restate the question, plus post to r.a.'s garbage heap, r.a.piloting. At 1000 rpm the prop produces some amount of thrust (lift), call it T[1000]. This thrust is only enough to move the plane in a moderate taxi. At double that rpm, 2000 rpm, the prop produces another amount of thrust, call it T[2000]. Now I'm not positive, but it sure seems that T[2000] T[1000] Why, if rpm only doubles, does thrust (seem to) much more than double? |
#4
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prop rpm question
"Bob Fry" wrote in message ... Uh, thanks for trying...I guess: snip Anyway I'll restate the question, plus post to r.a.'s garbage heap, r.a.piloting. At 1000 rpm the prop produces some amount of thrust (lift), call it T[1000]. This thrust is only enough to move the plane in a moderate taxi. At double that rpm, 2000 rpm, the prop produces another amount of thrust, call it T[2000]. Now I'm not positive, but it sure seems that T[2000] T[1000] Why, if rpm only doubles, does thrust (seem to) much more than double? The short way around is to look at your engine's power chart and evaluate how many excess HP it is developing at 1000 rpm vs 2,000 rpm. If we assume the plane in question is a C-152, the engine is making very little power at 1,000 rpm. I'd guess 15 hp, of which at least 5 hp is spent in friction inside the engine, leaving 10 hp for thrust. At 2,000 rpm, the engine is probably making 60 hp, of which 10 is spent on internal friction. Therefore, you have 50 hp for thrust. Another way to look at it is that your prop has an advance rate. Let's say it the advance rate is 4 feet per revolution. At 1,000 rpm, and no drag on the airplane (rolling or aerodynamic), the airplane would have a terminal velocity of 4,000 fpm, or about 48 mph. Of course, there is rolling and aerodynamic drag, and there is prop drag too, so the engine can only drag the plane along at, say, 30 mph, assuming a flat smooth runway. At 2,000 rpm, with no drag, the terminal velocity would be 8,000 fpm, or about 85 mph. Of course, there is still aerodynamic and prop drag, but there is no rolling resistance, so you get more bang for your RPM buck. Of course, it helps that your engine is delivering 60 hp, as opposed to 15 hp when it turns 1,000 rpm. Yet another way to look at it is that when your prop spins at 2x the speed, it requires 4x the power to turn it... KE=1/2MV^2.. And yes, all of this stuff is ideal world, no prop efficiency losses, etc... KB KB |
#5
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prop rpm question
I wish to leave the engine out of the discussion, but let's
continue... "KB" == Kyle Boatright writes: KB If we assume the plane in question is a C-152, Close enough, it's an Aircoupe with a C90. But let's look just at the prop. Why does a prop produce so much more thrust, much more than double, when it's turned at only twice the rate? KB Another way to look at it is that your prop has an advance KB rate. Let's say it the advance rate is 4 feet per KB revolution. Yep, 48" pitch. KB At 1,000 rpm, and no drag on the airplane (rolling KB or aerodynamic), the airplane would have a terminal velocity KB of 4,000 fpm, or about 48 mph. Of course, there is rolling and KB aerodynamic drag, and there is prop drag too, so the engine KB can only drag the plane along at, say, 30 mph, assuming a flat KB smooth runway. KB At 2,000 rpm, with no drag, the terminal velocity would be KB 8,000 fpm, or about 85 mph. Hmmmm...so prop thrust is indeed only twice at double the rpm?...ideally speaking of course. The idealized (no viscosity etc.) math seems to say that it is linear, but intuitive feel says not. |
#6
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prop rpm question
On Mon, 16 Jan 2006 21:15:49 -0800, Bob Fry
wrote: At 1000 rpm or so, my airplane will taxi and get up to, what, 15-20 kts? But at double the rpm it will fly at 80-90 kts, though it would take a long time to take off. Surely double the rpm produces more than double the propellor thrust...or does it? Anyway, it seems very nonlinear, that is, double the rpm and I get much more than double the performance. Why is that? http://www.allstar.fiu.edu/aero/Propulsion2.htm I'm not an engineer, nor do I play one on TV. I do find it interesting that the speed or velocity of the aircraft is a factor in figuring thrust in both the propeller and the jet propulsion formulas. Perhaps someone else can explain in it terms that you and I can understand. TC |
#7
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prop rpm question
This won't be a complete ( or maybe even correct answer) but ...
The prop is an airfoil. Here *lift* will be *thrust*. The thrust force generated will be T = C A d/2 V^2. V is the speed of the airfoil (prop). So the force is nonlinear, it goes as the square. For the rest of the parameters: C is coeff of lift (depends on angle of attack), A is airfoil area, d= air density. The prop will have induced and parasitic drag. The relative wind angle of attack on the prop foil will depend on the aircraft speed. At low speed it is at high angle of attack. The induced drag is large and the engine can't get up to full rpm. As the aircraft speed increases, the AOA decreases, the induced drag decreases and the can rev up to higher rpm (as constnt throttle). There will be an AOA where the drag is minimum, This is where the prop is most efficient. It is a narrow range, because, as you go faster, parasitic drag on the prop kicks in. Constant speed props are used to adjust the pitch to remian efficient over a wider raneg of airspeeds. Kershner states that the maximum thrust force occurs when the plane is standing still (at a fixed throttle setting, I guess), and decreases as you go faster. I do not understand this. Is it beacese AOA is largest? I am trying to see how this relates to power. Power would be force*distance/time or force*velocity. Maybe the thrust decreases slowly with airspeed, but the power still goes up as you go faster. This is just a hand waving argument. Please, anyone who knows more, feel free to correct this picture. Dave "Bob Fry" wrote in message ... I wish to leave the engine out of the discussion, but let's continue... "KB" == Kyle Boatright writes: KB If we assume the plane in question is a C-152, Close enough, it's an Aircoupe with a C90. But let's look just at the prop. Why does a prop produce so much more thrust, much more than double, when it's turned at only twice the rate? KB Another way to look at it is that your prop has an advance KB rate. Let's say it the advance rate is 4 feet per KB revolution. Yep, 48" pitch. KB At 1,000 rpm, and no drag on the airplane (rolling KB or aerodynamic), the airplane would have a terminal velocity KB of 4,000 fpm, or about 48 mph. Of course, there is rolling and KB aerodynamic drag, and there is prop drag too, so the engine KB can only drag the plane along at, say, 30 mph, assuming a flat KB smooth runway. KB At 2,000 rpm, with no drag, the terminal velocity would be KB 8,000 fpm, or about 85 mph. Hmmmm...so prop thrust is indeed only twice at double the rpm?...ideally speaking of course. The idealized (no viscosity etc.) math seems to say that it is linear, but intuitive feel says not. |
#8
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prop rpm question
"Bob Fry" wrote in message ... I wish to leave the engine out of the discussion, but let's continue... "KB" == Kyle Boatright writes: KB If we assume the plane in question is a C-152, Close enough, it's an Aircoupe with a C90. But let's look just at the prop. Why does a prop produce so much more thrust, much more than double, when it's turned at only twice the rate? KB Another way to look at it is that your prop has an advance KB rate. Let's say it the advance rate is 4 feet per KB revolution. Yep, 48" pitch. KB At 1,000 rpm, and no drag on the airplane (rolling KB or aerodynamic), the airplane would have a terminal velocity KB of 4,000 fpm, or about 48 mph. Of course, there is rolling and KB aerodynamic drag, and there is prop drag too, so the engine KB can only drag the plane along at, say, 30 mph, assuming a flat KB smooth runway. KB At 2,000 rpm, with no drag, the terminal velocity would be KB 8,000 fpm, or about 85 mph. Hmmmm...so prop thrust is indeed only twice at double the rpm?...ideally speaking of course. The thrust is probably 4x, like the engine's power. The advance rate makes the prop want to pull the airplane at 2x the speed, but again, 2x the speed requires 4x the power... The idealized (no viscosity etc.) math seems to say that it is linear, but intuitive feel says not. |
#9
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prop rpm question
: The thrust is probably 4x, like the engine's power. The advance rate makes
: the prop want to pull the airplane at 2x the speed, but again, 2x the speed : requires 4x the power... Check your equations. Drag force goes with the square of the velocity. Drag power (force*velocity) therefore goes with the *cube* of velocity. So, if that's the governing equation, 2x the speed requires 8x the power. If you look at the operator handbooks for identical airframes with different engine options (e.g. PA28-140/150/160/180/235, PA24-180/250/260/400), you'll see that they almost exactly follow this cubic (i.e. cube-root) equation. -Cory -- ************************************************** *********************** * Cory Papenfuss * * Electrical Engineering candidate Ph.D. graduate student * * Virginia Polytechnic Institute and State University * ************************************************** *********************** |
#10
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prop rpm question
Bob Fry wrote:
Uh, thanks for trying...I guess: DC "Bob Fry" wrote in message DC ... At 1000 rpm or so, my airplane will taxi and get up to, what, 15-20 kts? But at double the rpm it will fly at 80-90 kts, though it would take a long time to take off. Surely double the rpm produces more than double the propellor thrust...or does it? Anyway, it seems very nonlinear, that is, double the rpm and I get much more than double the performance. Why is that? DC The laws of physics (certainly those relating to mechanics - DC velocity, acceleration, thrust, drag and all that) are rarely DC linear. Eh? F=ma and many others are. DC On the ground your aircraft is probably not in an DC optimum attitude for drag reduction, so it's probably not fair DC to compare it with an aircraft in the sky. And will your DC aircraft fly straight and level at 1000rpm? If so, how fast? Kee-rist. Drag, and attitude (angle of attack, really) have little to do with the explanation I was looking for. No, of course it won't fly straight and level at 1000 rpm. That's nearly full idle landing rpm. DC The main thing dictating how your aircraft performs is DC drag. "Normal" drag increases with the square of the speed you DC fly at I think you mean parasitic drag. DC - so if you double the speed, you roughly quadruple the DC drag (hence everything has a terminal velocity when falling to DC earth - as you get faster, the drag increases faster than your DC speed increases, and you stop accelerating once drag equals DC the acceleration caused by gravity). Remember also that at low DC speeds you have induced drag, which is high at low speeds but DC vanishes as you get faster. Induced drag--drag caused by the wing producing lift at a vector not perpendicular to flight--never vanishes unless lift vanishes. Anyway I'll restate the question, plus post to r.a.'s garbage heap, r.a.piloting. At 1000 rpm the prop produces some amount of thrust (lift), call it T[1000]. This thrust is only enough to move the plane in a moderate taxi. At double that rpm, 2000 rpm, the prop produces another amount of thrust, call it T[2000]. Now I'm not positive, but it sure seems that T[2000] T[1000] Why, if rpm only doubles, does thrust (seem to) much more than double? Because prop thrust increases as the square of the RPM and thus you get four times more thrust at twice the RPM. There are other factors that come into play as well such as advance ratio, but I assume you can Google... Matt |
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