HP vs Torque vs Fuel Efficiency

[niky]

The prop thrust decreases as speed increases because the propeller is pushing against the air, and can't generate as much thrust at higher air speeds because it has to push harder to generate more thrust.

Much the same as with a driven wheel, except in this case, the variable pitch takes the place of gear ratios. Only thing similar for cars would be a CVT.

[mort]

Originally Posted by niky

The prop thrust decreases as speed increases because the propeller is pushing against the air, and can't generate as much thrust at higher air speeds because it has to push harder to generate more thrust.

Hi niky,
Hardly. For some propeller shapes it is almost the opposite: at some high propeller speed the air over the blade starts to stall. Lift, in this case in the forward direction, falls suddenly, but l/d doesn't disintegrate. The sudden drop in lift is accompanied by a drop in drag, so the engine revs higher (if not well governed) moving the prop further into stall. Above the critical tip speed the propeller fails like it was slipping on the drive shaft. - like the engine was suddenly unloaded. The prop is going faster, but loading the engine less.

A simple explanation of propeller theory.
Or you could have checked out the original source - from www.epi-eng.com

To find the speed which can be reached with a known engine HP, prop efficiency and airframe drag (thrust = drag in steady state level flight):

KTAS = ( HP * eff * 326 ) / Drag

It is clear from the relationship between power, thrust and speed, that if power and propeller efficiency are held constant, then propeller thrust decreases as true airspeed increases.

In case you were wondering, the hump in those curves is due to the fact that at low airspeeds, prop efficiency is very low. As airspeed increases, so does efficiency, quickly at first, then more slowly, up to it's max (about 85-87%).

And:

...the prop efficiency begins to decrease dramatically when the prop is operated at a helical tip velocity in excess of 0.85 Mach. That occurs because the local air velocity over the surface of the prop (near the point of maximum airfoil thickness) will reach Mach 1, and create a shock wave, separating the flow and dissipating prop energy.

At constant power thrust falls with increasing speed is basic physics. You won't be taken seriously about anything if you continue to deny the definition of power = speed times force.

from wiki.answers.com

Power is the time rate of change of energy, dE/dt= power= Force x velocity=Fv.

from wikidot.com

Mathematically we define work as:

W = F · s (1)

from physicsclassroom.com


from wikipedia

P(t) = F · v

For constant thrust power increases with speed.
It is plain that F = P/s: For constant power, as speed increases thrust will fall.
-mort

[sendler]

This has been very fun so I guess that makes me a serious Geek. I would much rather be spending my time studying physics than watching a movie even though my only formal training is a very rusty NYS High School algebra based physics class from 30+ years ago. I really appreciate the time and effort of everyone who is reading and or contributing to the discussion.
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The physics we need to understand are really very simple. Reciprocating engines make Torque and when looked at by multiplying that by the Revolutions (How many "Torques") in a certain amount of Time, you get the Power of the engine.
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Rear wheel Torque is Force on the road. But it is interesting to note that the highest rear wheel Torque will occur, ie with a constantly variable transmission, which is set to the Power peak of the engine. Not the Torque peak. Due to the fact that it will be using more torque multiplication (lower gear) at the higher rpm of the power peak.
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Force (from the rear wheel Torque) times the Distance the bike moved is Work.
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Power is the measure of Work vs Time. If you do the same amount of Work in a shorter Time, you have used more Power.
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P=W/T
Work=Force*Distance so
P=F*D/T and D/T is Velocity so
P=F*V
Given the same Power from the engine, The Force at the rear wheel must be less as the Speed between the road and the bike increases.
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F=M*A
Acceleration = Force/ Mass
F, and therefore A, must be less and less as the speed difference between the driven wheel and the road, or the airplane propeller and the air, is increasing.
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So this is the "Captain Obvious" over complicated way of saying the bike will pull harder in 1st than in 6th.
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It is fascinating to note also that the Propeller in air follows the same theory. It pulls harder at 100 mph than it does at 240 mph. Note also the the efficiency for this prop at 100 mph is only 64% versus 85% at 240 so the slope of the thrust drop off would be even steeper if real world efficiency were left out.

[sendler]

Now for the rocket engine. I'm having a tough time with this and I must admit that I intuitively think that for vehicles propelled by ejected mass, rather than a driven member that is coupled to a stationary medium like the ground or air, or water, that the speed that is valid for the calculation of Force to the frame of reference of the passenger, is the speed difference between the gases and the nozzle. Which might simply indicate that the term constant Power doesn't apply to such engines which will have fairly constant force at the nozzle. I'm having a hard time finding anything to contradict this. I was hoping to find some model rocketry pages with measured thrust data to compare to the well documented bench testing of the various engines which could show if speed comes into play.
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ThrustCurve Hobby Rocket Motor Data
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The efficiency of a rocket engine changes, starting at near 0% on lift off because the only thing moving is the ejected matter. And increases to the best efficiency when the speed of the exhaust gas referenced to the nozzle is equal and opposite to the speed of the rocket referenced to the Inertial Frame of Reference. The rocket is flying away from the ejected mass at mach 10 or whatever and leaving the mass which is exiting the nozzle at mach 10, essentially motionless in space behind it. And a rocket engine can continue to supply force to a vehicle that is traveling faster than the exhaust speed at the nozzle. But with decreasing efficiency as now the exhaust will have speed and momentum of it's own left over and in the same direction as the vehicle.
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Rocket engine - Wikipedia, the free encyclopedia
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Rocket - Wikipedia, the free encyclopedia

[sendler]

Here is the last piece we needed tying Power to Energy as proven by DieselMaxPower and mentioned way back when by Mort:
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"I'll add some info, even though you've basically got it. So a simplified equation for your last post is

Power = M*A*V

It took several relationships, and if you actually did it through torque you would have several factors of 2*pi and the wheel radius to carry around and eventually cancel. Now I know most people don't know calculus, so you will have to trust me on this:

E = 1/2 M V^2
d(E)/dt = M V (d(V)/dt)

d(E)/dt is power, and d(V)/dt is acceleration. So check this out
P = M V A

You get the EXACT same equation using calculus. And acceleration is:

A = P/(MV),

so as you go faster, your acceleration decreases (assuming constant P). This doesn't really add too much to your current understanding, but if you want to more advanced stuff it will really help. For instance, we may want to take into account the force it takes to spin the wheels. This gets so ridiculously tedious looking at forces. Instead, add it to the energy equation

E = 1/2 M*V^2 + 1/2 I omega^2
d(E)/dt = M*V*(d(V)/dt) + I omega (d(omega)/dt)

I is the mass moment of inertia and omega is the angular velocity of the wheel, and now d(omega)/dt is the angular acceleration of the wheel. In the simplest case (tires aren't slipping on the road, or slip by a constant percentage), d(omega)/dt is proportional to A:

P = M*V*A + I*omega*c*A
P = A*(M*V+I*omega*c)
A = P/(M*V+I*omega*c)
A = P/(M*V+I*V*d)

where c and d are constants. We can change omega to V by the same argument that we could change d(omega)/dt to A. So acceleration decreases with increased velocity (assuming constant power), but now we get a more realistic estimate. Some of the power produced goes into spinning the wheels.

Rockets get into a whole new mess, which I would actually say makes more sense by looking at momentum instead of energy. If you're interested, the Reynolds transport theorem for momentum should give you some good information. Most simple problems don't need calculus, even though the theorem in its purest form needs it. As a bonus, looking at model rockets is one of the most common simple problems."

[niky]

Power = F * V = MAV?

But the engine isn't putting out power. The engine's ability to cause acceleration is determined by the force at the wheels. Not power at the wheels. (Try looking for a chart... any chart... that shows horsepower at the wheels in different gears).

If force at the wheels is constant, then acceleration is constant, or as constant as can be dependent upon slippage, drag, etcetra. Constant application of power equals constant acceleration

Acceleration is directly linked to wheel torque, which gets lower at higher speeds because of lower gear multiplication at higher gears, which makes more acceleration at the engine result in less and less acceleration at the wheels. Not because of the speed of the vehicle itself.

[sendler]

C*V*T = Obvious

[niky]

C being? Or are you describing the change of velocity over time, which is delta-V over T? Which is acceleration?

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The acceleration tapers off (in the absence of drivetrain losses / frictional losses / air resistance and whatnot) not because of veloctiy, but because the time required to apply the force increases due to the lowering gear ratio. If the force could be applied directly to acceleration without having to match the speed of the road, then there's no such problem. It's in relation to the wheel speed that we start to lose acceleration. As wheel speed increases, we apply less and less force over time.

Remember: Power in relation the acceleration of the vehicle itself describes what is happening because of the application of Force. It is NOT what the engine is putting out.

Some further reading:

http://forums.xkcd.com/viewtopic.php?f=18&t=42952

http://www.daviddarling.info/encyclopedia/T/thrust.html

[niky]

Actually, looking back at it all, I think we are just approaching this from two different angles.

Power being the application of force through the wheels DOES go down as velocity goes up... I was describing it as force over time, which, yes, the more velocity, the less force. My bad.

But gears still matter because while gearing can never let you apply more force to the road than is theoretically available, improper gearing will certainly reduce it.

[sendler]

Continuously variable transmission - Wikipedia, the free encyclopedia
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A continuously variable transmission makes the concept of combining the equations for power and energy very obvious.
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Power = Mass * Acceleration * Velocity
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When you grab full throttle with a CVT, the transmission is controlled to change down to a lower ratio and wind the engine up to the power peak rpm. Where the rpm stays for the whole run until you let off the throttle. The CVT then continuously varies the gear ratio longer and longer to keep the same engine rpm while the vehicle Velocity is increasing from the Acceleration.
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The the throttle is held fully open by the rider and the engine will be kept at the same rpm by the transmission so the power from the engine is the same throughout the run even though the Velocity is increasing. The transmission is trading torque multiplication for speed as it moves up through it's range of gearing by squeezing the belt up in the front to make the engine pulley bigger and the wheel pulley smaller. The Power at the wheel is the same the whole time but there is less and less rear wheel Torque as Velocity increases.
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P = M*A*V
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Power stayed the same. And Mass stayed the same. So as the Velocity increases, the Acceleration and the Velocity must trade. One goes down if the other one goes up. Because the Energy that is added by the engine Power over time is linear. 1, 2, 3, 4. Going back to the Energy equation
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E = 1/2M*V^2
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Energy would have to increase exponentially to keep the same Acceleration as Velocity increases. 1, 4, 9, 16.