Subject: Shifting gears. Prop efficiency definition.
From: Paul
Date: 2/7/2000, 11:01 PM

Byron Brandon Ward wrote:

At 50 fps we have available out of the 120 HP 2200 RPM engine 1320  pounds of thrust
or a prop eff of 13%. (thrust in/thrust out. HP in/HP out).  Out of the 150
HP 2700 RPM constant speed engine we have 1650 pounds of thrust or a prop eff. of
32%.

Paul if this Lancair could produce 1650 lbs of thrust at 50 fps it could
climb straight up at 3000 fpm (you said it weighed 1600 lbs). Maybe the
prop efficiency is even lower at 50 fps? The prop efficiency
must be very low at low airspeeds. Otherwise, of course at 1 fps you
would have 66,000 lbs of thrust (120 hp at prop x 550 ft-lbs/s-hp)/1 fps.

Byron

Yes it is. The numbers quoted above are out of the engine and not the prop.

In theory a 150 HP engine could generate 8250 pounds of thrust at 10 fps.
That is why helicopters can go straight up. As Vance pointed out as the
disk loading gets lower the max efficiency of the prop gets higher at low speeds.
In our case the disk loading only varies between 5.5 HP per square foot
and 4.3 HP per square foot. This is with a six foot prop and 120 HP to
150 HP. A helicopter with a 30 foot dia. rotor on the other hand has a disk 
loading of only .2 HP per square foot so the prop efficiency can be much higher
and the helicopter can climb straight up with only 150 HP.

If we put longer landing gear on the Lancair and a larger prop with a higher
gear reduction we could make it climb straight up.

The fundamental thing to keep in mind at all times is the definition of
HP. HP is a force moving at a speed. If you slow the speed the force goes
up. If you increase the speed the force goes down. One HP is 550 pounds
of force moving at one foot per second. One HP is also 55 pounds of force
moving at 10 feet per second and 5.5 pounds of force moving at 100 feet per second.
  
Paul Lamar
 
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