Drum dia and torque

"Bob Johnson" wrote in message
news:XtUnd.87541$%x.66152@okepread04...
Hi Bill --

Today we towed in light wind and so turned 2300 engine RPM in the climb.
The
454 c.i. torque/hp curves show the engine was generating about 430 lb-ft
torque and 200 hp at these revs.

Were you at wide open throttle? Those curves are for WOT and say nothing
about part throttle operation. They are also for a bare engine on a
dynomometer corrected to sea level at standard atmosphere.

Now here is where I tend to go off the rails. Just 75 hp is required to
lift
a 1100 lb sailplane 1700 ft in 45 sec.

I know there are some aerodynamic and mechanical losses but it's hard to
believe they amount to some 125 hp.

The horsepower at the winch drum is easy. It's the cable speed in FPS times
the tension in pounds divided by the constant 550. (or if you have drum
torque data, RPM x torque in ft lbs. divided by 5252) Drum torque is tension
times effective drum radius. Say, 1100 x 1.5 or 1650 ft. lbs.

Let me guess that the cable speed is 90 FPS at the fastest point and the
tension equals the weight of the glider or 1100 pounds. That computes to
180 HP at the drum for just that moment when the cable speed peaks. Since
power = speed x force, the power will drop off quickly as the cable speed
drops even as the cable tension remains constant.

Your figure of 75HP is the power required to lift the glider. Still more
power is required to accelerate it to flying speed. The power delivered to
the glider increases until the peak cable speed is reached and then
decreases quickly.

The new German winches seem to have turbo diesels in excess of 350HP with
some as high as 800HP. The old Opel Diplomat V8 (Think Chevy 350) powered
winches are on the used market cheap. They must know something. Maybe it's
launching those water logged ASH-25's.

Bill Daniels