**More thrust with autogiros?**

More thrust with autogiros?

In a Popular Rotorcraft Flying article by Chuck Beatty, published in October 1977, Beatty gave a formula to calculate thrust. He stated that if a propeller had no losses, static thrust would be calculated as:

Thrust = (Horsepower X Diameter)2/3 X 10.4

Beatty added that allowing for profile drag, tip vortices, and other losses, the 10.4 would become 6.5 (a propeller of 62.5 percent efficiency). Beatty noted that in an aircraft speed range, the 10.4 would become 7, because with a 5- 5.5-foot propeller the tip speed is slower.

In 1999, George Pate, a now-deceased autogiro designer, calculated thrust requirements for some of the old-time 1930s autogiros. In Pate's calculations, the 6.5 factor was used, because the classic autogiros in the examples have propeller diameters larger than 5.5 feet.

So, using the equation "Thrust = (Horsepower X Diameter)2/3 X 6.5," Pate compared classic tractor autogyros of the 1930s to the "1/2 pound thrust for 1 pound of gross weight" rule of thumb for modern pusher gyroplanes.

Autogyro.............Horse-..Propeller....Gross.....Thrust......Pusher.....Difference

Make and............power...Diameter...Weight...Produced...Thrust....(Pusher thrust

Model..............................(Feet).......(Lbs).......(Lbs).....Needed.....- Tractor ")

Cierva C.19 Mk 1...100.......6.75..........1,300.......498.........650......-152 (-23%)

Pitcairn PA-24.......160.......8.00..........1,800.......763.........900......-137 (-15%)

Kellett D-1............245......8.80...........2,400....1,079......1,200......-122 (-10%)

Pitcairn PA-19.......420.....10.00...........4,200....1,683......2,100......-417 (-20%)

Pate concluded that classic tractor autogyros were more efficient, because they had been flown with 10 percent to 23 percent less thrust than "1/2 pound thrust for 1 pound of weight" rule of thumb for pusher gyroplanes.

What do you think of that?