11 Degree Rotor Thrust Vector

[rehler]

Chuck Beaty,

Normally we expect the RTV to be tilted back about 11 degrees during cruise flight in a typical gyro (9 degree rotor head tilt + 2 degrees "blow back" or "flap").

If a gyro weight 500 pounds (down force) and the RTV is at 11 degrees, then the prop thrust would only be about 97 pounds to cause the 11 degree tilt. But, the gyro would have a prop thrust around 350 pounds. So, is the remaining 253 pounds needed just to overcome drag?

Also, about how much of the total drag would be from the rotor and how much from the gyro?

[C. Beaty]

The lift/drag ratio of a rotor, Ken, is 1/(tan of disc angle of attack). If a rotor flew at 11º at 50 mph (draggy), its L/D would be 1/(tan11º) = 5.14:1.

If AUW was 600 lb, the rotor drag would be 600/5.14 = 117 lb.

The power required would be (117 * 50 * 1.47)/550 = 15.6 HP. If the propeller efficiency was 60%, the engine would have to supply 26 HP. The airframe drag would probably be something approaching that and you need a little left over for climb.

The climb lift/drag ratio is simply: (horizontal velocity)/(vertical velocity)

If you wanted to climb at 1000 fpm (16.666 fps) while flying at 50 mph (73.5 fps), the L/D is 73.5/16.666 = 4.41:1 and drag would be 600/4.41 = 136 lb.

With the same 60% propeller, the engine would have to supply another 30 hp.

There are trailer queens and flying machines. Flying machines require good power to weight ratio, low drag rotors and low drag fuselages.

[ymmv]

The power required would be (117 * 50 * 1.47)/550 = 15.6 HP.

Sorry, I'm missing something. Whence comes the 1.47 in that equation?


There are trailer queens and flying machines. Flying machines require good power to weight ratio, low drag rotors and low drag fuselages.

Care to comment on which fuselages are relatively low in drag? (Loaded question, yes. I think Little Wings are cool. It's not an informed opinion, but there it is.)

[C. Beaty]

MPH*1.47 = ft/sec

For an open airframe, a ballpark guesstimate would be to measure the frontal area and say it's equivalent to a flat plate with Cd = 1.

A hemispherical cup with open end forward has a normal Cd = 1.35.

A hemispherical cup turned the other way has Cd = 0.41

A fuselage similar to an RAF-2000 would be a a bit lower than a hemispherical cup, say 0.35.

If its frontal area (I don't know; just a rough guess) is 15 ft^2, then its drag at 50 mph would be:

0.5 * 0.0023 * 15 * .73.5 * 73.5 * 0.35 = 33 lb. Then there are wheels, axles, mast, struts, etc.

Fuselages such as a Little Wing aren't too bad; Cd of perhaps 0.25. Most of the drag will come from undercarriage, mast and struts.

[ymmv]

MPH*1.47 = ft/sec

Ah. Okay.


Fuselages such as a Little Wing aren't too bad; Cd of perhaps 0.25. Most of the drag will come from undercarriage, mast and struts.

Hmm. That's better than I had expected. I had assumed that in addition to the frontal area there would be pretty significant drag down those long sides. (In naval architecture they talk about wetted surface area and skin friction. What it's called in the air I don't know.)

Again, thanks for the math.

[eruttan]

I doubt it is as low as .25.
The typical jelly bean car is pretty slippery, and its .26-.28.

But doubting what Beaty says usualy means I am wrong. But I learn something.

ymmnv; It is called wetted area, and skin friction, in aircraft too.

[Doug Riley]

Wetted area and skin friction are a big deal with fast, slick planes like Glasairs and such. With a flying bed frame like most gyros, and at the speeds we fly, skin friction is not worth worrying about.

PRA Chapter 5 many years ago tested the drag of a typical Bensen frame by measuring towline pull and subtracting out the rotor drag. It was on the order of 60 lb. at 50 mph for a completely naked frame and down around 35 lb. with the "Dix" body. (The "Dix" body was a fiberglass unit sold by Marion Springer and her husband. It fit over the Bensen frame, tow boom and all. It had a windshield and enclosed the pilot up to just short of shoulder height.)

Where most gyro pods fall short is in the streamlining of the back end. Almost all of them end in a blank wall just behind the pilot's seat.

Enclosing an engine is more complex than making a nose cone. It also tends to cause the engine to run hotter. It makes engine access more difficult. Nevertheless, you get more performance bang for the buck (or pound) of streamlining by dealing with the back end than you do dealing with the front end. The McCulloch J-2 and Air and Space both enclose the engine and taper the rest of the body back to a fine edge.

[C. Beaty]

Any object moving through the air has both skin (wetted surface) and form (pressure) drag.

Abbott & Von Doenhoff give an approximate formula for turbulent skin friction drag coefficient as:

Cf = 0.072* R^-0.2

Form drag depends upon the size of the wake. No matter how streamlined the nose of an object, if it’s chopped off without tapering to a fine point, the wake will be large.

The dimples of a golf ball force a turbulent boundary layer that sticks better and causes the airstream to remain attached farther around the ball and leave a smaller wake. A golf ball is a case where form drag is of more concern than skin drag.

Skin friction drag is the greater portion of drag for a streamlined shape that leaves a small wake.

Marks’ has a table from data extracted from NACA reports that gives some measured drag numbers for various fuselages at 100 mph:


A, ft^2 ---------10---------------15---------------20
D/A, average---5.5--------------4.5--------------4.0


That’s drag per square foot for various frontal areas.

At 100 mph, the dynamic pressure is 24.85 lb/ft^2

The drag coefficient for a 15 ft^2 fuselage would be 4.5/24.85 = 0.18

The LW has a well tapered tail cone so a Cd guess of 0.25 may not be too optimistic.

[Udi]

Chuck, on the old forum you had some graphs that showed the rotor drag and the airframe drag on the same chart. I remember the airframe drag goes up (exponentially?) with airspeed, and the rotor drag goes down with airspeed (up to a point). You showed three or four curves for different kinds of enclosures.

Do you still have that graph handy?

Thanks,

Udi-

[C. Beaty]

I no longer have those graphs, Udi, but it's no trouble to replot them.

My equation plotting program is on another HD but when I have occasion to swap HDs, I'll replot and attach to this thread.

[raghu]

An obvious way to go about estimating body drag is from the best climb speed of a gyro. At Vz the drag due to the airframe is exactly equal to the rotor drag. Given that we have resonable estimates for rotor drag working out the drag of the airframe is straight forward.

[C. Beaty]

The graph below illustrates the general trend of rotor and airframe drags at various airspeeds.

[Udi]

Thanks, Chuck. Interesting to note that the minimum drag airspeed (best glide speed? Best L/D?) is not at the same airspeed where the two graphs intersect. I am not sure why I expected it would...

Udi-

[C. Beaty]

One picture is worth 1,000 words, Udi.

The most accurate way, by far, of quantifying aircraft performance is by towline pull testing; an ideal dynamometer.

Bensen made extensive use of towline testing to evaluate changes but it wouldn't be of much use to those modern gyroplane "designers" who don't understand the difference between power and torque.