I am building a 100" E-2C Hawkeye (AWACS with radar dome on top). I need some assistance in calculating my wing loading. Using a wing loading calculator, I determined the following:
wing area= 1150 SQ.IN with flaps retracted, 1225 with flaps extended.
Radome area=660 SQ.IN
Engines= 2ea Saito 125
Props= Zinger 14x6, 4-blade

If I factor the estimated fliying weight of 25-26lbs and use only the wing area, I come up with an approximate wing loading of 53 oz/sq.ft.

If I add the radome area, the wing loading drops to approximately 34oz/sq.ft.

If the radome is set with an incidence of zero, does this get factored in as wing area? COuld the radome incidence be adusted to assist in the total lift of the aircraft.

Any help is appreciated.
Mike

Wing loading is not a very good indicator of how well a model will perform. A much better indicator IMHO is the formula that Mr. Reynold proposed in the September 1989 issue of Model Builder. It applys favorably to any size model, all the way from the little rubber powered jobs to the giant scale. It is called "Wing Volume Loading (WVL)" and is defined as:
Weight of model in ounces divided by wing area of the model raised to the 1.5 power.
If the value falls below 12 the model will fly very well, if 7 or less it will be a real floater. If much above 12 it becomes a real lead sled. If the radom is providing lift just as a second wing on the biplanes, I'd suspect you could safely add its area to the wing area for the purpose of this calculation.

Would this be the formula you mentione? If so, this doesnt come anywhere near 7 or 12.


416 OZ/ (1150)^1.5=

416 OZ/ (1810)^1.5=

Mike

The formula SEEMS to work better if you divide the area by the weight, then times 1.5

wing area in sq inches/model weight in oz x 1.5 =

However, that's not the way the formula works.

When it's stated as........Weight of model in ounces divided by wing area of the model raised to the 1.5 power. .......... it means that the theory behind the formula depends on your figuring the area of the wing raised to the 1.5 power and then dividing that into the weight of the model.

It's considered a "volume" theory because you're getting something like the volume of the wing (3 dimensional), not just using it's area (2 dimensional)

OK, puting the formula discussion aside, how does my project come out? What are the results of the volume formula? Any way I do it, I cant get a number within the range of 7-12. Have I totally forgot my math skills?

Mike

The area in the formula should be in square feet, i.e. (1810/144)^1.5 to get square feet raised to the 1.5 power. My fault for not stateing that the area was in square feet, not square inches. Sorry about that. In your case the WVL would be just over 9 and should result in a very flyable plane.

AAHHH, that makes a big difference. The numbers I was gettting were not even close.

Thanks Rodney.

Mike

BTW, the radome is part of the airplane and will provide lift just like a wing of that shape would. Matter of fact, have been wings of that shape.