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The following question came from Frank fnasta1@tampabay.rr.com" fnasta1@tampabay.rr.com )


How do you figure sq inches of wing area?


From : Don Stackhouse

For a rectangular wing, it's just chord times span. For a single-tapered (i.e.: trapezoidal) wing, it's (root chord + tip chord)/2 times the span.

For something more complicated than a single-tapered trapezoid, just break up the shape into a series of trapezoids that approximate the shape well, figure the area of each trapezoid and add them all up. The accuracy of this method depends on how well the trapezoids approximate the actual shape.

CAD systems on computers use a variation of this method. They divide the shape into a series of rectangular strips and add up the areas of the rectangles. The reason it works is because they can divide the shape up into MANY rectangles. For exampler, if I was finding the area of an elliptical wing with highly rounded tips, I might tell the CAD system to use strips only .005" or even .001" wide. Thousands of strips, but together they make a very close approximation of the true shape, with the calculated area accurate within a tiny fraction of a square inch. I sure wouldn't want to try it that way with just pencil and paper though!

    and how much will each sq inch lift?

That depends on a great many other factors. The formula for lift of an infinite-span wing is:

L = 1/2 Cl rho V^2 S

where L = lift the 1/2 is leftover from the equation that this one originally came from Cl = "lift coefficient", which depends on airfoil and angle of attack rho = air density V^2 = velocity squared S = wing area

As you can see, if you fly higher where the air is thinner (and therefore the density is lower), you don't get as much lift. OTOH, more area, a different airfoil or angle of attack, or especially an increase in airspeed can make a lot more lift.

Cl depends primarily on the shape of the airfoil, the angle of attack it's flying at, and on the Reynolds number or "Re", which is a measure of what we call scale effect. It depends on chord length, speed, air density and on air viscosity (the stickyness or slipperyness of the air; maple syrup and peanut butter have high viscosity, water has lower viscosity, and air's viscosity is even lower). To calculate Re at sea level, multiply 778 times the chord in inches times the sped in miles per hour. Smaller airfoils like the ones on our models make more drag and less lift than the same airfoils would on something larger or faster. In addition, each airfoil shape will have a certain range of Re's where it works best. The airfoils that work well on full-scale usually do not work well on models, and model airfoils usually do not work well on full scale aircraft. The airfoils that do well on "conventional" R/C models are usually inappropriate for our slow fliers. The details of why and how this is are very complex, and can fill books. I won't go much further into it here, but there is some more discussion of this in some of the articles in the "Ask Joe and Don" section of our website: http://www.bright.net/~djwerks/

For typical airfoils for indoor models, an efficient Cl is likely to be somewhere around .2 to .6, maybe .8 to .9 for a good airfoil design working close to stall. Yes, it is possible to get somewhat higher than that, but with few exceptions, the drag is going to get very high. More drag means more power required, which means bigger batteries, which negates the increase in lift. In addition, the turns we make are so tight that the inside wingtip gets into stall problems if we don't leave it some reserve capability. For preliminary design purposes, I'd recommend trying to stay under about .6 to .8 maximum unless there's some solid data indicating it's safe to go higher. I have designs that can do much better than that, but it takes a lot of very good design tools plus a LOT of experience using them to achieve that.

Then there's the effects of finite span. Most of us don't have room in our storage area to keep an infinite span wing, so we have to opt for a span that's somewhat less than that. The problem with a finite span wing is that the air spills around the tips, so the lift is reduced. How much it's reduced depends on the details of the each specific wing design, but for typical indoor models we could reasonably assume it's about 75% to 85% of the lift of an infinite span wing of the same area. That means that our Cl limits of .6 to .8 that I discussed above become only .45 to .68 when you throw in the effects of a finite span wing. Once again, the right design tools and the experience to back them up can do much better, but very few folks have access to that sort of design capability, nor the obsessiveness to go through all the work required to make it happen. If you want to stick to just rules of thumb and cookbook airfoils and planforms, you'd better assume conservative values for Cl.

Now let's talk about rho, the air density. In English units it's measured in slugs, and at sea level standard conditions it's about .002378 slugs. At 2500 feet msl ("above Mean Sea Level") it drops to .002209 slugs, and it's down to .002049 on a standard day at 5000 msl in Denver. Meanwhile, the viscosity stays nearly constant, so as you gain altitude you also get lower Reynolds numbers, which in turn reduces Cl. It's a double whammy, we lose both Cl and rho as we go higher, so it's no wonder that models that are marginal performers at sea level in Santa Barbara sometimes become flightless R/C cars in Denver or Reno. Also, a hot day reduces the air density as well. For example, here in Ohio at about 1000 msl, the air density on a hot day in August can approximate the standard day density of a 3000 msl altitude.

For the formula and numbers above, you should use feet per second for velocity and square feet for wing area. To convert miles per hour to feet per second, multiply by 22 and divide by 15.

Frank, we're now finally ready to answer your question!

Let's assume we have an effective Cl for the whole wing of .6, and we're flying at a density altitude of 2500 feet msl.

At 5 mph at these conditions, a square foot (144 square inches) of wing will support .0356 pounds, which is .57 ounces.

At 6 mph, that increases to .82 ounces.

At 7 mph, it can support 1.11 ounces.

At 10 mph, it can support 2.28 ounces.

At 15 mph, it can support 5.13 ounces.

At 20 mph, it can support 9.12 ounces.

Those advertising claims of indoor models with 4 mph flying speeds are starting to look a bit questionable, aren't they! It is possible to fly that slowly with a heavier-than-air fixed-wing model, but it's extremely difficult. I'd say in most cases, about 7 mph or more is probably more reasonable. That's still pretty dog-gone slow.

Don Stackhouse
DJ Aerotech



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