I wrote a blog a little while ago entitled “One Form of Lift” which talked about the generation of lift and how there was, from an engineering standpoint, no such thing as “impact lift”. Classical engineering analysis only uses lift (which is primarily generated by pressure flows around the wing) and drag. My assertion is that most or all of the force people are calling “impact lift” is really drag, but was there a way to prove it by using a real world example? To take a look at that, I decided to analyze what happens if I stick a set of flat plate wings on my light sport aircraft, a Flight Design CTSW.

We will assume that the wings on the CTSW are mounted so that they are at zero degrees angle of attack (and angle of incidence) when the airplane is sitting level. The wing area of a CTSW is 107.4 square feet and the airplane’s max gross weight is 1320 pounds.

The equation for lift is: Lift = ½ D V2 S CL where D= air density, S = wing area, V2= the velocity squared, and CL = the coefficient of lift.

The equation for drag is of the same form, i.e., Drag = ½ V2 S CD where = air density, S = area being analyzed for drag, and CD = coefficient of drag.

For these formulas, lift=weight, the density of air at sea level is used, and S= 107.4 sq ft.

A flat plate has a lift coefficient, and the curve maximum hits a CL = 0.7 at an angle of attack of approximately 12 degrees.

SOURCE:http://home.planet.nl/~kpt9/The%20boundary%20layer.htm

At twelve degrees’ angle of attack, then:

Lift = ½ (.002378 slugs/ft3) V2 (107.4) (0.7)

For the airplane to fly, lift must equal to weight, so:

1320 = ½ (.002378) V2 (107.4) (0.7)

2640/(.002378*107.4*.7) = V2

2640/0.17877804 = V2

14,767 = V2

121.5 = V (feet per second) or 72 knots

This would get my CTSW with a flat plate wing off the ground at 72 knots while nibbling at the stall. Let’s say you understand this will be the case and elect to back off on the liftoff a bit, flying off at approximately 8 degrees angle of attack. (I believe you will either strike the tail of the CTSW at 12 degrees or be very much at risk of it.) So, when would you want to rotate? At that 8 degrees, you have a lift coefficient of 0.6. That gives you:

2640/.153238 = V2

17,228.1 = V2

131.25 =V (fps) or 77 knots

Since the CT actually gets airborne as low as 42 knots with flaps and about 50 without, you get an idea of the work a good airfoil design (using Bernoulli’s) is doing for you. You can “reverse engineer” the lift coefficient at those speeds by substituting them in for V and solving for CL.

Doing so for the no flap configuration and 50 knots yields a CL = 1.45; in other words, the shaping of the airfoil increases the “flat plate” lift coefficient by slightly over 100%. (The use of flaps yields a lift coefficient of 2.06…which is part of the reason why it leaps off the runway and feels like you’re in an elevator going straight up…)

But, let’s go back to flying by the flat plate alone since our real target is to examine the idea of “impact lift”. As I said in an earlier blog, the only way to do this from an engineering standpoint is to use drag as the “impact lift” force. There is no accounting for such a thing in conventional performance analysis (which should tell you something in and of itself about the rigor of the idea, despite some folks trying to discount how engineers do it).

Let’s see how we might be able to generate enough “impact lift” (drag) to fly. A flat plate has a drag coefficient of 1.28. Since the equations for lift and drag are the same and we’re considering the same area in both equation sets (i.e., only the area of the wings), then we can shortcut the calculation of the total drag force the wings can generate by simply taking a ratio of 1.28/0.7 = 1.83. Then, that same 1320 lbs of total force…now as drag instead of lift…would be generated at 72/1.83 = 39 knots. But this would be for a CTSW wing deflected at 90 degrees to the windstream (and the fuselage). Since we’re just trying to use “impact lift” (i.e., drag) to get airborne, the max lift would occur at a wing inclined 45 degrees to the windstream. The “lift” would be equal to the sine of 45 degrees times the total force. The sine of 45 degrees is 0.707. So, then the speed at which you’d have just enough drag acting vertically to lift off would be: 39 knots/0.707 = 56 knots.

However, for this to occur, the drag acting in the horizontal direction would also be 1320 lbs (sine and cosine of 45 degrees…the horizontal and vertical force components would be the same). Let’s see if the Rotax 912 can get there.

Using Thrust = (HP*efficiency*326)/KTAS and assuming we have a 0.8 propeller efficiency:

1320 = (100*.8*326)/KTS

1320 KTS = 26,080

KTS = 19.75

This is the speed the Rotax could attain in this “lift/drag” configuration. This is a conservative answer since the airplane will have more drag area than this, i.e., we did not account for the frontal area of the fuselage or other components. This puts us well short of the approximate 56 knots it would take to get airborne. To see how big an engine it would take, let’s use the same formula and solve for HP.

1320= (HP*.8*326) / (56)

1320 = (260.8HP)/ (56)

1320*56 = 260.8 HP

73920 = 260.8 HP

283.4 = HP

So, it would take an engine producing almost three times the horsepower the CT has available which would increase the weight and cause us to do all these calculations over again. Keep in mind, too, that we are only talking about the power it would take to get the CT into the air since climb rate is a function of excess horsepower. The engine would have to be bigger still. (At least a 300HP IO-540…) What the numbers are telling you is that the idea of impact lift doesn’t work, and that’s before we try to consider the flying characteristics and design of the flight controls for such a beast. We know that increasing camber (curvature) of an airfoil causes an increase in lift and that’s what makes ailerons and elevators work; how do you explain what’s happening if “impact lift” (drag) is the only force?

“Impact lift” is a misnomer and a popular myth, like saying you can discount Bernoulli’s principles and just use Newton’s laws to easily explain how airfoils work. Good luck with either. From both an engineering and piloting standpoint, it’s better and proper to see lift as a result of Bernoulli’s (i.e., pressure distributions) principles and drag as something that needs to be overcome and is not your friend (except when trying to slow down). You’ll not only stay safer that way but you’ll teach your kids the right concepts, making sure it is “as simple as possible but no simpler”.

*AUTHOR’S NOTE: Many thanks to Dave Witwer, Jim Gardner, and Matt Zwack for their review of the drafts of this blog. They’re all pilots and aerospace engineers (Dave, Jim and I worked at Johnson Space Center and Matt works out at Marshall Spaceflight Center.) Great guys I’m honored to know as colleagues and friends! *