During Saturday's sport gold heat race I rememebered that Jeff Lavelle's Glasair had had problems with its landing gear during the final race of 2018, and I started checking my pictures to see if the tell-tale shadow appeared. Sure enough there it was. My pictures aren't as clear as some of those posted on the forum but they were clear enough to show Jeff and his team. (I borrowed WJ Pearces photo for this post, I hope that's OK.)

Here's one of my pics:

Attachment 24495

I got to wondering how much of an effect the gear being out had.

My first cut at it is based on the assumption that both the qualifying run and Saturday's race were run flat out. (Footnote 1)

The power required to for an airplane to fly in a given condition is simply Drag times Velocity:

Pr = D * V

If Pr is the same for qualifying as it was for the heat race then:

Pr = D_qual * V_qual = D_race * V_race

Since V_race is slower than V_qual, we know the D_race must have been bigger than D_qual.

Drag can be calculated from speed (V) , air density (rho) , wing area (S) and a drag coefficient (Cd):

D = 1/2 * rho * V^2 * Cd * S

Pr is just D*V so it is:

Pr = 1/2 * rho * V^3 * Cd *S

Now, we don't know Cd for the airplane. But we can say that during the heat race Cd was increased by some amount by the gear leg hanging out, which we'll call deltaCd.

Since we assume Pr is the same in both conditions we have:

1/2 * rho * V_qual^3 * Cd * S = 1/2 * rho * V_race ^3 * Cd * S + 1/2 * rho * V_race^3 * deltaCd * S

With a little algebra we can find the ratio of the change in drag to the basic drag:

deltaCd/Cd = (V_qual^3 - V_race^3) / V_race^3

V_qual = 396mph = 578 ft/sec

V_race = 373mph = 544 ft/sec

deltaCd = 19.6% of the basic drag. Pretty big!

Another way to think about this is to consider the "Drag Area" of the airplane. This is the area of something with the same drag as the airplane, but that has Cd = 1.

We can write Drag the same way as above, but use a shorthand for 1/2 * rho * V^2 above.

We'll call that 'q'

So:

Drag = q * Cd * S

If we pretend Cd = 1, then

Drag = q * S

and the drag area is:

S = Drag / q.

We'll call Drag Area Ds for short.

Using the same idea of letting Pr be the same for both qual and racing we get

Ds_qual / Ds_race = V_qual^3/V_race^2 = 1.196 - which not too surprisingly is 19.6% more than the basic drag area.

But what *is* the basic drag area?

From the Glasair website we learn the following:

Glasair III HP Max speed at Sea Level Wing Area

300 290mph 81sq ft.

We can rearrange our power equation to solve for Cd based on wing area

Cd = 2 * Pr / (rho * V^3 * S)

We have to adjust the engine power by the efficiency of the propeller get Pr - Pr is probably 85% of 300hp, or 255hp.

which gives us and estimate of Cd = 0.0192.

We can then use Cd to calculate Pr at 393mph: 650 hp

And adjust that by the prop efficiency to get the engine power: 765 hp. (Note that I did this at sea level)

Remember that Pr = Drag * V. And we can get the drag area with Pr / (V * q) = Ds

Ds = 1.56 ft^2.

Now I'm not all that great at estimating height by shadows, but I remember that if you can measure two shadows, and know the height of one of the things casting a shadow, the height of the other thing is in proportion to the ratio of the shadows. The flap bracket casts a shadow roughly the same length as the gear door. The flap bracket looks a lot taller in the shadow photo than it does from the wingtip (WJ Pearces picture is much sharper than mine), and with

a little kentucky windage, I'm going to guess that the gear leg as up about an inch along about 36 inches of wing.

Attachment 24496

Attachment 24497

This gives a frontal area of 36 sq inches.

I'm going to estimate that the drag coefficient is about that of a bluff body or Cd = 1, giving a drag area of 36 sq in, or 0.25 sq ft

0.25sq ft / 1.56 sq ft = 16%.

Close enough. =)

Footnote 1:

We assumed that Pr was the same for both qualifying and racing.

One might assume that means that the engine was producing the same horsepower, but that's not quite true.

At a constant airspeed, the Drag is equal to the thrust, so we can also think of Pr like this:

Pr = T * V

The engine produces power at the crankshaft; the propeller applies that power via thrust. If V is lower, then T must be higher to keep Pr the same. This could be done by spinning the engine faster or through prop pitch being a little different. Without knowing the engine parameters I can't say for sure, but I think for this estimate assuming that Pr is constant is reasonable.