I’ll Pound Y’all

Let’s get wingfoil technical, OK? The photo shows a lift analysis of a#SkyDock airfoil at an attack angle of 4 degrees. At this angle, we’re generating a coefficient of lift of 1.39 with a drag of .0144. This produces a Lift / Drag ratio of 96.73.


In order to generate a lift of 660 pounds (#SkyDock gross weight); the induced drag from the entire wing is 7 pounds.

(Real world values vary substantially… topic for another post…)

Now let’s get practical and estimate in-flight speed to exactly carry the airplane at a 4 degree angle of attack. How fast does the airplane have to go to balance lift & aircraft weight exactly?

Let’s do this in English units, not metric.

It’s easy. Using the lift formula, which looks complicated but isn’t:

Lift = (0.5) ρ v² A Cl

ρ = air density;
v = velocity;
A = Area of wing;
Cl = Coefficient of lift;

We want to solve this equation for velocity. So the equation is rearranged as follows:

v = ((Lift * 2) / (ρ * A * Cl)) ^ 0.5

And now we solve.

Lift = 660 pounds (SkyDock @ gross weight)
ρ = 0.074887 pounds per cubic foot
A = approximately 120 square feet
Cl = 1.39

The answer is … 10.28 feet per second or 7 mph.

Oh My! that’s not correct! NOT EVEN CLOSE!

The problem is the we have solved without taking into account the conversion of force into pounds. Our result should have been intermediately calculated using the relatively obscure unit of ‘poundal’, with a consequent conversion to pounds.

It kind of reminds me of my youth, when I’d be avoiding some bully in the hallway, and he’d ‘pound y’all’ if I got in his way. Different situation… I digress.

A poundal is a unit of force, unlike the bully, who was a unit of terror. At acceleration of one gravity, it is equivalent to a pound using this formula:

1 poundal = 0.031081 pounds.

Fixing the formula from above,

v = ((Lift * 2) / (ρ * A * Cl * 0.031081)) ^ 0.5

And solving one more time using the same variables

Lift = 660 pounds (SkyDock @ gross weight)
ρ = 0.074887 pounds per cubic foot
A = approximately 120 square feet
Cl = 1.39

The corrected answer is … 58.31 feet per second or 40 mph.

Now we know: A SkyDock flying at 40mph with zero flaps and angle of attack of 4 degrees will generate exactly the lift required to balance offsetting gross weight of the airplane.

The author of this post, James Wiebe, has a degree in Math from Tabor College, Hillsboro, KS.

6 thoughts on “I’ll Pound Y’all

  1. I see a laminar flow stream line. I think the analysis results with different Re numbers in the range of 10^5~10^6 may be not much difference. Considering Aspect ratio may reduce lift coefficient 5% or so. However, SkyDock now contain technical component with your aerodynamic analysis. applaud.

  2. Oh my! James you are a truely a scientist! Your calculations reminds me of what Sheldon and Leonard did when the encountered a math problem.

    Respected and liked!

    Happy for y’all

  3. Hey Jim, it looks like the real impediment to slow flight is the gross weight of 660 lbs. I guess I understand that for marketing reasons, designs need to target a gross weight that covers the (gross!) weight of the average overweight American buyer. On the other hand, If a plane was to be designed for, say, a 150 lb pilot, then the increased performance of an ultralight would be truly significant. Any thoughts?

  4. Vis my prior reply, of course the gross weight cannot be the only change, I was thinking more of an 80% dynamic scaling, so that the wing area gets smaller also, avoiding the “leaf-in-the-wind” effect of simply reducing weight while maintaining wing area, etc. Bonus is that with scaling, power requirement is reduced as well as airframe weight, landing gear, and everything else.

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