5
$\begingroup$

I'm not an aeronautical engineer, but I am making a brave attempt to perform a preliminary design for an 4-metre wingspan radio controlled glider. With special thanks to @mns I have now learned a formula for calculating the coefficient of induced drag. The question now is how do I use it to calculate total drag.

Considering the formula below, how do I make a reasonable estimation of efficiency factor (e)

enter image description here

I've selected a prismatic inner wing with tapered outer wing to assimilate an ellipse? I understand that they state that (e) for a perfect ellipse is 1. What is 0 then, a square? Here is my wing planform:

enter image description here

They then state that I should add CDi to CD0 to offer me a total CD.

The question I ask now is why add CD0 (That is CD at 0 degrees AoA if my angle of incidence (setting angle of the wing) is 6.25 degrees?

I assume that I then run this through the lift formula to get drag in newtons.

The curve for CD / Alpha can be seen here

Improve this question
$\endgroup$

1 Answer 1

Reset to default
3
$\begingroup$

Typical values of e are between 0.7 and 0.98 in subsonic flow, and 0.3 to 0.5 in supersonic flow. Since it is in the denominator, e cannot become 0 or drag would be infinite. A square wing has an e anywhere between 1 (with aspect ratio close to zero) and 0.8 (with high aspect ratio and away from the design point for the twist distribution)

As a rough first-order assumption, the zero-lift drag coefficient is constant over the moderate AoA range when the flow is attached over the full chord. This drag is the sum of friction (shear forces tangential to the wing surface) and pressure drag (caused by pressure on the backwards-facing sections being lower than on the forward-facing sections of the airfoil, again due to viscosity).

Of course, once you have a full airfoil polar, your results will be more precise if you use the particular value of the drag coefficient at the specified angle of attack. However, sometimes you will need to calculate lift and drag without knowing the angle of attack, and then you fall back to the first-order approximation. For an example where this assumption is useful please read this answer.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Problem is first approx is wrong, because at higher AoA pressure drag is higher(even lower pressures at backward facing sections). I never understand why theory neglect pressure drag as AoA rise, and keep zero lift pressure drag through all AoA range.. If increase AoA of hand out of car window, you feel huge increase in drag/resistance, is this "pressure drag" or "induced drag"? $\endgroup$
    – user207141
    Apr 30 at 5:42
  • $\begingroup$ @user207141 No, this approximation is useful and - as I said - something to fall back to if no other data is available. Your example with the hand is pressure drag from flow separation. There you are outside of the linear AOA range and that area is of no interest for flight performance. $\endgroup$
    – Peter Kämpf
    Apr 30 at 17:44
  • $\begingroup$ I just say pressure drag when wing produce lift is higher then on zero lift AoA, so total drag must be higher than formula predict... In other words formula is too simplified.. $\endgroup$
    – user207141
    May 1 at 8:29
  • $\begingroup$ @user207141 Look here for a counterexample. $\endgroup$
    – Peter Kämpf
    May 1 at 13:05

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .