# Induced drag confusion - More than one cause?

1. Mar 8, 2013

### scanwinder

Hi,

I'm a bit confused by induced drag on an aerofoil and hoping somebody can clear it up for me.

I understand that as air slips around the edges of the wing due to the pressure differential between the top and the bottom surfaces, the angle of attack at the ends of the wings is effectively reduced(as the air is being deflected downward).

What i'm unclear about is, some sources seem to state that component of lift opposing motion is the induced drag(i.e. if the wing is at an angle with respect to the airflow, the wing generates a force perpendicular to it, and a component of that is lift and a component of that is drag)

Below is an image illustrating the latter:
http://selair.selkirk.bc.ca/training/aerodynamics/images/induced-definition.gif [Broken]

So my question is, is induced drag due to downwash from wingtip vortices, or the rearward component of lift since the force the wings generate isn't directly upward. Or is it a combination of both of these things?

Last edited by a moderator: May 6, 2017
2. Mar 9, 2013

Induced drag is sort of a catchall term for the sources of drag that are unavoidable on any body producing lift, for example the drag as a result of the wing being at an angle of attack and the drag associated the wingtip vortices. In short, the answer to your question is "both of these things".

3. Mar 9, 2013

### scanwinder

Thanks for that, I thought that would be the case. It just seems that most text books explain one or the other but not that it's the combination of both. The derivation in the text i'm reading makes a whole lot more sense now that I realise it's a combination.

4. Mar 9, 2013

### rcgldr

For an ideal or perfect wing, using the wing as a frame of reference, this ideal wing diverts the relative air flow by only changing direction and not speed, so that there is no change in energy of the affected air. The change in direction means the "forwards" component of the momentum of the affected air is reduced, while the downwards component is increased from zero. The decrease in the forward component of momentum corresponds to the drag component of impulse, which is force x time. Induced drag can be calculated based on the angle of diversion and lift. If I did the math correctly, and defining θ as the angle of diversion, I get induced drag = lift (1 - cos(θ)) / sin(θ).

My math (induced drag calculated so that it's a postive number):

f = force
i = impulse
m = mass
v = velocity
t = time
θ = angle of deflection

i = f Δt = m Δv
f = m Δv / Δt

lift = m Δv / Δt = m v (sin(θ) - 0) / Δt
induced drag = m (-Δv) / Δt = m v (1 - cos(θ)) / Δt
lift / sin(θ) = m v / Δt
induced drag = lift (1 - cos(θ)) / sin(θ)

Last edited: Mar 9, 2013