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I Just wanted to see if my understanding of drag is correct, as I am unsure. I understand that during straight and level flight, the drag can be calculated as:

enter image description here

With the first term being the parasitic drag component and the second term being the induced drag component.

However, in a coordinated turn with the same angle of attack at the same airspeed, I have been told that the total drag can be expressed as:

enter image description here

And because the angle of attack is the same, the lift to drag ratio is the same. In this scenario the load factor is 2.

However, since in straight and level flight L = W. So does it follow that the total drag in this turn is double the total drag in the straight and level flight?

i.e.

enter image description here

I don't know why but I feel that this is wrong.

If this is correct, however, the question says that the parasitic drag of the turn is equal to the parasitic drag of the straight and level flight; so would it correct to follow that the induced drag is equal to the following equation?

enter image description here

Ginger
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8packgrandma
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    You should clarify whether you mean the same airspeed in both cases. If you do, there is an error in your application of the "same angle of attack" idea. – quiet flyer Oct 26 '20 at 15:44
  • Hi, I've edited the question to reflect that the airspeed is the same in both cases. I thought that the same angle of attack maintains the same lift to drag ratio in turning and straight level flight – 8packgrandma Oct 26 '20 at 23:10
  • In "a turn" with a 10 degree bank angle, the drag is probably well less than that of level flight. If you're in an F-16 with a 9-G turn at a bank angle close to 90 degrees, then your drag is well beyond double the level flight case. Somewhere between those two extremes, you'll find "a turn" (quoting your question's title) that gives you 2x the drag of level flight. Is your question if that always occurs at 60 degrees bank & 2 G's? – Ralph J Oct 27 '20 at 05:02
  • yes in this question its 60 degrees bank with load factor 2 G's – 8packgrandma Oct 27 '20 at 08:40
  • @Ralph J in "a turn" with a 10 degree bank angle, the drag is probably well less than that of level flight??? – Robert DiGiovanni Oct 27 '20 at 09:33
  • What would really neat is to hold the aircraft level and turn using rudder and fuselage as a symmetrical airfoil, generating sideways lift. But, a little more drag there too. (My original thought was to turn just using the wheel spats). – Robert DiGiovanni Oct 27 '20 at 10:23
  • Sorry, 10 degree banked turn, drag well less than double the S&L case. The 9G turn has drag well above double the S&L case. – Ralph J Oct 27 '20 at 13:25
  • While airborn lift always equals weight, doesn't matter what sort of contortions or altitude changes are happening. Weight is a shorthand for net force, it is not mass, don't confuse mass and force. – Max Power Jan 28 '23 at 01:07

2 Answers2

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Could drag in a turn be twice the drag in straight and level flight?

Yes, but this will be a rater steep turn.

… in a coordinated turn with the same angle of attack at the same airspeed …

No, this is impossible. Either angle of attack or speed have to be higher so the higher lift required for turning can be generated.

… angle of attack is the same, the lift to drag ratio is the same. In this scenario the load factor is 2

And in order to create twice the lift of straight flight, the speed has to be higher by a factor of √2. In this case, L/D is indeed the same, but speed is higher so twice the dynamic pressure is available for creating the required lift. Since both lift and drag are proportional to dynamic pressure, creating twice the lift at the same L/D means that also drag will double.

In order to turn with a load factor of 2g, the bank angle has to be 60°. That is rather steep.

Peter Kämpf
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For a set air density, a given airspeed and angle of attack will always produce the same lift and drag. lift = weight in all cases and Weight=Mass*Acceleration(gravitational and mechanical combined. Vector sum.) If lift=weight and weight=M*A then lift=M*A and if M=1 and A=1 then L=1, if A=2 then L=2. If we know L=1 based on air density speed and AoA then W=1 and if M still =1 then then A must =1. The other option is that A=2 and so M=1/2.

Although load factor is tossed around quite often in aviation, even by many very competent instructors, I don't feel it is used appropriately in most cases. "Design load factor" is a circuitous method of specifying a target stress condition of an airframe for baseline engineering purposes. G units of acceleration is really more accurate terminology for lift vector calculations in my opinion. one G unit = about 9.8 in kg/M/S units or 32.2 in Slug/foot/S units

Alternately you could just use weight regardless of its source for your calculations. That is the lift needed for 2*mass at 1G is the same as 1*mass at 2G, and aerodynamically equivalent. With the possible exception of an extremely small radius turn where you would get effects from the tail and nose experiencing significantly different angles of attack, but this would be a radius of only a few fuselage lengths.

Max Power
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