I understand that the total drag includes the parasitic and induced component, with the parasitic component being the same as with straight and level flight.

How do I calculate the induced component?

The end goal is to produce a graph of drag versus airspeed.

  • $\begingroup$ I took the liberty of adding a clarifying sentence at the end based on content in your comments; feel free to delete if you think it detracts from the question. $\endgroup$ – quiet flyer Oct 26 '20 at 10:52

Induced drag $D_i$ is proportional to the square of lift $L$: $$D_i = \frac{2\cdot L^2}{\rho\cdot v^2\cdot\pi\cdot b^2\cdot\epsilon}$$

where $b$ is wingspan, $v$ is flight speed and $\rho$ is air density. $\epsilon$ is an efficiency factor which tends to be between 0.8 and 1. Compared to straight and level flight, induced drag will quadruple when turning at 2g and the same flight speed.

In order to know how much total drag goes up you need to know how much of your total drag is due to induced drag. Only then can the question be answered.

If your straight and level polar point is that for best range (when zero-lift and induced drag are equal), 2g flight at the same speed will increase total drag by a factor of 2.5.


Do you know the L/D ratio at whatever angle-of-attack is being used in the turn? If so, in a steady-state constant-altitude turn, Drag = (D/L) * Weight * Load Factor.

Note that the ratio of Lift / Drag is also the ratio of Lift coefficient / Drag coefficient.

Note also that the L/D ratio can easily be computed, if the Thrust or Drag for the same angle-of-attack is known for straight-and-level flight.

In a turn, the airspeed must be increased according to the square root of the increase in wing loading, to allow steady-state flight at the same angle-of-attack as we had in wings-level flight.

This answer treats the drag due to increased loading due to turning as being exactly equivalent to the drag due to increased loading that we would get simply by adding weight to the aircraft. This is not exactly true, because in curving flight, the relative wind is actually curved, which has aerodynamic consequences. But it is a very good first approximation for aircraft that have reasonably high flight speeds, and it is really the only practical way to tackle the problem short of an elaborate computer model.

  • $\begingroup$ for this particular question I do not have the Lift or Drag Coefficient. Is there any other way to calculate or a different formula for that D/L ratio? $\endgroup$ – 8packgrandma Oct 25 '20 at 14:54
  • $\begingroup$ I do know drag, but I am unsure how to compute the ratio, given that I do not know the lift $\endgroup$ – 8packgrandma Oct 25 '20 at 15:07
  • $\begingroup$ no we've been asked to plot drag against airspeed for different cases, nothing said about angle of attack, I guess I will write that the angle of attack is constant $\endgroup$ – 8packgrandma Oct 25 '20 at 23:51
  • $\begingroup$ does that mean that the drag to lift ratio is constant for both turning and level flight? $\endgroup$ – 8packgrandma Oct 25 '20 at 23:52

Turning with a load factor of 2 is a steep turn of 60 degrees, so your entry speed will be faster to avoid an "accelerated" stall. Vbg of a Cessna 172 is around 65 knots, but steep turns are performed at 90 knots. The turn is entered by rolling, increasing angle of attack, and increasing throttle.

A practical way to determine increase in total drag would be to monitor needed rpm increase (fuel flow) during the manuever. One could also measure the actual AOA in the turn.

Theoretically, the difference in AOA at level flight and AOA in the turn could be compared with the coefficient of lift and coefficient of drag polars for that airfoil for those respective angles (at the same speed) to derive a calculated solution.

Comparison with increased fuel flow in the turn should be close, but you may find the parasitic component increases too if the turn is not coordinated.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.