How does maneuver g-load affect drag?

Question

High-performance jets can use high-g turns to scrub airspeed and energy (e.g., in run-and-break landing patterns).

What is the relationship between the aggressiveness of a turn and the rate of decrease in airspeed?

For example, does the rate of decrease of airspeed in a turn depend only on the g-load, or is it also a function of the no-load airspeed/drag of the aircraft? Is there some aircraft characteristic (e.g., something like "form factor") that describes the relationship? Or is this a unique characteristic of each aircraft type?

Answer 1

The decrease in Airspeed is a result from lower Thrust than Drag (T<D). It is convenient to use the Specific Excess Power (SEP) in this case. \begin{equation} SEP = \frac{dh}{dt} +\frac{V}{g} \frac{dV}{dt} = \frac{(T - D)V}{mg} \end{equation} Solving for the decreasing speed: \begin{equation} \frac{dV}{dt} = \frac{T - D}{m} - \frac{g}{V}\frac{dh}{dt} \end{equation} Now we can see the possibilities to reduce the aircrafts Speed (making dV/dt negative).

1. Lowering the Thrust (turning off the engines / using reverse thrust)
2. Increasing Drag
3. Climbing Number 3 is not interesting in this case, since we want to land the aircraft. Therefore Number 1 and 2 are the most practical Solutions. The Drag can be split up into the Zero Lift Drag D_0 ("no load") and your Induced Drag D_i ("load") \begin{equation} D = D_0 + D_i \end{equation} The Zero Lift Drag for a specific Configuration usually depends only your Flight Speed V_TAS and the density rho, which depends on your altitude. (Note: Speedbrakes, Opening up Weapon bays etc. are increasing C_D0). (I would say C_D0 is one of the important "form factors" but has only a small effect on subsonic flight compared to the following form factor) \begin{equation} D_0 = 0.5 \cdot rho \cdot C_{D0} \cdot V_{TAS}^2 \cdot S_{ref} \end{equation} On the other hand, the induced drag is dependent on the current load factor n_z. \begin{equation} D_i = 0.5 \cdot rho \cdot C_{Di}(n_z) \cdot V_{TAS}^2 \cdot S_{ref} \end{equation} So the needed Lift L can be calculated with the load factor by multplicating the loadfactor with the mass of the aircraft and the gravitational constant \begin{equation} L = 0.5 \cdot rho \cdot C_L \cdot V_{TAS}^2 \cdot S_{ref} = n_z \cdot m \cdot g \end{equation} We can solve for the lift coefficient: \begin{equation} C_L = \frac{2 \cdot n_z\cdot m \cdot g}{rho \cdot V_{TAS}^2 \cdot S_{ref}} \end{equation} (Side Note: So, to achieve higher load factors we need higher C_L -> unstable Configurations) One important thing to see here is, that C_L ~ n_z. This Lift Coefficient is needed for modelling your Drag. I am assuming we have a quadratic Dragpolar (not correct, but good enough for explanations). For this case: \begin{equation} C_{Di} = k \cdot C_L^2 \end{equation} Since C_L ~ n_z -> C_Di ~ n_z^2. So by increasing the load factor from 1 to 6 will increase our induced Drag by 35 times! (Therefore high g maneuvers are used) But getting this factor k (I would call this the main "form factor") right is quiet difficult, so I am going to mention only some dependencies on it (More about the influences on the induced Drag can be read in Raymers or Roskams books). This factor highly depends on the form of your leading edge. For Round airfoils you can get a net suction force (=Thrust) on the Leading Edge (Commercial). This suction force can not be created that effectivly on Fighter Aircraft, since they usually have quiet sharp leading edges. This factor can also be influenced by:
4. Higher Aspect Ratio -> smaller induced Drag
5. Taper
6. Flaps
7. Airfoil
8. Lift slope

Some other ideas to increase the Drag:

1. Moving the Center of gravity so aircraft gets too stable and therefore a lot of trim drag will be created
2. Post stall maneuvers

But the easiest and most effective way is just to fly high g's.