There’s been a lot of discussion about the impact of flaps on turn performance. Specifically, how does the increased drag of flaps impact the rate of turn of an aircraft?
My previous responses to this question relied on a series of calculations to explain the physics at work in a sustained turn and the resulting turn rate performance. This explanation is harder to intuitively understand because of the stages of calculation needed. I’ve since found a more mathematically intuitive method to explain sustained turn rate performance.
The ProblemIt’s widely stated that sustained turn performance is a function of an airplane’s specific excess power. As already discussed specific excess power is

Specific excess power is the sum of the energy usage in maneuvering the airplane. It gives us insight into the maneuvering loads an airplane can sustain bound by it’s energy gain or loss as a function of thrust and drag.
A difficulty, however, lies in how to translate specific excess power to turn performance. Turn performance is usually expressed as load factor, turn radius, and turn rate. Specific excess power does not give us direct visibility to these performance variables. They must be derived from key aerodynamic coefficients embedded in Eq. (1.1) in order to make them visible. For instance an aircraft’s rate of turn is

To arrive at a value for turn rate in Eq. (1.2) as a function of specific excess power from Eq (1.1) requires several stages of calculations. Therefore, it is non-intuitive to arrive at an aircraft’s turn performance through specific excess power.
Is there a more intuitive way of understanding the relationship between specific excess power and turn rate without having to go through a series of calculations to gain understanding? The answer is yes.
An aircraft in a sustained turn neither gains or loses altitude and airspeed while holding it’s turn. This means that the aerodynamic forces are in equilibrium while in a maneuver. In mathematical terms specific excess power equals zero (PS = 0). Bypassing the mathematical derivation, knowing this relationship leads to the following

where n is the load factor of the aircraft in a sustained turn as a function of the basic aerodynamic variables of lift, drag, thrust, and weight.
The Key Equation: Turn Rate as a Function of L/D, T/W, & VelocitySubstituting Eq. (1.3) for load factor in Eq. (1.2)

We now have an expression of an aircraft’s sustained turn rate in key aerodynamic ratios derived from specific excess power. Eq. (1.4) gives us a more intuitive way to understand how key aerodynamic factors affect an airplane’s sustained turn rate.
We can now intuitively evaluate the affect of deploying flaps on turn rate as a function of the flap’s lift and drag. From Eq. (1.4) we see that turn rate is a function of lift-to-drag ratio, thrust-to-weight ratio, and velocity. Rate of turn is maximized when lift-to-drag and thrust-to-weight are maximized while velocity is minimized.
We know that deploying flaps reduces an aircraft’s lift-to-drag ratio. Flaps increase the lift of aircraft but at a cost of increased drag as well. If L/D ratio reduces with flaps, then so should turn rate right? So what’s going on if turn rate remains the same or is actually better with flaps deployed?
The reason is we can’t forget about velocity in Eq. (1.4). Flaps also reduce the minimum maneuvering speed of an airplane too so while L/D decreases, so does velocity. Secondly for piston-propeller aircraft thrust varies with velocity as well increasing as velocity decreases.
The bottom line is sustained turn rate is a function of lift-to-drag ratio, thrust-to-weight ratio, and velocity which all need to be factored in because they all vary depending on the configuration and airspeed of the airplane.
Illustration Using the EquationLet’s put some data to this to demonstrate the dependence of turn rate on the combined ratio of lift-to-drag and thrust-to-weight divided by velocity. For the below I used the same static variables and propeller efficiency curve from my previous calculations. The calculations represent sustained turns at maximum lift coefficient (Clmax), or the maximum lift-limit achievable by an aircraft.
Here’s a table with variation in Cl/Cd for comparison to evaluate Eq (1.4) above. At the point of best sustained turn rate (Ps=0) Cl/cd, V, T, and W were plugged into Eq (1.4). These are the results for turn rate.
Plane Config Clmax Cd0 Cd cl/cd V Ps=0 T W n rate (dps)
1 clean 1.6 0.02 0.2 8 160 3949 12400 2.55 18.4
2 flaps 2.8 0.1 0.6 4.3 100 5164 12400 1.79 18.7
3 flaps 2.2 0.1 0.4 5 120 4755 12400 1.92 17.1
4 flaps 2.8 0.18 0.7 3.8 97 5281 12400 1.64 17.2
Plane #1 and #2 were the same ones in my previous calculations (same results as before). Note that they are both approximately the same turn rate despite L/D ratio being lower for Plane #2. The reason is the thrust is greater while the velocity where thrust=drag is also lower. The net result is Plane #1 and #2 have approximately the same turn rate despite Plane #2 being "draggier" because of the flaps.
Plane #3 I hypothetically decreased Clmax. This results in a better lift-to-drag ratio (remember CD includes both the parasite and induced components of drag) compared to Plane #2, yet Plane #3 has a worse turn rate. The reason is the velocity where thrust=drag (Ps=0) has increased which also reduces the amount of thrust available. The combination of both the increase in airspeed for Ps=0 and reduction in thrust results in a lower turn rate.
Plane #4 I left Clmax the same but increased the profile drag from .1 to .18. This results in a lower velocity where thrust=drag (Ps=0) and higher thrust but in combination with the lower lift-to-drag ratio results in a lower turn rate compared to that of Plane #1 or Plane #2.
What happens if we changed the engine BHP available for Planes #1 and #2? It was assumed to be 2380 HP. Let’s reduce it to 2000 HP and see what the result is.
Plane Config Clmax Cd0 Cd cl/cd V Ps=0 T W n rate (dps)
1 clean 1.6 0.02 0.2 8 151 3520 12400 2.27 17
2 flaps 2.8 0.1 0.6 4.3 96 4625 12400 1.6 16.4
We see now that Plane #2 has a lower sustained turn rate compared to Plane #1. Changing the available horsepower changed the thrust and also the velocity where thrust=drag. The net result is a lower turn rate for Plane #2.
All these examples demonstrate that sustained turn rate is function of the combination of lift-to-drag, thrust-to-weight, and airspeed as expressed in Eq. (1.4).
A key learning is that we should be careful with making conclusions regarding airplane performance that rely on generalizations. Equation (1.4) demonstrates why conclusions such as increased drag of flaps reduces sustained turn performance or extrapolating steady climb performance to turn performance are inaccurate because they oversimplify what occurs in a sustained turn.
I hope this helps to illuminate the topic.
Tango, XO
412th FS Braunco Mustangs