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According to an answer to this question, at best L/D the the drag coefficient is evenly split between Cd0 and Cdi.

I'm going to limit the question to the Cessna 172 and its NACA 2412 wing with an estimated Reynolds number of 2.5 million at a rotation speed of 55 knots.

Cd = Cd0 + Cdi

Cdi can be replaced with Cl$^2$/(3.14 × AR × epsilon) or Cl$^2$/K. K = around 20.

Cl values of 0.8 at 5 degrees AoA yield a Cdi of 0.032, which is indeed close to the Cd0 of 0.031.

My quandary is that if all this is true, there should be a noticeable deceleration when the aircraft is rotated at 55 knots while taking off, resulting from a doubling of drag. This is not observed.

Furthermore, while lift to drag ratios of the NACA 2412 airfoil are commonly well over 50 at Re > 2.5 million, the L/D (based on glide ratio) of the entire aircraft is only around 10.

Is it possible the mathematical modeling is overestimating Cdi? How did they arrive at K = 3.14 × AR × epsilon?

Is it possible differences in Re number (when testing models in wind tunnels) has skewed the Cdi calculation too high for full scale aircraft?

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  • $\begingroup$ Robert, at 55 knots the dynamic pressure is lower. While the induced coefficient of drag doubles when going from 65 knots to 55, the drag itself is only a bit higher and propeller thrust (being proportional to speed inverted) also is higher. The airplane will accelerate just fine when rotated at 55 knots. $\endgroup$ Feb 20 at 9:03
  • $\begingroup$ @PeterKämpf yes, that is observed. My query was why (a tricycle gear at much lower AoA) does not experience a loss of acceleration upon rotation if the calculated Cdi increases to that near Cdo (a doubling in drag). But it may be that, as Rob McDonald and Thomas Perry pointed out, the ratio of lift generating drag compared to the form drag of the entire aircraft may not be expressed in K, which may be reflecting only the wing (airfoil)(3 dimensionally). $\endgroup$ Feb 20 at 10:39

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Answering the question: Why are induced and parasite drag equal at best L/D?

It is easy enough to do the math...

Starting with a simple parabolic polar

$C_D=C_{D,0}+K\,{C_L}^2$

Divide both sides by $C_L$. We will observe that extremum of D/L will occur at the same place as extremum of L/D.

$\frac{C_D}{C_L}=C_{D,0}\,{C_L}^{-1}+K\,{C_L}$

Take the derivative of both sides with respect to $C_L$.

$\frac{d}{d\,C_L}\left(\frac{C_D}{C_L}\right)=-C_{D,0}\,{C_L}^{-2}+K$

Set this equal to zero. This condition will be true at any extremum of $D/L$.

$0=-C_{D,0}\,{C_L}^{-2}+K$

Re-arrange

$C_{D,0} = K\,{C_L}^2$

Which demonstrates that parasite drag and induced drag are equal at the extremum of $L/D$.

This is simply a property of a parabola -- that gets applied to an airplane. It is a mathematical consequence.

We can take this some steps further...

At best $L/D$, what is the $C_L$? Start with our last equation and solve for $C_L$

${C_L}^* = \sqrt{ \frac{C_{D,0}}{K}}$

At best $L/D$! Note that we call this ${C_L}^*$ -- pronounced See-Ell-Star.

If we want to know the value of best $L/D$, we can divide this equation by twice the parasite drag coefficient (since we know they are equal our desired condition)

${\frac{C_L}{C_D}}_\mathrm{Best} = \left(\sqrt{ \frac{C_{D,0}}{K}}\right)/\left(2\,C_{D,0}\right)$

Square the denominator and combine with the numerator.

${\frac{L}{D}}_\mathrm{Best} = \sqrt{ \frac{C_{D,0}}{K\,4\,{C_{D,0}}^2}}$

Cancel the extra $C_{D,0}$..

${\frac{L}{D}}_\mathrm{Best} = \sqrt{ \frac{1}{K\,4\,{C_{D,0}}}}$

You can pull out the $1/4$ if you want to...

${\frac{L}{D}}_\mathrm{Best} = 0.5 \sqrt{ \frac{1}{K\,{C_{D,0}}}}$

So, for the example you cited...

$C_{D,0}=0.031$ and $K=0.054$

${\frac{L}{D}}_\mathrm{Best} = 12.22$

And, it occurs at ${C_L}^*$

${C_L}^* = 0.758$

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  • $\begingroup$ So, it boils down to how K is derived, and what it's value will be for various aircraft types. But thanks for this (far more mathematicly satisfying) answer. $\endgroup$ Feb 20 at 8:36
  • $\begingroup$ The derivation I've given above works for any two-term parabola. This definition of K is the conventional one. Much of the above derivation also works for a three-term parabola $C_D=C_{D,0}+K_1\,C_L+K_2\,{C_L}^2$, but it gets a lot messier and some of the details don't work out the same. As you can see, the relative values of $C_{D,0}$ and $K$ are what determine the shape of the drag polar. $\endgroup$ Feb 20 at 16:48
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An airfoil does not have $c_{d,i}$.

Induced drag is a 3D effect. An airfoil is a 2D entity. When you look at airfoil drag data in a book, the non-constant value (that sometimes looks like a parabola) is actually the variation of form drag with lift coefficient. It is not induced drag.

An airfoil can readily have an L/D of 50 to 100 -- as you pointed out, the max L/D of a Cessna is about 11.

When you are rolling down the runway, you do not have zero induced drag. The wing is at some angle of attack, it is producing some amount of lift. You then rotate to higher angle of attack (increasing the induced drag) and lift off.

The lift will increase and induced drag will increase -- but you were accelerating on the ground -- i.e. there was already an imbalance of forces, your thrust was enough greater than the sum of retarding forces that you were accelerating.

If you had not rotated (and the runway was long enough), you would have continued accelerating to higher speeds. At some point, you might exceed the speed that the wheels and landing gear can handle.

So perhaps when you rotate, your rate of acceleration is slightly reduced. Of course, you're also losing the rolling resistance of the tires.

The speed for rotation is determined as a speed that is fast enough to allow you to take off safely beyond the stall speed. Immediately after takeoff, you are still in ground effect and will continue accelerating as you climb out. When you reach pattern altitude and level off, you will be traveling much faster than your rotation speed.

In fact, although the power setting is nearly constant for this entire period, your thrust is decreasing as you go faster (prop thrust is greatest at static conditions and drops as you accelerate). So it is clear that at rotation speed, you have thrust substantially greater than drag.

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  • $\begingroup$ "So perhaps when you rotate, your rate of acceleration is slightly reduced". My point exactly. So why does it mathematicly calculate to a coefficient similar in value to Cd0? I suspect for very slick aircraft it might, but for something draggier like a biplane, perhaps not. Makes me think about winglets, strakes, and fairings. $\endgroup$ Feb 20 at 0:30
  • $\begingroup$ No no no. Do not use my quote against me like that. What even is your question at this point? Your title question makes no sense, an airfoil does not have induced drag. Your point about sudden deceleration during takeoff was addressed by my post -- there is plenty of thrust to keep accelerating. Do you want to know why at best L/D, an aircraft has half induced and half parasite drag? This has nothing to do with takeoff. $\endgroup$ Feb 20 at 5:35
  • $\begingroup$ Not against you at all. The observations are in agreement. I'm trying to figure out how the calculated Cdi is so high. $\endgroup$ Feb 20 at 8:12
  • $\begingroup$ I believe I see your error. K is usually defined as $K\,{C_L}^2$. You're defining it as $K=\pi\,e\,AR$. Which means your value of about 20 should really be about .050 -- which matches. $\endgroup$ Feb 20 at 16:45
  • $\begingroup$ that's 0.05 in the numerator or 20 in the denominator. Same. I suspect these calculations are for the wing (3D) only. $\endgroup$ Feb 20 at 17:46
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Is it possible the mathematical modeling is overestimating Cdi?

Maybe they used the wrong data!

For the answer, below, a simple approach is taken that converges on the desired result. In completing some back of the envelope modeling of the 172 (in Excel), supporting background developments in my notes show the following...

We can start with the general proposition

$$T = \frac{T_{Hp}\cdot 550}{V} = \frac{V^2\rho C_DS}{2}$$

where

$$C_D = C_{D_0} + \frac{{C_L}^2}{\pi eAR} .$$

We can estimate $e$ and $C_{D_0}$ or assume a polar.

The value of $$e = \frac{1}{1+\delta} .$$

The variable coefficient $k$ in the above is $(\pi eAR)^{-1}$. Therefore,

$$C_D = C_{D_0} + k {C_L}^2 .$$

Plotting $C_D$ as y vs ${C_L}^2$ as x gives a fitted-line of slope $k$ and intercept $C_{D_0}$. Given slope $k = (\pi eAR)^{-1}$, then $e = (k\pi AR)^{-1}$. We can do this just for the wing alone, or for the whole airplane.

But how does one do this?

Given data in the POH, the whole-airplane approach must necessarily involve something close to a PIW or VIW assessment by using power surrogates for the aircraft for which the POH data are overlaid. In the individual analysis being completed for the 172S or for the 172M, the POH evaluation for each plane was overlayed on a surrogate hodograph that could be independently adjusted for power and velocity to match the POH data. From the surrogate, various elements were extracted as necessary. In other words, a polar-like hodograph was assumed and fitted to the corresponding measures for the aircraft. This is actually a hodographic plot of aircraft $H_p$ available (at the blade) vs $H_p$ required for flight, both as a function of velocity in knots. The analysis was completed based on fitting the derived polar to real aircraft data. Of particular note, data were rectified to velocity in knots and elevation at sea level for standard temperature. Using this process, and an evaluation of data in the POH for each plane, some very interesting perspectives on aircraft performance can be ascertained. The results appear to be surprisingly representative or satisfactory, and are easily validated.

Here are the resulting polars -

enter image description here

Of particular note, data given for for the earlier version POH for the 172 M are not presented as velocities in knots. This was noted as a possible source of error in calculations. The later 172 M POH provides velocities in knots. For the 172 M polar, above, velocities are in knots.

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  • $\begingroup$ So at cruise of 100 knots the Cl need only be half of what it is at 70 knots and 1/4 of what it need be at 50 knots, showing Cd0 and Cdi to be dependent on airspeed. Thanks. Interesting the Cl$^2$ (therefor Cdi) is linear WRT Cd. $\endgroup$ Feb 20 at 8:28
  • $\begingroup$ @RobertDiGiovanni Keep in mind that the induced drag is over there in that $k{C_L}^2$ thingy. Of course $C_{D_0}$ is the drag from all that other stuff catching air... but you probably know all this already... $\endgroup$ Feb 21 at 4:24
  • $\begingroup$ I suppose this is why flying wings were all the rage years ago, but modern gliders seem to be very high AR wings with smooth fairings at attachment points. $\endgroup$ Feb 21 at 13:01
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My quandary is that if all this is true, there should be a noticeable deceleration when the aircraft is rotated at 55 knots while taking off, resulting from a doubling of drag. This is not observed.

Right, this is not observed because this is not the whole story.

Just before taking off, the engine has to win the aerodynamic drag and the rolling friction of the wheels. The aerodynamic drag can be expressed via the usual $D= qS(C_{D_0}+K\,{C_L}^2)$ while the rolling friction is simply proportional to the weight of the airplane via a coefficient $\mu$ which depends on the surface of the landing field: it goes from some 0.03 for dry aspahlt to some 0.1 for unprepared wet field. So, the total force before taking off is:

$$F= qS(C_{D_0}+K\,{C_L}^2) + \mu mg$$

Just after takeoff, the rolling friction disappears and the induced drag becomes the one given by lift equals weight:

$$F= qSC_{D_0}+K\frac{(mg)²}{qS}$$

Let's do the math with the values you supplied plus the ones from Wikipedia:

$C_{D_0}=0.031$, $K=0.054$, $mg=1'111kg \times 9.81≈11'000N$, $S=16.2m²$, $C_L$ before takeoff $≈0.1$, $\mu$ of wet grass $≈0.08$ and $V=55kts≈30m/s$.

We get (just before and after takeoff respectively):

$qS≈ 8'200$

$\begin{cases} F= 8'200(0.031+0.054 \times 0.1²) + 0.08 \times 11'000=1'138N\\ F= 8'200 \times 0.031 + 0.054\frac{11'000²}{8'200}=1'051N \end{cases}$

As you can see there's no much of a difference between before and after rotation and actually after rotation there is a reduction of the needed thrust/power (at least in the conditions used in this example).

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