Does increase in the thickness of a camber airfoil mean more drag will be produced? Why do slow flying planes have thick airfoils when it will just slow them down even more?
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asked Jun 16, 2018 at 14:51
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2 Answers
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There are several reasons for using thick$^*$ airfoils.
- Slow planes need high lift coefficients in order to fly slow. The increased drag coefficient is the prize to pay for a higher $C_{L,max}$. Take a look at this diagram:
Source: H. Schlichting, E. Truckenbrodt, Aerodynamik des Flugzeuges. Colors added.
At low Reynolds-Numbers the maximum lift coefficient is achieved by the thickest airfoil (18%). At slightly larger Re-Numbers the 12% thickness airfoil gains the upper hand. However, the thinnest airfoil (9%) has on average about 20% less maximum lift capability than the thicker ones.
The following diagram shows polars of symmetric airfoils with different thicknesses.
Source: B. Rögner, Flugwissen. I couldn't find the original source Rögner took the diagram from, if anybody knows it feel free to comment.
The quality of image is bad, but you can recognize both the higher $C_{L,max}$ and drag coefficient of thick airfoils.
The drag coefficient depends on many factors and, as shown in the formula given in Peter's answer, is not proportional to the thickness. The drag coefficient's thickness dependency however, can be assumed to be proportional to the thickness: the $\delta^4$ term is one order of magnitude smaller than the term multiplied with $\delta$.
$$\frac{c_{D,\text{normal airfoil}}}{c_f}=2+4\cdot\delta+120\cdot\delta^4$$
$$\frac{c_{D,\text{laminar airfoil}}}{c_f}=2+2.4\cdot\delta+140\cdot\delta^4$$
In Hoerner's Fluid Dynamic Drag, Chapter 6.A.2, you can find more formulas to approximate the drag coefficient and take other parameters into account.
A good resource for understanding the influence of airfoil parameters on their aerodynamic properties is NACA TM 824.
Stall behavior of thicker airfoils is usually more forgiving.
Another reason for using thick airfoils is that it decreases the wing's weight, as thick structures can carry better the bending loads in a wing. If a wing tank is used, thicker airfoils allow for a higher tank volume.
$^*$: I interpret thick as >12%, e.g. 15% as found in the wings of many GA airplanes.
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$\begingroup$ At low Reynolds-Numbers the maximum lift coefficient is achieved by the thickest airfoil (18%) . My R.number is 1million+,I ll be using a 12hp engine for the ultralight...that means I don't have all the thrust in the world to compensate drag,do you still advice the use of a thick airfoil? @Gypaets $\endgroup$ Commented Jun 18, 2018 at 6:38
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1$\begingroup$ @DavidTeahay At Re>10^6 I'd pick an airfoil with 12% to 15% relative thickness. Compared with a 9% thick airfoil you are increasing $c_{A,max}$ by 10-20%, while the drag increase is only a single digit percentage. $\endgroup$– GypaetsCommented Jun 18, 2018 at 21:17
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$\begingroup$ understood!....I last question.@Gypaets..."slow planes need high Cl to fly".....The only airfoil with a desirable Cl for my ultralight is a Goe 518 with thickness of 18%,should I look for another with lesser thickness? $\endgroup$ Commented Jun 19, 2018 at 2:21
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$\begingroup$ @DavidTeahay I can't really tell without knowing more details about the aircraft and how you intend to use it. I'd tend to something thinner, but the Goe is probably a valid choice too. $\endgroup$– GypaetsCommented Jun 19, 2018 at 17:54
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$\begingroup$ Alright😊... .......appreciate! $\endgroup$ Commented Jun 19, 2018 at 18:35
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Yes, and even more precisely it should be worded a bit differently: The drag coefficient is grows linearly with airfoil thickness. Airfoil thickness means that the air has to flow around the airfoil. This displacement effect causes the flow around a thick airfoil to speed up more than around an equivalent but thinner airfoil. The thicker airfoil pushes the air aside and around itself more, causing the flow to accelerate and create more friction than the slower flow around a thinner airfoil. This effect is normally approximated with an additional term in the friction drag formula which is proportional to relative thickness.
Starting from the friction coefficient along a straight wall $c_f$, this additional friction drag has been captured in an empirical formula which gave the best fit to a wealth of airfoil drag data, cambered and uncambered. This is the formula for the zero-lift drag coefficient $c_{d0}$ of an airfoil: $$c_{d0} = c_f\cdot \left(2 + 4\cdot\delta + 120\cdot\left(\frac{1}{\sqrt{1-Ma^2}}\right)^3\cdot\delta^4 - 0.09\cdot Ma^2\right)$$ where $\delta$ is the relative thickness of your airfoil and $Ma$ the Mach number.
That slow flying aircraft use thick airfoils is not generally true. However, larger aircraft want to use thicker airfoils in their wing root in order to make the wing spar lighter. By using a wider distance between the lower and upper spar caps, smaller caps can be used for the same bending strength.
In order to maximise lift, leading and trailing edge flaps are used. Thicker airfoils make their integration easier, and they allow to carry more fuel due to the wing's higher internal volume. However, beyond 20% thickness at subsonic speed and 14% at transsonic speed thickness becomes a liability - the flow will separate too early to make thicker airfoils practical.
answered Jun 16, 2018 at 18:21
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$\begingroup$ Insightful! as always......One question: do you advice the use of a thick airfoil in an ultralight with low speed cos of the higher Cl that thick airfoils odten have $\endgroup$ Commented Jun 16, 2018 at 19:54
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$\begingroup$ @DavidTeahay: No. The structural advantage is low and the lift coefficient is higher for airfoils of 12%. More thickness doesn't help (look at the data in Gypaets' answer - your Reynolds number is safely above 1 million). The nose shape and especially camber are more important than thickness. $\endgroup$ Commented Jun 16, 2018 at 21:40
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$\begingroup$ @Gypaets: There is a constant (to account for both sides of the wing) and then a linear summand which is proportional with thickness. What is not proportional about that? The fact that the line will not run through the origin? That doesn't make it non-proportional. $\endgroup$ Commented Jun 16, 2018 at 21:42
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$\begingroup$ @DavidTeahay: Please accept Gypaets' answer. Maybe he will then stop holding a grudge about my answer. $\endgroup$ Commented Jun 17, 2018 at 17:55
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$\begingroup$ @PeterKämpf The answer is covering the zero-lift range (question is wider). I think will be useful to include the validity of this formulay (I guess not useful getting close to transonic) and also point out that proportionality might not be there when including lifting effects. $\endgroup$ Commented Jun 18, 2018 at 22:14