# How can I calculate the TOTAL drag of the ENTIRE airplane?

I have an aircraft similar to a Cessna 172 and am using a virtual wind tunnel software (flowdesign) to get a drag estimate. I import the airplanes 3D model and I get around 2240 lbs of drag with a drag coefficient of 0.33 at a speed of 253fps, which I think is way too much to be trusted.

I then tried to calculate it using an equation (although the equation is supposedly only for wing drag): D=q∗S(Cd0+CL^2/(π∗AR))

Where q is dynamic pressure, S is the wing surface area (208sqft in my case) and AR is the wing aspect ratio (6). I used a D/q value of 7 (to be conservative) and a Cd0 of 0.033. With that, I got a result of 560lbs of drag (remember, this is supposed to be just for the wing) at a speed of 253fps (which is the speed at which the best L/D ratio is supposed to be according to the equation - and is the same speed I used in the virtual wind tunnel)

With that result, I then went to calculate how much thrust I would need for a climb rate of 1250fpm. The equation is: T=W*(Vv/V+D/L)

Where W is the weight of the airplane, Vv is the targeted vertical speed (1250fpm), V and D are the speed and drag for the best L/D ratio (so 560lbs and 253fps), L is probably the lift and I used the same value as the weight there. With that, I got the result of 923lbs of thrust needed. That is around 723 horsepower.

Now if I go look up the horsepower of a Cessna 172, google says it is 180. The climb rate is 720fpm, which is half of what I targeted for, but the difference is still too big for my calculations to be right. The wing area is 175sqft, only about 30sqft less than I have.

I then tried to calculate the drag with another equation, in which I additioned the induced and parasitic drag. The equation for the induced drag is: Di=Cdi*S*q

Where Cdi=Cl^2/(π∗AR) where Cl=W/(q*S) and I got a result of about 118lbs of induced drag. For parasitic drag, I used Dp=D/q*q and got around 284lbs of parasitic drag. The total drag is 403lbs, which I believe is still way too much to be true.

I then tried using the wright equation, which is: D=Cd*q*S, and I got a result of 3160 lbs of drag using a Cd of 0.033 (now I know I probably should have used something closer to what I got from the virtual wind tunnel, but the result would have been even greater) and a total area of 3160sqft (that is the total wetted area of the whole airplane).

All of the above results could be countered with enough horsepower (and that is somewhat realistically doable), but when I compare the horsepower required, the climb rate and the wing area to other already existing planes, my calculations seem off.

So what is a simple way of calculating the total drag of the entire airplane accurately?

Sorry for the wall of text.

P.S.: if anyone is interested, I am using these videos (or series of videos) to conceptualize an airplane design:

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• So, you are basically calculating the drag with 3 equations which are the same and you get 3 different results. I suggest you to stick with a consistent system of units (best SI) and recheck the calculation. And get some standard book about general airplane design (Raymer for example) instead of online tutorials: they are really easy to read and full of (correct) numbers for any aerodynamic coefficient 🖖 Sep 18 at 17:17
• Have also a look to this questions: aviation.stackexchange.com/questions/43410/… aviation.stackexchange.com/questions/56629/… Sep 19 at 9:26
• Does this answer your question? How is the zero-lift drag coefficient calculated? Sep 19 at 19:18

EDIT

Doing the math at a speed of 149kts (as suggested by @RobertDiGiovanni) and a weight halfway between EW and MTOW, gives a total drag of some 308lbs which should be a reasonable value.

Try with:

$$C_{D_{total}}=0.02+0.05C^2_L$$

this should give you a reasonable target for your simulations.

How can I calculate the TOTAL drag of the ENTIRE airplane?

Total drag polar for an aircraft can be calculated as the sum of 1. the polar of the wing plus 2. the drag of all the other components:

1. $$C_{D_{wing}}=C_{D_{min}}+ k(C_L-C_{L@C_{D_{min}}})^2$$
2. $$C_{D_{rest}}= C_{D_{fuselage}}+C_{D_{landing gear}}+C_{D_{nacelle}}+…$$

where:

1. $$C_{D_{min}}$$, $$k$$ and $$C_{L@C_{D_{min}}}$$ depend on the particular airfoil(s) used for the wing and its geometry (sweep angle, aspect ratio, twist, ...).
2. each term of $$C_{D_{rest}}$$ is normally estimated using known standard historical values, scaled via the basic dimension of each component.

Each and every term depends also on the speed i.e. on Reynolds and Mach numbers. Drag terms of 2. may also depend on lift (landing gear for example).

The sum of 1. and 2. gives the following general form for the drag coefficient:

$$C_{D_{total}}=a+bC_L+cC^2_L$$

Normally the second term $$b$$ is neglected and the total drag coefficient assumes the well known form:

$$C_{D_{total}}=C_{D_0}+KC^2_L$$

For a single-engine light-aircraft with symmetrical airfoil and flying at subsonic speed, plausible rough values for $$C_{D_0}$$ and $$K$$ based on historical values are:

$$C_{D_{total}}=0.02+0.05C^2_L$$

Anyway a better approximation (within a ±10% error) can be achieved only evaluating each term of 1. and 2. I suppose that doing the math here for one particular case is beyond the scope of this Stackexchange, also because it would take some time and space. The values to be used and the general theory behind it can be found in any standard book of general airplane design, like the one by D. P. Raymer, Dr. J. Roskam, E. Torenbeek or L. Nicolai. All this books are really easy to read, fluent and with almost no equation.

• Running the calcs at high cruise finds us squaring a Cl value of less than one. Why don't you show one calc at 149 knots. It'd be worth an upvote perhaps. Sep 20 at 16:31
• @RobertDiGiovanni: I don't believe in doing the homework of somebody else (not referring to you), I prefer to give the tools and show the path ;-) Anyway I'm going to do it, I just need some time Sep 20 at 17:16
• Thanks for the answer. Also, you were right in your comment under my question. My dumbass forgot to convert from knots to fps (should have been 150 fps, not knots). I have redone my calculations and now the result seem more reasonable. I also want to thank @RobertDiGiovanni for his answer, having something to compare my results to is very useful as a first rough "reality" check. Sep 22 at 21:03
• :-D glad I could help 🖖 And seriously consider getting one of those books, older version can be found almost for free Sep 23 at 6:46

Drag of the $$entire$$ airplane is readily available by gliding, or monitoring fuel consumption.

Checking fuel consumption is a bit more complicated because what you really need is thrust.

in steady state flight, thrust = drag

But just to get "in the ball park", a 172 weighing 2400 lbs gliding 8:1 generates 300 lbs of drag at Vbg. That is about how much thrust you need for level flight, with around 300 lbs more available to climb.

Of course these numbers will vary depending on airspeed and configuration but we can see even 923 lbs of thrust is a bit high.

253 feet per second comes out to 149 knots$$^1$$! Best L/D for a 172 is around 65 knots, or around 110 fps. Try that number and see what you get.

$$^1$$ keep in mind "cruise" numbers in POH are true airspeed, where higher altitude will help. 403 pounds of total drag here seems to be closest to reality, especially considering decrease in fixed prop thrust at that airspeed.