Goal of your design

Basic wingdesign

L= lift; W= weight; D= drag; T= thrust.
D is much less then L.
The smaller D/L, the better.

The picture shows what forces are present when flying at cruisespeed ( constant speed, constant height). All forces are compensated.


Weight is a constant, ...unless your passenger chooses the safety of a descent by parachute above a landing with your flight-skill.

I do not take in count the fuelconsumption. Could be it will be taking in count in a further stage of the theory.


At levelflight at a constant speed lift compensates weight. So L = W (formula 1). Main formula for lift is:

L = 0,5 . . v2 . Cl . S

(formula 2)
  L = lift
rho.jpg (829 bytes)= density air
v = velocity (speed)
Cl = lift coefficient
S = wingarea

Let's simplify this formula. The density of air is 0,3125 kg/m3 (0.0195 lb/cubic ft or 0.52675 lb/cubic yd) at sealevel. So it is a constant at a certain height. Cl does not change at levelflight and at a certain speed. Wingarea is also a constant. What's left?

L= constant . v2 (formula 3)

So lift is related to the second power of speed. But most important is another relation. If we put formulas 1 and 2 together you see:

th_004.jpg (2630 bytes) (formula 4)

Speed needed to keep the airplane in the air is related to the wingloading. Don't make the mistake thinking that this is the stallspeed! For stallspeed you need to look further. This makes that the ultralights for Belgium have a fixed speedrange, because the wingloading is limited by law. Another conclussion could be that unswepted flying wings fly slower than a swepted flying wing. Unswepted flying wings use a reflexed airfoil and these airfoils have a lower Cl. There is more wingarea needed to create the same lift needed to keep the weight in the air. More wingarea for the same weight means lower wingloading. Formula 4 says that then the needed speed gets lower too.


Most would think now that if you increase speed all will be well. That would be true if drag did not excist. But it does. Drag has two different components. Parasite drag and induced drag. Parasite drag is the commonly known frictiondrag. Induced drag is more difficult. I have found a explanation that says: "energy lost to create lift is induced drag" ("Composite construction for homebuilt aircraft" by Jack Lambie,  Aviation Publishers, Hummelstown, PA, USA).

D = Dp + Di (formula 5)

I also found in nearly every book the following formulas. Even if you know these formulas, make sure to read further in this dragsection, because some people forget one detail.

Dp = 0,5 . rho.jpg (829 bytes) . v2 . Cd . S (formula 6)

Di = W2 / rho.jpg (829 bytes) . v2 . b2 (formula 7)

A important relation can be seen between Di and v. The higher the speed, the smaller the induced drag! Now we introduce ar.jpg (732 bytes) or AR (aspect ratio) to see another relation. There are several ways to calculate aspect ratio.

ar.jpg (732 bytes) = b2 / S (formula 8)

ar.jpg (732 bytes) = S / k2main (formula 9)

ar.jpg (732 bytes) = b / kmain (formula 10)

Let's put formula 7 and 8 together.

Di = W2 / rho.jpg (829 bytes) . v2 . S . ar.jpg (732 bytes) (formula 11)

The higher ar.jpg (732 bytes), the smaller Di. That is why most gliders have slender, long wings. These wings need to be extra strong, because at such length they are very fragile. Extra strong, means extra weight. Another formula I found shows another relation. First you need to know that:

D = 0,5 . rho.jpg (829 bytes). v2 . (CD + Cdi) . S

(formula 12)
  CD = drag coefficient total airplane
Cdi = induced drag coefficient

Cdi =


(formula 13)

pi.jpg (794 bytes) . ar.jpg (732 bytes)

remark_small.jpg (1234 bytes) I received this remark from Don Stackhouse. "I know about the famous equation that relates induced drag to aspect ratio. If you look closely at that equation you will notice it also includes Cl. In most real-world applications there is a relationship between aspect ratio and Cl, and when you factor that relationship into the equation, you will find that aspect ratio and Cl drop out, and leave you with only span and span efficiency. Most discussions of induced drag I've seen lately seem to overlook this."

Thanks, Don. I didn't know that. Could somebody quote me the relation between aspect ratio and Cl. I would like to state it in this section.

This curve shows that the total drag is the lowest at a certain speed and not at the lowest speed.

Here you have two important things to see. In formula 13 you see that the the lift-coefficient is related to the induced drag. If you read the explanation of Jack Lambie about induced drag it isn't hard to understand. The more lift created, the more energy lost, the higher the induced drag.

Q: One thing I don't understand is that if you use formula 13, the induced drag is related again to v2 and not 1/v2 (you need to combine formula 13 and 12 to understand this). Isn't the induced drag normally reducing when the speed increases? Can someone clear this out for me? Did I forget a certain condition or something else? A: (Ivan G. Becuwe) "Formula 13 contains Cl2 and Cl is related to 1/v2 "(crude translation from original text). So putting Cdi in formula 12 creates crudely the relation:

D= something . (1/v2)2 . v2
And that can be transformed into
D = something . 1/v2
And this is what we wanted to reach.

Secondly you see in formula 12 that CD stands for the parasite drag coefficient of the WHOLE airplane. Many forget to keep the drag of the fuselage, wheels, struts, tail and so in mind while designing a model. Here follows a typical conversation with a modelmaker about his first own design.

Modelmaker: "My flying wing model does't perform like it should."
Other person: "What did you do?"
Modelmaker: "Well, I designed according to the book my wing. You know ... airfoil, wingarea, control areas and so. It all looked nice on paper."
Other person: "What did you do then?"
Modelmaker: "Then I added a fuselagepod, wheels and some fancy dummy rockets."
Other person: ...speechless...

You have to keep in mind that drag not only works on the wing. It also works on the fuselage. If you work on a ultralight, you will see that the fuselage is rather large in reference to the wing. The frontal area of the fuselagepod can be used to estimate the parasite drag.

Dp = cd_pi.jpg (952 bytes) . a_pi.jpg (945 bytes) . q

(formula 14)
  cd_pi.jpg (952 bytes) = drag coefficient based upon a_pi.jpg (952 bytes)
a_pi.jpg (952 bytes) = frontal or cross sectional area
q = 0,5 . rho.jpg (829 bytes) . v2

In the book "The design of the aeroplane" of Daroll Stinton I found a table with some crude estimates of some airplaneparts. A enclosed cockpit of a light airplane with tractorpropeller had a cd_pi.jpg (952 bytes)-range between 0.1 and 0.2. If you imagine a small cockpit of 0.8 m wide and 1.1 m high, you will see that the drag created by this airplanepart may not be forgotten in a complete design. Not to mention the drag of the gear!

Part of airplane

cd_pi.jpg (952 bytes)

Cockpit of small airplane 0.1 to 0.2
Cockpit of large airplane 0.07 to 0.12
Nacelles for propeller above wing 0.2
below wing 0.06
Jet 0.05 to 0.07
Wingtanks under wing 0.2 to 0.25
at tip 0.06
Struts 0.12 to 0.2
Bracing 0.22
Hull for flying boat 0.11 to 0.15
Float for seaplane 0.15 to 0.2
Parachute (extended) 1.33

(multiply the sum of the above drag areas by the landing gear factors shown for each type of arrangament

fixed tailgear unfaired 1.3
faired 1.2<
fixed nosegear unfaired 1.35
faired 1.25
Retracting into fuselage 1.08
nacelles 1.03
wings 1.0 <

We hope that you could swallow this chapter before we do mention the factor e or K. The former description of lift is true when using a wing with infinite span. Since we don't have such wings, we need to view the problems which are created at the tip of a wing. What happens? Under the wing you have a positive pressure, above the wing you have a negative pressure. At the tip there is a "leak" of the positive pressure zone towards the negetive pressure zone. This leak creates a vortex (little tornado at the end of the wing). The larger this vortex is, the more energy is lost when creating lift. So Cdi is related to this vortex. Lift is also related to this vortex, because there is little lift in the wing area where the leak passes by. Not all the wingspan creates lift. A certain part of the wing is lost. e is a factor that shows how much of the wing is usefull to create lift. So if e = 0,9 then 10% of the wingspan is lost due to the vortex.

hlp_small.jpg (1417 bytes) Can someone give me some true e -figures of some known aircrafts. I would like to use them as a example. And if you might have a picture of this aircraft (without copyright), could you please send me a scanned copy (JPG-file).


K = (1 + f_haak.jpg (760 bytes)) = 1/e

(formula 15)

These symbols are used in different books, but are all symbols related to efficienty

Formula 2 and 13 get transformated into:

L = 0,5 . rho.jpg (829 bytes) . v2 . Cl . S . e

(formula 16)

Cdi =


(formula 17)

pi.jpg (794 bytes) . ar.jpg (732 bytes) . e

As you can see, as e gets less lift gets less, but Cdi gets larger.


T = D (formula 18)
In levelflight at a constant speed

I found the following formulas.

v0 = n.jpg (781 bytes) .

haak_l.jpg (891 bytes)

2 . P

haak_r.jpg (896 bytes) 1/3 (formula 19)
pi.jpg (794 bytes) . rho.jpg (829 bytes) . D2 . (1-n.jpg (781 bytes))
v0 = velocity of incoming flow
n.jpg (781 bytes) = efficiency factor propeller
P = given engine power
rho.jpg (829 bytes)= density of air
D = diameter propeller

T =

Pengine . n.jpg (781 bytes)

(formula 20)


  T = thrust

You need to know that friction and rotational losses are not included in these formulas. Our classic propellers have in reality a efficiency of 10 to 15% less than the efficienty of the ideal propeller (stated here). Keep this in mind while making estimated calculations!

Another thing you need to know is the fact that a airplane not only has to fly level, but it also needs to be able to climb. If two airplane have two same engine and prop, which one is going to climb the best. Well, it is going to be the one with the lowest wingloading. Why? It will have the lowest sinkrate and will need less power to fly level (=at the same height). So all the remaining power can be used to climb. The other airplane with the higher wingloading needs more power to fly level and has less reserve to climb.

Once a member of the Nurflugel-mailinglist did ask what power he did need to make his glider self-launching and he have the data of his glider.

The glider weighs 180 lbs, and with me its gross is 330.
Wingspan 42ft
Empty Weight 185 lbs
Aspect ratio 9.5
Wing area 185 sq. ft
Pilot Weight 100-230 lbs
Glide Ratio 19 to 1 at 40 mph
Min sink rate 162 ft/min at 30 mph
Speed Range 35 to 70 mph

Al Bowers did answer: "The important ones are L/D and min sink and weight. Make sure the units are the same (everything is "per second" and "pounds-force", etc; or do it all SI and remember
that kilograms is NOT weight!)...

Power = weight * min-sink = 891 ft-lbs/sec = 1.62 HP

That's after the prop losses and electrical losses, and that is also the absolute MINIMUM required for level flight. Small props with high RPM tend to be pretty bad efficiency-wise, so 4 HP is probably the minimum for level flight sustainer. If you want to climb with it, then 12 HP might get about 500 fpm climb rate. I'm assuming a prop efficiency of 0.5 (which might be high for some of the "lawnmower" engines being used in the ultralight community)." (L/D and glide ratio are the same)

What Al Bowers didn't mention in the text, but he did use it to get his result, is the fact that the formula (slightly modified) also can be used to get the needed power to not only fly level but to climb as well.

Needed power = weight * (minimum sink + climb rate) (formula 21)

Be sure to use all the right dimensions. Weight is Newton (or lbs), minimum sink rate in m/sec (or ft/sec) and climb rate in the same dimension. Now try to find the data of a motorized glider and see if your calculations fit with the data of the glider.

In the sector "Basic wingdesign"  we will show you some other formulas (mainly related to wingform and performance).

If you have a problem understanding this section, please state us your question. We will try to search for a possible answer.

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