I found this equation online but it is not from a reputable source and has no derivation. Could anyone help me by deriving this equation or even verifying its accuracy? I would love to cite and use this equation in my assignment so urgent help would be most appreciated. enter image description here

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    $\begingroup$ There is no such simple relationship. The wings profile shape also influences the Lift-to-drag ratio. $\endgroup$ Commented Apr 16, 2018 at 7:59
  • $\begingroup$ Merely the fact that the text and example discusses paper airplanes, wing spans the tens of cm range, and a launch height of 2 m, should hint that the formula probably isn't applicable to the general case, or even to all gliders. $\endgroup$
    – user
    Commented Apr 16, 2018 at 11:08

2 Answers 2


Your mystery equation is $$\text{Distance} = \text{Height}\cdot0.5\cdot\sqrt{\pi\cdot e\cdot\frac{b^2}{f}}$$ Since the flight distance is the product of the glide ratio (or lift-to-drag ratio) L/D and height, the equation can be written like this: $$\frac{L}{D} = \frac{c_L}{c_D} = 0.5\cdot\sqrt{\pi\cdot e\cdot\frac{b^2}{f}}$$ if we use dimensionless coefficients instead of the dimensionful forces. Now that does look familiar, right? Only something seems to be missing. And that drag area $f$ should be the wing area.

Now you asked for the derivation, and I've done that only with wind in this answer, so let's do it for the simpler case of a glider and without wind right here. You start with the question: What polar point gives the best distance for a given height? Or, in other words, when does the lift-to-drag ratio reach its optimum? To get anywhere, you need a simple equation for the lift-to-drag ratio, so I recommend to start with the equation for the drag polar $$D = \frac{\rho\cdot v^2}{2}\cdot S\cdot\left(c_{D0}+\frac{c_L^2}{\pi\cdot AR\cdot\epsilon}\right)$$ The terms inside the big bracket are the sum of the parasitic and the induced drag coefficient and the term in front of the bracket is the product of dynamic pressure and wing area. It is needed to turn the coefficients into a force. The symbols are:
$\kern{5mm} \rho\:\:\:\:\:$ air density
$\kern{5mm} v\:\:\:\:\:$ velocity
$\kern{5mm} S\:\:\:\:\:$ wing surface area
$\kern{5mm} c_{D0} \:$ zero-lift drag coefficient
$\kern{5mm} c_L \:\:\:$ lift coefficient
$\kern{5mm} \pi \:\:\:\:\:$ 3.14159$\dots$
$\kern{5mm} AR \:\:$ aspect ratio of the wing
$\kern{5mm} \epsilon \:\:\:\:\:\:$ the wing's Oswald factor, normally between 0.8 and 1.0

Since lift should be constant over speed for a glider, the optimum lift-to-drag ratio is when drag reaches a minimum. We express the lift coefficient by what it stands for and differentiate drag with respect to speed: $$D = \frac{\rho\cdot v^2}{2}\cdot S\cdot\ c_{D0}+\frac{(m\cdot g)^2}{\frac{\rho\cdot v^2}{2}\cdot S\cdot\pi\cdot AR\cdot\epsilon}$$ $$\Rightarrow\:\:\frac{dD}{dv} = \frac{2\cdot \rho\cdot v}{2}\cdot S\cdot c_{D0} - \frac{2}{v^3}\cdot\frac{2\cdot(m\cdot g)^2}{\rho\cdot S\cdot\pi\cdot AR\cdot\epsilon}$$

Drag should reach its minimum when the derivative equals zero. We try to isolate the zero lift drag by dividing the equation by all what is collected around it: $$0 = c_{D0} - \frac{2}{v^4}\cdot\frac{2\cdot(m\cdot g)^2}{\rho^2\cdot S^2\cdot\pi\cdot AR\cdot\epsilon}$$ Now bring $c_{D0}$ to the other side and re-insert the lift coefficient. The result is: $$c_{D0} = \frac{c_L^2}{\pi\cdot AR\cdot\epsilon}$$ This means that the optimum lift-to-drag ratio is reached when the zero-lift drag equals induced drag. Total drag is the sum of both and in this case equal to twice the induced drag. Written in coefficients, this is: $$c_D = \frac{2\cdot c_L^2}{\pi\cdot AR\cdot\epsilon}$$ Written as the lift-to-drag ratio, this becomes: $$\left(\frac{L}{D}\right)_{opt} = \frac{\pi\cdot AR\cdot\epsilon}{2\cdot c_L}$$ Since the optimum lift coefficient can be expressed as $$c_{L_{opt}} = \sqrt{c_{D0}\cdot\pi\cdot AR\cdot\epsilon}$$ this can be inserted and the lift-to-drag ratio can be written as $$\left(\frac{L}{D}\right)_{opt} = 0.5\cdot\sqrt{\frac{\pi\cdot AR\cdot\epsilon}{c_{D0}}}$$ The wing's aspect ratio is $AR = \frac{b^2}{S}$, and the mysterious drag surface must be $f = c_{D0}\cdot S$, and voilá, the strange equation is proven right. When the correct values are used, it is valid from mini paper airplanes up to Jumbo Jets and above. Airfoil drag is included in the zero-lift drag coefficient.

  • $\begingroup$ You may add links and references to provide further reading (I m curious about where this equarion cones from and who uses it for the first time) $\endgroup$
    – Manu H
    Commented Apr 18, 2018 at 6:06
  • $\begingroup$ Dear down voters, if you don't understand the answer, please consider asking in the comments instead of blindly down voting what is a correct answer. $\endgroup$ Commented Apr 18, 2018 at 11:45
  • $\begingroup$ @ManuH: Sheesh - are you guys never pleased? It’s the sum of parasitic and induced drag, plain and simple, and was most likely proposed first by Prandtl or one of his pupils. $\endgroup$ Commented Apr 18, 2018 at 17:44
  • $\begingroup$ it is not a question of being pleased but of having more references not only for further readings but also to ease navigation in external resources (I do believe it improve quality) $\endgroup$
    – Manu H
    Commented Apr 19, 2018 at 5:53
  • $\begingroup$ @ManuH: It is really the most simple and most often used of equations, so I was not expecting to defend its use. I added more explanation, but cannot really prove who has used it first. $\endgroup$ Commented Apr 19, 2018 at 6:17

You may be confusing wingspan with aspect ratio, roughly span over chord. More accurately, AR=square of wingspan divided by wing area. Still not the whole story on glide ratio, but it's most of it I think.

The FAA's Glider Handbook is a good starting reference.


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