The fact: Gliders have ballast tanks that can be filled with water. The addition of ballast increases weight, and this allows the glider to fly at faster airspeeds while maintaining the same glide ratio. This means that if two gliders P1 and P2, being both of the same model, start at some point A, and P2 weights more than P1, they will both end at point B, but P2 will arrive before P1. The movement is rectilinear, and during the flight, each plane is able to maintain constant airspeed and vertical speed, being P2's speed greater than P1's. Angle of attack might be different too. This is a little diagram I made:

enter image description here

To simplify things, let's consider the movement is made inside uniform air, with no density variation with height, and winds are calm. I've read some aviation manuals, and the most frequent explanation provided is that the glider retains the same Lift/Drag (L/D) ratio, but nowhere in the books I've read so far includes the proof. I'm trying with my very limited skills to reach an expression relating speed with weight, but almost always get lost in the process.

This second diagram I made shows the different forces involved. As gliders don't (usually) have an engine, there's no thrust, and the only forces present are Lift (L), Drag (D) and Weight (W). (Sorry, arrow lengths are not proportional to vectors magnitudes): enter image description here

As the movement is uniform, the acceleration is null. Then, Newton's Second law tell us that:

${\vec F = m·\vec a}$

As the mass cannot be null, then the resulting force F should be null. So we have:

${\vec F = \vec F_x + \vec F_y}$
${\vec F_x = \vec L_x + \vec D_x = \vec 0}$
${\vec F_y = \vec L_y + \vec D_y + \vec W_y= \vec 0}$

The lift formula is:

${L = \frac{C_L(\alpha)V^2 \rho S}{2} }$


  • ${C_L(\alpha)}$ Is the coefficient of lift, which is proportional to the angle of attack below the stall.
  • ${V}$ is the speed. Shame on FAA handbook for not detailing which speed it was. According to Wikipedia, it is IAS (Indicated Airspeed) in the direction of movement, which makes sense.
  • ${\rho}$ Is the air density
  • S is the wing surface area.

Drag formula is:

${D = C_D(\alpha) q S }$


  • ${C_D(\alpha)}$ Is the drag coefficient, which is a function of the angle of attack, roughly quadratic.
  • ${q}$ is the dynamic pressure (???).
  • S is the wing surface area.

Again, sorry for not including units. I continued transforming the expressions but couldn't get a clear W/V equation, instead I ended up with a messy trigonometric expression. There might be an easier way, perhaps approximating some coefficients? Anyway, my questions:

  1. Are all assumptions above correct?
  2. In the real thing, airspeed may be constant, but ground speed is a different thing. Is the movement really uniform?

(Sorry for my awful English, it is not my native language)

Thanks in advance.

Here's my attempt to express speed as function of the mass:

${\vec F_y = \vec L_y + \vec D_y + \vec W_y= \vec 0}$

${|\vec L|\cos(\varphi) = -|\vec D|\sin(\varphi) -gm}$

${\frac{C_L(\alpha)V^2 \rho S}{2}\cos(\varphi) = -C_D(\alpha) q S \sin(\varphi) -gm}$

As lift and drag coefficients are constant for a given angle of attack, ${\rho}$ and q are constants because the air is uniform, and S is constant because the glider flies without modifying the wing area (no flaps added), we can group the constants and this is what we get:

${k V^2 = -g m + k'}$

  • 3
    $\begingroup$ For a non-native English speaker, your English is very good. $\endgroup$ – Mike Dunlavey Jul 4 '12 at 18:24
  • $\begingroup$ v is true air speed (TAS), or the addition of $\rho$ to the formula would not be necessary. $\endgroup$ – Peter Kämpf Aug 12 '18 at 21:47
  • $\begingroup$ The result I calculated is $V^2 = 2mg\cos \phi/ (CρS)$ $\endgroup$ – enbin Mar 12 '20 at 11:20

When a glider is trimmed for maximum glide range, for which Lift over Drag (L/D, also called glide ratio) is maximized, it descends at an angle whose tangent is the inverse of that ratio. For example, a Cessna 172 has a glide ratio of about 9 at roughly 65 knots. A good glider can have a glide ratio of 30.

The glide ratio (within reason) is not a function of weight, because lift and drag vary together, proportional to velocity squared.

Velocity itself is a function of weight, because lift has to equal weight. When an aircraft is balanced in a normal way (slightly nose-heavy) it automatically seeks the speed necessary to support its weight.

So adding weight increases velocity, so as to generate the additional lift needed to support the weight. That additional velocity increases the drag by the same proportion.

So, the glider is flying faster, and descending faster (because sink rate is proportional to drag), but since the drag and lift are in the same proportion, the angle of descent is the same.

As an aside, aircraft that are "slippery", or have low drag, are difficult to descend in. You can see the airport just fine, but you can't get to it, because you can't get down. To remedy that, you need a way to either reduce lift (with spoilers) or increase drag (with speed brakes). Some spoilers double as speed brakes. If you find yourself in a commercial airliner approaching an airport, and you see the spoilers on the top of the wing being raised, and you hear a roaring sound, that's what's happening. The pilot is trying to lose altitude without gaining additional speed. Other aircraft do it by "slipping" - that is, using the rudder to put the plane sideways to the wind, increasing drag.

  • $\begingroup$ +1. As ${\vec L_x = - \vec D_x}$; then ${|\vec L|\sin(\varphi) = - |\vec D| cos(\varphi)}$; thus ${\tan(\varphi) = -|\vec D| / |\vec L|}$. $\endgroup$ – Mister Smith Jul 5 '12 at 15:09
  • $\begingroup$ I didn't noticed it at first, but the rate of sink vs airspeed is depicted in the so called polars graphs. Minimum sink rate is determined from this graphic by drawing the tangent to the curve starting at point (0,0). this is why sink rate is expressed in knots in gauges instead of feet per minute as it is usual in other aircraft. $\endgroup$ – Mister Smith Jul 5 '12 at 15:18

There is an omission in input data: glider's "fuel" is a sum of its potential and kinetic energies. So heavier glider (towed to same height and velocity) requires more work from the towing plane and starts with more energy than lighter glider.


enter image description here

As shown, $AB$ is the wing of the glider. The yellow arrow is the speed direction $V$ of the glider. $G$ is the weight of the glider. $G$ has two components, one is a component $G_t$ parallel to the velocity direction, and one is a component $G_n$ perpendicular to the velocity direction. There are also two components of the air force applied to the wing, one is the component $D$ parallel to the speed direction, and the other is the component $L$ perpendicular to the speed direction.

If the speed $V$ of the glider is low, then $G_t> D$ and $G_n> L$. Thus, under the thrust of $G_t$, the speed $V$ of the glider increases. As the speed $V$ increases, $D$ will increase, and $L$ will increase. Finally, $G_t = D$, $L = G_n$, and the glider will glide along the yellow line at a uniform speed.

From this analysis of the glider, we can see that the flight of the glider is a thrust flight. Without gravity as the thrust, the glider cannot fly and cannot generate lift.

Increasing the weight $G$ of the glider will get a larger $G_t$, so the glider can fly further.

enter image description here


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.