# Polar, Performance, and Water Ballast

Have you ever wondered what differentiates the 'high-performance' gliders from the normal two-seaters, apart from the price tag and the rather fragile appearance? What are the ridiculously long wings of a Duo-Discus good for? Or more importantly, after you have paid the launch fee, what can you do to stay in the air for longer? The answers to these questions require an understanding of the performance metrics of the glider.

You might have heard of more experienced pilots talking about 'polars', or you might have seen the convex curve which is confusing to get started with. You might also have seen people adding or dumping water into and out of their gliders. Building on the knowledge of glider performance, we can have a closer look at how these tools help cross-country pilots to fly faster and further.

There is a wealth of text, published or online, discussing the topics mentioned above. However, some of these are rather scattered pieces of discussions on the forums, or they can be written in another system of conventions than what is adopted in Cambridge. Some are sloppy about their assumptions and approximations, and some dive straight into the calculus making it impossible to follow. This work aims to present the derivations of the governing equations and the polar functions in a clear and detailed manner, and summarises the implications for those who would rather not follow the mathematics.

1. Consider the forces acting on the glider in unaccelerated flight: lift, drag, and glide ratio.
2. Aerodynamic coefficients: definitions and meanings.
3. Relationship between lift and drag.
4. General method of solution, and assumptions necessary to simplify it.
5. Analytical form of the glide polar.
6. Implications of the polar: minimum sink speed, and best glide.
7. Implications of the polar: water ballast
9. The non-dimensional polar and recommended readings.

## Glider in Unaccelerated Flight in Still Air

We start by considering a glider in unaccelerated flight in still air. We shall assume the following:

1. The aeroplane in question is a glider, i.e. it creates no thrust.
2. The flight is unaccelerated, i.e. the glider is flying straight and level without changing its airspeed.
3. The air is still, i.e. there is no macroscopic movement of air in forms such as wind, thermals, ridge lift, etc.
4. The air is homogeneous in its thermodynamic properties, especially, it has a uniform density $$\rho$$.

### Governing equations from a force perspective

Hopefully you already understand how a glider can remain airborne, but just in case you are in confusion, consider an unpowered glider in unaccelerated flight in still air: three forces act on the glider, namely:

1. Gravity (weight), pointing vertically downwards.
2. Lift, pointing upwards and perpendicular to the flight path.
3. Drag, pointing backwards and along the flight path.

By Newton's first law, in order for the glider to stay unaccelerated, these three forces must balance. Imagine the glider is flying horizontally. If this is the case, then the lift force must point vertically upwards. We then have a drag force pointing horizontally backwards with no force balancing it, because the other two are both in the vertical direction.

Therefore, the only way for the forces to balance is that, the glider cannot be flying in the horizontal direction. The flight path must be at an angle to horizontal. We shall denote this angle as $$\theta$$. By experience, a glider in unaccelerated flight in still air keeps descending, rather than climbing. Therefore, we know the flight path is inclined downwards. We shall define this direction as positive $$\theta$$.

With this made clear, the gravity ($$W$$) can be decomposed into two components, one to balance the lift ($$L$$), and one to balance the drag($$D$$). The following relationship holds:

$W \sin(\theta) = D$ $W \cos(\theta) = L$

Dividing these two expressions, $$W$$ can be eliminated, giving:

$\frac{L}{D} = \frac{1}{\tan(\theta)}$

The quantity $$\frac{L}{D}$$ is referred to as the Lift-to-Drag Ratio.

### Governing equations from an energy perspective

An alternative way to think about this is from an energy perspective. Because the drag force wants to slow down the glider and take its kinetic energy away, the glider must keep descending, so that it releases its gravitational potential to make up for the loss, otherwise it cannot remain at the same speed. Consider riding a bicycle: if you stop pedalling on level ground, you will gradually slow down and eventually stop, this is because drag force steals your kinetic energy away and you have no means of replenishing it. However, if you cycle downhill, you will not stop even if you do not pedal.

Therefore, we conclude that a glider flies downhill. This is in agreement with the conclusion of the previous section. We can borrow the notation and call the slope angle of this imaginary hill $$\theta$$. Geometrically, if we travel for a unit distance on the face of the hill, then in the horizontal direction the distance travelled will be $$\cos(\theta)$$ and in the vertical direction the height drop will be $$\sin(\theta)$$.

From an energy conservation point of view, the following expression holds (it means the energy that the drag force uses up equals to the energy the gravity must provide):

$D \times 1 = W \times \sin(\theta)$

This is the same result as in the last section.

It is worth noting that the energy approach tells us nothing about the lift force directly.

### Glide ratio

The glide ratio is a measurement of the efficiency of the glider. It means 'how many feet can the glider travel forward for every foot of altitude drop?' If a glider has a glide ratio of 50:1 (fifty-to-one), it means the glider is capable of travelling 50 feet forward for every foot of altitude drop, the same thing applies in meters, etc. We want the glider to travel as far as possible while losing minimum altitude, therefore, a larger glide ratio is favourable over a smaller one.

This measurement we can vaguely call 'performance', although strictly speaking this is only one aspect of it. We can say glider A (50:1) has a higher performance than glider B (30:1), for example.

In the last section, we concluded that, for each unit distance travelled on the hill, the horizontal distance covered is $$\cos(\theta)$$ and the vertical distance covered is $$\sin(\theta)$$. Therefore, the glide ratio is:

$\textrm{Glide ratio} = \frac{\cos(\theta)}{\sin(\theta)} = \frac{1}{\tan(\theta)} = \frac{L}{D}$

This is a very important result. It is also worth noting that, up until now, we have made no approximations.

### Glide ratio: influential factors

If you are familiar with calculus you will already have noticed that everything derived above is valid in an instantaneous sense. If you are not, take a minute to appreciate that the scale of time does not play a role in the process explained above: a glider with a glide ratio of 50:1 can travel fifty feet while dropping by one foot, it can also travel for fifty miles while dropping by one mile (slightly more than 5000 ft). If we extend this to the other extreme of the length scale such that the time associated is very small, we can see that the glide ratio is defined for any instant of the flight process.

There is no restriction on the glide ratio changing from one instant to another either, otherwise we will have no need to consider the polar if the glider flies the same at whatever speeds. Before proceeding into the more detailed discussions, some of the most important factors are presented here.

The glide ratio is mainly affected by:

1. The aerodynamic design of the glider. The more streamlined, sleek, and aerodynamic-looking a glider is, with long slender wings and smooth gel coating, the more likely it is to have a larger glide ratio.
2. The configuration of the glider. The glide ratio is almost always the highest in the clean configuration, i.e. with nothing sticking out or deployed. Lowered undercarriage, extended brakes and spoilers, deployed and windmilling propellers, opened or lost canopies, attached ropes are things that will reduce the glide ratio. Generally speaking, having the flaps set to other angles than neutral is not good for the glide ratio, but this very much depends on other factors.
3. The way the glider is flown. For a fixed glider mass (which is generally the case with the exception of jettisoning water), the glide ratio is a function of indicated airspeed, giving rise to the polar which is the immediate next topic. Moreover, if the glider is not flown straight (with sideslip), or flown otherwise than normal (e.g. stalled, inverted), the glide ratio can decrease drastically.

## Lift and Drag Coefficients

In the discussions that follow, only the incompressible flow regime is considered. This is justified by the low speed that gliders fly at.

### Definitions

In aerodynamics, the lift coefficient ($$C_L$$) and drag coefficient ($$C_D$$) are defined as follows:

$C_L = \frac{L}{\frac{1}{2} \rho V^2 S}$ $C_D = \frac{D}{\frac{1}{2} \rho V^2 S}$

Where:

• $$\rho$$ is the true density of air.
• $$V$$ is the true airspeed of the aeroplane.
• $$S$$ is the area of the wing (projected onto the ground), a fixed value for a given glider.
• $$\frac{1}{2} \rho V^2$$ is collectively known as the dynamic pressure, or dynamic head.

### A comment on the dimension

You should notice that, for both coefficients, the unit of both the numerator and the denominator is the unit of force (Newton in SI units). The denominator comprises $$\frac{1}{2} \rho V^2$$ which has the unit of pressure, and $$S$$ which has the unit of area, so the product yields a force.

Consequently, both $$C_L$$ and $$C_D$$ are non-dimensional. These quantities have no units. Non-dimensional quantities are the language of aerodynamics: it allows us to study the underlying physics without being distracted by how things are measured. A K-21 is heavier than a Junior, therefore, in unaccelerated glide with the same angle-of-attack, the wings of the K-21 produces more lift than the Junior wings. What causes this? Is it because the design of the K-21 is aerodynamically superior? The answer is not necessarily, as the K-21 can be flying faster, for instance, or has larger wings. The comparison only becomes meaningful when the lift is non-dimensionalised into the lift coefficient.

### A comment on dynamic pressure

The true air density and the true airspeed always appear together as a compound quantity $$\frac{1}{2} \rho V^2$$ which is referred to as the dynamic pressure or dynamic head.

The density of air is not a constant: it depends on pressure (which most notably depends on altitude) and temperature. This causes a major inconvenience as we have to assert a value to it in order to arrive at any numerical results directly useful for flying: e.g. you do not check any non-dimensional quantities in the cockpit, you read the instruments instead which tells you the airspeed in knots or the altitude in feet.

To overcome this problem, we notice that density only appears within $$\frac{1}{2} \rho V^2$$. Therefore, we can define an equivalent density $$\rho_e$$ and an equivalent airspeed $$V_e$$ such that:

$\frac{1}{2} \rho V^2 = \frac{1}{2} \rho_e V_e^2$

We can assert a value to $$\rho_e$$ and arrive at a value of $$V_e$$ such that, when used together, they produce the same amount of dynamic head, therefore, the aerodynamic effect is exactly the same.

The most reasonable value to assign to $$\rho_e$$ would be the density of air at some standard conditions. This can then be implemented into some instrument that tells you $$V_e$$ (all this instrument has to do is to measure the dynamic head). So long as all the manuals and polar charts express airspeed in $$V_e$$ assuming the same value of $$\rho_e$$, the change in true air density will not cause these performance guidelines to vary.

In practise, the instrument that tells you $$V_e$$ is the air speed indicator (ASI), and $$V_e$$ is known as indicated airspeed. Based on the discussions above, you should realise that:

1. Indicated airspeed is directly related to the dynamic head.
2. The dynamic head is the only way true airspeed affects glider aerodynamics (before it disintegrates by overspeeding).
3. Therefore, the glider's aerodynamics is affected only by indicated airspeed, not true airspeed (apart from the never-exceed speed).
4. We should tabulate performance figures and draw polar graphs using indicated airspeed.
5. We do not need to adjust the performance tables or polar graphs to compensate for non-standard atmospheric conditions.

### Components of lift and drag coefficients

To proceed with the discussions, it is necessary to quote these without proof. Indeed, these formulae cannot be proven. There are complicated aerodynamic theories that derives these, however, while the success in doing so is remarkable, the theories themselves rely on rigorous assumptions and extensive modelling, so the derivations cannot really be called proofs. You are advised to understand the following as experimental correlations.

$C_L = C_{L0} + C_1 \alpha$ $C_D = C_{D0} + \frac{k}{\pi A} C_L^2$

It is, however, necessary to explain the physical rationale in detail.

The lift coefficient $$C_L$$ can be decomposed as follows:

1. $$\alpha$$ is the angle-of-attack.
2. $$C_{L0}$$ is the lift coefficient at zero angle-of-attack. This term equals to zero if the aerofoil is symmetric, greater than zero if the aerofoil is cambered, and smaller than zero if the aerofoil is cambered the wrong way.
3. $$C_1$$ can be thought as an empirical factor. It is rather close to $$2\pi$$.
4. The lift coefficient increases proportionally with the angle-of-attack up to the point where the wing stalls.

The drag is more complex: the drag on an aeroplane has three components:

1. Friction drag, this is the drag caused by the air sticking onto the glider and trying to slow it down. Imagine flying a glider in honey which is rather sticky. The friction drag coefficient $$C_{DF}$$ is approximately a constant for a given glider.
2. Pressure drag, this is the drag associated with the glider trailing a wake. This is also known as the form drag because it is related to the form of the glider being not fully aerodynamic. You would intuitively think that a Land Rover Discovery has more drag than a Jaguar fastback: because the Discovery is not streamlined while the fastback is, and this is what pressure drag is about. The pressure drag coefficient $$C_{DP}$$ is approximately a constant for a given glider, because its form does not change in flight. Were this approximation not to be made, the following derivation can remain unaltered by pretending this variation is a part of the induced drag.
3. Induced drag, this is the drag caused by having lift. There is no free lunch in aerodynamics and wherever you have lift you must have drag, no matter how good your design is. The induced drag coefficient $$C_{DI}$$ takes the following form:

$C_{DI} = \frac{k}{\pi A} C_L^2$ Where $$A$$ is the aspect ratio of the wings (how slender the wings are), and $$k$$ is a factor that depends on the wing design. This drag component increases quadratically with $$C_L$$.

By the explanations above, it should be evident that:

$C_{D0} = C_{DF} + C_{DP}$

## Relationship Between Lift and Drag: the Parabolic Polar

The following relationship between $$C_D$$ and $$C_L$$ is fundamental to the discussions that follow:

$C_D = C_{D0} + \frac{k}{\pi A} C_L^2$

This is a parabolic function. It is this function that is referred to when talking about a 'parabolic polar': the actual (and more useful) polar curve that we shall derive is not a parabola.

### A statement of the task that follows

From the relationship presented above, and making use of the following facts or assumptions:

1. Mass of the glider remains constant.
2. Energy is conserved.
3. The air is still.
4. The density of air is uniform and known, or rather and better, we work in the corrected (indicated) airspeed system.

We will be deriving a one-to-one relationship between indicated airspeed and sink rate.

## A General Method of Solution

Before making further assumptions and simplifications, a general method of solution is worth presenting. The algebraic difficulties, as we shall see, is formidable, but it lends itself nicely to numerical methods.

Re-arranging the definitions of $$C_L$$ and $$C_D$$:

$W \cos(\theta) = C_L \times \frac{1}{2} \rho V^2 S$ $W \sin(\theta) = C_D \times \frac{1}{2} \rho V^2 S$

The two expressions can both be squared and added together. Notice that $$\cos^2(\theta) + \sin^2(\theta) =1$$, the following is arrived at:

$W^2 = (C_L^2 + C_D^2) \times (\frac{1}{2} \rho V^2 S)^2$

Or rather, in the more insightful form:

$C_L^2 + C_D^2 = \frac{W^2}{(\frac{1}{2} \rho V^2 S)^2}$

The right hand side of the expression above is a function of indicated airspeed only, because air density is a constant for it to be compatible with the indicated airspeed.

Substitute the parabolic relationship between $$C_D$$ and $$C_L$$ into the expression above, we have:

$f(C_L) = g(V)$

Where $$f(x)$$ and $$g(x)$$ are functions that are too cumbersome to typeset. Keep in mind that $$C_{D0}$$ is embedded in $$g(x)$$.

Solving the above which the author does not believe is analytically possible:

$C_L = \frac{W \cos(\theta)}{\frac{1}{2} \rho V^2 S} = h(V)$

In words, a relationship between the lift coefficient and the indicated airspeed can be arrived at.

The above can be further re-arranged, such that:

$\cos(\theta) = \frac{h(V) \rho V^2 S}{2W}$

This is a relationship between the glide slope and the airspeed. From here on, determining the sink rate from the glide slope and airspeed is a trivial geometrical task, so the required relationship between airspeed and sink rate is essentially derived.

## An Approximate Method of Solution: The Analytical Polar Curve

We shall attempt a derivation of the analytical polar curve again but using a slightly different algebraic approach than what is used in the previous section. We will see that, by adopting this approach, and by making a simple approximation, the algebra becomes simple enough for us to explicitly express the analytical form of the polar equation (an equation relating sink rate to indicated airspeed).

Firstly, the definitions of $$C_L$$ and $$C_D$$ shall be substituted into the parabolic relationship between $$C_L$$ and $$C_D$$, giving:

$\frac{D}{\frac{1}{2} \rho V^2 S} = C_{D0} + \frac{k}{\pi A} \frac{L^2}{\frac{1}{4}\rho^2 V^4 S^2}$

Both sides of the equation above need to be multiplied by $$\frac{1}{4} \rho V^4 S^2$$ (notice that this is the square of $$\frac{1}{2} \rho V^2 S$$), then divided by $$\frac{1}{2} \rho V S$$, giving:

$DV = \frac{1}{2} \rho V^3 S C_{D0} + \frac{2kL^2}{\pi A \rho V S}$

To proceed, the conservation of energy must be invoked. We realise that the kinetic energy of the glider is not changing because the glider is flying unaccelerated. Therefore, the release of gravitational potential, the rate of which equals to the power of the gravitational force, must balance the rate at which the mechanical energy of the glider is being dissipated by aerodynamic drag, which is the power of the drag force.

By definition, the power of a force is given by:

$P = F \times V$

In words: the power is the product of the magnitude of the force and the speed of the subject in the direction of the force. By using this relationship, we realise that: the power of the gravitational force is given by $$W \times V_S$$ (weight times the sink rate), and the power of the drag force is given by $$D \times V$$ (drag times the airspeed). This relationship can also be obtained by a geometrical argument using basic trigonometry.

It should be noticed that the above argument is not watertight: this is because $$V$$ is the indicated airspeed which generally differs from the true airspeed, and it is the latter that must be used to calculate the drag power. There are two ways to think around this:

1. You can think of this as an approximation that is being made: true airspeed is being approximated with indicated airspeed. As a result, some systematic error will be introduced into the results.
2. If you can understand the relationship between $$W \times V_S$$ and $$D \times V$$ from a geometrical perspective, you can think the following: because we are working in the indicated system where $$V$$ is the indicated airspeed, the corresponding $$V_S$$ obtained geometrically is the indicated sink rate. It needs to be converted to the true sink rate via the compound quantity $$\frac{1}{2} \rho V^2$$.

By using $$D \times V = W \times V_S$$, the last equation becomes:

$WV_S = \frac{1}{2} \rho V^3 S C_{D0} + \frac{2kL^2}{\pi A \rho V S}$

This expression should be examined in detail. The following quantities are known (either set, from design, or can be measured):

1. $$W$$, weight of the glider, depends on the design, cockpit loading, and amount of water carried, but can be known and usually does not change midway in flight.
2. $$\rho$$, density of air, because we work in the indicated system, this becomes the air density value used in the ASI, which is a fixed number.
3. $$S$$, wing area of the glider, known and stays constant (we shall not consider the effects of deploying flaps, etc. on the performance).
4. $$C_{D0}$$, this depends on the aerodynamic design of the glider.
5. $$k$$, this depends on the wing design of the glider, a highly complex series expansion to obtain a numeric value exists, but for all practical purposes this is a constant.
6. $$\pi$$, 3.1415926...
7. $$A$$, aspect ratio of the wing, depends on the glider design and a known constant.

Therefore, there are three changing quantities in this equation:

1. $$V_S$$, this is the quantity we are interested in, the y.
2. $$V$$, this is the quantity we can control, the x.
3. $$L$$, lift on the glider, what is it?

You should realise that, the existence of $$L$$ in the equation above prevents us from obtaining a deterministic relationship between $$V_S$$ and $$V$$ which is the polar equation we desire. $$L$$ can be related to $$W$$ by using $$V_S$$ and $$V$$ and geometrical arguments, but this will complicate the equation and prevent us from arriving at an explicit relationship. In other words, doing so is the equivalence of reverting to the method in the last section.

Instead, we shall introduce the following approximation: the weight of the glider equals to the lift force acting on the glider. This sounds intuitively true, but there is an error associated with it, whose magnitude is given by $$1-\cos{\theta}$$. Fortunately, this error is gracefully small at typical glide angles. If the glide ratio is 30:1, the error is 0.056%, which becomes even smaller if the glide ratio is higher.

We have shown that this is a good approximation. Therefore, we can replace the $$L$$ in the existing equation with $$W$$, and the analytical polar curve is arrived at:

$V_S = \frac{1}{2W} \rho S C_{D0} V^3 + \frac{2kW}{\pi A \rho V S}$

## Implications of the Analytical Polar Curve: Minimum Sink and Best Glide

### Shape and General Features of the Analytical Polar

The polar equation is a combination of a third order term which is monotonically increasing throughout the domain of definition, and a hyperbolic term which, in the domain of $$V>0$$, decreases monotonically. Therefore, a global minimum is expected. This will be (confusingly) referred to as the global maximum because conventionally, the Y-axis ($$V_S$$) is turned upside-down, such that going down means higher sink rate.

The analytical form of the polar curve applies to the speed range from several knots above stall to $$V_{NE}$$ ($$V_{NE}$$ must be converted to its indicated value). It does not apply close to stall, which is because the aerodynamics of the glider changes considerably before the onset of stall such that the drag ceases to be a parabolic function of the lift.

The analytical polar given above can be plotted by any computer code, or a plot can be found in any gliding textbook. You can also ask an instructor to draw you one.

### Minimum Sink Airspeed

We seek an indicated airspeed that will give us the minimum sink rate. This is the indicated airspeed to fly at only if you want to stay in the air for as long as possible with a certain amount of altitude drop. It is useful when thermalling. Flying at this speed may not be able to get you anywhere (in extreme cases you can go backwards rather quickly), so caution and thought is needed.

To find this airspeed, the analytical polar is differentiated to reveal the maximum:

$\frac{d V_S}{dV} = 0$

This gives:

$V_{MS} = (\frac{4kW^2}{3 \rho^2 S^2 C_{D0} \pi A})^{\frac{1}{4}}$

It is common practice to define a quantity called wing loading as $$\omega = \frac{W}{S}$$ which quantifies how much weight each meter squared wing area is carrying. With this definition in place, notice that:

$V_{MS} \propto \sqrt{\omega}$

The implication is, the minimum sink airspeed is not fixed: with the glider loaded heavier it will become higher.

It is worth noting that, by flying at this airspeed:

$V_S(V_{MS}) \propto \sqrt{\omega}$

Therefore, by loading the glider heavier, the minimum sink rate possible is also higher. This implies that gliders with low wing loading can make use of weaker thermals with a limited rising speed.

If a glider thermals at the minimum sink airspeed, carrying water ballast will enable the glider to fly faster and likely at a larger radius. This can prove to be beneficial as some experienced pilots will say, but a mathematical proof is not possible in the absence of a model to characterise the behaviour of the thermal. Water ballast is usually carried on good thermal days but not on days with marginal conditions. You will sometimes hear pilots say that the water doesn't work, the author's interpretation to which is that, because the thermals are not strong and big enough, the increased minimum sink by carrying water ballast outweighs the possible benefits if any.

### Best Glide

The best glide ratio achievable for a given glider in a particular loading and configuration can be determined from the analytical polar. Recall that the glide ratio is given by the horizontal distance covered over the vertical altitude drop. For analytical purposes, it is necessary to make a small angle approximation such that the horizontal distance is approximately given by $$V \times t$$ where $$t$$ is time. The accuracy of this approximation is shown in previous sections.

Using the small angle approximation, the inverse of the glide ratio is given by:

$\frac{V_S}{V} = \frac{\rho}{2 \omega} C_{D0} V^2 + \frac{2 k \omega}{\pi A \rho V^2}$

This compound quantity is to be differentiated with respect to $$V$$ to reveal the minimum:

$\frac{d}{dV}(\frac{V_S}{V}) = \frac{\rho}{\omega} C_{D0} V - \frac{4k \omega}{\pi A \rho V^3} =0$

This yields:

$V_{BG}=(\frac{4k \omega^2}{\pi A \rho^2 C_{D0}})^{\frac{1}{4}}=(3)^{\frac{1}{4}} V_{MS}=1.3161 V_{MS}$

Therefore, the best glide speed is always 31.6% higher than the minimum sink airspeed according to our analysis. Slight discrepancies may arise in reality due to the approximations we have made, mainly the aerodynamic ones.

From the expression for $$V_{BG}$$ given above, it is evident that:

1. $$V_{BG} \propto \sqrt{\omega}$$, such that the best glide speed will increase as the wing loading increases, by means such as using water ballast.
2. Increasing the aspect ratio can reduce the best glide speed.
3. Decreasing $$C_{D0}$$ can increase the best glide speed.

It is also of interest to calculate the best possible performance of the glider, which, by definition, happens at the best glide speed. The algebra proceeds as follows:

$V_{BG}^2 = \sqrt{\frac{4k}{\pi A C_{D0}}} \frac{\omega}{\rho}$

This is to be substituted into:

$(\frac{V_S}{V})_{\text{best}} = \frac{\rho}{2 \omega} C_{D0} V_{BG}^2 + \frac{2 k \omega}{\pi A \rho V_{BG}^2}$

To yield:

$(\frac{V_S}{V})_{\text{best}} = 2 \sqrt{\frac{k C_{D0}}{\pi A}}$

Or, alternatively (to give the large number like 40 or 50 that we are familiar with):

$\text{Best Glide Ratio} = 0.5 \times \sqrt{\frac{\pi A}{k C_{D0}}}$

This is a very important result, as it gives all the factors underpinning the best performance of a glider (in a particular configuration):

1. The wing loading does not change the best performance. Therefore, a 50:1 glider will be 50:1 with a light pilot or a heavy pilot, or with or without water ballast. This is, however, based on our model, and in reality more factors may come into play. For example, if the wing loading is high, then the best glide speed increases accordingly and the change in Reynolds number may have some effect. Alternatively, the different structural deflections of the wings may produce subtle differences in the aerodynamic geometry. Nevertheless, this is the rationale underpinning the use of water ballast: it does not degrade aerodynamic performance.
2. Increasing the aspect ratio of the wing is an effective (and, in fact, easiest) way to improve the best performance, as the best glide ratio scales with $$\sqrt{A}$$. This is the reason why high performance gliders have slender wings.
3. Improving aerodynamic design, such that $$C_{D0}$$ or $$k$$ is reduced, can improve the best glide ratio as we would intuitively expect. However, modern advancement in aerodynamics has been agonisingly slow and you realise that there is not much potential to be released by comparing a fibre glass glider built in the 1980s with a modern one. What differences do you spot?

From a geometric point of view, the above solution process is equivalent to finding a ray from the origin that is tangent to the polar curve. You should ask an instructor to demonstrate this to you to reinforce the understanding. This geometric method is useful when more factors are taken into account, such that an analytical solution cannot be obtained easily.

### Water Ballast

Water ballast has no effect on the glider best performance, but it makes the best glide speed faster, so the pilot can cover a certain amount of cross-country distance faster. This is the first reason for using water ballast.

In fact, the use of water ballast does not change the shape of the polar at all, not only for the best performance point. To see this, please read the next section on non-dimensional polar. The shape of the polar is dictated only by the best glide speed and the sink rate at best glide, but, as shown previously, both quantities are proportional to $$\sqrt(\omega)$$. Therefore, as the wing loading changes, the polar curve scales around the origin with $$\sqrt(\omega)$$ but keeps its shape. Because the best glide is a tangent to the polar, and that the polar is scaled around the origin of the ray, the slope of the ray (best performance) is invariant.

The second reason for using water ballast is to improve the performance in headwind and sinking air. This is difficult to prove mathematically as the workings in the next section will show, but geometrically this can easily be demonstrated. Because the polar curve is scaled to be larger, any shift in origin due to headwind and sinking air is comparatively smaller. This makes the new tangent to the polar closer to the best glide line in stationary air, such that the degradation of performance is less.

Conversely, it can be demonstrated graphically that water ballast is detrimental to performance (in terms of covering ground distance) when there is tailwind or rising air. However, gliders are not usually flown downwind for meaningful distances, and when rising air is present, a pilot will attempt to stay in it and soar, rather than moving to another place, so these effects are unimportant.

Experienced pilots sometimes argue that carrying water ballast improves thermalling performance. A mathematical establishment cannot be made unless a model exists to characterise the behaviour of a thermal (which indeed exists, but the validity is questionable in the author's opinion). It is worth pointing out that, very hand-wavingly we can say, if there is any benefit in carrying water ballast when thermalling, it will come from thermalling at a larger radius, rather than at a higher speed.

You may have been instructed that you need to fly faster in headwind or sinking air to cover ground efficiently. This section briefly demonstrates the underlying mathematics, but you are encouraged to use the geometric method to prove to yourself that it is indeed the case.

We firstly approximate true airspeed with indicated airspeed. By doing this, the following construction is possible:

$V = V_g + V_w$

Where $$V_g$$ is ground speed and $$V_w$$ is the headwind speed. This expression is to be substituted into the analytical polar. We should also bear in mind that it is the most efficient covering of ground distance that is of interest, therefore:

$\frac{V_S}{V_g} = \frac{\rho C_{D0}}{2 \omega} \frac{V^3}{V-V_w} + \frac{2k \omega}{\pi A \rho}\frac{1}{V(V-V_w)}$

This quantity must be differentiated with respect to $$V$$ to find the optimum. Performing the differentiation and, after considerable simplifications:

$2V^5 - 3V^4 V_w - 2CV + CV_w = 0$

Where:

$C = V_{BG}^4$

This is a fifth-order polynomial. It is a fact established by Abel in 1820 that a fifth-order polynomial cannot, in general, be solved by radicals. However, we can 'prove' that the solution lies in $$V > V_{BG}$$ by considering the following:

1. Substitute $$V=V_{BG}$$ into the equation above, and show that the value is negative.
2. Differentiate the expression and substitute $$V=V_{BG}$$ into the differentiated expression, and show that the value is positive.
3. Hence, we have a function that is presently negative, but it is increasing, so we would expect a root (the required solution) at a larger $$V$$ than the present $$V$$ which is $$V_{BG}$$

The above arguments are far from watertight: The differentiated expression only gives a positive value if $$V_w \leq \frac{2}{3} V$$. While this is usually the case, the above cannot constitute a proof. A more rigorous analysis on the locations of the maxima and minima is required.

The problem is much simpler if the geometric method is used: to use the geometric method, imagine setting up a ground speed zero which is different from the airspeed zero. The polar is plotted with respect to the airspeed zero but the tangent ray needs to start from the ground speed zero. Because the ground zero is located in $$V>0$$, the tangent is steeper and intersects the polar at a larger $$V$$.

### Effects of Sinking Air

If the glider is flying in some air that is sinking with a uniform downward speed $$V_{SA}$$, then the polar equation should be adapted into the following form:

$V_S = \frac{\rho C_{D0}}{2 \omega} V^3 + \frac{2k \omega}{\pi A \rho} \frac{1}{V} + V_{SA}$

Using the differentiation method to find the optimum airspeed for covering ground (notice that, because there is no headwind or tailwind, the indicated airspeed is equivalent to ground speed. This is not to say we approximate TAS with IAS, but there is a monotonic relationship between the two which is dictated by the altitude, which is a free variable in our problem.)

$\frac{d}{dV}(\frac{V_S}{V}) = 0$

$\frac{\rho C_{D0}}{\omega} V^4 - V_{SA} V - \frac{4k \omega}{\pi A \rho} = 0$

This equation has the solution of $$V=V_{BG}$$ if $$V_{SA} = 0$$ as expected, but if $$V_{SA} > 0$$, then the solution is $$V>V_{BG}$$. The proof of this is left as an exercise for the reader.

### Effects on the Minimum Sink

It should be obvious by now that the above adjustments to the polar have no effect on the minimum sink speed: the difference only arises when $$V$$ is divided over to the left side, i.e. a glide ratio is sought after. Physically this makes sense: the minimum sink speed is purely an interaction between the glider and the surrounding air, and if we disregard all relativity to the ground, then the air in which the glider flies can move in whichever possible way (so long as it is not accelerating) and the glider can perform the same macroscopic motion with it without altering the detailed aerodynamics.

## The Non-Dimensional Polar and the Determination of the Polar in Practice

The polar equation can be abstracted into the following form:

$V_S = AV^3 + \frac{B}{V}$

With $$A=\frac{\rho C_{D0}}{2 \omega}$$ and $$B=\frac{2k \omega}{\pi A \rho}$$.

At best glide, by our calculations from the previous sections, the best glide speed is given by (notice the change in notation for reading convenience):

$V_i = (\frac{B}{A})^{\frac{1}{4}}$

$V_{Si} = 2(AB^3)^{\frac{1}{4}}$

These can be substituted into the abstract polar equation, such that:

$2 (\frac{V_S}{V_{Si}}) = \frac{AV^3}{(AB^3)^{\frac{1}{4}}} + \frac{B}{(AB^3)^{\frac{1}{4}}V}$

This can be simplified into:

$2\frac{V_S}{V_{Si}} = (\frac{V}{V_i})^3 + (\frac{V_i}{V})$

This is the non-dimensional polar. It tells us that the polar curve is deterministic from only two quantities: the best glide speed, and the sink rate at the best glide speed. These two quantities both depend on the wing loading, so the additional requirement is that they be measured with the same level of wing loading.

Aerodynamic coefficients such as $$C_{D0}$$ are difficult to determine and measuring such quantities require sophisticated equipment and techniques include wind tunnel testing and flight tests. However, the polar can be determined simply by test flying the glider and plugging the measured airspeed and sink rate into the non-dimensional polar as coefficients. This is a useful method to determine a polar of a glider which you may not have a manual for.

The above, nevertheless, assumes that the parabolic relationship between $$C_L$$ and $$C_D$$ holds true, which is something we have been doing throughout this article. This relationship has its limitations and such limitations lead to most of the deviations from the analytical polar as observed in flight.