Climbing, Ceiling & Gliding

J. Gordon Leishman

doi:https://doi.org/10.15394/eaglepub.2022.1066.n38

39 Climbing, Ceiling & Gliding

Introduction

After the takeoff roll, all aircraft need to be able to climb quickly away from the runway into flight and also be able to climb up to a cruising altitude at a reasonable rate. The process looks almost effortless when watching a modern airliner climb away from the runway, as shown in the photograph below. However, substantial amounts of thrust and power are required to perform this aeronautical feat. Also, the airplane must be flown at a reasonably precise airspeed to maximize its angle and/or rate of climb. As will be explained, the climbing process takes more power over and above that for level flight at the same airspeed. Sufficient excess engine thrust (and hence power) must be incorporated into the design of the aircraft to allow for adequate climb performance over a wide range of weights and airspeeds.

An airliner in the initial climbout condition before the undercarriage is retracted.

As the airplane climbs higher and higher, its rate of climb will diminish, and the thrust (or power) required for flight and the thrust (or power) available begin to approach each other. Eventually, the rate of climb will reduce to the point that the aircraft can climb no higher for all practical purposes. This condition is called the operational flight ceiling. The estimation of the ceiling is also an essential part of the aircraft’s design, and sufficient excess power must be available to allow it to reach and cruise at an efficient operational altitude.

Learning Objectives

Understand the factors that will influence an airplane’s climb rate and the time it takes to climb to altitude.
Use the basic equations for aircraft performance to set up and estimate an airplane’s climb rate under different flight conditions.
Appreciate the concept of a flight ceiling and the factors that will affect the attainable ceiling of an airplane.
Understand the principles of gliding flight and the significance of the lift-to-drag ratio in determining an aircraft’s gliding performance.

Steady Climbing Flight

A schematic of an airplane in a climb is shown in the figure below. The angle $\theta$ is the climb or flight path angle. Notice that the angles are exaggerated here for clarity compared to what a normal climb would look like. For descending or gliding flight, the value of $\theta$ would be negative, a condition considered later.

An airplane in a steady climb along a rectilinear flight path is typical of a standard operational climb.

The equations of motion are

(1) $\begin{equation*} \left( \frac{W}{g} \right) \frac{d V_{\infty}}{dt} = T \cos \epsilon - D - W \sin \theta \end{equation*}$

and

(2) $\begin{equation*} \left(\frac{W}{g}\right) \frac{V_{\infty}^2}{r} = L \cos \phi + T \sin \epsilon \cos \phi - W \cos \theta \end{equation*}$

For the steady, unaccelerated flight along a rectilinear flight path, then $dV_{\infty}/dt = 0$ and $V_{\infty}^2/r = 0$ (i.e., no radius of curvature), so the equations of motion reduce to

(3) $\begin{eqnarray*} T \cos \epsilon - D - W \sin \theta & = & 0 \\ L + T \sin \epsilon - W \cos \theta & = & 0 \end{eqnarray*}$

It can be further assumed that $\epsilon = 0$ , which means that the line of action of the thrust vector is along the direction of flight; this is a reasonable assumption. Therefore, the equations of static equilibrium further simplify to

(4) $\begin{eqnarray*} T - D - W \sin \theta & = & 0 \\ L - W \cos \theta & = & 0 \end{eqnarray*}$

The preceding equations for the climb condition are analogous to the results obtained for straight-and-level flight. However, notice that in a climb, for vertical equilibrium, the lift on the airplane’s wings is now slightly smaller than its weight because some of the airplane’s weight is supported by a vertical component of the thrust from the propulsion system.

By multiplying the first of the preceding two equations by $V_{\infty}$ , then

(5) $\begin{equation*} T V_{\infty} = D V_{\infty} + W V_{\infty} \sin \theta \end{equation*}$

and with rearrangement, then

(6) $\begin{equation*} \frac{T V_{\infty} - D V_{\infty}}{W} = V_{\infty} \sin \theta \end{equation*}$

Notice that the vertical rate of climb $V_c$ is equal to $V_{\infty} \sin \theta$ so that the rate of climb (ROC) is

(7) $\begin{equation*} \mbox{ROC} = V_c = V_{\infty} \sin \theta = \frac{T V_{\infty} - D V_{\infty}}{W} \end{equation*}$

The abbreviations “ROC” or “R/C” are used variously in the literature on aircraft performance to denote the rate of climb.

The term $T V_{\infty}$ is the power available for flight $P_{\rm A}$ , and $D V_{\infty}$ is the power required $P_{\rm{req}}$ to overcome the drag of the aircraft. Therefore, the excess power available, $\Delta P$ , is

(8) $\begin{equation*} \Delta P = T V_{\infty} - D V_{\infty} = P_A - P_{\rm req} \end{equation*}$

so the ROC is then

(9) $\begin{equation*} \mbox{ROC} = V_c = \frac{\Delta P}{W} \end{equation*}$

It can be deduced then that the best climbing performance for an aircraft will be obtained at a low weight, with low drag, and with a large amount of excess power available. In this latter regard, excess power means the power available over and above required for straight-and-level flight at the same weight, airspeed, and density altitude. The former observations are certainly not unexpected based on physical reasoning.

Some aerodynamics can now be introduced into the analysis. In steady climbing flight, then

(10) $\begin{equation*} C_L = \frac{L}{\frac{1}{2} \varrho_{\infty} V_{\infty}^2 S} = \frac{W \cos \theta}{\frac{1}{2} \varrho_{\infty} V_{\infty}^2 S} \end{equation*}$

and the standard drag polar for an airplane can be assumed, i.e.,

(11) $\begin{equation*} C_D = C_{D_{0}} + \frac{{C_L}^2}{\pi A\!R e} \end{equation*}$

Therefore, the total drag is

(12) $\begin{equation*} D = \frac{1}{2} \varrho_{\infty} V_{\infty}^2 S \left( C_{D_{0}} + \frac{{C_L}^2}{\pi A\!R e} \right) \end{equation*}$

Substituting into the equation for the drag polar gives

(13) $\begin{equation*} D = \frac{1}{2} \varrho_{\infty} V_{\infty}^2 S \left( C_{D_{0}} + \frac{1}{\pi A\!R e} \right) \frac{(W \cos \theta)^2}{\frac{1}{2} \varrho_{\infty} V_{\infty}^2 S} \end{equation*}$

After some rearrangement using the preceding results, the rate of climb is

(14) $\begin{equation*} \mbox{ROC} = V_c = V_{\infty} \left[ \frac{T}{W} - \frac{1}{2} \varrho_{\infty} V_{\infty}^2 \left( \frac{W}{S} \right)^{-1} C_{D_{0}} - \frac{W}{S} \left( \frac{2 \cos^2\theta}{\varrho_{\infty} V_{\infty}^2 S \pi A\!R e} \right) \right] \end{equation*}$

Two critical parameters appear again, namely the wing loading of the airplane $W/S$ and its thrust-to-weight ratio $T/W$ , both of which have primary dependencies on the airplane’s climb rate. Notice also that the rate of climb is improved with improved aerodynamic efficiency, i.e., with a lower value of $C_{D_{0}}$ and a higher value of $e$ .

There are no exact solutions to Eq. 14, but numerical and graphical solutions are possible. In part, closed-form solutions are impossible because engine performance (e.,g., thrust and power) is often difficult to generalize through simple equations. However, under certain assumptions, it is possible to estimate the climb rates for weight and altitude variations. For example, because the climb angle is typically small for many classes of airplane, then $\cos^2\theta \rightarrow 1$ , so the governing equation for the ROC becomes

(15) $\begin{equation*} V_c = V_{\infty} \left[ \frac{T}{W} - \frac{1}{2} \varrho_{\infty} V_{\infty}^2 \left( \frac{W}{S} \right)^{-1} C_{D_{0}} - \frac{W}{S} \left( \frac{2}{\varrho_{\infty} V_{\infty}^2 \pi A\!R e} \right) \right] \end{equation*}$

In the solution to this latter equation, the outcome is either:

The value of $V_c$ (the ROC) for a given value of $\theta$ .
The value of $\theta$ for a given value of $V_{\infty}$ .

Rate of Climb & Excess Power

It has been previously shown that the ROC of an airplane is proportional to the excess power available over and above that required for level flight, i.e.,

(16) $\begin{equation*} {\mbox ROC} = V_c = \frac{\Delta P}{W} \end{equation*}$

What this latter equation means, however, is that the ROC is related to the excess power available over and above that required for straight-and-level flight at the same weight $W$ , altitude, and airspeed $V_{\infty}$ . This outcome can be seen easily when the power required versus available power curves are plotted, which are shown in the figure below for a representative propeller-driven airplane.

Representative power required versus power available curves for a propeller-driven airplane. The excess power available allows the aircraft to climb.

Remember that for a propeller-driven airplane, the power available is relatively constant over a wide range of airspeeds, but for a jet where $T$ is nominally constant, then the power available (i.e., the useful power or $T V_{\infty}$ ) increases linearly with airspeed, as shown in the figure below. Therefore, it can be concluded that the propeller and jet airplanes will have a markedly different rates of climb characteristics.

Representative power required versus power available curves for a jet airplane. Notice that compared to a propeller-driven airplane there is significantly less excess power available at low airspeeds.

In each case, however, the ROC depends on the excess power available, for example, the resulting climb characteristics shown in the figure below, both for propeller-powered and jet airplanes. Jets have relatively low amounts of excess power available at lower airspeeds so they can achieve only relatively low rates of climb at low airspeeds, i.e., just after takeoff.

Representative rate of climb curves for a propeller-driven airplane versus a jet airplane.

Again, it is apparent that the maximum rate of climb, ROC $_{\rm max}$ , occurs where there is the largest excess power available $\Delta P$ . Because of the larger excess power available with propeller-driven airplanes at low airspeeds, they generally have excellent climb characteristics just after takeoff. They are ideally suited for operations from shorter runways or runways surrounded by mountainous terrain.

Jet aircraft, however, exhibit characteristics where the largest rates of climb are obtained at higher airspeeds. For example, when observing an airliner taking off, it will be noted that it first climbs relatively slowly until airspeed has built sufficiently and the undercarriage and flaps have been retracted to reduce drag before the best rates of climb are possible. Long-haul commercial airliners, which are laden with fuel, often have rates of climb that are less than 1,000 ft/min until airspeed builds and the aircraft exits the terminal airspace. The resulting climb to the cruise altitude is usually performed in stages, where fuel must be burned off to decrease the in-flight weight and allow a climb to a higher altitude.

In the event of an engine failure called one engine inoperative (OEI), the aircraft also needs to be able to climb, albeit at reduced rates. Therefore, OEI performance considerations are always an essential part of the design of a multi-engine airplane. OEI climb is particularly critical for twin-engine airliners. They are designed to have much larger amounts of excess thrust and power available from any of the engines than would be used for three or four-engined airplanes.

Hodograph Method

A graphical method often used for solving for the rate of climb is called the hodograph method. A hodograph is a diagram, as shown below, which gives a vectorial visual representation of a point in space; it is the locus of one end of a variable vector with the other end fixed at the origin (0,0) of the graph. In this method, the vertical speed or ROC or $V_c = V_{\infty} \sin \theta$ is plotted versus the horizontal speed $V_H = V_{\infty} \cos \theta$ , which gives a locus of points that is called a hodograph. As shown in the figure, a horizontal tangent to the hodograph gives the maximum rate of climb. Also, any line from the origin of the graph to a point on the hodograph gives the angle of climb, and the length of this line will be the corresponding airspeed $V_{\infty}$ .

A hodograph is used to give a visual depiction of the rates and angles of climb.

Moving along the hodograph in a continuous counterclockwise direction leads to the following observations:

The value of $V_c$ first increases and $V_{\infty}$ decreases.
There comes a point where $V_c$ reaches a maximum (point 2). $V_c$ begins to decrease and reaches a tangent to the curve (point 3), which is the steepest angle of climb.
The maximum or steepest climb angle does not occur at the same airspeed as the maximum rate of climb.

Why is both the maximum angle of climb and maximum rate of climb important?

The airspeed for the best climb angle, also known as $V_x$ , is the airspeed that allows an aircraft to gain the most altitude in the shortest possible distance over the ground, e.g., for obstacle or terrain avoidance. The best angle of climb airspeed occurs when the difference between the power available and the power required is greatest. The best rate of climb, also known as $V_y$ , is beneficial when the aircraft needs to achieve the highest climb rate possible. A high climb rate is beneficial because it allows the aircraft to reach more efficient higher operational altitudes quicker.

Effects of Altitude

Both weight and altitude (i.e., density altitude) will affect the rate of climb of an aircraft. Because the effects of changes in altitude are more important than changes in weight after the aircraft first takes off and burns off much fuel weight, it is logical to examine altitude effects on climb performance. Recall that the possible rate of climb is proportional to the excess power available over and above that required for straight-and-level flight at the same altitude. The figure below shows representative results for the power required for a propeller-driven airplane with increasing altitude.

The power required varies according to the relationship for a propeller-driven airplane, i.e.,

(17) $\begin{equation*} P_{\rm req_{\rm alt}} = P_{\rm req_{\rm MSL}} \left( \frac{\varrho_0}{\varrho} \right) \end{equation*}$

which shows that the power requirements increase significantly with altitude. However, the advantage of flying at a higher altitude is that the drag on the aircraft is less so for the same power then the aircraft will be able to fly at a higher airspeed.

The corresponding behavior for a jet aircraft is shown in the figure below. Again, the thrust and the power available are also affected by altitude, dropping off quickly above a certain altitude. The reduction or lapse in engine power depends, of course, on the particular characteristics of the engine and propulsion system. Therefore, the excess power available (to climb) diminishes with altitude because of these preceding effects, i.e., more power is required for flight, and less power is available for flight. When the power required is equal to the power available at a given altitude, then the aircraft reaches its maximum attainable airspeed. Furthermore, without any excess power available, the aircraft cannot climb at any airspeed.

The excess power available to climb diminishes at higher altitudes as the gap narrows between the power required and the power available.

Determining Rates of Climb – Analytical Approach

From previously, the governing equation for the ROC becomes

(18) $\begin{equation*} V_c = V_{\infty} \left[ \frac{T}{W} - \frac{1}{2} \varrho_{\infty} V_{\infty}^2 \left( \frac{W}{S} \right)^{-1} C_{D_{0}} - \frac{W}{S} \left( \frac{2 }{\varrho_{\infty} V_{\infty}^2 \pi A\!R e } \right) \right] \end{equation*}$

Given a value of airspeed $V_{\infty}$ , the corresponding $V_c$ and the climb angle is

(19) $\begin{equation*} \sin \theta = \frac{V_c}{V_{\infty}} \end{equation*}$

or from Eq. 18 simply by dividing by $V_{\infty}$ then

(20) $\begin{equation*} \sin \theta = \frac{T}{W} - \frac{1}{2} \varrho_{\infty} V_{\infty}^2 \left( \frac{W}{S} \right)^{-1} C_{D_{0}} - \frac{W}{S} \left( \frac{2}{\varrho_{\infty} V_{\infty}^2 \pi A\!R e} \right) \end{equation*}$

Also, it is clear that

(21) $\begin{equation*} V_{\infty} \sin \theta = \frac{T V_{\infty} - D V_{\infty}}{W} \end{equation*}$

or

(22) $\begin{equation*} V_{\infty} \sin \theta = V_{\infty} \left( \frac{T - D}{W} \right) \end{equation*}$

and

(23) $\begin{equation*} \sin \theta =\frac{T}{W} - \frac{D}{W} \end{equation*}$

so the rate of climb increases with higher values of thrust and also with a reduction of drag and weight, as would be expected.

To find the maximum angle of climb then the result that $L = W \cos \theta$ is used and so

(24) $\begin{equation*} \sin \theta = \frac{T}{W} - \frac{\cos \theta}{L/D} \end{equation*}$

Typically, the maximum angle of climb would be used on takeoff, perhaps to clear an obstacle at the end of the runway. Most airliners will be flown just after takeoff at the speed for the maximum angle of climb. When terrain clearance is assured, then the airspeed can be increased to give a better rate of climb.

Assuming again that $\theta$ is small such that $\cos \theta =1$ then

(25) $\begin{equation*} \sin \theta = \frac{T}{W} - \frac{1}{L/D} \end{equation*}$

i.e., the angle of climb depends on the lift-to-drag ratio $L/D$ of the airplane.

Consider first a jet airplane (for which thrust is basically constant at all airspeeds) then the maximum angle of climb $\theta_{\rm max}$ is achieved at the maximum value of lift to drag ratio, i.e.,

(26) $\begin{equation*} \sin \theta_{\rm max} = \frac{T}{W} - \frac{1}{(L/D)_{\rm max}} \end{equation*}$

For a jet aircraft it can be shown that the maximum lift-to-drag occurs at the condition

(27) $\begin{equation*} (L/D)_{\rm max} = \sqrt{\frac{1}{4 C_{D_{0}} K}} \end{equation*}$

where $K$ is given by

(28) $\begin{equation*} K = \frac{1}{\pi A\!R e} \end{equation*}$

so

(29) $\begin{equation*} \sin \theta_{\rm max} = \frac{T}{W} - \sqrt{ \frac{1}{4 C_{D_{0}} K}} \end{equation*}$

From these results then the airspeed corresponding to the maximum climb angle can be found, which is

(30) $\begin{equation*} V_{\theta_{\rm max}} = \sqrt{ \frac{2}{\varrho} \left( \frac{K}{C_{D_{0}} }\right)^{1/2} \left( \frac{W}{S} \right) \cos \theta_{\rm max}} \end{equation*}$

The conditions for the maximum rate of climb can also be estimated analytically.

For a propeller airplane then the power required is

(31) $\begin{equation*} P = \frac{T_A V_{\infty}}{\eta_p} \end{equation*}$

and the power available is equal to the power required so

(32) $\begin{equation*} P_A = \eta_p P \end{equation*}$

and this power is assumed to be constant with both airspeed and rate of climb. In this case, then

(33) $\begin{equation*} \sin \theta = \frac{\eta_p P}{V_{\infty} W} - \frac{1}{2} \varrho_{\infty} V_{\infty}^2 \left( \frac{W}{S} \right)^{-1} C_{D_{0}} - \frac{W}{S} \frac{2K}{\varrho_{\infty} V_{\infty}^2} \end{equation*}$

By differentiating this latter equation with respect to $V_{\infty}$ and setting equal to zero for a maximum then with some algebra it is possible to show that

(34) $\begin{equation*} V_{\theta_{\rm max}} = \frac{ 4 (W/S) K}{\eta_p \varrho_{\infty} (P/W)} \end{equation*}$

for a typical propeller airplane. Again, the conditions for the maximum rate of climb can also be estimated analytically.

Time to Climb

The time it takes to climb from one altitude to another is also of consideration, i.e., to climb from an altitude $h = h_1$ to altitude $h = h_2$ . This latter characteristic is important in operational service when an aircraft may be required to climb to a certain altitude in a given time, such as to comply with air traffic requirements, i.e., to fit in with flight routings or for collision avoidance. It is also more efficient for an aircraft to fly at higher altitudes (the drag is less to the thrust or power required is less), so the ability to climb quickly to an efficient cruising altitude is important. Just after takeoff, an airliner is laden with fuel and relatively heavy, affecting not only its best rate of climb but also the airspeed for the best rate of climb.

The rate of climb is the vertical component of the airspeed so

(35) $\begin{equation*} \frac{dh}{dt} = V_{\infty} \sin \theta = V_c \end{equation*}$

or

(36) $\begin{equation*} dt = \frac{dh}{V_c} \end{equation*}$

where $V_c$ would be a function of weight an altitude. This result means that the time to climb from $h_1$ to $h_2$ would be

(37) $\begin{equation*} t = \int_{h_1}^{h_2} \frac{dh}{V_c} \end{equation*}$

The minimum time to climb would be obtained at the maximum rate of climb so the time would be

(38) $\begin{equation*} t_{\rm min} = \int_{h_1}^{h_2} \frac{dh}{V_{c_{\rm max}}} \end{equation*}$

Flight Ceilings

The airplane’s absolute ceiling is defined as the altitude for which the rate of climb approaches zero, i.e., $V_c \rightarrow 0$ . In this regard, the excess power available over and above that required for straight-and-level flight at the same weight, airspeed, and altitude, diminishes to zero, as shown in the figure below. The service ceiling is defined as the altitude where the rate of $V_c$ reduces to 100 ft/min and is a more useful measure of the ceiling because it represents the upper limit for level flight.

The flight ceiling is reached whenever the aircraft runs out of excess power available over and above that required for level flight at the same weight, airspeed, and altitude.

The service and absolute ceilings can be determined as follows:

Calculate the rate of climb at a given weight for different altitudes, which can be done graphically using the hodograph method or numerically.
Plot the results as a curve in terms of altitude versus the achievable rate of climb, as shown in the figure below.
As the rate of climb diminishes toward a lower threshold, then cease the calculations and extrapolate the resulting curve to find the value of altitude when $V_c$ = 100 ft/min, which is the service ceiling.
Extrapolate the curve to find the height value at $V_c$ = 0, which is the aircraft’s absolute ceiling beyond which it cannot climb (at the given weight).

This latter approach is tedious but effective. Another approach is to set $V_c$ equal to zero in Eq. 18 and solve for the air density, which can then be related to pressure altitude using the equations of the ISA.

The determination of the maximum possible rate of climb at different altitudes allows for an estimation of both the service ceiling and the absolute ceiling for the airplane. The results will be a function of aircraft weight.

Notice that a structural pressurization limit may also set an aircraft’s ceiling because the fuselage is basically a large pressure vessel. For example, a commercial jet airliner is generally limited to altitudes less than 41,000 ft, even though it may have sufficient excess power to climb higher. This characteristic is because the pressure differential between the cabin and the external ambient pressure becomes large enough for the fuselage to reach structural stress limitations.

Gliding Flight

Consider now an airplane in power-off gliding flight, as shown in the figure below. In this case, the thrust from the powerplant(s), $T$ , is zero. However, in the zero thrust condition, the engine may produce drag (e.g., from a windmilling or feathered propeller). It must be factored into the total drag of the airplane and hence the resulting glide performance. The glide angle is denoted by $\theta$ . Again, the airspeed along the flight path can be decomposed into horizontal and vertical components, i.e., $V_c$ and $V_H$ , respectively.

In this case, the equations of motion reduce to

(39) $\begin{eqnarray*} D & = & W \sin \theta \\ L & = & W \cos \theta \end{eqnarray*}$

Dividing one equation by the other gives

(40) $\begin{equation*} \frac{\sin \theta}{\cos \theta} = \tan \theta = \frac{1}{L/D} \end{equation*}$

which shows that the glide angle $\theta$ will be a function only of the lift-to-drag ratio $L/D$ . Obviously, the higher the $L/D$ the shallower the glide angle and the further the glide distance from a given height. The shallowest glide ratio will be obtained at the maximum value of $L/D$ , i.e.,

(41) $\begin{equation*} \theta_{\rm min} = \tan^{-1} \left( \frac{1}{L/D_{\rm max}} \right) \end{equation*}$

Proceeding further by considering the aerodynamics in the glide gives for lift

(42) $\begin{equation*} L = W \cos \theta = \frac{1}{2} \varrho_{\infty} V_{\infty}^2 S C_L \end{equation*}$

and solving for the airspeed gives

(43) $\begin{equation*} V_{\infty} = \sqrt{ \frac{2 \cos \theta}{\varrho_{\infty} C_L} \left( \frac{W}{S} \right) } \end{equation*}$

where the familiar wing loading term $W/S$ appears once again in the results.

This forgoing result tells us that the higher the wing loading the higher the expected glide speed, which also depends on density altitude. The glide angle itself remains a function of the $L/D$ , but at a higher gliding airspeed, the rate of descent will be higher. This all means that the airplane will cover the same distance over the ground (i.e., the gliding distance) but simply reach the ground sooner with a higher airspeed. The range $R$ from a given altitude $h$ is easily calculated because

(44) $\begin{equation*} R = \frac{h}{\tan \theta} = h \left( \frac{L}{D} \right) \end{equation*}$

as shown in the figure below.

For example, if an airplane with a $L/D$ of 15 is initially at 30,000 ft altitude and suffers a complete power failure, then the approximate distance it can glide will be $R$ = 30,000 (15) = 450,000 ft = 82 miles. This latter result gives some comfort in that the loss of total power in an airliner (which is not unheard of) gives the pilot a considerable gliding distance to reach an airfield. However, to achieve a specific $L/D$ , the airplane must fly at a specific airspeed, called the equilibrium glide speed. For an airplane, both $C_L$ and hence $L/D$ (or $C_L/C_D$ ) depend on the angle of attack.

Sailplanes & Gliders

These preceding aerodynamic principles are also fundamental to the sport of gliding. The ability to glide long distances from a given altitude depends on the ability of a sailplane (a high-performance glider capable of efficient soaring flight) to have a high lift-to-drag ratio and shallow glide angles $\theta$ . To sustain its flight, a sailplane can take advantage of rising air currents from thermals, uplifted air when the wind blows against hills and ridges, sea-breeze frontal boundaries, or atmospheric waves.

As will be apparent from the figure below, for the shallow angles of glide typical of a sailplane then

(45) $\begin{equation*} \theta \approx \frac{1}{L/D} \end{equation*}$

Therefore, the shallowest glide ratio will be obtained at the maximum value of the lift-to-drag ratio, $L/D$ , While gliders and sailplanes are basically the same, sailplanes usually have $L/D$ values exceeding 30:1.

The balance of forces on a sailplane, showing that the higher the lift-to-drag ratio of the aircraft the shallower the glide angle.

High aerodynamic efficiency is essential to achieve good gliding performance, so gliders often have aerodynamic features that are not found in other aircraft. For example, the wings of a modern glider use smooth, low-drag laminar airfoils. The wings are made of composite materials with great geometric accuracy, and the surfaces are made glassy smooth by polishing. Sailplanes may also use winglets to decrease drag further and improve the wing’s aerodynamic efficiency by reducing induced drag. In addition, the hinge gaps of the ailerons, rudder, and elevator are carefully sealed on sailplanes to reduce drag further. The consequence of a better lift-to-drag ratio or $L/D$ is that a sailplane can glide further from a given altitude, as shown in the figure below.

A sailplane with a better lift-to-drag ratio $L/D$ will be able to glide further.

Not only is the maximization of the $L/D$ of a sailplane important, but in competitions where both distance and speed are essential, the best $L/D$ must be obtained at relatively high airspeeds. This latter result can be seen from the gliding polar; an example is shown below. In this case, the polar is a hodograph for descending flight. Notice that the best glide ratio is obtained when the straight line from the origin of the graph at (0,0) just touches the polar.

Notice also that the best glide ratio is obtained at a higher airspeed when the sailplane is flown at a higher weight. Because the gliding airspeed is a function of wing loading, all things being equal, a sailplane with a higher wing loading will be able to glide at a higher airspeed and travel further in a given time. In this regard, modern competition gliders carry jettisonable water ballast in the wings. The extra weight the water ballast provides is advantageous if the strong thermals and soaring conditions allow the sailplanes to climb in rising air. Although heavier gliders have a slight disadvantage when climbing in thermals, they can achieve a higher airspeed at any given glide angle. This latter characteristic is advantageous in strong conditions when sailplanes spend only a small amount of time climbing in thermals. The pilot can jettison the water ballast at any time and usually before landing to minimize landing loads on the airframe.

Summary & Closure

Establishing the climbing performance of an aircraft is fundamental to its overall flight capability. The ability to take off from a runway and establish a sufficient angle of climb or rate of climb to clear obstacles is an essential consideration in the design process. The rate of climb for any given aircraft is a function of its excess power available over and above that needed for level flight at a given weight and operating density altitude. Climb capability is crucial for one engine inoperative conditions, and sufficient power must still be available from the remaining engine(s) to permit a safe rate of climb.

Propeller-driven airplanes tend to have significant excess power available at low airspeeds and generally have impressive short takeoff and high initial rates of climb compared to jets, so they are ideal for use from short runways or those surrounded by high terrain. Jets, however, have relatively low amounts of excess power available at lower airspeeds and initial climb performance is much lower. The maximum attainable altitude of an aircraft will also be limited by the excess power available; power will diminish with increasing density altitude, so there comes the point where the excess power available diminishes so that the aircraft cannot climb any higher, which is referred to as the ceiling.

The ability to glide to a safe landing with complete power loss is also fundamental to the safe design of all aircraft. While powered aircraft rarely make good gliders, the ability to land without power is always a consideration for all aircraft types in regard to safety of flight.

5-Question Self-Assessment Quickquiz

For Further Thought or Discussion

Explain why the maximum rate of climb of an airplane depends on density altitude.
An airliner typically climbs to its maximum cruising altitude in a series of stages. Explain why this process is required.
After takeoff from a valley surrounded by mountainous terrain, an airplane will need to climb at its best angle or best rate? Discuss.
For operations out of smaller airports surrounded by mountains, would a propeller or jet aircraft be preferred and why?
In competition gliding, a sailplane may often carry jettisonable water ballast. Why?

Other Useful Online Resources

To dive further into the the climbing and gliding characteristics of aircraft, visit the following web sites:

Skybrary’s entry on gilding, including a link to a report on a successful glide descent to a landing in an Airbus A330 after it ran out of fuel.
How Stuff Works article on how far a plane can glide after engine failure.

License

Icon for the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License

Introduction to Aerospace Flight Vehicles by J. Gordon Leishman is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License, except where otherwise noted.

Digital Object Identifier (DOI)

https://doi.org/https://doi.org/10.15394/eaglepub.2022.1066.n38