Mastering the Glide: How Aircraft Fly Without Thrust

Introduction

Gliding flight demonstrates how an aircraft can remain airborne without engine thrust by efficiently balancing lift, drag, and weight. Whether in engine-out scenarios or in glider operations, understanding the principles of glide performance is essential for pilots and engineers alike. Factors such as glide ratio, angle of attack, airspeed, and aircraft configuration play a crucial role in determining how far and how efficiently an aircraft can travel without power. This article explores the aerodynamic principles behind gliding flight and explains how aircraft are able to sustain controlled, stable flight without thrust

Gliding is a term applied to the flight of an aircraft where there is no power, or thrust, delivered from the engines. If we consider the forces acting upon the aircraft below, we can neglect the thrust.

Generating expressions for lift and drag, by balancing the forces above, gives:

D = W sin γ

L = W cos γ

Where γ is the glide angle.

Dividing these 2 expressions gives us a very useful parameter ^D/_L.

^D/_L = ^{W sin γ} / _{W cos γ} = tan γ

tan γ = ^D/_L = ^C_D / _CL

Example 1: Calculating Glide Angle from Aerodynamic Coefficients

Given: Lift coefficient, C_L= 0.6,  Drag coefficient, C_D= 0.03

Calculation:

Tan⁡ γ = ^C_D/ _CL = ^0.03 / _0.6 = 0.05

γ = tan⁡⁻¹(0.05) ≈ 2.86º

The glide angle γ  is approximately 2.86°.

Example 2: Finding Drag-to-Lift Ratio and Glide Angle

Given: Lift L = 20 000 N, Drag D = 1 000 N
Calculation:

^D/_L = ^1 000 / _20 000 = 0.05

tan⁡ γ = 0.05 ⇒ γ =tan⁡−1(0.05) ≈ 2.86º

 The aircraft descends at a glide angle of 2.86°.

Example 3: Calculating Lift Coefficient for a Given Glide Angle

Given: Drag coefficient C_D = 0.04, Glide angle γ = 4º
Calculation:

tan⁡4º ≈ 0.0699

C_L = ^C_D / _{tan ⁡γ} = 0.04 / 0.0699 ≈ 0.57

 Required lift coefficient is C_L ≈ 0.57

This is a useful equation which tells us that as drag increases and lift decreases the ratio ^D/_L gets bigger – this also corresponds to an increase in the glide angle. Another way of looking at it is, as the lift increases and drag decreases for an aircraft, the Lift to drag ratio ^L/_D increases and the glide angle minimises. 

Minimising the glide angle is important for an aircraft as it determines how far the aircraft will glide with the absence of engine power. It can also be thought of as the efficiency of the aircraft, where in this sense the efficiency is maximising the lift produced compared to the drag produced. 

It will be noticed that this is the same criterion as for maximum range, so that an aeroplane that has a flat gliding angle should also be efficient at flying for range, neglecting the influence of the propulsion efficiency.

If an aeroplane is to glide as far as possible, the angle of attack during the glide must be such that the lift/drag is a maximum. If the pilot attempts to glide at an angle of attack either greater or less than that which gives the best ^L/_D, then in each case the path of descent will be steeper. 

To summarise:

• We want to know how far the aircraft can travel in gliding flight.
• For maximum distance the glide angle is minimised.
• Aircraft must fly at velocity and angle of attack to minimise glide angle and hence minimum ^C_D / _CL

We can also find the glide range of an aircraft by considering the diagram below.

The glide range can be calculated as follows:

tan γ = ^h / _S

S = ^(h₁-h₂) / _{tan γ}

Subbing in for ^L/_D:

S = ^L/_D ( h₁-h₂)

Example: Find the glide range for an aircraft that has an altitude of 2.5km. The aircraft has a lift to drag ratio ^L/_D of 15. 

Solution:

S = ^L/_D ( h₁-h₂)

S = 15 ( 2500 – 0)

S = 37500m

We can see here from this answer that for every 1000m the aircraft descends it is covering 15000m horizontally or 15 times as much. 

Minimum Glide Angle

Since the mathematical relationship between glide angle and lift and drag coefficients is:

tan γ = ^C_D / _CL

If we express ^C_D / _CL = ^{(C_DO + kC_L2 )} / _CL  

We can differentiate ^C_D/_CL   and make this equal to zero, giving the minimum value of this function. After some algebraic manipulation the minimum value of ^C_D/_CL  

is given by 2 √kC_DO  Hence the minimum glide angle is given by:

γ _min = tan ^-1 ( 2 √kC_DO  )

Where;

K = ¹ / _πARe (1 over the effective aspect ratio)

C_DO is the drag coefficient at zero lift

It can then be seen to minimise the glide angle one must increase the effective aspect ratio (as this will reduce the value of k) 

The diagram below  shows how aspect ratio changes for various wing designs. It can be seen that the longer wings have a much higher aspect ratio and hence a smaller minimum glide angle. 

The zero-lift drag coefficient is reflective of parasitic drag which makes it very useful in understanding how “clean” or streamlined an aircraft’s aerodynamics are. The zero-lift drag coefficient can be minimised by streamlining the design of an aircraft. The removal or redesign of drag-inducing structures such as landing gear, engines and radiators, struts and wires, open cockpits, and armament could greatly increase the performance of both civilian and military aircraft.

Example: A sailplane with a mass of 500kg has a drag polar equation C_D = C_DO + kC_L² where the value of C_DO is 0.01 , the value of k is 0.022 and the lift coefficient is 0.75. Find the lift to drag ratio and the minimum glide angle. 

Solution:

C_D = 0.01 + 0.022 x 0.75²

C_D = 0.0233

The lift to drag ratio is therefore:

^L/_D = C_L/D_C = 0.75 / 0.0233 = 33.6

We can also find the minimum glide angle with:

γ _min = tan ^-1 ( 2 √kC_DO  )

γ _min = tan ^-1 ( 2 √0.022 x 0.01 )

γ _min = 1.7º

Speed for Minimum Rate of Descent

With some consideration to the forces applied on an aircraft in glide and algebraic manipulation we can determine the velocity required for the minimum rate of descent, v_min

 V_min = 4(^2w/_ρ)^½ ( ^k3C_DO / ₂₇ ) ^1/4

In this equation one must consider: ρ,the air density. wing loading, w. the constant k and the drag at zero lift C_DO 

wing loading is the total mass of an aircraft divided by the area of its wing. 

Example: Worked example (sea level)

Given: Weight: W = 150,000 N, Air density: ρ = 1.225 kg m^−3, Induced-drag factor: k = 0.05, Zero-lift drag coefficient: C_D0 = 0.025

Calculating the first bracket

(^2w/_ρ)^½ = ( ^2×150000 / _1.225) ^1/2

(^2w/_ρ)^½ = (³⁰⁰⁰⁰⁰/_1.225) ^½ 

(^2w/_ρ)^½ = ( 244897.96)^½ = 494.87

Calculating the second bracket

( ^k3C_DO / ₂₇ ) ^¼ =( ^{(0.05)3 ( 0.025)} / ₂₇ )^¼

 ( ^k3C_DO / ₂₇ ) ^¼= ( ^{3.125 x 10-6} / ₂₇ ) ¼

( ^k3C_DO / ₂₇ ) ^¼ = 0.01844

So the final calculation 

V _min = 4 x 494.87 x 001844

V _min = 36.51ms^-1

Converting to Knots 

36.51 x 1.94 = 79.97knots

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