8/13/2019 AerE 355 - Lecture 2

1/25

Lecture 2

Transition from Considerations ofPerformance to Stability & Control:

Steady Glider Performance

8/13/2019 AerE 355 - Lecture 2

2/25

(3) Quality of equilibrium (trim) associated withthe flight path. (stability)

Flying Wing Problem

A flying wing problem is used to demonstratethe transition from performance considerations tostability and control considerations. It includesthe following elements:

(1) Establishment of a flight path fromPoint A to Point B. (flight dynamics/performance)

(2) Control requirements for flight pathimplementation. (control)

8/13/2019 AerE 355 - Lecture 2

3/25

A Flying Wing Example

3

8/13/2019 AerE 355 - Lecture 2

4/25

Cliff

2000 ft

A flying wing glider configuration rests on the flat topof a mountain 2000 ft (MSL) high. The mountain issurrounded with water (sea level) and has steepcliffs on all sides. The nearest land mass is located20,000 ft in the distance at sea level. Can the gliderconfiguration fly from the mountain top to the land masswithout touching water?

Sea level water surface

20,000 ftB

A

Land mass

(Additional information on next slide)

8/13/2019 AerE 355 - Lecture 2

5/25

Additional Information about the glider configuration

Cd o = zero lift drag coefficient = 0.035

W = glider + pilot weight = 200 lb

AR = glider aspect ratio ( length to breadth ) = 8

e = glider span efficiency factor = 0.85

S = glider wing area = 80 sq ft

r = atmospheric density = 0.002378 slug/ft 3

(assume constant density)

8/13/2019 AerE 355 - Lecture 2

6/25

This is a performance problem. We need only beconcerned with the motion of the glider cg.That is, the wing/pilot can be considered as a point

mass. The ensuing motion is governed by Newtons laws. They provide the basis for the simulation ofthe admissible flight paths.

Fundamental question to be answered -

Is there an admissible flight path that spansthe 20,000 ft distance from point A to point B?

8/13/2019 AerE 355 - Lecture 2

7/25

Lift, L

Drag, D

Weight, W

Free stream velocity, V

Local Horizon

g

g = Flight Path Angle

g

g

8/13/2019 AerE 355 - Lecture 2

8/25

Lift, L

Drag, D

Weight, W

Free stream velocity, V

Local Horizon

g

g = Flight Path Angle

g

g

8/13/2019 AerE 355 - Lecture 2

9/25

Vx

Vh

V

Velocity Components

x

h

Coordinate System

x = range coordinateh = altitude coordinate

8/13/2019 AerE 355 - Lecture 2

10/25

Note: These ordinary differential equations

are 1st

order in velocity components.

Apply Newtons 2 nd Law

Note: g = accelerationof gravity

L

D

W

g

g

V

Fx = -D cos( g) L sin( g) = (W/g) d(V x)/dt+ (1)

+ Fh = L cos( g) - D sin( g) - W = (W/g) d(V h)/dt (2)

8/13/2019 AerE 355 - Lecture 2

11/25

Relationship between velocity components, rangecoordinate, and altitude coordinate.

Vh = d(h)/dt

Vx = d(x)/dt

Note that these equations are kinematic inNature and are essentially used to reducethe previous equations (Newtons 2nd law)

to first order equations.

(3)

(4)

8/13/2019 AerE 355 - Lecture 2

12/25

Lift and Drag Force Definitions

L = C L (1/2) r V2 S (5)

D = C D (1/2) r V2 S (6)

where CL = Lift CoefficientCD = Drag Coefficientr = Atmospheric density

S = Configuration wing area

Lift and Drag are traditionally defined in thefollowing terms:

Drag components sum to produce total drag Skin friction Base pressure differential Shock-induced pressure differential (M > 1)

Lift components sum to produce total lift Pressure differential between upper and

lower surfaces Wing

Fuselage Horizontal tail

8/13/2019 AerE 355 - Lecture 2

13/25

Lift/Drag Relationship

CD = C do + C L2 /(p e AR) (7)

where

C do = zero lift drag coefficient andCL2 /(p e AR) = induced drag coefficient

A 2 nd order relationship termed parabolic drag polar exists between lift and drag for streamlinedconfigurations. This relationship is generally usedin a performance analysis and is given by:

8/13/2019 AerE 355 - Lecture 2

14/25

Velocity Components Defined

The relationship between the wing/pilotVelocity ( V) and the horizontal/verticalvelocity components ( Vx and Vh ) must bedefined and included in the final set ofequations.

V2 = V x2 + V h 2 (8)

8/13/2019 AerE 355 - Lecture 2

15/25

Flight Path Definition

Finally, the flight path angle must be definedin terms of the velocity components.

g = tan -1(Vh /V

x) (9)

All equations required to establish the set ofadmissible flight paths from point A to point B

have been developed in the previousdiscussions. The equations can now becollected.

8/13/2019 AerE 355 - Lecture 2

16/25

-D cos( g) - L sin( g) = (W/g) d(V x)/dt

L cos( g) - D sin( g) - W = (W/g) d(V h)/dt

Vh = d(h)/dt Vx = d(x)/dt

L = C L (1/2) r V2 S

D = C D (1/2) r V2 S

CD = C do + C L2 /(p e AR )

V2 = V x2 + V h 2

g = tan -1(Vh /Vx)

Equation Set

Variable Set (D, g, L, V x, Vh , h, x, C L, C D, V)(If sufficient boundary conditions are

specified.)

8/13/2019 AerE 355 - Lecture 2

17/25

Reasonable result???

Equation Set Closure

Question Is there a unique solution to thesystem of equations?

Answer One must determine if the systemhas closure. Count the numberof equations and compare to thenumber of variables.9 equations - - - 10 variables(1 excess variable)

Therefore a unique solution does not exist.In fact, the system has infinite solutions.

8/13/2019 AerE 355 - Lecture 2

18/25

Sounds like control! Its only reasonable to believethat the interaction between the pilot andthe wing during the flight process can, anddoes, impact the flight path characteristics.That is, the pilot can control the flight.

Reasonable Result!!!

One shouldnt expect a unique flight path. Thatcould occur only if the excess variablein the set is specified. That would provideclosure for the system and a unique flightpath would result.

8/13/2019 AerE 355 - Lecture 2

19/25

It would appear that if you want the best chance toglide from cliff to land mass, you must generate theCL function that will maximize the range of theconfiguration.

What Variable Should be Specified?

That is, what should be used as the control variable?

Looks as if CL (lift coefficient) would be mostappropriate. In fact, the pilot can control theattitude of the wing during the flight process toset instantaneous values for CL.

8/13/2019 AerE 355 - Lecture 2

20/25

How Does One Determinethe Best C L Function?

How Does One Solve theSystem of Equations?

8/13/2019 AerE 355 - Lecture 2

21/25

This can be accomplished by merely settingthe accelera t ions to zero and solving the systemof equations for all remaining variables .

Lets assume a constant value for C L throughoutthe flight. Furthermore lets investigate only the equi l ibr ium f l igh t pa ths that occur between pointsA and B.

8/13/2019 AerE 355 - Lecture 2

22/25

-D cos( g) - L sin( g) = 0 (1)

L cos( g) - D sin( g) - W = 0 (2)

Vh = d(h)/dt (3)

Vx = d(x)/dt (4)L = C L (1/2) r V2 S (5)

D = C D (1/2) r V2 S (6)

CD = C do + C L2 /(p e AR) (7)

V2 = V x2 + V h 2 (8)

g = tan -1(Vh /Vx) (9)

8/13/2019 AerE 355 - Lecture 2

23/25

Set CL value

Evaluate C D from (2)Evaluate g from (1)Evaluate V from (3)

Evaluate Range from (4)

Reduced equation set:

g = tan -1(-C D /C L) (1)

CD = C do + C L2 /(p e AR) (2)

V2

= 2.0/( r S)[ W/(C L cos( g) - C D sin( g)) ] (3)range = -2000.0/tan( g) (4)

Procedure

8/13/2019 AerE 355 - Lecture 2

24/25

CL Gamma Velocity Range0.2 -10.5 102 10,8500.3 -7.5 83 15,300

0.4 -6.0 72 18,8000.5 -5.3 65 21,4100.6 -4.9 59 23,1400.7 -4.6 55 241600.8 -4.6 51 24,6300.9 -4.7 48 24,6901.0 -4.8 46 24,000

8/13/2019 AerE 355 - Lecture 2

25/25

Unsteady Flight Paths Considered Next Period