Summary of Results from Flight

Summary of Results from Flight Sep 3 – Dec 6, 2006 Notes of Merit 1. Argument and output of all transcendental functions must be dimensionless (or of non-linear scale). 2. Always check your units you fucking idiot 3. We are deep in the quicksand of subscript hell

Constants J ft 2 ft lb R  287  1716  1716 KgK s 2 R slug  R c P    1.4 cv kg Viscosity of Air:   1.79 105 m  s kg Density of Kerosene:   799.242 k m3

Coefficient of Friction for a paved runway: r  0.02

Standard Atmosphere Ideal Gasses: P  RT dP Pressure Equation:  g dz g  az  aR For lapse rate a:   Pz  P0 1   T0  Basic Aerodynamics In steady flight, T=D, L=W

Conservation of mass on a streamtube: 1V1 A1   2V2 A2 Conservation of momentum: the forces on a fluid are composed of body forces and surface forces. In general body forces are negligible. Surface forces are composed of normal forces (generally pressure) and tangential forces (shear forces, always friction from fluid viscosity).

dV dP Euler’s Equation for a differential cube of fluid encountering pressure:    dt dx For steady motion, dx  vdt Which gives us: VdV  dP By integration along a streamline (with constant density) we get Bernouli’s Equation: 1 1 P  V 2  P  V 2 1 2 1 2 2 2 Brief Review of Thermodynamics  1  P  1   1  e  w  q , dW  PdV  Pd  , h  e  , dh  de  Pd    dP            1  dP And thus q  e  Pd   dh      de dh Specific Heats: c  , c  and thus for ideal gasses, e  c T , h  c T v dT P dT v P

And by definition: h  e  cP  cv T  RT R a 2 c   P  1  1 dP 1 For isentropic flow on a streamline (no friction, no dissipative processes) h    V 2  C   2 1 giving h  V 2  Const 2

More isentropic relations: de P P  giving de  d d  2  2 P dP Thus: de  cv dT  d and similarly: dh  c dT   2 P  1 P 1 P dP d dT  2  Which produces   and thus from the ideal gas law cv  cP  P    1  P2   2  T2   2  P  T   1       2   2  P    T      1  1  1  1  P1  T1  Note that these relationships apply even when density is no constant!

Speed of Sound Stuff Mass is conserved across a sound wave (with velocity a):   a    d a  da  a  ad  da da (a second order differential drops out), giving a   d dP Momentum is conserved: dP  ada =0 and thus da   which combines to a dP P a 2    RT from the ideal gas law, or: a  RT d  Some Stuff about Wings 1 If there is no heat transfer on a streamline, we combine above relations to get: h  V 2  Const , 2 which applies at a stagnation point giving: 2 2 1 2 T0 V   1 V   1 2 cPT  V  cPT0  1  1    1  M  2 T 2cPT  2  RT  2 

The isentropic relations can be applied to this formula to get some helpful properties

In flight, wings produce a downwash in the rear. This effectively increases the angle of attack α 2 cL b by αi where αi is given by:    , this induces an additional drag: i AR S 2 cL cL Di  L i  q S , and thus c  (with some given efficiency e (e can be 1 for elliptic  AR Di ARe wings)

Note that a finite aspect ratio reduces the slope of the lift vs. α curve: a0 for infinite wings, a for finite wings:

Infinite Wings a   const dc  c  c  0 a L L L 0  and also cL  a0     const , a0 d  i   ARe  1 ARe Finite Wings dc a L  0  a d a 0 1 0 ARe

Wind Tunnel Relations

By combing Bernouli’s equation and incompressible mass flow (V1 A1  V2 A2 ), we get a formula 2 1 P  P 2  1 2 for low speed wind tunnels: V2  2 A2 1 2 A1 U-Tube Manometer

With fluid density  m we have P1  P2   m gh 1    2  m  For low speed flow on a Pitot Tube: P  V  P0 and thus, V  2 gh when it is 2   a  connected to a U-tube manometer Subsonic Compressible Flow

T0  1 2 Using the enthalpy and energy relations,  1 M  , by using the isentropic relations, T 2   1  2a 2  P   and definition of the speed of sound: V     1      1  P    Supersonic Duct Flow

dA dV d dV dA AV  C  ln(A)  ln(V )  ln()  C     1 M 2    A V  V A Thus, for M < 1, decrease area to increase speed, for M > 1, increase area to increase speed. Thus you can only accelerate past M=1 by going through a minimum section area.

The relationships can be fleshed out to generate:   2 dT   1MdM dA M 1dM  dV dM  1   T  1 2    A   1 2  1 M V M  1 2 M 1 M  2 1 M  2  2    In a supersonic wind tunnel, M=1 at the minimum throat section (denoted *) 1 1 * * * * A  V  a a * * 2 * *  1   a  T    T  *       A V aM a  T    T 

1 1    1   1 2 1 M 2 A 1     2  A* M   1     2 

Drag Effects On a sphere with very low skin friction, only pressure drag remains: 1  d 2  2   D  V   2  4 

 dV  Viscosity creates a shear stress on a surface:  w    Thus, the steeper the  dY  y0 boundary layer plot slope, the greater the skin friction

D  PdS Pressure drag can be found as P  S D   dS Skin friction drag can be found as f  w S Boundary Layers Reynold’s number: Large Reynold’s number means small viscous effects, except in very thin layer of fast variation. A small Reynold’s number means that viscous effects are important throughout the flow field. VL For flow on a flat plate, Re  - Re close to 1 in the leading edge region, Re >> 1 in laminar  flow region, Re = 105 in turbulence layer

Vext Vy We define δ as the height of a boundary layer where  0.01 Vext ky2 For a laminar boundary layer on a flat plate, V Vy  e where V is the flow speed external to the boundary layer

5x For laminar flow on the flat plate:   Re x 1  For turbulent flow on the flat plate:   0.37x(Re) 5

 w 0.664 The skin friction coefficient: c f  , For a flat plate in laminar flow: c f  q Re x 0.74 c  For turbulent flow on a flat plate: f 1 Re 5 This can be integrated over a plate to find the net friction force. Turbulent boundary layers produce more drag than laminar boundary layers. Note that turbulent boundary layers have increased momentum towards a surface, so by increasing turbulence a boundary layer can be maintained further down a surface, lowering wake drag (and perhaps the overall drag in blunt bodies).

Note: Stalling occurs when the angle of attack raises to high, and a streamlined body experiences blunt body drag.

Airfoils and Wings Camber line bisects thickness distribution (the camber at a point of the boundary of an airfoil is the distance from the chord to camber line at that point)

Note that most wings generate a moment which is balanced by the tail of a plane

Lift is the force perpendicular toV , Drag is the force parallel to V It is most convenient to describe wings with the dimensionless coefficients: L, D, M

Overall, Force on a wing is a function of many properties: F  f L, D, M ,  ,V , S,  ,c,a ,  (with c as the chord length, S is the wing projected onto a ground plane) Dimensional Analysis produces: L D M C  C  C  L 1 D 1 M 1  V 2 S  V 2 S  V 2 Sc 2   2   2   The pressure on a point of an airfoil is described with the pressure coefficient: 2 V P  P cP  1 2  V 1 2  V 2  For finite Mach number, and isentropic flow, we get the stagnation pressure related to a general pressure   1 P0   1 2   1 M  , by related the stagnation pressure with P and some other point: P  2 

   1 2  1 1 M P      2  P   1 2    1 M   2  Using the definition of the coefficient of pressure, this can be used to find:    1 2   1 1 M   2  2  cP  2 1 M   1 2    1 M   2  TE TE  P cos( )dS  P cos( )dS The net lift per unit span of an airfoil can be found as  u  l LE LE

Where θ is the angle between the flight path and the horizontal

For infinite wings, we have drag per unit span as D  D f  DP  DW With wave drag only factored in for transonic/supersonic flow

For finite wings, we get wing tip vortices with cause induced drag proportional to cL

Dimensionless Quantities Dimensional Analysis tells us that givenQ   quantities and N   dimensions, you can make Q-N dimensionless quantities (counting inverse pairs only once). Only multiplication and division may be used to relate these quantities. In addition to the above wing coefficients, we define:  V c S b S Re    AR   b   c S c Shock Waves Wings have a point of maximum velocity which creates a shockwave once the wing reaches a critical Mach number. This creates a sharp increase in pressure and disturbs the boundary layer. A shock wave is a discontinuous transition in a supersonic flow, and is described by a shock 1 angle: sin()  M Shockwaves only occur in supersonic flows when dictated by the boundary conditions.

From some extensive analysis with the conservation laws:

  1  1 2 P2  P1 1    2  1  a* where a* is some constant  2  2

P2  P1 For a constant diameter stream tube traveling through a shockwave:  V1V2  a*  2  1

We can thus solve for M2 (the velocity of a flow after passing through a shockwave) 1 2  2 1   1 2   2 1 M 1  M  1 M  2  2  1  2  M 1  2  1 M 1 1   1 

For Oblique shocks, the component of flow velocity tangential to the shockwave is unaffected, only the normal component is changed.

Shock waves introduce wave drag to a vehicle. Thus with shock angle β and deflection angle θ:

V1 cos   V2 cos  

The change in flow angle can be found as

1  2 1   1 2 2     tan  1 M 1 sin    1 2 2  M 1 sin cos   For weak shocks, θ goes to 0, and β goes to μ and our above velocity relations go to V V  2 1  V1  V2 1 tan  2 when analyzed with some differential V2  1 dP M 2 d dP M 2 d  u   calculus, 2 this is more usefully described as: 2 and P P  1 M 1 1 dP M 2 d l  2 P M 1 1 4 cD  cL  By integrating around the surface of a wing, we get: 2 M  1 Range and Endurance For jets, we define the thrust specific fuel consumption (ct) as the weight of fuel/(force of thrust- time) The endurance of a jet is found as: 1 1 C W  2 1 1 L 0 2 1 CL   E j  ln  Its range is found as: R  2 W 2 W 2 c C  W  j  0 l  t D  l    S ct CD   For propeller aircraft, we use the specific fuel consumption (c) as the weight of fuel / (unit of power-time). Also, η denotes the propeller efficiency. 3 2 1 1 1  C W   CL     L 0 E  2 S 2 W 2 W 2 R p  ln  p     1 0  c C  W  c CD   D  1 

For optimal performance, the CL/CD are maximized 1  1  4 1 1 C eAR 2    D0  CL CD0eAR  C 2  3  L    C 2C   D max D0  CD  4   CD0 max 3  3  3 2 4  CL  3CD0eAR     CD  4CD0  max

Takeoff and Landing Performance For takeoff, maximum lift coefficient is employed (that is allowed). 2W In general for takeoff: VLO  1.2   SCL,TO 2W Where VSTALL   SCL,TO 1  C 2  D   V 2 SC   L  Drag becomes more complicated for takeoff:    D,0  2  eAR  Here, Φ is a ground effect coefficient, and can be approximated (for h = height of wing, and b is wingspan) 2 16h     b    2 16h  1    b 

By combining takeoff forces: 2 2 1 VLO 1.44W S LO  M  2 T  D   R W  Lavg g  SCL,TO T  D   R W  Lavg  Sometimes, the resistive forces are negligible in comparison to the thrust. Generally, the average resistive forces are taken at 0.7VLO

2W For landing, we generally use: VTD  1.3  SCL,TD 1.69W 2 Thus: STD  g  SCL,MAX Davg  Trev   R W  Lavg 

Again, average values are found at 0.7VTD

Turning Flight mV 2 For the plane body rolled into a level turn by an angle Φ, W  L cos and Lsin   F . r r 2 2 Not that the lift can be related to the radial force and weight as: L  Fr W L We define a load factor n: n  , Up to some structural limit, n can be maximized with: W 1 C S n   V 2 L,Max max 2   W For maximum maneuverability maximize the load factor and cornering velocity, the corner 2n W velocity is defined as: V   max  CL,max S Here, W/S is called the wing loading 2 2 2 WV V which produces (for turning radius R): Fr  W n 1 , and thus: R  gR g n 2 1 g n 2 1 Similarly, the turn rate is defined as:   V

The pull-up maneuver rotates the plane tip upwards, and has turn radius and rate as: V 2 gn 1 WV 2     R  2 , here the radial force becomes: Fr  gn 1 V gR

For the pull down maneuver in inverted flight (assuming n>>1 for high performance military craft that perform these maneuvers)

2 gn n C V 2L    g  L R    and V W gn  SC  2  L S

Note that the minimum radius appears at CL,MAX Control Surfaces Ailerons - control roll, located on trailing edges of wings Flaps – controls trim, can affect pitch Elevator – Horizontal, Located on tail, controls pitch Rudder – Vertical on tail, controls yaw Slats – on leading edge of wings, limits flow separation from large angles of attack Spoilers – Act as airbrakes, located on tops of wings

Stability and Control Static Stability: rapid return to initial state Dynamic Stabilty: Motion is tracked over some time – equilibrium is returned slowly.

Aerodynamic center – point on chord where the moment does not vary with angle of attack.

For longitudinal static stability, the slop of the CMCG vs. Angle of Attack is negative.

Note that a positive moment rotates the wing tip upward

For positively cambered wing sections, CMAC <0, the tail produces a positive moment to allow for level flight.

Moments about CG:

M cgw  M acw  Lw cos w h  hacw c  Dw sin w h  hacw c  Lw sin w zc  Dw cos w zc

Here h is length from wing tip to the center of gravity of the wing body, hac is the length from the wing’s aerodynamic center to the center of gravity, and z is the height from the wing chord line to the center of gravity.

We often assume that the angle of attack is very small, so:

M cgw  M acw  Lw  Dw w h  hacw c  Lw  Dw w zc

This leads directly to a formula for the moment coefficient

CMcg  CMacw  CLw  CDw w h  hacw c  CLw  CDw w zc

Z is generally quite small, so we get: CMcg  CMacw  CLw h  hacw 

For lift curve slope aw, the wing’s lift coefficient can be approximated: CLw  aw w

For the wing and fuselage, simply replace all terms referring specifically to the wing (subscript w) with terms referring to the wing and body (subscript wb)

CMcgwb  CMacwb  awb wb h  hacwb 

A tail can now be added to this model with horizontal moment arm lt, and vertical moment arm zt (assume small). For typical wings (positive camber) the tail must produce downward lift to balance the pitch of the plane. Wihout a tail, most planes would tend to dip down (dive).

Lt M cgt  Lt lt where CLt  q St lt St By defining the volume ratio: V  , we can define: CMcgt  VH Clt H cS

The effective angle of attack of the tail can be defined as:  t   wb    it , where ε is an effect from the downwash of the wing, an it is the absolute angle of attack of the tail measured from the zero lift line of the wing body.

Thus, for lift curve slope of the tail at, we can get     CLt  at t  at  wb    it   at wb 1   at it   0  where    0   wb      0

 at    Thus: CMcg  CMacwb   wb awb h  hacwb VH 1  VH at it   0   awb    For stability, the slope of the moment coefficient vs. angle of angle should be negative. This slope goes to zero when h = ha, where ha is found as:

at    ha  hacwb VH 1  a    Static Margin: the rate at which the equilibrium condition is restored (below):

CMcg  aha  h, The steeper this slope, the faster the equilibrium condition can be restored.  a

Pitch can be controlled by deflecting the rear of the tail (we assume the rear tail to be

CL symmetric), this leads to: CLt  at t   e for deflection angle δe  e This leads to a new equation for the moment coefficient around the center of gravity of the wing  C   Lt  body: CMcg  CMacwb  CLwb h  hacwb VH CLt  CMacwb  CLwb h  hacwb VH at t   e    e 

CLt and thus CMcgwb  VH  e  e (ignoring z components of moment) By deflecting the tail, the total CMcg vs. angle of attack slope can be shifted downwards, allowing for higher speeds as the wing’s lift coefficient can be lowered.

The load on a tail can often be large, and it can be analyzed as follows: for He – Elevator Hinge Moment CT – Tail Chord Length Cb – Distance from the leading edge of the elevator to the hinge Ce Distance from the trailing edge of the elevator to the hinge.

H e Che Che Che    t   e 1 2    V S C t e 2   e e For stable planes (and in general), both partial derivatives are negative Note that a positive δe or αt pulls the elevator up, inducing a negative pitching moment, which brings the plane tip up.

When no control is provided to the tail and elevator, the elevator will reach the stick free

Che   t  t deflection:  e  This leads to an equation for the stick free lift of the tail: Che

 e

 Che   Che      CLt  t 1 CLt  t C'Lt  at t    t  at t F where F is found as: F  1    e  Che  at  e  Che        e    e 

With no stick forces, CLt goes to C’Lt, and the moment coefficient about the center of mass becomes: C'Mcgwb  CM ac  CLwb h  hacwb VH at t F

at    The leads to a similar expression: ha  hacwb  FVH 1  , thus a stick free elevator a   

CMcg reduces the overall static margin as:  ah'a h  e Rudder For the vertical tail with a deflection moving its center lvt to a side of the plane: 1 L   V 2 SC , and it produces a yawing moment: L l vt 2   Lvt vt vt Without deflection, the vertical tail provides some restoring capability

Lateral Static Stability If the airplane is banked by some angle: Φ with wings inclined by a dihedral angle Γ (angle between ground plane and wing plane) The downwash velocities become:

Vni  V sin  : inner wing to turn

Vni  V sin  : outer wing to turn Thus, the lift on the inner wing becomes greater than the lift on the outer wing, generating a restoring moment.

Propulsion For a propeller, β is the angle between the plane of blade rotation and the propeller chord lines. Β is a function of r, approaching 0 at the prop tips and approaching 90 near the hub. Relative wind   speed: V, is defined as: V  V  r This angle between this velocity and the propeller is called α – the angle of attack. The angle between the chord line and rω is equal to α + φ where φ is found as:  V   V    tan 1    This reveals:     and     tan 1     r   r  T is the thrust, and is defined by D (the drag, parallel to the relative wind V), and L (Lift, normal to the relative wind). T  L cos Dsin, the available power thus becomes PA  TAV , T V because of skin friction drag and tip vortices, P  P ,   A  A P TA depends on the angle of attack, φ, and d – the propeller diameter. For revolutions per second n, and angular velocity:   2n For a fixed pitch propeller angle β, the advance ratio (J) can be defined. V J   ` dn 1 β β β 3 > 2 > 1 β β η 2 3 β 1

J For large values of J, V becomes large, and thus φ becomes large. φ -> β sets α -> 0 and T -> 0. For φ > β, the thrust becomes negative

Reciprocating Piston Engine 1. Piston is at top dead center, v is minimal (V1). 2. Piston decends to bottom dead center, the intake value open, fuel/air mixture is injected at constant P to maximum volume (V2) 3. Intake valve closes and piston moves to top dead center, P3 and V3 = Vmin =V1 4. Spark ignites the mixture at constant volume V4 = V1, Pressure is increased to P4  5. Power stroke: Piston is moved down isentropically ( PV  C ) to BDC, V5 = Vmax = V2

6. Exhuast valve opens, reduces pressure to P2 = P6

5 The work produced in the power stroke is found as:  PdV 4 3 The work consumed in the compression stroke is found as:  PdV 2 W The net work is the difference between these two. The power output is net . For n turns per unit t n time, and N cylinders, the indicated power IP is: IP  NW 2 net

The shaft power is found as Ps   mech IP n The available power with a propeller thus becomes P    NW A prop mech 2 net An alternative description degines the piston diameter b (bore), the piston stroke (s). The b 2 displacement d is thus found as: d  N s . The effective pressure is defined as Pe such that 4 b 2 n W  P s and PA can be found as P    P d net e 4 A prop mech 2 e

Jet Engine The turbo jet engine operates in stages 1. Air Inlet 2. Diffuser – drops speed, bumps up pressure 3. staged compressor, output fluid is slowed to allow steady combustion. From 1-3 PV   C 4. Combustion, kerosene is sprayed into air and ignited, constant pressure 5. Turbine, extracts work from engine. Isentropic work extraction: PV   C 6. exhaust

Thrust can be found as the integral of the surface forces on the duct. We want the component along the axis of revolution (Anderson 9.14) T  P ds  P A  A T  m˙ V V  A P  P  s   e i  this distilss to:  e   e  e  

Rocket Engines Fuel and oxidizer combust in chamber to produce stagnation temperatures and pressures (T0, P0). The product gassesare expelled through a convergent-divergent duct to produce supersonic speeds. 1    1  2 ˙ ˙  2   Pe    T  m fuel  mox Ve  Pe  P Ae for an isentropic nozzle, V  RT 1   e  0      1   P0      The specific impulse of a fuel mixture is found as: 1    1  2 T T 1 2RT   Pe    I     0 1    This specific impulse is proportional to sp      w˙ g 0 m˙ fuel  m˙ ox  g0  1   P0      R the square root of the gas constant: R  where M-bar is the molecular weight, and R-bar is M the universal gas constant  1   1 2   P0  2  The exit mass rate is solved as: m˙  A   R T0   1 V m b For flight the initial and final masses are related by: i  e g0I sp m f Thus, for ML – mass of payload, MS – mass of structure, MP – mass of payload m M  M  M M i  L S P  1 P m f M L  M S M L  M S

Astronautics 2V For a projectile launched from the ground, the time of flight without drag is found as: t  0 g GmM Lagrangian: L  T   where Ke is T, and Φ is gravitational potential energy (  ) r P A solution of the Lagrangian gives all possible orbits as: r  1 ecos  c h 2 Ah 2 Where P  , h  r 2˙ (angular momentum divided by mass), k 2  GM , and e  k 2 k 2 Note that h, A, and C are defined by the initial conditions. For 01 the orbit is hyperbolic. e >= 1 is an escape velocity

From the Lagrangian solution, we can arrive at: 1 k 4 k 4 T  m e 2  2ecos  c1   m 1 ecos  c 2 h 2 h 2 The difference between the absolute values of the kinetic and potential energies is called H. 1 k 4 H   m 1 e2  2 h 2 k 2k For circular orbit, Vb  For minimum escape velocity, Vb  r r Ballistic Entry through Isothermal Atmosphere During reentry the stagnation temperature on a vehicle can reach extreme values: T  1 0  1 M 2 . During reentry, the velocity at the maximum deceleration can be found as: T 2 2 V entry dV VE Z sin Vmax accel  ,  . For a general point along the flight path, e dt max 2e 2 dV D V CD S  CD S      and thus V  VE exp  where ρ is the density and Z is the dt m 2m  2mz sin altitude at the point of inquiry

Entry Heating dQ 1 CH  For a vessel entering the atmosphere, we define a heat transfer coefficient: dt 1 3  V S 2   dQ 1 3 mCH   V SCH From some analysis we arrive at: dQ   V dV dt 2 CD The total heat on the vessel can be found by integrating from the entry velocity to zero.  C V 2  H  E Qtot  m   CD  2 Reynold’s analogy relates the heat transfer coefficient and the coefficient of skin friction 2 1 mVE CF CH  CF , thus Qtot  2 2 2CD  CF 

Lifting Entry Lifting entry assumes flight at a constant angle θ to the horizon. We define rc, the local radius of curvature, to be approximately equal to the radius of the earth: re. g mV 2 V 2  L W cos   1 SC 1 By substituting in the definition of L, this leads to:  L  re 2 2 re This can be directly used to find the velocity as a function of altitude since g and ρ are known functions of altitude. However, a solution of the differential equation gives:  C S   h D  m  V  VE exp   2z sin   