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Transcript of Gravity Turn Concept - · PDF file Gravity Turn Concept Curvilinear Coordinate System Gravity...

  • Gravity Turn Concept

    Curvilinear Coordinate System

    Gravity Turn Manoeuvre concept

    Solutions for Constant Pitch Rate

  • Inclined Motion Concept

    In reality, vertical motion is used only for a very small

    part of the overall ascent mission and for the most part,

    ascent trajectory is inclined & curvilinear in nature.

    This is mainly because one of the terminal constraint is

    that the inclination of the velocity vector with respect tothat the inclination of the velocity vector with respect to

    the local horizon is required to be close to zero.

    Considering Earth’s curvature, the rocket needs to

    undergo large flight path angle changes (~1100) during

    the ascent mission.

    This requirement calls for a different methodology of

    trajectory design & solution.

  • Effect of Inclination

    A curvilinear flight path requires motion in a plane and

    therefore, needs models for a planar motion.

    Also, thrust is used mainly for velocity increments and

    is always along the flight path, so that a normal force is

    needed to produce the curvilinear path. Consider a

    rocket having inclination with the vertical.rocket having inclination with the vertical.

    us, vs

    un, vnmgT

    L

    Fc

    θ

    X

    Y

  • Consider the following schematic of a planar motion.

    Curvilinear Motion Model

    0 ˆlim s

    t

    s ds V V s u

    t dt∆ →

    ∆ = = → = ⋅

    � ɺ

    ( ) ( )

    Acceleration:

    ˆ

    ˆ

    s

    s s s n

    s n

    d d a V Vu

    dt dt

    du a Vu V Vu V u

    dt

    a a a

    θ

    = =

    = + = +

    = +

    ��

    � � � �ɺɺ ɺ

    � � �

  • Planar Motion Equations

    The equations of planar motion are as follows.

    0 cos

    sin

    s sp

    dV ma m mg I mg

    dt

    d ma mV mg

    θ

    θ θ

    = = − −

    = =

    ɺ

    In this case, the resulting trajectory is called ‘gravity

    turn’ trajectory, as ‘g’ alone is responsible for (dθ/dt).

    The above non-linear time-varying differential equations

    contains three unknowns i.e. V, θ and m(t) (the design input), for which no general solutions exist.

    sinnma mV mg dt

    θ= =

  • Special Analytical Solutions

    Special analytical solutions to the gravity turn equations

    are possible which, while restricting the overall degree of

    freedom, provide immense practical utility.

    In addition, lot of insight can be obtained by analyzing

    the equations themselves. In a kinematic sense, the

    can be rewritten as,gravity turn equations can be rewritten as,

    Also, it is possible to obtain V and θθθθ, if m(t) is specified, or vice versa. This is the basis for ‘Pitch Program’ in

    launch vehicle mission design.

    0 sin ( ) cos ( );

    ( ) ( )

    spmg I g t V g t

    m t V t

    θ θ θ= − − =

    ɺ ɶ ɺɺ ɶ

  • Case – 1: Constant Pitch Rate

    In this case, rocket is commanded to track a specified

    pitch rate i.e. (dθ/dt), which is achieved through an independent pitch rate tracking control system.

    This results in the second equation providing the velocity

    solution, which is then used in the first equation to obtainsolution, which is then used in the first equation to obtain

    the required burn profile {m(t)} or ‘pitch program’.

    0 0 0

    0

    0 0

    0 0 0

    sin ( ) ; ( ) ; cos

    2 cos 2 ln (sin sin )

    sp sp

    g q t q t V t V g

    q

    dm g dt m g

    m g I m q g I

    θ θ θ θ θ

    θ θ θ

    = → = + = =

    = − → = −

    ɶ ɺ ɺ ɶ

    ɶ ɶ

  • Case – 1: Constant Pitch Rate

    As q0 is constant at all times including the initial time,

    we can write,

    0 0 0 0

    0

    sin 0 or 0, =0 not admissible.

    g q V

    V

    θ θ= ≠ ∞ → =

    ɶ

    This means that gravity turn manoeuvre can be started

    only from a non-zero pitch down angle, after it acquires

    a minimum forward speed.

    This requirement is usually met by giving a ‘pitch kick’

    to the vehicle at appropriate time to initiate manoeuvre

    and usually happens after acquiring some altitude.

    0

  • Case – 1: Constant Pitch Rate

    The altitude profile can be obtained by resolving the

    velocity V in vertical direction as follows.

    2

    0 0

    cos sin cos cos

    dh dh V g V

    dt d q q

    θ θ θ θ

    θ = → = =

    ɶ

    Can θθθθ(t) be more than 900? What would such a condition represent? What is the impact on the burn rate and total

    propellant mass?

    0 0

    0 02 2

    0 0

    sin 2 ( ) (cos 2 cos 2 )

    2 4

    dt d q q

    dh g g h h

    d q q

    θ

    θ θ θ θ

    θ = → = − + ɶ ɶ

  • Case – 1: Constant Pitch Rate

    Another trajectory parameter of interest is the final flight

    path angle, which can be evaluated as,

    0 01 0 0sin ln sin

    2

    sp

    b

    b

    g q I m

    g m θ θ−

        = +  

       ɶ

    Burnout time & horizontal distance are as follows.

    ( )

    ( ) ( )

    0

    0 0 2

    0 0

    0

    0 02

    0

    ( ) 1 ; sin 1 cos 2

    2

    sin 2 sin 2 ( ) ( )

    2 2

    b

    b b

    b

    b

    dx t V x x g d

    q dt q

    g x x

    q

    θ

    θ

    θ θ θ θ θ

    θ θ θ θ θ θ

    − = = → − = −

     − = − − + 

     

    ∫ ɶ

    ɶ

  • Constant Pitch Rate Example

    First stage of the Chinese Long March rocket has the

    following lift – off parameters. m0 = 79.4 Tons, mp = 60

    Tons, Isp = 241 s, g0 = 9.81m/s 2, Payload mass = 9.4 Tons,

    β0 = 600 kg/s (until ti), ti = 10s,

    (1) Determine the trajectory parameters at end of 10s.(1) Determine the trajectory parameters at end of 10s.

    Vi = 0.0876 km/s, hi = 0.426 km,

    (2) Determine terminal parameters in case the rocket

    executes the gravity turn for a further 90s. θi = 5 o.

    Vt = 0.827 km/s, ht = 35.0 km, mt = 39.3 Tons, θt = 55.3o, q0 = 0.559

    o/s, tb = 90 s, xt = 25.6 km.

  • Constant Pitch Rate Example

    (3) Also, determine if all the propellant can be burnt to

    reach 90o? If yes, give final burnout parameters. If no,

    give reasons as well as the final burnout mass.give reasons as well as the final burnout mass.

    No. mt = 36.5 Tons.

    (4) What should be θi if all fuel is to be burnt? (θb = 90 o)

    q0 = 0.32 o/s, θi = 2.87

    o, tb = 272.3 s.

  • Summary

    Gravity turn trajectories take much longer time, but

    result in lower velocities in denser atmosphere and also

    reduce the energy loss due to gravity.reduce the energy loss due to gravity.

    Constant pitch rate solution is simple to obtain in closed

    form, though requiring initial conditions consistent with

    the amount of propellant to be burnt.