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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
0ˆlim s
t
s dsV V s u
t dt∆ →
∆= = → = ⋅
∆
�ɺ
( ) ( )
Acceleration:
ˆ
ˆ
s
ss s n
s n
d da V Vu
dt dt
dua 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
dVma m mg I mg
dt
dma mV mg
θ
θθ
= = − −
= =
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 mgdt
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 tV g t
m t V t
θθ θ= − − =
ɺ ɶɺɺ 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
00
0 0 0
sin( ) ; ( ) ; cos
2 cos 2ln (sin sin )
sp sp
gq t q t V t V g
q
dm g dt m g
m g I m q g I
θθ θ θ θ
θθ θ
= → = + = =
= − → = −
ɶɺ ɺ ɶ
ɶ ɶ
Case – 1: Constant Pitch Rate
As q0 is constant at all times including the initial time,
we can write,
00 0 0
0
sin0 or 0, =0 not admissible.
gq V
V
θθ= ≠ ∞ → =
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 coscos
dh dh V gV
dt d q q
θ θ θθ
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 gh h
d q q
θ
θθ 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 00sin ln sin
2
sp
b
b
g q I m
g mθ θ−
= +
ɶ
Burnout time & horizontal distance are as follows.
( )
( )( )
0
00 2
0 0
0
0 02
0
( ) 1; sin 1 cos 2
2
sin 2 sin 2( ) ( )
2 2
b
bb
b
b
dxt V x x g d
q dt q
gx x
q
θ
θ
θ θθ θ θ
θ θθ 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/s2, 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 = 5o.
Vt = 0.827 km/s, ht = 35.0 km, mt = 39.3 Tons, θt =
55.3o, q0 = 0.559o/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 = 90o)
q0 = 0.32o/s, θi = 2.87o, 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.