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.