Game Development Reference
In-Depth Information
The first term in Equation (11.3) is called the
momentum thrust
and is caused by the
rocket engine ejecting exhaust gases from the rocket nozzle. The momentum thrust is equal to
the
exhaust velocity
of the combustion gases,
v
ex
, multiplied by the mass flow rate. The
l
term
in the momentum thrust expression is a correction factor that accounts for losses due to the
shape of the nozzle. It generally has a value less than but close to 1.
The second term in Equation (11.3) is called the
pressure thrust
and is due to the pressure
difference at the exit plane of the nozzle. The quantities
p
e
and
p
a
are the exit and ambient air
pressures respectively, and
A
e
is the area of the exit plane of the nozzle. The pressure thrust can
be positive, negative, or zero depending on the relative values of the exit and atmospheric pressure.
The value of pressure thrust will also change as the rocket climbs into the air because atmospheric
pressure decreases with increasing altitude.
The thrust generated by a rocket engine is often written in terms of an
effective exhaust
velocity
,
v
e
, which incorporates the effects due to the correction factor and pressure thrust.
dm
Fv
dt
=
(11.4)
T
e
For most rockets running at a constant throttle setting at a constant altitude, the effective
exhaust velocity and mass flow rate are assumed to be constant. The effective exhaust velocity
will increase with increasing altitude because the pressure thrust component increases with
increasing altitude.
The thrust shown in Equation (11.4) is the force imparted to the exhaust gases. The thrust
force applied to the rocket is equal and opposite to the force applied to the exhaust gases.
dm
Fv
=−
(11.5)
T
e
dt
The Rocket Equation
The force applied to the rocket from the engines causes the rocket to accelerate according to
Newton's second law. The thrust force on the rocket,
F
T
, is equal to the mass of the rocket,
m
,
times the acceleration of the rocket, and is also equal to the effective exhaust velocity times the
change in mass with respect to time.
dv
dm
Fm
=
= −
v
(11.6)
T
e
dt
dt
Multiplying the two sides of Equation (11.6) by the time increment,
dt,
and dividing by
mass results in a differential equation that is a function of mass and velocity only.
dm
dv
=−
v
(11.7)
e
m
The two sides of Equation (11.7) can be integrated, yielding an expression for the velocity
at any point in time as a function of the mass at that point in time.
⎛
m
⎞
vt
()
=+
v
v
ln
0
(11.8)
⎜
⎟
0
e
mt
()
⎝
⎠
