Sub.: Range vs. difficulty for ballistic missile
development.
Date: March 27, 1998
From: R.L. Garwin (914) 945-2555
26-234 Yorktown Heights, NY
IBM Fellow Emeritus
Email: RLG2@watson.ibm.com
RSA@watson.ibm.com
(Jean A. Hernandez, Secretary)
The question frequently arises as to the difficulty or ease
of extending the range of a ballistic missile from the
common Scud B of perhaps 300 km range to a range of 9000 km
or 10,000 km.
One way to answer this question is simply to point to the
various missiles that exist in the inventory of the various
nations, to show that ICBM range is indeed possible and
conventional. But this conceals the very substantial
difference between the technology of a short-range ballistic
missile or medium-range ballistic missile and that of an
ICBM.
Another approach is to go back to first principles-- namely,
the rocket equation-- to see what is involved actually in
reaching such range for a ballistic missile. The reader
should bear with us while we explain some simple concepts,
probably familiar to just about everybody.
A "ballistic missile" is a vehicle or object that continues
on its path under the force of the Earth's gravity. It
continues because of Newton's laws of motion. More
specifically, it exchanges its kinetic energy of radial
motion for potential energy as it rises to apogee and then
recovers kinetic energy as it loses potential energy as it
comes closer to the center of the earth. That applies to a
rock just as well as it does to an ICBM.
But how does the missile "go ballistic"? The artillery
projectile follows a ballistic trajectory after having been
accelerated by high pressure gas in a gun barrel to its
final velocity. But a bazooka or other shoulder-fired
rocket does not need a heavy gun barrel. It does not push
on the gun barrel to be accelerated in a short time; rather,
it pushes on the exhaust gases of the combustion of rocket
fuel.
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Of course, since the manufacturer and owner of the gun
barrel probably believes that it is more important (and
certainly costs more) than the individual projectile
that it shoots, this is usually presented from the
point of view of the gun barrel's pushing on the
projectile.
A given rocket stage operates ordinarily at constant chamber
pressure. Hot gas is produced by the combustion of liquid
or solid fuel, exchanging chemical energy for thermal
energy. The hot gas creates some chamber pressure in the
combustion chamber, which is connected to space by the
throat of a rocket nozzle. If there were just an aperture
in the combustion chamber, one would still have a rocket,
but the gas would be emitted from that aperture in a
hemisphere, rather than in a directed stream. So the
momentum associated with the thermal energy would be lower
than if all the gas went in the same direction.
Furthermore, even if the gas were all to go in the same
direction, but were to remain hot, the gas stream would have
less momentum than if the gas were cooled by expansion and
still kept largely uni-directional.
In fact, the function of the rocket nozzle is to expand the
combustion gas in such a fashion that it cools and
constitutes a largely uni0directional jet. At the throat,
one has hot gas with the molecules going in every direction.
As the gas expands in the rocket nozzle, it repeatedly
pushes on the material of the nozzle (except in the
direction of the exit circle), and as the rocket nozzle
diameter increases with distance from the throat, the gas
expands and cools itself. So in the rocket jet that is
ejected into space, the gas is quite cold, but moving
extremely rapidly. The best that can be done in obtaining
exhaust velocity is to convert all of the thermal energy of
the fuel into kinetic energy of the exhaust (leaving nothing
left over for internal thermal energy of the exhaust plume).
For rocket exhaust into space (above the atmosphere), that
condition is closely approached. Certainly billions of
dollars have gone into wringing the last bit of performance
out of rocket nozzles. Eventually one gets to a point at
which the size and mass of additional nozzle outweighs the
small benefit that can be obtained by reducing the already
small residual thermal energy in the exhaust plume.
For exhaust into the atmosphere in the boost phase of a
rocket, such large expansion ratios are not available,
because the atmospheric density and pressure is finite. So
first-stage rocket nozzles lack the long skirt of a
deep-space nozzle.
Often if one's goal is synthesis the best or only way to
proceed is through analysis of various specific examples.
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For instance, it is a simple matter to follow the trajectory
of a specific choice of rocket design, launched at a
specific angle to the horizontal; one can determine the
entire course of propulsion, burnout, and the range to
impact. One would, of course, like to know the elevation
angle corresponding to maximum range, which is a simple
problem for a flat Earth (the result is 45 deg, without
taking into account gravity loss or atmospheric drag), but
even that could be obtained through multiple computer
trials. Even when a complex synthesis technique-- based on
experience, complex computer models, and the like-- purports
to provide a solution to the problem, it is a good idea to
check it by simple analysis and by elementary step-by-step
following of the trajectory.
Quantitatively one can have an excellent approximation to
rocket performance by the assumption that all of the thermal
energy of the fuel emerges as uni-directional kinetic energy
of the exhaust. Then Newton's third law states that for
every action there is an equal and opposite reaction. This
means that in a coordinate frame centered on the rocket at
any given time, the momentum associated with the exhaust
emitted in a very short interval (say, one second), is
countered by the momentum given to the remaining rocket plus
un-exhausted fuel. So one has an equation relating the two,
and specifically giving the rate of increase of rocket
momentum as determined by the rate m (in grams/second) at
which mass is exhausted, together with the velocity Ve of
the exhausted mass.
Eq. 1: mVe = M dV/dt
In this Equation, m is the mass ejected per second in the
rocket plume; Ve is the exhaust velocity of the jet; M is
the remaining mass of the rocket (decreasing with time); V
is the instantaneous velocity of the rocket, and t is the
time.
This equation has a simple solution which can be written as:
Eq. 2: Mf/Mo = e**-(Vg/Ve)
Here Mf is the final mass of the rocket stage-- the payload
plus the dry weight of the rocket; M0 is the gross launch
overall weight (GLOW); Vg is the velocity gain overall; and
Ve is again the rocket exhaust speed. The reader is asked
to pardon the rather gross appearance of the simple
mathematics; expedience may be worth more than elegance.
"**" stands for exponentiation; "r x s" or "r*s" is the
product of r and s, as is "r s".
Note that this simple Eq. 2 has its ultimate simplicity
because it ignores the time at which the combustion occurs,
and this means that for rockets operating in space and not
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captive in the Earth's gravitational field, a tiny rocket
nozzle achieving a certain exhaust velocity and ultimately
expelling the same total mass will ultimately give the
residual rocket mass the same speed as would a large rocket
nozzle. And the smaller nozzle and combustion chamber and
pumps have less parasitic mass and allow more mass for
payload or a higher velocity of that payload. So deep-space
systems have low-thrust rockets and long burn times.
Obviously one cannot push this too far in the ballistic
missile application, because once the rocket engine is lit
for the missile on the ground, the missile must produce a
thrust greater than the Earth's gravity, or else the missile
will continue to sit on the ground, despite the furious
expenditure of fuel.
It is quite remarkable that large liquid-fuel space
boosters, for instance, have an excess acceleration of only
about 0.3 g, so that most (1.0/1.3) of their thrust and
their initial fuel expenditure simply replaces the upward
thrust of the launch stand. Ballistic missiles as weapons,
however, typically have a larger initial thrust and have in
this way less "gravity loss" associated with their burn
rate.
In fact, it is a useful approximation to ignore gravity loss
in the first estimate of ballistic missile performance.
We have now demonstrated how to calculate the ratio between
gross launch overall weight (GLOW) and payload. That is, we
know how for a single stage to relate the GLOW, the fuel
mass, the dry weight, and the payload. There is a direct
trade-off between dry weight (structure, fuel tanks, rocket
nozzle, pumps, etc.) on the one hand, and payload on the
other. Indeed, because there is a maximum speed even with
zero payload, and because some of that dry weight is
associated with the initial thrust and the initial nozzle,
there is benefit or even necessity to "staging" rocket
propulsion.
For a given technology level, the idea is to throw away as
much of the inert mass as possible as the fuel is being
expended, and to end up with a smaller rocket, with unburned
fuel, as the second "stage". The trick can be repeated, so
that in principle a given level of technology could
accelerate a second stage to a certain velocity, which would
then burn and separate from the third stage, which would
then burn and separate from the fourth stage, etc.
It is of interest to ask what can be done with "continuous
staging" in which one assumes that not only rocket fuel is
exhausted, but at the same time the associated weight of
tankage and motors. In this approximation of continuous
staging, with a small enough payload Mf, one is able to
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calculate a sufficient initial mass Mo, no matter what the
exhaust velocity. Of course, the payload can be vanishingly
small, but whatever the required velocity gain, there is
such a payload.
For continuous staging, the payload fraction is given by
Eq. 2, but with an equivalent exhaust velocity Vee < Ve, in
view of the assumption of zero dry weight that goes along
with Vee.
In terms of the exhaust velocity, the payload, and the gross
launch overall weight, one can now calculate some examples
for a single stage. But first we want to relate the
velocity to the range.
There is an old story about the physicist looking for a
job who with the confidence common to physicists and
mathematicians felt that he could revolutionize the
dairy industry. So he was hired on speculation and sat
down at his brand new, shiny, empty desk with a pen and
paper and began to write, "Consider a spherical cow."
It is sometimes instructive to begin with a simple, if
artificial problem, for which the assumptions and the
mode of calculation are more transparent.
So we consider first a flat Earth, which in our common,
local experience is a reasonable approximation.
If one sketches the optimum trajectory for a stone to go the
greatest distance for given initial speed (or a baseball
hit) one finds that the optimum launch angle is 45 degrees
from the horizontal, and one can obtain the range simply by
noting the time required for the vertical velocity Vz to
reverse. During flight without thrust or drag, the rocket's
horizontal velocity Vh is constant. So we have the set of
equations:
Voz = Vo/sr2
(For brevity, I write "sr2" for the square root of two).
T = 2Voz/g; R = VhT = (2/g)(Vo/sr2)(Vo/sr2) = Vo**2/g
g = 9.8
m/s**2 R = 300 km Vo = sr(gR) = sr(9.8x3x10**5) =1.7146 km/s.
Vo km/s R km(Flat Earth) Mo/Mb Reality
1.7146 300 2.140 2.69 Scud B
2.425 600 2.933 3.80 Al Hussein
3.1305 1000 4.01
9.900 10,000 80.82
(Round Earth)
7.1 10,000 23.33
PAGE 6
Table 1 required initial velocity Vo for range R in the
flat, non-rotating Earth approximation. Mo is the initial
launch weight and Mb is the burn-out mass of the rocket.
It should be noted that an exhaust velocity Ve = 2.254 km/s
corresponds to a "specific impulse" (Isp) of 230 sec = Ve/g.
The specific impulse is a time that characterizes the energy
per unit mass of the propellant. Equipped with an
appropriate nozzle, it is equal to the time that an initial
mass of propellant could provide a thrust equal to the
Earth's gravitational force on that initial mass.
From the MIT group, we have data on the Al Hussein. The
empty weight is 1785 kg with a warhead of 300 kg for a dry
weight of 1485 kg. Similarly the Scud B has an empty weight
of 2370 kg and a warhead of 965 kg for a dry weight of
1405 kg. The Iraqis did not add very much structure in
lengthening the tanks of the Scud B to make the Al Hussein.
The Al Hussein is supposed to have a gross GLOW of 6785 kg,
so that Mb/Mo = 1485/6785 = 0.2189. From Eq. 2, this
corresponds to a velocity gain of 1.5193 x Ve. And for
Ve = 2.254 km/s for an Isp of 230 s (given by the MIT group
as the Isp for the Scud B and for the Al Hussein), we find
that the Vo = 2.254 km/s x 1.5193 = 3.4245 km/s. From our
Table above, this corresponds to a range of 1200 km with
zero payload.
If one wants to have a greater range than 1200 km, even with
zero payload, one must either change the technology or go to
a rocket with more than one stage. First we consider adding
a second stage. By "technology" we mean here only two
things-- the structural weight of the rocket and the motors
(the dry weight or structural fraction SF) and the Isp.
First we consider the use of staging for a given technology,
taking Isp = 230 s and a structural fraction of 22% (SF).
We assume that the SF is constant for a payload less than or
equal to the SF, assuming that not much structure is
required to support the payload.
Then we have:
c Eq. 3: V2 = Ve ln M2/(M2*SF +PL)
wherefore the second stage, we have written the burnout
c weight explicitly as (M2*SF + PL), where PL is the payload
c in kg. By "ln" we mean the base-e logarithm.
And for the first stage we have
Eq. 4: V1 = Ve ln M1/(M1*SF + M2)
PAGE 7
The entire mass of the second stage rocket-- fuel and all--
is the payload of the first stage.
Now we want to minimize M1 for a given total velocity gain
(V1+V2):
(V1+V2)/Ve = ln M1/(M1*SF+M2) + ln M2/(M2*SF+PL)
e**(V1+V2)/Ve = M1 x M2/(M1*SF+M2)x(M2*SF+PL)
Eq. 5: e**Vg/Ve = 1/(SF+M2/M1)(SF+PL/M2)
Let PL' = PL/(SF*M2) SR = M2/(SF*M1), where we have
introduced the "staging ratio" SR.
Eq. 6: e**(V1+V2)/Ve = (SF)**-2 /(1+SR)(1+PL') =
constant
Note that for a given M1
Vg is maximized for the minimum of Q = (1+SR)(1+PL') with
respect to M2. So
c dQ/dM2 = 0 for -PL*SF*M1 = PL*M2 + M2**2(SF+PL/M2) = 0
or
-PL*SF*M1 + M2**2 SF = 0
or M1*PL = M2**2.
Therefore M2 = sr(M1*PL). This means that the velocity gain
is the same for each stage of the multi-stage rocket with
the same technology for each stage.
We can now solve for M1 to obtain
(SF + sr(M1*PL)/M1)(SF + PL/sr(M1*PL)) = e**-Vg/Ve
(SF + sr(PL/M1))(SF + sr(PL/M1)) = e**-Vg/Ve
or
(SF + sr(PL/M1)) = e**-(Vg/2)/Ve
Now take the Vg = 7.1 km/s; Vg/2 = 3.55 km/s;
Ve = 2.254 km/s
sr(PL/M1) = e**-(Vg/2)/Ve - SF
Eq. 7: (M1/PL) = 1/(e**(-Vg/2)Ve - SF)**2 = 1/(0.2070-0.22)**2
PAGE 8
Unfortunately, this denominator is negative, and there is no
launch mass that will propel even a two-stage rocket with Al
Hussein technology to a range of 10,000 km on a round Earth.
We should have noted that the launch velocity requirements
for a round Earth are less than those for a flat Earth at
long range simply because the Earth falls away from the
rocket as it travels-- a phenomenon noted by Newton in his
comparison between the speed of a falling apple and the
speed of the falling moon. For instance, for a range beyond
20,000 km, there is no additional velocity requirement,
since the rocket has gone into orbit and can have any range
with no additional velocity.
So maybe one wants to go to better technology rather than to
a third stage. For instance, we could imagine a structural
fraction SF = 0.15 instead of the structural fraction of
0.22.
Better structural fraction can be obtained by use of higher
strength materials, by chemical milling or other tailoring
of parts, and so on. Assuming a 30% reduction in SF (to
0.15), the denominator in the previous equation is
D = 0.2070 - 0.15 = 0.0570, and the launch mass of a
two-stage rocket would be given by M1/PL = (1/D)**2 = 308.
So for a payload of 300 kg , this would be a launch weight
of 90 tons.
If this seems high, and we want a 30 ton launch weight for
600 kg payload; (M1/PL) = 50, then we need D = 0.1414. So
that SF = 0.2070 - 0.1414 = 0.0656. This is an
astonishingly low structural fraction compared with the
0.22.
Another option is to move to higher Isp-- for instance 300
seconds. Then Ve = 2.94 km/s and e**(-Vg/2Ve) = 0.271. So
we would have the desired denominator of D = 0.1414 for a
structural fraction SF = 0.130, allowing a 30 ton launch
weight for a 600 kg payload. Or we could use three stages.
Now we discuss "gravity loss". We have noted early on that
a rocket launched vertically with a thrust less than its
weight will just sit on the pad, exhausting its fuel,
without any acceleration at all. Successful rockets thus
have acceleration greater than 1 g, but they are still
penalized by the Earth's gravity during boost. Because the
burn time of a Scud B is 70 s, gravity loss during burn
imposes a requirement on the rocket to exert a vertical
component of thrust that would have sufficed to provide
additional vertical velocity of 70 x 9.8 m/s, or about
0.69 km/s. In the first line of Table 1, the initial
velocity for 300 km is 1.7146 at 45 degrees elevation. This
has a horizontal component of 1.212 km/s, and a vertical
component of 1.212 km/s. Added to the vertical component
PAGE 9
requirement to be produced by the rocket is the 0.69 km/s
that we have just calculated, for an effective vertical
velocity gain of 1.802 km/s. These two components of the
right triangle thus correspond to a total velocity gain of
2.172 km/s, and with the Isp of 230 s, to a required
Mo/Mb = 2.621. This accounts for a good deal of the
difference between the instantaneous burn calculation of
Mo/Mb = 2.140 and the "Reality" of 2.69 for the Scud B.
This rather lengthy discussion indicates the changes of
technology toward reduced structural fraction and improved
Isp that are mandatory in producing a rocket of ICBM range.