Not So Fast:  Comments on ?Estimates of Performance and Cost for Boost Phase 
Intercept? presented to the Marshall Institute?s Washington Roundtable on Science 
and Public Policy by Greg Canavan on 24 September 2004. 
 
 
Dr. Greg Canavan?s paper, ?Estimates of Performance and Cost for Boost Phase 
Intercept,? [http://www.marshall.org/article.php?id=262] examines some implications of 
constellation size and interceptor cost and weight for the total costs and feasibility of a 
space-based boost-phase interceptor (SBI) system.  The paper argues, in general, that a 
?concentrated? system, that is, one that is tailored to defend against missiles launched 
from a small geographic area, can be substantially cheaper than is currently believed.  
North Korea might be considered ?small.?  The paper states that reductions in expected 
cost come about from a combination of lower estimates of SBIs mass, lower estimates of 
individual satellite cost, and a constellation that requires fewer interceptors because it 
covers only a restricted range of latitudes.   
 
We believe that mass and cost estimates are wrong and the simple model of 
satellite coverage exaggerates the effect of concentration.  All the errors together lead to 
an extreme underestimation of the cost.  The paper?s SBI masses are based on unproven 
and very optimistic estimates of kill vehicle masses;  its per satellite costs are based on 
unrealistic learning curve performance;  and a more accurate model of satellite orbits 
shows that the benefits of concentration are somewhat smaller than the paper?s simple 
model suggests. 
 
 The total mass of the system in orbit is critical because launching mass into orbit 
is expensive.  The mass of the constellation is just the number of required interceptors 
times the mass of each interceptor.  The interceptor consists of a ?kill vehicle,? which 
actually maneuvers and collides with the enemy rocket, the booster rocket that propels 
the kill vehicle from its orbit to the general area where the intercept will occur, and a ?life 
jacket? that provides the maintenance functions for the interceptor while it waits in orbit 
but is not required for intercept, so is left in orbit when the intercept begins. 
 
 Defense system designers do not have to worry about making the kill vehicle too 
small and too light.  Even a one kilogram projectile, closing with a thrusting booster 
rocket at several kilometers per second, will disable the rocket if it scores a hit.  So the 
goal is to get the kill vehicle mass as low as technically feasible while still being able to 
maneuver well enough to hit it.  Several estimates have been made of kill vehicle mass.  
Canavan uses masses reported in the American Physical Society (APS), Boost Phase 
Intercept Systems for National Missile Defense  
[http://www.aps.org/public_affairs/popa/reports/nmd03.cfm] and the Congressional 
Budget Office (CBO) Alternatives for Boost-Phase Missile Defense 
]http://www.cbo.gov/showdoc.cfm?index=5679&sequence=0]  as baselines for 
comparison with emphasis on the latter because they have the smallest mass. 
 
One must keep in mind that when the paper repeatedly refers to the ?CBO? kill 
vehicle, these are not designs that CBO developed or endorses as feasible.  Like any other 
 2
CBO study, their space-based interceptor study takes the form of ?if one believes X, then 
the costs will be Y?? with consideration of a range of Xs.  The lowest kill vehicle 
masses used by CBO were provided by Lawrence Livermore National Laboratory 
(LLNL).  Thus, every time the paper refers to the ?CBO interceptor,? it really means the 
interceptor proposed by LLNL. 
 
CBO cites a briefing from LLNL in November of 2003 that used a kill vehicle 
mass of 30 kg.  [See footnote 4 on p. 24 of CBO report.]  While the CBO report does not 
explicitly distinguish between fueled and unfueled masses (often called ?wet? and ?dry? 
weights), it is clear from the context that they consider the Livermore kill vehicle to be 30 
kg fully fueled.  CBO notes [p. 25] that ??producing a 30-kg kill vehicle with BPI 
performance would require a technological leap in miniaturization.?  The APS study also 
states that the Livermore Advanced Technology Kill Vehicle ??remains largely a 
conceptual design at this point;  it has not been built or flight-tested? [APS p. 250] and 
might be ready for testing in a decade.  More importantly, however, the mass of the 
Livermore interceptor, reported to APS in a briefing in September of 2001 [see APS 
footnote 160, cited on p. 250, appearing on p. 255], is 23 kg dry.  With fuel, the total 
mass is 67 kg [APS Table 14.2, p. 253]?or more than twice what was reported to CBO 
just over two years later?and it is the heavier kill vehicle that the APS believes is 
challenging.  Because launch costs to low Earth orbit are about $20,000/kg, launch costs 
are a substantial fraction of total system cost and the weight of the kill vehicle is critical. 
 
 The paper also introduces much lower costs than CBO for building the kill 
vehicles, or ?KVs.?  It points out that ?For CBO data base costs, KV costs dominate SBI 
cost,?  [p. 2] and argues that the CBO study errs by basing its kill vehicle costs estimates 
on individual military satellites.  The paper states that a better model is the Iridium 
system of telephone satellites that were ?mass produced.?  Specifically, ?IRIDIUM does 
not compute the cost of individual satellites because production showed significant 
learning, which caused later satellites to cost a small fraction of the early ones.?  [p. 18]  
Seventy two Iridium satellites were produced for a cost per pound of satellite that is about 
12% of the cost of the military satellites used as the CBO baseline. 
 
By mass-producing the satellites (or anything, for that matter), the per unit costs 
should come down.  This is often referred to as a ?learning curve.?  The literature on 
learning curves is extensive and complex.  Obviously a subject of much interest to 
industry, a great deal of effort has gone into uncovering the determinants of learning 
curves.  Some studies show that learning curves depend on the type of industry and others 
show they also depend on management skill.  Some reduction in cost results from true 
learning, that is, just avoiding mistakes and developing some tricks of the trade.  Other 
reductions in cost come about because large production runs allow investment in capital 
equipment that reduces unit costs if the capital costs can be shared over a large enough 
number of units. 
 
There are several ways to express reductions in unit cost as the number of units 
increases.  The most common way is to express the reduction in average cost with a 
second lot of the same size as a first lot.  For example, if the average cost of 200 widgets 
 3
is 90% as much as with a production of 100 widgets, then the widget production is said to 
have a 90% learning curve.  To a good approximation, for many types of manufacture, 
the average cost after the next lot of 200 will be 90% of the first two hundred, and so on 
(this is just to say that the ?learning? is a logarithmic function of the number of units).  
Most observed learning curves are in the range of 85-95% (100% represents no learning).  
[See NASA?s tutorial on learning curves, including a learning curve calculator, at 
http://www1.jsc.nasa.gov/bu2/learn.html .] 
 
For learning to account for the reduction in cost between a single military satellite 
and the 72 Iridium satellites, the learning curve would have to be 72%.  This learning 
curve is at the extreme range observed in any type of production and is significantly 
better than the 85% estimated by NASA for aerospace industry.  Therefore, even if this 
high learning curve occurred during Iridium production, it is an anomaly, not a general 
example.  In fact, there are few truly unique, one-of-a-kind military satellites?they 
typically are part of a series?thus the basis costs for military satellites used in the paper 
already include some degree of ?learning,? meaning that the paper is further exaggerating 
the learning effects in Iridium manufacture.  It is far more likely that the difference in 
cost represents fundamental differences between the performance, specifications, and 
missions of commercial and military satellites. 
 
Finally, the paper points out that the system does not need uniform satellite 
coverage over the Earth but can concentrate the coverage over a small area, reducing the 
total number of interceptors.  Specifically, if we focus only on North Korea, a 
constellation with peak concentration at that latitude will be much smaller than one 
covering a larger area.  This is true, but at several points the paper uses approximations or 
assumptions that overestimate the effect.   
 
 Figure 10 in the Canavan paper shows that SBIs in a constellation inclined at 42.5 
degrees (the northern latitude of North Korea) will spend 11% of their orbital period 
within a latitude ?band? 3.3 degrees wide, with 42.5 degrees as its northern edge.  (We 
cannot reproduce the author?s numbers from the paper alone but it appears that he uses 
the 3.3 degree band as covering North Korea. Figure 10 shows the fraction of SBI from 
constellations inclined at 43 and 30 degrees ?in 3.3-degree bands, which are roughly the 
width needed to cover trajectories of missiles from North Korea.?  [p.25]  North Korea 
runs from 38 to 42.5 degrees, or 4.5 degrees, not 3.3.  This is important because later 
Canavan seems to reduce the mass of the interceptor based on its being able to just barely 
cover a 3.3 degree band from edge to edge.)  Note this concentration effect is feasible 
only against missiles launched from a small country like North Korea.  Iran is much 
larger, running between 25 and 40 degrees latitude and a constellation that covered Iran 
would clearly be much larger than one covering North Korea. 
 
The APS effort started out considering only land-based boost-phase interceptors 
and later added analysis of space-based systems.  The APS study team was well aware of 
the concentration effect, noting that ?For any given orbit, satellite coverage will be 
concentrated at the same latitude as the inclination of the orbit, leaving the equatorial 
region underpopulated, as discussed in [80]. (The inclination of an orbit is the angle that 
 4
it makes with the equator.) Although that concentration would be beneficial if the defense 
needs only to intercept missiles from a very narrow latitude band, such as from North 
Korea, it makes complete coverage over a wide range of latitudes more difficult to 
achieve.?  [APS p. 108]  The APS study team also noticed that complete coverage at the 
equator gave fairly constant double coverage at the latitude of North Korea.  [David 
Mosher, personal communication.] 
 
Comparing APS Fig 6.3 [p. 109] and Canavan?s  Fig 11 [p. 26] shows that 
Canavan calculates substantially fewer interceptors even without concentration, but half 
of this reduction is due to seeking only single coverage, that is, only one interceptor is 
within range at any moment. That is, the system described in the paper is effective 
against a single North Korean missile.  An obvious countermeasure would be to build a 
second missile that could be launched at the same time.  The APS study does not 
concentrate over the latitude of North Korea and calculates a constellation ?sufficient to 
cover any point in space and time in the region between approximately 30 and 45 degrees 
North latitude with an average of two interceptors and a minimum of one.?  [APS p. 110]   
 
 The Canavan paper uses the effective radius, r, of the area where the interceptor 
can engage the missile to be half the width of the band 3.3 degree band, so it just reaches 
from edge to edge.  It then points out that ?At their northmost latitude all SBI are headed 
due east, which produces a ring of eastward moving SBI on a line of latitude ?  equal to 
the constellation inclination.?  Thus, all that is needed for complete coverage over any 
point in the latitude band is to have SBIs at that latitude separated by 2r.  We can 
calculate the circumference of the Earth at that latitude and divide by 2r to get the number 
of satellites needed in that latitude band.  If 11% of all the satellites are in the band, we 
can divide the number in the latitude band by 0.11 to get the total number of satellites.  
This approximation assumes that the coverage of the SBIs is a square moving around the 
band when it is in fact a circle.  The circle covers only about three quarters of a square it 
fits in.  This is not a big problem but shows the way to another source of error with this 
model.  The 11% value is the fraction of time the SBI is anywhere within the band but the 
approximation assumes that the SBI is traveling along the center of the band at all times 
that it is within the band.  In fact, the SBI will have some coverage over the band even 
before it enters the band because of the range of the interceptor will reach into the band 
before the SBI arrives.  On the other hand, when the interceptor first barely enters the 
band, only half of its coverage is within the band, and half is lying south of the band.  
That is, it is covering only about 40% of the ?square? assumed by the simple coverage 
model.  At the very northernmost point of its orbit, the SBI is still ?inside the band,? but 
just grazing the top and again only half its effective coverage lies within the band, with 
the other half now north of the band. 
 
 The qualitative critique above suggested the need for more rigorous numerical 
calculations.  The details of a MathCad worksheet for calculating satellite coverage are 
shown in the appendix.  These calculations are based on equations found in ?Earth 
Coverage by Satellites in Circular Orbits,? by Alan R. Washburn, Department of 
Operations Research, Naval Postgraduate School.  
[http://spica.or.nps.navy.mil/searchdocs/classnotes/Coverage.pdf] 
 5
Washburn?s paper considers coverage of the Earth by low orbit communications 
satellites, but the mathematics applies directly to the problem of coverage by space-based 
interceptors.  In particular, Fig 1 of Washburn defines the geometry used in the appendix.  
The MathCad calculation finds solutions to Washburn equation 8, 9, and 10.  These 
equations express the distance between the satellite and the target.  (Since the Earth and 
the satellite orbits are both assumed circular, it is convenient to measure distances in 
angles, using the center of the Earth as the origin of the coordinate system.)  Once the 
formula for satellite-target separation is known, it is easy to find the fraction of the time 
that any one interceptor is within range of the target.  The inverse of that number is the 
number of satellites needed to have one interceptor always within range. 
 
 The number of satellites needed depends on the range of each interceptor.  If each 
one can cover a greater area, fewer are needed in total to cover any given target.  The 
range that the interceptor can reach during the boost time of the targeted missile is almost 
directly proportional to the change in speed, or delta V, available to the interceptor.  (The 
?almost? is because the time taken to get up to speed, depending on the acceleration, 
must also be taken into account.)  But the higher delta V comes with greater fuel weight, 
and thus greater total interceptor weight.  There is, therefore, a direct tradeoff between 
the number of interceptors and the weight of each individual interceptor and this 
relationship between the interceptor speed and the number of required interceptors is 
critical because it determines, along with the interceptor mass relationship, the minimum 
total weight for the system. 
 
 Using his simple model, Canavan calculated the relationship between interceptor 
speed and total interceptor number and shows his results in Fig 11 [p. 26].  We 
recalculated the same relationship with the more rigorous model and found that the 
simple model works fairly well, but there are two significant differences. 
 
 First, our calculated result shows that the number of required interceptors is more 
sensitive to the booster burn time than the simple model suggests. Canavan writes, ?Thus, 
uniform and concentrated coverage require roughly equal numbers of SBI for liquid 
missiles, but uniform coverage requires about twice as many SBI for solid missiles.  That 
is because concentrated coverage scales on missile burn time as 1/T, while uniform 
coverage scales as 1/T
2
.  The ratio of uniform to concentrated coverage scales as 1/T 
which is a significant penalty for solid missiles.?  [p. 25]  This result of Canavan?s 
calculation would be critical, because it suggests that using concentrated coverage 
overthrows the widely-held assumption that a key countermeasures to any boost-phase 
interceptor defense is to reduce the vulnerable boost time by moving from liquid to solid 
boosters.  It appears, however, that the concentrated constellation?s apparently reduced 
sensitivity to ICBM booster burn time is an artifact of simplifications in the Canavan 
model.  This means that fast-burning solid-propellant boosters remain a challenging 
countermeasure even to concentrated space-based boost-phase interceptor defense. 
 
 Second, at the low delta Vs that Canavan determines are optimal, the simple 
model underestimates the number of interceptors required.  At higher delta V, the simple 
model actually overestimates.  That is, the curve describing the required number of 
 6
interceptors as a function of the interceptor delta V is steeper than predicted by the simple 
model.  The net effect of this different sensitivity is that the optimal delta V is not the 2 to 
2.5 km/sec (depending on whether liquid or solid fuel boosters are being attacked) that 
Canavan calculates but 3 to 3.5 km/sec.  [The paper?s data on total mass is apparently 
contained in fig. 12, which seems to be accidentally omitted, but the text and fig. 13 make 
clear where the minima lie.  Note also, that the calculations in the Appendix do not 
include the weight of the in-orbit ?lifejacket.?]  At the higher optimal delta V, the reach 
of the interceptors increases and the benefit of concentrating the interceptors is less.  This 
particular difference between the simple model and the analytical model is not the most 
significant one because the minima are quite flat and the precise point of the minimum 
does not affect total mass on orbit much. 
 
 Canavan?s discussion of reducing the number of interceptors in a constellation of 
space-based missile defense interceptors compared to previous studies relies heavily on 
the well-known principle that giving up on coverage near the equator allows a 
constellation to concentrate coverage at a higher latitude.  The simplifications of the 
Canavan coverage model exaggerate the benefit.  More accurate calculations show the 
benefits of concentrations, of course, but they are somewhat less than predicted by the 
simple model.  In any case, the differences between the simple model and the analytical 
model are small compared to the difference between single and double coverage and 
much smaller than the effects of hopeful estimates of kill vehicle weight and cost. 
 
In summary:  With a combination of (1) extremely optimistic kill vehicle mass 
estimates, (2) SBI cost estimates based on unrealistic learning curve values, (3) a 
somewhat exaggerated benefit of interceptor concentration based on a simplified model 
of satellite coverage, and (4) single rather than double coverage of North Korea, the 
author shows large reductions in the cost of a defensive space-based interceptor 
constellation defending against North Korea compared to previous estimates.  
 
Ivan Oelrich, Federation of American Scientists 
Steve Fetter, University of Maryland 
 
Q 15.857= ground trace parameter (orbits per day)
P
S
e
2004()
Q
:= P 5434= orbital period (s)
a
3
?
P
2 ??
?
?
?
?
?
2
?:= a 6680= semi-major axis of circular orbit with period P (km)
haR
e
?:= h 302= altitude of satellite orbit (km)
Calculate the maximum SBI range r (km), given SBI burn-out velocity v (km/s) and ICBM burnout time 
t
bo
 (s), assuming constant SBI acceleration of c (g) and a SBI launch delay of t
d
 (s):
rv t
bo
, t
d
, c,()vt
bo
t
d
?
v
2c? g?
?
?
?
?
?
?
?:= maximum SBI flyout distance (km)
Calculate the maximum intercept angle ? from satellite to an ICBM at burnout altitude h
bo
 (the angle 
formed by the intercept point, the center of the Earth, and the satellite at time of SBI launch):
? vt
bo
, t
d
, c, h, h
bo
,()acos
a
2
R
e
h
bo
+()
2
+ rv t
bo
, t
d
, c,()
2
?
2a? R
e
h
bo
+()?
?
?
?
?
?
?
:=
Calculate the actual angle A between the satellite and the potential intercept point as a function of 
time:
A ??,?, i,()acos cos ?()cos ?
?
Q
+
?
?
?
?
?
? cos ?()? cos ?()sin ?
?
Q
+
?
?
?
?
?
? cos i()? sin ?()sin i()?+
?
?
?
?
?
sin ?()?+
?
?
?
?
?
?
:=
where ? measures time by angle in satellite orbit, ? is the latitude of the target, ? measures the initial 
longitude of the target on the fixed celestial sphere, and i is the inclination of the orbit.
Below is a plot of A as a function of time (the number of SBI orbits) together with illustrative value of ?; 
when A is less than ?, the SBI is within range of the ICBM burnout point:
Absentee Ratio for Fixed ICBM Burn-out Latitude and SBI Inclination
First define some basic parameters:
G 6.672 10
11?
?:= M
e
5.9736 10
24
?:= ? GM
e
?()10
9?
?:= ? 398559=
R
e
6378.1:= equatorial radius (km) R
m
6371.0:= mean radius (km)
g
?
R
m
2
:= g 0.00982=
S
e
t( ) 86164.09055 0.00015 t 1900?()?+:= S
e
2004( ) 86164.106= seconds per day
Define a frozen satellite orbit (i.e., one with a repeating ground trace) with altitude of about 300 km:
m 111:= number of orbits before ground trace is repeated
n7:= number of days before ground trace is repeated
Q
m
n
:=
? 0 0.01, 2 ?? 20?..:=
0 5 10 15 20
0
2
A ?
?
4
, 0,
?
4
,
?
?
?
?
?
? 4 300, 60, 10, h, 300,()
?
2 ??
A proper calculation of the fraction of time that satellite is within range of target would integrate the 
function if(A(?) < ?,1,0) from ? = 0 to 2?m, where m is the number of unique orbits:
f ? i, v, t
bo
, t
d
, c, h
bo
,()
0
2 ?? m?
?if A ??, 0, i,()? vt
bo
, t
d
, c, h, h
bo
,()? 1, 0,
?
?
?
d
2 ?? m?
:= TOL 0.001?
Unfortunately, the function is too poorly behaved to be evaluated numerically with MathCad, so we'll have 
to use a discrete function:
? 0 0.01, 2 ?? m?..:=
f ? i, v, t
bo
, t
d
, c, h
bo
,()
?
if A ??, 0, i,()? vt
bo
, t
d
, c, h, h
bo
,()? 1, 0,
?
?
?
?
?
?
?
1
?
:=
f
?
4
?
4
, 4, 300, 60, 10, 300,
?
?
?
?
?
0.0086316=
The minimum number of satellites required to provide continuous coverage of a point on Earth is 1/f:
Nvt
bo
, t
d
, c, h
bo
,()ceil f
?
4
?
4
, v, t
bo
, t
d
, c, h
bo
,
?
?
?
?
?
1?
?
?
?
?
?
:= N 4.5 300, 60, 10, 300,()9=
v 1 1.5, 6..:=
0123456
10
100
1
.
10
3
1
.
10
4
N v 300, 60, 10, 300,()
N v 180, 60, 10, 200,
v
Canavan's Eq. 6 for 2-stage 
SBI with penalties for payload, 
fuel, and thrust, assuming 
constant acceleration c and 
KV mass of 30 kg; fig. 3 for his 
option 5:
M
sbi
vv
e
, c,()30
1.05 e
v
2v
e
?
?
1 0.087 0.1+
0.0009 v
e
? c? g?
v
+
?
?
?
?
?
e
v
2v
e
?
1?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
2
?:=
123456
10
100
1
.
10
3
M
sbi
v 2.94, 10,()
v
Mv v
e
, t
bo
, t
d
, c, h
bo
,()Nvt
bo
, t
d
, c, h
bo
,()M
sbi
vv
e
, c,()?:= Total constellation mass
123456
1
.
10
4
1
.
10
5
1
.
10
6
M v 2.94, 300, 60, 10, 300,()
M v 2.94, 180, 60, 10, 200,
v
M v 2.94, 180, 60, 10, 200,()
1000
347.9
126.4
87.5
74.8
72.7
73.4
76.2
85.8
98.1
115.3
142.0
=
M v 2.94, 300, 60, 10, 300,()
1000
41.9
29.6
24.1
21.7
21.1
21.3
22.4
24.6
28.1
31.9
38.8
=
v
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
=