5
re
.
nas a
ree stentee
,
16) The transformation to the principal axes, o, and r,., is accomplished by means of the
equations
(oe + 7)? = 03 + oF + 20.0,V1 — p*,
(o, — t)? = o2 + of? — Qar0yV'l— p?.
In solving for ¢, and 7,, take ¢, — t, > 0 when p > 0 and o, — r, < 0 when p < 0.
17) The angle a through which the principal axes are inclined is given by
tan 2a = [2pe,0,/(02 — 02)}(c. ¥ o,)
and
:
a@ = 45° (oz = ay).
It is to be noted that 90° — a is also a solution of this equation; i.e., the equation alone
does not tell which of the axes the angle is measured toward.
This procedure tacitly assumes that the city is not markedly bimodal. If it is, it is
probably safe to assumethat it is because of some geographical barrier — twin cities sepa-
rated by a river, or a city like San Francisco, on opposite sides of a bay — since otherwise
reasons of commerce and convenience would tend to make the halves coalesce. This
barrier can then be used to dissect the city into a pair of unimodal subcities, each of which
may be fitted according to the method used here.
BOMBS AIMED AT CENTER OF CITY
ony yaa
:
ve
Tl ae eae NSA
nee Cal eee eeee ee ee ne
ieee eaee ee
For the sake of simplicity it is assumed from now on that the bombs arecircularly
normally distributed, i.e., that gg = 7g. When n bombs of the same characteristics are
aimed at the center of the city, the expected level of casualties is given by
1
+o
f+o
2
y?
c
_ NE + aT
+
210-7. [ f exP (- 207 y(t ~~ [1 ~ T+x08 exp
d&du,
(A12)
which works out to be:
Result 4. For n bombs aimedat the center of the city, the expected level of casualties
is given by
d (—)e*(7)Low/(L + oB)H—T](LL + Co + mod)]L] + Mow + urd).
unl
This is relatively easy to compute from, although the increasing number of terms as n
increases is unfortunate. A rather rough one-term approximation can be made asfollows.
If expected levels of survival rather than casualties are considered, the curve tends to 0
asm — o, and momentsof all orders exist. From Eq. Al2 the zeroeth moment is simply
+a
216eT<
+o
at dv
§ EA
2a? — ga)
Q7? EAE exp 21+
ro}
[cf exp (= $5
(1 + of)?
(ATS)
~ eV[1 + A(o2 — 02) [1 + A(oR — 72)]
Nowfit this by exp(— an), whose zeroeth momentis 1/[1 — exp (— a)], by equating these
moments.
Thus
exp (— a) = 1~ {cV1 + X(o% — o2)LL + M(oR — THI/(L + Ao$)*},
(A114)
so that the expected level of survival of n bombs is approximately
{1 —[e/(1 + Aog)"]V[1 + Mok — 02)IL + AOR — 73) ]}”
78
(A15)
ORO-R-17 (App B)