Project Cyclops, A Design... - Department of Earth and Planetary ...

L-4 we see that the length of IF cable to an antenna
whose coordinates are x,y is
sin y0
x
1 - cos 0
-
sin 0
y+x
0
= y tan- + x
2
Y
(L16)
The maximum cable length is found by setting x equal
to _ in equation (L16) and differentiating. The
result is
a
£max cos(O/2) cos(rr/2m) (L17)
Y
a
2 O = SECTOR
SUPPLIED
BY
EACH
_./ ._ MAIN
_ I_ ,, TUNNEL
2 /9
= --
zsin 0
dy
Lx/_ _y2
tv tan -- + x/dx
Oa2 y/tan 0 2
2a tan (0/2) 4ma n
= -- - tan --
3 (0/2) 3_ 2m (LI 9)
For the tunnel patterns we have been considering the
data given in Table L-6
TABLE L-6
O m _max/a "_/a fl/9.o
_r/2 2 1.4142 0.8488 1.2732
lr/4 4 1.0824 0.7032 1.0548
rr/6 6 1.0353 0.6823 1.0235
0 oo 1.0000 0.6666 1.0000
The case 0 = 0 corresponds to having an individual
straight tunnel for each IF line, and gives the minimum
possible lengtla. We see that the hexagonal lattice with 6
main tunnels (pattern 2b) has an average length only
2.35% greater, that the square lattice with 4 main tunnels
(pattern lb) has an average length only 5.5% greater,
but that the patterns with only two main tunnels (the a
patterns) increase the average length by over 27%.
/ O /'
The total length of IF cabling can now be expressed
in terms of the number of antennas n, and thek spacing
s. For our purposes we can omit the term 1/2 in
equations (L4) and (LS). Substitution into equation
(LI 9) then gives for pattern la:
Figure L-4. Distance through tunnels to an antenna
The average cable length is very nearly that for a
continuous distribution of end points (antennas); that is,
for lb:
8 n 312
Lla - 3 rr s = 0.4789 na/2s (L20)
_ f£dA
f dA
(L18)
16 n 3/2
Llb
=
-_- (V_'- 1)
_
n
s = 0.3967na/2s (L21)
for 2a:
where dA is an element of the array area. Taking the
integrals over the area between the main feeder tunnel
and the x-axis (see Figure L.4), we have
L2a ="3- 8 _ /'!3/2m n
s = 0.4457 na/2s (L22)
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