and Equation 2-23 reduces to
H
2
B =
Thus,
10
1
2 (0.6430)
7.78 feet ,
B = — = — = 5.0 feet .
2
2
When z = - d,
2 sinh (2îrd/L)
2 (0.8394)
and, B = 0.
(b) With
= 10.45 feet, and z - - 25, evaluate the exponent of e
for use in Equation 2-24, noting that L = Lo,
2n(- 25)
512
- 0.307
thus
e"0-307 = 0 736 .
Therefore,
H
27Tz/
10.45 z
A = B = — e /l= ------ (0.736) = 3.85 feet .
2
2
The maximum displacement or diameter of the orbit circle would
be 2(3.85) = 7.70 feet.
, X
Lo ~512
(c) z = — — = ----------------------------- = — 256 feet ,
2
2
2ttz _ 2tt(— 256) _ _ 3 J42
~L~ ”
512
Using Table C-4 of Appendix C,
e-3 142 = 0.043.
H
27Tz4
10.45
, A = B = — e 7L = ------- (0.043) = 0.225 feet .
2
2
2-21