Table 17. Computation of the Itcoth by Iteration 
2 arL 
Let H = 9 ; ot = ae = aan = 5 = .785 
Number 
of 
Iterations 
a) eoth( .785) = 1.524 
2 coth(.785) (1.524) = coth(1.12) = 1.238 
3 coth(.785) (1.238) = coth(.972) = 1.333 
4 coth(.785) (1.333) = coth(1.05) = 1.282 
5 coth(.785) (1.282) = coth(1.006) = 1.309 
6 coth(.785) (1.309) = coth(1.028) = 1.294 
Y, coth(.785) (1.294) = coth(1.0158) = 1.302 
8 coth(.785) (1.302) = coth(1.022) = 1.297 
Therefore 1.297 <1(.785) <1.302; 
ab L, were equal to 1000 feet and 
if H were 125 feet, then L would be 
1000/1.30 or L = 769.2 feet. 
of the depth water wave length. Then » “H/g equals 2H/L, which in 
turn yields 2rL,/8L, or the number 7/4. 
The value of 7/4 to three figures is given by 0.785. The 
hyperbolic cotangent has the value 1.524 as shown in the first row. 
The second row gives the hyperbolic cotangent of 0.785 times 1.524. 
The true value of the Itcoth lies between the two numbers given by 
2.5924 and 1.238. 
Eight iterations then yield the values given by row 7 and row 
8. Within an error of one half of one percent the true value of 
the Itcoth for p 2u/e equal to 7/4 is 1.30. Given that the wave 
length in deep water is 1000 feet, the depth would then be 125 feet 
Su 
