1. Find: db [ft] Ho’ db Ho’ = 10 [ft] A) 7.75 B) 11.25 C) 14.75
Find the breaking depth, d b, in feet. [pause] In this problem we are provided ---
T = 13 [s]

2. Find: db [ft] Ho’ db Ho’ = 10 [ft] A) 7.75 B) 11.25 C) 14.75
the unrefracted deepwater wave height, H knot prime, and we are also provided, ---
T = 13 [s]

3. Find: db [ft] Ho’ db Ho’ = 10 [ft] A) 7.75 B) 11.25 C) 14.75
the period of the wave, T. [fin] We are asked to find, d b, ---
T = 13 [s]

4. Find: db [ft] Ho’ db Ho’ = 10 [ft] A) 7.75 B) 11.25 C) 14.75
which is the depth of water which causes the wave to break. [pause] The depth of water, d, ---
T = 13 [s]

5. Find: db [ft] Ho’ db T = 13 [s] Ho’ = 10 [ft] d = db depth breaking
equals the breaking depth, d b, when the actual height of the wave, H A, ---
depth
breaking
depth

6. Find: db [ft] Ho’ db T = 13 [s] Ho’ = 10 [ft] d = db HA = Hb depth
equals the breaking height of the wave, H b. [pause] Therefore, we can solve for the breaking depth ---
depth
breaking
actual
breaking
depth
height
height

7. Find: db [ft] Ho’ db T = 13 [s] Ho’ = 10 [ft] d = db HA = Hb depth
by guessing values for the depth, ---
depth
breaking
actual
breaking
depth
height
height

8. Find: db [ft] Ho’ db T = 13 [s] Ho’ = 10 [ft] guess d = db HA = Hb
d, and then checking, whether or not, the actual height, equals, ---
depth
breaking
actual
breaking
depth
height
height

9. ? Find: db [ft] Ho’ db T = 13 [s] Ho’ = 10 [ft] guess d = db HA = Hb
the breaking height. [pause] For computational convenience sake, we’ll divide these height terms by, ---
depth
breaking
actual
breaking
depth
height
height

10. ? Find: db [ft] = Ho’ db T = 13 [s] Ho’ = 10 [ft] breaking guess
actual
height
d = db HA Hb
the length of the wave, L. [pause] When looking back at the possible solutions, ---
height = L L
depth
breaking
depth
wavelength

11. ? Find: db [ft] = Ho’ db T = 13 [s] Ho’ = 10 [ft] breaking actual
height HA Hb
for the breaking depth, we’ll begin by guessing a depth of 13 feet,
height =
possible L L
breaking
wavelength
depths

12. ? Find: db [ft] = Ho’ db T = 13 [s] Ho’ = 10 [ft] breaking
d1= [ft]
A) 7.75
B) 11.25
C) 14.75
D) 18.25
height HA Hb
for our first iteration, because 2 possible answers are shallower and 2 possible answers are deeper. To enter the table of wave values, --- =
possible L L
breaking
wavelength
depths

13. ? Find: db [ft] = Ho’ db T = 13 [s] Ho’ = 10 [ft] breaking
d1= [ft]
A) 7.75
B) 11.25
C) 14.75
D) 18.25
height HA Hb
we next need to compute d i over L knot. The letter i simply represents the iteration number, so we’ll change this variable i --- = di L L Lo
wavelength

14. ? Find: db [ft] = Ho’ db T = 13 [s] Ho’ = 10 [ft] breaking
d1= [ft]
A) 7.75
B) 11.25
C) 14.75
D) 18.25
height HA Hb
to a 1, for our first iteration, and plug in, --- = di L L
i=1 Lo
wavelength

15. ? Find: db [ft] = Ho’ db T = 13 [s] Ho’ = 10 [ft] breaking
d1= [ft]
A) 7.75
B) 11.25
C) 14.75
D) 18.25
height HA Hb
13 feet, into the numerator. [pause] L knot is computed as g T squared, --- = d1 L L
i=1 Lo
wavelength

16. ? ? Find: db [ft] = Ho’ db T = 13 [s] Ho’ = 10 [ft] breaking
d1= [ft]
height HA Hb
divided by 2 PI. [pause] After plugging in the known variables, --- =
g * T2 d1 L L
Lo =
2 * π
i=1 Lo
wavelength ?

17. ? Find: db [ft] = Ho’ db T = 13 [s] Ho’ = 10 [ft] g=32.16 [ft/s2]
breaking
d1= [ft]
height HA Hb
L knot equals, feet. This value of L knot is a constant because variables g, T and PI are all constants. This makes d1 over L knot, --- =
g * T2 d1 L L
Lo =
2 * π
i=1 Lo
wavelength
Lo = [ft]

18. ? Find: db [ft] = =0.01503 Ho’ db T = 13 [s] Ho’ = 10 [ft]
g=32.16 [ft/s2]
breaking
d1= [ft]
height HA Hb
equal to, With the d over L knot value, --- =
g * T2 d1 L L
Lo = =
2 * π Lo
wavelength
Lo = [ft]

19. Find: db [ft] =0.01503 Ho’ db d1=13.000 [ft] Ho’ = 10 [ft] d/Lo d/L d1
we have an entry point into the tables, and we can linearly interpolate, ---
d/Lo
d/L d1
0.015 = Lo
0.016

20. Find: db [ft] +0.03 * 0.05132 =0.01503 Ho’ db d1/L = 0.97 * 0.04964
d1= [ft]
+0.03 *
Ho’ = 10 [ft]
to calculate the values of d1 over L, which equals, ---
d/Lo
d/L d1
0.015 = Lo
0.016

21. Find: db [ft] +0.03 * 0.05132 =0.01503 Ho’ db d1/L = 0.97 * 0.04964
d1= [ft]
+0.03 *
Ho’ = 10 [ft]
d1/L =
[pause] Next, we’ll compute the wavelength, L, ---
d/Lo
d/L d1
0.015 = Lo
0.016

22. Find: db [ft] +0.03 * 0.05132 =0.01503 Ho’ db d1/L = 0.97 * 0.04964
d1= [ft]
+0.03 *
Ho’ = 10 [ft]
d1/L =
by dividing 13 feet, by, the value of ---- d1 d1 L= =
(d1/L) Lo

23. Find: db [ft] +0.03 * 0.05132 =0.01503 Ho’ db d1/L = 0.97 * 0.04964
d1= [ft]
+0.03 *
Ho’ = 10 [ft]
d1/L =
, and the wavelength for this first iteration, equals, --- d1 d1 L= =
(d1/L) Lo

24. Find: db [ft] +0.03 * 0.05132 = 261.622 [ft] =0.01503 Ho’ db
d1/L = 0.97 *
d1= [ft]
+0.03 *
Ho’ = 10 [ft]
d1/L =
feet. [pause] Since this wavelength depends on the depth, we’ll add a subscript, --- d1 d1 L=
= [ft] =
(d1/L) Lo

25. Find: db [ft] +0.03 * 0.05132 = 261.622 [ft] =0.01503 Ho’ db
d1/L = 0.97 *
d1= [ft]
+0.03 *
Ho’ = 10 [ft]
d1/L =
1, to signify the first iteration. [pause] Next, we’ll return to the tables, and calculate, --- d1 d1
L1=
= [ft] =
(d1/L) Lo

26. Find: db [ft] =0.01503 Ho’ db d1=13.000 [ft] Ho’ = 10 [ft]
L1= [ft]
the hyperbolic tangent of k times d, and, the actual height divided by the unrefracted deepwater height. [pause] This second term, H A over ---
d/Lo
tanh(k*d)
HA/Ho’ d1
0.015
0.3022
1.307 = Lo
0.016
0.3117
1.288

27. Find: db [ft] =0.01503 Ho’ db d1=13.000 [ft] shoaling Ho’ = 10 [ft]
coefficient
L1= [ft]
H knot prime, is sometimes called the shoaling coefficient. [pause] Using linear interpolation, ---
d/Lo
tanh(k*d)
HA/Ho’ d1
0.015
0.3022
1.307 = Lo
0.016
0.3117
1.288

28. Find: db [ft] +0.03 * 0.3117 =0.01503 d1=13.000 [ft] Ho’ = 10 [ft] db
L1= [ft]
tanh1(k*d)= 0.97 *
+0.03 *
tanh k d for a depth of 13 feet, equals, ---
d/Lo
tanh(k*d)
HA/Ho’ d1
0.015
0.3022
1.307 = Lo
0.016
0.3117
1.288

29. Find: db [ft] +0.03 * 0.3117 =0.01503 d1=13.000 [ft] Ho’ = 10 [ft] db
L1= [ft]
tanh1(k*d)= 0.97 *
+0.03 *
tanh1(k*d)=
[pause] Similarly, we can compute, ---
d/Lo
tanh(k*d)
HA/Ho’ d1
0.015
0.3022
1.307 = Lo
0.016
0.3117
1.288

30. Find: db [ft] +0.03 * 1.288 =0.01503 d1=13.000 [ft] Ho’ = 10 [ft] db
L1= [ft]
tanh1(k*d)=
HA,1/Ho’= 0.97 * 1.307
+0.03 * 1.288
H A 1, over, H knot prime, and we get, ----
d/Lo
tanh(k*d)
HA/Ho’ d1
0.015
0.3022
1.307 = Lo
0.016
0.3117
1.288

31. Find: db [ft] = 1.3064 +0.03 * 1.288 =0.01503 d1=13.000 [ft]
Ho’ = 10 [ft] db
L1= [ft]
tanh1(k*d)=
HA,1/Ho’= 0.97 * 1.307
+0.03 * 1.288 =
[pause] Keep in mind, we want to compare ---
d/Lo
tanh(k*d)
HA/Ho’ d1
0.015
0.3022
1.307 = Lo
0.016
0.3117
1.288

32. Find: db [ft] =0.01503 HA d1=13.000 [ft] L Ho’ = 10 [ft]
L1= [ft]
tanh1(k*d)=
HA,1/Ho’=
the actual height of the wave divided by the wavelength, and the breaking height ---
d/Lo
tanh(k*d)
HA/Ho’ d1
0.015
0.3022
1.307 = Lo
0.016
0.3117
1.288

33. Find: db [ft] =0.01503 HA Hb d1=13.000 [ft] L L Ho’ = 10 [ft]
L1= [ft]
tanh1(k*d)=
HA,1/Ho’=
of the wave, divided by the wavelength. [pause] We can compute H A over L, for this first iteration, ---
d/Lo
tanh(k*d)
HA/Ho’ d1
0.015
0.3022
1.307 = Lo
0.016
0.3117
1.288

34. Find: db [ft] = =0.01503 HA,1 HA,1 Ho’ d1=13.000 [ft] * L1 Ho’ L1
Ho’ = 10 [ft]
L1= [ft]
tanh1(k*d)=
HA,1/Ho’=
by using a few of the parameters we already calculated. And H A 1, over L 1, equals, ---
d/Lo
tanh(k*d)
HA/Ho’ d1
0.015
0.3022
1.307 = Lo
0.016
0.3117
1.288

35. Find: db [ft] = = 0.04993 =0.01503 HA,1 HA,1 Ho’ d1=13.000 [ft] * L1
Ho’ = 10 [ft]
L1= [ft]
HA,1
tanh1(k*d)= = L1
HA,1/Ho’=
[pause] Next, we can calculate H B 1, ---
d/Lo
tanh(k*d)
HA/Ho’ d1
0.015
0.3022
1.307 = Lo
0.016
0.3117
1.288

36. Find: db [ft] = 0.142 * tanh1 (k*d) = 0.04993 =0.01503 Hb,1
d1= [ft]
= * tanh1 (k*d) L1
Ho’ = 10 [ft]
L1= [ft]
tanh1(k*d)=
HA,1/Ho’=
HA,1
divided by L 1, by plugging in tanh 1 K D, we get, --- = L1
d/Lo
tanh(k*d)
HA/Ho’ d1
0.015
0.3022
1.307 = Lo
0.016
0.3117
1.288

37. Find: db [ft] = 0.142 * tanh1 (k*d) = 0.04296 = 0.04993 =0.01503 Hb,1
d1= [ft]
= * tanh1 (k*d) L1
Ho’ = 10 [ft]
L1= [ft]
Hb,1 =
tanh1(k*d)= L1
HA,1/Ho’=
HA,1
[pause] Now we can compare H A over L, --- = L1
d/Lo
tanh(k*d)
HA/Ho’ d1
0.015
0.3022
1.307 = Lo
0.016
0.3117
1.288

38. Find: db [ft] = 0.142 * tanh1 (k*d) = 0.04296 = 0.04993 =0.01503 Hb,1
d1= [ft]
= * tanh1 (k*d) L1
Ho’ = 10 [ft]
L1= [ft]
Hb,1 =
tanh1(k*d)= L1
HA,1/Ho’=
HA,1
and H b over L, for this first iteration. [pause] If H A over L is less than, --- = L1
d/Lo
tanh(k*d)
HA/Ho’ d1
0.015
0.3022
1.307 = Lo
0.016
0.3117
1.288

39. < Find: db [ft] = 0.04993 = 0.04296 HA Hb d1=13.000 [ft] if L L
Ho’ = 10 [ft]
L1= [ft]
tanh1(k*d)=
HA,1/Ho’=
HA,1
and H b over L, then wave has not yet broken, --- = L1
tanh(k*d)
HA/Ho’
Hb,1
0.3022
1.307 = L1
0.3117
1.288

40. < Find: db [ft] = 0.04993 = 0.04296 HA Hb d1=13.000 [ft] if L L
Ho’ = 10 [ft]
then d > db
L1= [ft]
tanh1(k*d)=
HA,1/Ho’=
HA,1
and the breaking depth is less than the depth we used for that iteration. In this case, --- = L1
tanh(k*d)
HA/Ho’
Hb,1
0.3022
1.307 = L1
0.3117
1.288

41. < Find: db [ft] = 0.04993 = 0.04296 HA Hb d1=13.000 [ft] if L L
Ho’ = 10 [ft]
then d > db
L1= [ft]
use smaller d
tanh1(k*d)=
HA,1/Ho’=
HA,1
we should use a smaller depth, d, in the following iteration. [pause] If H A over L is greater than, --- = L1
tanh(k*d)
HA/Ho’
Hb,1
0.3022
1.307 = L1
0.3117
1.288

42. < > Find: db [ft] = 0.04993 = 0.04296 HA Hb d1=13.000 [ft] if L
Ho’ = 10 [ft]
then d > db
L1= [ft]
use smaller d
tanh1(k*d)= HA Hb
>
HA,1/Ho’= if L L
HA,1
H b over L, then the wave doesn’t exist because it already broke at a deeper depth of water, --- = L1
tanh(k*d)
HA/Ho’
Hb,1
0.3022
1.307 = L1
0.3117
1.288

43. < > Find: db [ft] = 0.04993 = 0.04296 HA Hb d1=13.000 [ft] if L
Ho’ = 10 [ft]
then d > db
L1= [ft]
use smaller d
tanh1(k*d)= HA Hb
>
HA,1/Ho’= if L L
HA,1
then d < db
and we should guess a larger value for the depth, d, for the following iteration. Once these two heights are equal, --- = L1
use larger d
Hb,1 = L1

44. < > = Find: db [ft] = 0.04993 = 0.04296 HA Hb d1=13.000 [ft] ifL L
Ho’ = 10 [ft]
then d > db
L1= [ft]
use smaller d
tanh1(k*d)= HA Hb
>
HA,1/Ho’= if L L
HA,1
then d < db
then the depth, d, equals, --- = L1
use larger d
Hb,1 HA Hb = = if L1 L L

45. < > = Find: db [ft] = 0.04993 = 0.04296 HA Hb d1=13.000 [ft] ifL L
Ho’ = 10 [ft]
then d > db
L1= [ft]
use smaller d
tanh1(k*d)= HA Hb
>
HA,1/Ho’= if L L
HA,1
then d < db
the breaking depth d b. [pause] For our first iteration, the depth of 13 feet resulted in --- = L1
use larger d
Hb,1 HA Hb = = if
then d = db L1 L L

46. < > = Find: db [ft] = 0.04993 = 0.04296 HA Hb d1=13.000 [ft] ifL L
Ho’ = 10 [ft]
then d > db
L1= [ft]
use smaller d
tanh1(k*d)= HA Hb
>
HA,1/Ho’= if L L
HA,1
then d < db
H A over L greater than H b over L, therefore, ---- = L1
use larger d
Hb,1 HA Hb = = if
then d = db L1 L L

47. < > = Find: db [ft] = 0.04993 = 0.04296 HA Hb d1=13.000 [ft] ifL L
Ho’ = 10 [ft]
then d > db
L1= [ft]
use smaller d
tanh1(k*d)= HA Hb
>
HA,1/Ho’= if L L
HA,1
then d < db
we need to select a larger depth, d, for our second iteration. Our second depth is computed using --- = L1
use larger d
Hb,1 HA Hb = = if
then d = db L1 L L

48. > Find: db [ft] = 0.04993 = 0.04296 HA Hb d1=13.000 [ft] if L L
Ho’ = 10 [ft]
then d < db
L1= [ft]
use larger d
tanh1(k*d)=
HA,1/Ho’=
( HA,i / Li )
di+1= di *
HA,1
( Hb,i / Li )
the first depth and the values of H A over L and H b over L. For i equal to 1, --- = L1
Hb,1 = L1

49. > Find: db [ft] = 0.04993 = 0.04296 HA Hb d1=13.000 [ft] if L L
Ho’ = 10 [ft]
then d < db
L1= [ft]
use larger d
tanh1(k*d)=
HA,1/Ho’=
( HA,i / Li )
di+1= di *
HA,1
( Hb,i / Li )
we plug in the appropriate variables, and our next depth, d 2, equals, --- = L1
Hb,1 = L1

50. > Find: db [ft] = 0.04993 = 0.04296 HA Hb d1=13.000 [ft] if L L
Ho’ = 10 [ft]
then d < db
L1= [ft]
use larger d
tanh1(k*d)=
HA,1/Ho’=
( HA,i / Li )
di+1= di *
HA,1
( Hb,i / Li )
feet. [pause] Next, we’ll begin the second iteration, --- = L1
d2= [ft]
Hb,1 = L1

51. Find: db [ft] = ? = ? d2=15.109 [ft] Ho’ = 10 [ft] L2=? tanh2(k*d)= ?
HA,2/Ho’= ?
HA,2
using a depth of feet. [pause] First, we’ll compute d2, divided by, ---
= ? L1
Hb,2
= ? L1

52. Find: db [ft] = ? = ? d2 d2=15.109 [ft] Lo Ho’ = 10 [ft] L2=?
tanh2(k*d)= ?
HA,2/Ho’= ?
HA,2
L knot. Recall, we computed L knot during the first iteration, ---
= ? L1
Hb,2
= ? L1

53. Find: db [ft] = ? = ? d2 d2=15.109 [ft] Lo Ho’ = 10 [ft] L2=?
Lo = [ft]
tanh2(k*d)= ?
HA,2/Ho’= ?
HA,2
and after plugging in the known variables, d 2 over L knot, equals, ---
= ? L1
Hb,2
= ? L1

54. Find: db [ft] = 0.01747 = ? = ? d2 d2=15.109 [ft] Lo Ho’ = 10 [ft]
Lo = [ft]
tanh2(k*d)= ?
HA,2/Ho’= ?
HA,2
[pause] Next we’ll calculate the value for ---
= ? L1
Hb,2
= ? L1

55. Find: db [ft] = 0.01747 =0.01747 d2 d2=15.109 [ft] Lo Ho’ = 10 [ft]
Lo = [ft]
tanh2(k*d)= ?
HA,2/Ho’= ?
d over L for this second depth, by using linear interpolation between the values of ---
d/Lo
d/L d2
0.017 = Lo
0.018

56. Find: db [ft] = 0.01747 +0.47 * 0.05455 =0.01747 d2 d2=15.109 [ft] Lo
Ho’ = 10 [ft]
L2=?
Lo = [ft]
tanh2(k*d)= ?
d2/L = 0.53 *
HA,2/Ho’= ?
+0.47 *
0.017 and 0.018, for d over L knot, we get d 2 over L, equals, ---
d/Lo
d/L d2
0.017 = Lo
0.018

57. Find: db [ft] = 0.01747 +0.47 * 0.05455 =0.01747 d2 d2=15.109 [ft] Lo
Ho’ = 10 [ft]
L2=?
Lo = [ft]
tanh2(k*d)= ?
d2/L = 0.53 *
HA,2/Ho’= ?
+0.47 *
d2/L =
[pause] Again, we’ll add the subscript 2, ---
d/Lo
d/L d2
0.017 = Lo
0.018

58. Find: db [ft] = 0.01747 +0.47 * 0.05455 =0.01747 d2 d2=15.109 [ft] Lo
Ho’ = 10 [ft]
L2=?
Lo = [ft]
tanh2(k*d)= ?
d2/L = 0.53 *
HA,2/Ho’= ?
+0.47 *
d2/L2 =
since we’re on the second iteration. [pause] And to compute L 2, ----
d/Lo
d/L d2
0.017 = Lo
0.018

59. Find: db [ft] = 0.01747 +0.47 * 0.05455 d2 d2=15.109 [ft] Lo
Ho’ = 10 [ft]
L2=?
Lo = [ft]
tanh2(k*d)= ?
d2/L = 0.53 *
HA,2/Ho’= ?
+0.47 *
d2/L2 =
we divide feet by , and we get, --- d2
L2=
(d2/L2)

60. Find: db [ft] = 0.01747 +0.47 * 0.05455 = 281.359 [ft] d2
d2= [ft] = Lo
Ho’ = 10 [ft]
L2= [ft]
Lo = [ft]
tanh2(k*d)= ?
d2/L = 0.53 *
HA,2/Ho’= ?
+0.47 *
d2/L2 =
feet. [pause] Next we’ll compute the hyperbolic tangent of k d, and, --- d2
L2=
= [ft]
(d2/L2)

61. Find: db [ft] = 0.01747 +0.47 * 0.05455 =0.01747 d2 d2=15.109 [ft] Lo
Ho’ = 10 [ft]
L2= [ft]
Lo = [ft]
tanh2(k*d)= ?
d2/L = 0.53 *
HA,2/Ho’= ?
+0.47 *
d2/L2 =
the shoaling coefficient, H A or H knot prime, by using ---
d/Lo
tanh(k*d)
HA/Ho’ d2
0.017
0.3209
1.271 = Lo
0.018
0.3298
1.255

62. Find: db [ft] +0.47 * 1.255 +0.47* 0.3298 =0.01747 d2=15.109 [ft]
HA,2/Ho’= 0.53 * 1.271
+0.47 * 1.255
Ho’ = 10 [ft]
L2= [ft]
tanh2(k*d)= 0.53 *
+0.47*
interpolation again. [pause] And tanh 2 k d equals, ---
d/Lo
tanh(k*d)
HA/Ho’ d2
0.017
0.3209
1.271 = Lo
0.018
0.3298
1.255

63. Find: db [ft] +0.47 * 1.255 +0.47* 0.3298 =0.01747 d2=15.109 [ft]
HA,2/Ho’= 0.53 * 1.271
+0.47 * 1.255
Ho’ = 10 [ft]
L2= [ft]
tanh2(k*d)= 0.53 *
+0.47*
tanh2(k*d)=
0.3251, and the soaling coefficient for this second iteration, equals, ---
d/Lo
tanh(k*d)
HA/Ho’ d2
0.017
0.3209
1.271 = Lo
0.018
0.3298
1.255

64. Find: db [ft] +0.47 * 1.255 +0.47* 0.3298 =0.01747 d2=15.109 [ft]
HA,2/Ho’= 0.53 * 1.271
+0.47 * 1.255
Ho’ = 10 [ft]
L2= [ft]
HA,2/Ho’ =
tanh2(k*d)= 0.53 *
+0.47*
tanh2(k*d)=
[pause] Next, we’ll compute H A 2, over L 2, ---
d/Lo
tanh(k*d)
HA/Ho’ d2
0.017
0.3209
1.271 = Lo
0.018
0.3298
1.255

65. Find: db [ft] = =0.01747 HA,2 HA,2 Ho’ d2=15.109 [ft] * L2 Ho’ L2
Ho’ = 10 [ft]
L2= [ft]
tanh2(k*d)=
HA,2/Ho’=
and after plugging in the known variables, we get, ---
d/Lo
tanh(k*d)
HA/Ho’ d2
0.017
0.3209
1.271 = Lo
0.018
0.3298
1.255

66. Find: db [ft] = = 0.04491 =0.01747 HA,2 HA,2 Ho’ d2=15.109 [ft] * L2
Ho’ = 10 [ft]
L2= [ft]
HA,2
tanh2(k*d)= = L2
HA,2/Ho’=
[pause] And to solve for H b 2, over L 2, ---
d/Lo
tanh(k*d)
HA/Ho’ d2
0.017
0.3209
1.271 = Lo
0.018
0.3298
1.255

67. Find: db [ft] = 0.142 * tanh2 (k*d) = 0.04491 =0.01747 Hb,2
d2= [ft]
= * tanh2 (k*d) L2
Ho’ = 10 [ft]
L2= [ft]
tanh2(k*d)=
HA,2/Ho’=
HA,2
we’ll plug in the hyperbolic tangent of k d, and compute, --- = L2
d/Lo
tanh(k*d)
HA/Ho’ d2
0.017
0.3209
1.271 = Lo
0.018
0.3298
1.255

68. Find: db [ft] = 0.142 * tanh2 (k*d) = 0.04616 = 0.04491 =0.01747 Hb,2
d2= [ft]
= * tanh2 (k*d) L2
Ho’ = 10 [ft]
L2= [ft]
Hb,2 =
tanh2(k*d)= L2
HA,2/Ho’=
HA,2
[pause] As before, we’ll compare --- = L2
d/Lo
tanh(k*d)
HA/Ho’ d2
0.017
0.3209
1.271 = Lo
0.018
0.3298
1.255

69. Find: db [ft] = 0.142 * tanh2 (k*d) = 0.04616 = 0.04491 =0.01747 Hb,2
d2= [ft]
= * tanh2 (k*d) L2
Ho’ = 10 [ft]
L2= [ft]
Hb,2 =
tanh2(k*d)= L2
HA,2/Ho’=
HA,2
the computed height of the wave and the breaking height of the wave, --- = L2
d/Lo
tanh(k*d)
HA/Ho’ d2
0.017
0.3209
1.271 = Lo
0.018
0.3298
1.255

70. < > Find: db [ft] = 0.04491 = 0.04616 HA Hb d2=15.109 [ft] if L
Ho’ = 10 [ft]
then d > db
L2= [ft]
use smaller d
tanh2(k*d)= HA Hb
>
HA,2/Ho’= if L L
HA,2
then d < db
and this time, we find out that the wave breaks at a shallow depth than d 2, --- = L2
use larger d
Hb,2 = L2

71. < > Find: db [ft] = 0.04491 = 0.04616 HA Hb d2=15.109 [ft] if L
Ho’ = 10 [ft]
then d > db
L2= [ft]
use smaller d
tanh2(k*d)= HA Hb
>
HA,2/Ho’= if L L
HA,2
then d < db
and we should test a smaller depth d. For the next iteration, --- = L2
use larger d
Hb,2 = L2

72. < Find: db [ft] = 0.04491 = 0.04616 HA Hb d2=15.109 [ft] if L L
Ho’ = 10 [ft]
then d > db
L2= [ft]
use smaller d
tanh2(k*d)=
HA,2/Ho’=
( HA,i / Li )
di+1= di *
( Hb,i / Li )
HA,2
we’ll compute the next depth to evaluate, d 3, which equals, --- = L2
Hb,2 = L2

73. < Find: db [ft] = 0.04491 = 0.04616 HA Hb d2=15.109 [ft] if L L
Ho’ = 10 [ft]
then d > db
L2= [ft]
use smaller d
tanh2(k*d)=
HA,2/Ho’=
( HA,i / Li )
di+1= di *
( Hb,i / Li )
HA,2
feet. [pause] If we went through another entire iteration, our fourth depth, d 4, --- = L2
d3= [ft]
Hb,2 = L2

74. > Find: db [ft] = 0.04578 = 0.04555 HA Hb d3=14.700 [ft] if L L
Ho’ = 10 [ft]
then d < db
L3= [ft]
use larger d
tanh3(k*d)=
HA,3/Ho’=
( HA,i / Li )
di+1= di *
( Hb,i / Li )
HA,3
would equal feet. [pause] Next, if we look at our successive approximations --- = L3
d4= [ft]
Hb,3 = L3

75. > Find: db [ft] HA Hb if L L then d < db d1=13.000 [ft]
use larger d
d2= [ft]
d3= [ft]
( HA,i / Li )
di+1= di *
d4= [ft]
( Hb,i / Li )
for the breaking depth, d b, we notice the depth is converging on a value close to ---

76. > Find: db [ft] HA Hb if L L then d < db d1=13.000 [ft]
use larger d
d2= [ft]
d3= [ft]
( HA,i / Li )
di+1= di *
d4= [ft]
( Hb,i / Li )
14.7 something feet. [pause]
14.7 [ft] < db < 14.8 [ft]

77. > Find: db [ft] HA Hb if L L then d < db d1=13.000 [ft]
use larger d
d2= [ft]
d3= [ft]
A) 7.75
B) 11.25
C) 14.75
D) 18.25
d4= [ft]
When reviewing the possible solutions, ---
14.7 [ft] < db < 14.8 [ft]

78. > Find: db [ft] HA Hb if L L then d < db d1=13.000 [ft]
use larger d
d2= [ft]
d3= [ft]
A) 7.75
B) 11.25
C) 14.75
D) 18.25
d4= [ft]
the answer is C. [fin]
14.7 [ft] < db < 14.8 [ft]
answerC