1. Find: αb wave direction breaking αo wave contours αb beach A) 8.9
Ho = 20 [ft]
Find alpha b, the apporach angle of the wave when the wave breaks. [pause] In this problem we are provided, --- o
T = 15 [s] o
αo = 30 o o

2. Find: αb wave direction breaking αo wave contours αb beach A) 8.9
Ho = 20 [ft]
the deepwater waveheight, H knot, the period of the wave, --- o
T = 15 [s] o
αo = 30 o o

3. Find: αb wave direction breaking αo wave contours αb beach A) 8.9
Ho = 20 [ft]
T, and alpha knot, which is, --- o
T = 15 [s] o
αo = 30 o o

4. Find: αb wave direction breaking αo wave contours αb beach A) 8.9
Ho = 20 [ft]
the deepwater approach angle. [pause] The problem asks to find alpha b, --- o
T = 15 [s] o
αo = 30 o o

5. Find: αb wave direction breaking αo wave contours αb beach A) 8.9
Ho = 20 [ft]
which is the angle between the crestline of the wave and the ocean bottom contours, when the wave breaks. To find alpha b, --- o
T = 15 [s] o
αo = 30 o o

6. Find: αb = wave direction breaking αo wave contours αb beach sin (αb)
sin (αo) =
we’ll use the relationship which relates alpha b to alpha knot, L b and L knot. Where L b and L knot --- Lb Lo

7. Find: αb = wave direction breaking αo wave contours αb beach sin (αb)
sin (αo)
deep water =
correspond to the wavelength for the breaking condition and deepwater condition, respectively. [pause] After solving for alpha b, ---
wavelength Lb Lo
wavelength at the
breaking depth

8. Find: αb = wave direction breaking αo wave contours αb beach sin (αb)
sin (αo)
deep water =
we can plug in the value of alpha knot, which is 30 degrees, ---
wavelength Lb Lo Lb
wavelength at the
αb = sin-1
sin(αo) * Lo
breaking depth

9. Find: αb = wave direction breaking αo wave contours αb beach sin (αb)
sin (αo)
deep water =
and was provided in the problem statement. [pause] But we still have to find ---
wavelength Lb Lo o 30 Lb
wavelength at the
αb = sin-1
sin(αo) * Lo
breaking depth

10. Find: αb = wave direction breaking αo wave contours αb beach sin (αb)
sin (αo)
deep water =
L b, and, L knot. [pause] L knot, equals, ---
wavelength Lb Lo o 30 Lb
wavelength at the
αb = sin-1
sin(αo) * Lo
breaking depth

11. Find: αb = wave direction breaking αo wave contours αb beach sin (αb)
sin (αo)
g * T2 =
g times T squared, divided by 2 PI. [pause] Variable g represents the ---
Lo = Lb Lo
2 * π o 30 Lb
wavelength at the
αb = sin-1
sin(αo) * Lo
breaking depth

12. Find: αb = wave direction breaking αo wave contours αb beach
32.16 [ft/s2]
sin (αb)
sin (αo)
g * T2 =
gravitational acceleration constant, and variable T is the wave period, ---
Lo = Lb Lo
2 * π o 30 Lb
wavelength at the
αb = sin-1
sin(αo) * Lo
breaking depth

13. Find: αb = wave direction breaking αo wave contours αb beach
32.16 [ft/s2]
T = 15 [s]
sin (αb)
sin (αo)
g * T2 =
which was provided as 15 seconds. [pause] This makes L knot, equal to, ---
Lo = Lb Lo
2 * π o 30 Lb
wavelength at the
αb = sin-1
sin(αo) * Lo
breaking depth

14. Find: αb = =1,151.645 [ft] wave direction breaking αo wave contours αb
beach
32.16 [ft/s2]
T = 15 [s]
sin (αb)
sin (αo)
g * T2 =
1, feet. [pause] Lastly, we need to determine, L b, ---
Lo =
=1, [ft] Lb Lo
2 * π o 30 Lb
wavelength at the
αb = sin-1
sin(αo) * Lo
breaking depth

15. Find: αb ? = =1,151.645 [ft] wave direction breaking αo wave contours
beach
32.16 [ft/s2]
T = 15 [s]
sin (αb)
sin (αo)
g * T2 =
the wavelength of the wave, when the wave breaks. [pause] We know we have found L b, when we select a wave depth, d, ---
Lo =
=1, [ft] Lb Lo
2 * π o 30 ? Lb
wavelength at the
αb = sin-1
sin(αo) * Lo
breaking depth

16. Find: αb ? =1,151.645 [ft] wave direction breaking αo wave contours αb
beach
32.16 [ft/s2]
T = 15 [s]
d=?
g * T2
which results in a wave condition, where the actual height of the wave, H A, ---
Lo =
=1, [ft]
2 * π
wave o 30 ?
depth Lb
αb = sin-1
sin(αo) * Lo

17. Find: αb ? = Lo=1,151.645 [ft] wave direction breaking αo wave
contours
actual
wave
wave height at
height
breaking depth Ha = Hb
d=?
equals the breaking height of the wave, H b. [pause] This is the same comparison, as if we happen to divide ---
Lo=1, [ft] o
wave 30 ?
depth Lb
αb = sin-1
sin(αo) * Lo

18. Find: αb ? = Lo=1,151.645 [ft] wave direction breaking αo wave
contours
actual
wave
wave height at
height
breaking depth Ha Hb
d=? =
both sides of the equation by the wavelength, L. [pause] To begin, we’ll arbitrarily choose a wave depth of ---
Lo=1, [ft] L L o
wave 30 ?
wavelength
depth Lb
αb = sin-1
sin(αo) * Lo

19. Find: αb ? = Lo=1,151.645 [ft] wave direction breaking αo wave
contours
actual
wave
wave height at
height
breaking depth Ha Hb
d=20 [ft] =
20 feet for our first iteration. [pause] Next, we’ll compute d over L knot, ---
Lo=1, [ft] L L o
wave 30 ?
wavelength
depth Lb
αb = sin-1
sin(αo) * Lo

20. Find: αb ? = Lo=1,151.645 [ft] wave direction αo αb beach d Lo
d=20 [ft]
and after plugging the appropriate variables, d over L knot, ---
Lo=1, [ft] o
wave 30 ?
depth Lb
αb = sin-1
sin(αo) * Lo

21. Find: αb ? =0.01736 Lo=1,151.645 [ft] wave direction αo αb beach d Lo
d=20 [ft]
equals, [pause] This value of d over L knot corresponds to a d over L value of, ---
Lo=1, [ft] o
wave 30 ?
depth Lb
αb = sin-1
sin(αo) * Lo

22. Find: αb ? =0.05353 =0.01736 Lo=1,151.645 [ft] wave direction αo d L
beach d = Lo
d=20 [ft]
Therefore, the wavelength for this wave, ---
Lo=1, [ft] o
wave 30 ?
depth Lb
αb = sin-1
sin(αo) * Lo

23. Find: αb ? = 0.05353 =0.01736 Lo=1,151.645 [ft] wave direction αo d L
beach d = Lo
d=20 [ft]
at a depth of 20 feet, equals, --
Lo=1, [ft] o
wave 30 ?
depth Lb
αb = sin-1
sin(αo) * Lo

24. Find: αb ? = 0.05353 = 373.622 [ft] =0.01736 Lo=1,151.645 [ft] wave
direction αo d = L αb d L=
= [ft]
d/L
beach d = Lo
d=20 [ft]
feet. [pause] Also, by knowing the value of d over L knot, ---
Lo=1, [ft] o
wave 30 ?
depth Lb
αb = sin-1
sin(αo) * Lo

25. Find: αb ? = 0.05353 =0.01736 Lo=1,151.645 [ft] wave direction αo d L
L= [ft]
beach
d=20 [ft] d = Lo
we can interpolate to find the hyperbolic tangent of k d, and, ---
Lo=1, [ft] o 30 ? Lb
αb = sin-1
sin(αo) * Lo

26. Find: αb ? = 0.05353 =0.01736 Lo=1,151.645 [ft] wave direction αo d L
L= [ft]
beach
d=20 [ft] d = Lo
the shoaling coeffcient, H over H knot prime, where the variable H, ---
Lo=1, [ft] o
tanh (k*d) = 30 ? Lb
H/Ho = ‘
αb = sin-1
sin(αo) * Lo

27. Find: αb ? = 0.05353 =0.01736 Lo=1,151.645 [ft] wave direction αo d L
L= [ft]
beach
d=20 [ft] d = Lo
represents the actual height of the wave, H A, at the tested depth of, ---
Lo=1, [ft] o
tanh (k*d) = 30 ? Lb
H/Ho = ‘
αb = sin-1
sin(αo) * Lo

28. Find: αb ? = 0.05353 =0.01736 Lo=1,151.645 [ft] wave direction αo d L
L= [ft]
beach
d=20 [ft] d = Lo
20 feet, and that variable H, is what we’ll use to compare the actual wave, H a, ---
Lo=1, [ft] o
tanh (k*d) = 30 ? Lb
H/Ho = ‘
αb = sin-1
sin(αo) * Lo

29. Find: αb ? = 0.05353 = =0.01736 Lo=1,151.645 [ft] actual wave d heightHa Hb
L= [ft] = L L
d=20 [ft] d = Lo
with the wave at its breaking condition, H b. [pause] The right hand side of this equation, H b over L, ---
Lo=1, [ft] o
tanh (k*d) = 30 ? Lb
H/Ho = ‘
αb = sin-1
sin(αo) * Lo

30. Find: αb = 0.05353 = =0.01736 = 0.142 * tanh (k*d) actual wave d
height L Ha Hb
L= [ft] = L L
d=20 [ft] d = Hb Lo
equals, times the hyperbolic tangent of k d, ---
= * tanh (k*d) L
tanh (k*d) =
H/Ho = ‘

31. Find: αb = 0.05353 = = 0.0460 =0.01736 = 0.142 * tanh (k*d) actual
wave d =
height L Ha Hb
L= [ft] = L L
d=20 [ft] Hb = d L = Hb Lo
which equals, [pause] Solving for the actual wave height, H a, ---
= * tanh (k*d) L
tanh (k*d) =
H/Ho = ‘

32. Find: αb = 0.05353 = = 0.0460 =0.01736 = 0.142 * tanh (k*d) actual
wave d =
height L Ha Hb
L= [ft] = L L
d=20 [ft] Hb = d L = Hb Lo
we must first compute the approach angle the wave would make, ---
= * tanh (k*d) L
tanh (k*d) =
H/Ho = ‘

33. Find: αb ? = 0.05353 = Lo=1,151.645 [ft] actual wave d height 0.0460 LHa Hb
L= [ft] = L L
d=20 [ft] L
Lo=1, [ft]
α = sin-1
sin(αo) * Lo
tanh (k*d) =
at a depth of 20 feet. And after plugging in the known variables of, ---
H/Ho = ‘

34. Find: αb ? = 0.05353 = Lo=1,151.645 [ft] actual wave d height 0.0460 LHa Hb
L= [ft] = L L o 30
d=20 [ft] L
Lo=1, [ft]
α = sin-1
sin(αo) * Lo
tanh (k*d) =
L, L knot, and alpha knot, the crestline of the wave would be ----
H/Ho = ‘

35. Find: αb ? = 0.05353 = Lo=1,151.645 [ft] actual wave d height 0.0460 LHa Hb
L= [ft] = L L o 30
d=20 [ft] L
Lo=1, [ft]
α = sin-1
sin(αo) * Lo
tanh (k*d) =
9.319 degrees offset from the ocean bottom contours. With the value of alpha, we can compute ---
α = 9.319 o
H/Ho = ‘

36. Find: αb ? = 0.05353 = Lo=1,151.645 [ft] actual wave d height 0.0460 LHa Hb
L= [ft] = L L o 30
d=20 [ft] L
Lo=1, [ft]
α = sin-1
sin(αo) * Lo
tanh (k*d) =
the wave refraction index, K r, which equals, ---
α = 9.319 o
H/Ho = ‘
cos (αo)
KR =
cos (α)

37. Find: αb ? = 0.05353 = Lo=1,151.645 [ft] actual wave d height 0.0460 LHa Hb
L= [ft] = L L o 30
d=20 [ft] L
Lo=1, [ft]
α = sin-1
sin(αo) * Lo
tanh (k*d) =
[pause] After that, we’ll compute the unrefracted deepwater wave height, ---
α = 9.319 o
H/Ho = ‘ o
cos (αo) 30
KR =
cos (α)
KR =0.9368

38. Find: αb ? = 0.05353 = Lo=1,151.645 [ft] actual wave d height 0.0460 LHa Hb
L= [ft] = L L
d=20 [ft]
Lo=1, [ft]
Ho = KR * Ho ‘
tanh (k*d) =
H knot prime, by multipliying the refraction coefficient by the deepwater wave height. We just computed ---
unrefracted
H/Ho = ‘
deepwater
α = 9.319 o
wave height
KR =0.9368

39. Find: αb ? = 0.05353 = Lo=1,151.645 [ft] actual wave d height 0.0460 LHa Hb
L= [ft] = L L
d=20 [ft]
Lo=1, [ft]
Ho = KR * Ho ‘
tanh (k*d) =
the refraction coefficient, K r, and the problem statement provided the deepwater ---
unrefracted
H/Ho = ‘
deepwater
α = 9.319 o
wave height
KR =0.9368

40. Find: αb ? = 0.05353 = Lo=1,151.645 [ft] actual wave d height 0.0460 LHa Hb
L= [ft] = L L
d=20 [ft]
Ho = 20 [ft]
Lo=1, [ft]
Ho = KR * Ho ‘
tanh (k*d) =
wave height as 20 feet. This makes the unrefracted deepwater wave height, equal to, ---
unrefracted
H/Ho = ‘
deepwater
α = 9.319 o
wave height
KR =0.9368

41. Find: αb ? = 0.05353 = Lo=1,151.645 [ft] actual wave d height 0.0460 LHa Hb
L= [ft] = L L
d=20 [ft]
Ho = 20 [ft]
Lo=1, [ft]
Ho = KR * Ho ‘
tanh (k*d) =
feet. [pause] Now, to find the actual height of the wave at this depth, ---
Ho = [ft] ‘
H/Ho = ‘
α = 9.319 o
KR =0.9368

42. Find: αb ? = 0.05353 = Lo=1,151.645 [ft] actual wave d height 0.0460 LHa Hb
L= [ft] = L L
d=20 [ft]
Ho = 20 [ft]
Lo=1, [ft]
Ho = KR * Ho ‘
tanh (k*d) =
we’ll multiply H knot prime by the shoaling coefficient, and the actual height, equals, ---
Ho = [ft] ‘
H/Ho = ‘
α = 9.319 o
H = Ho * (H/Ho ) ‘ ‘
KR =0.9368

43. Find: αb ? = 0.05353 = Lo=1,151.645 [ft] actual wave d height 0.0460 LHa Hb
L= [ft] = L L
d=20 [ft]
Ho = 20 [ft]
Lo=1, [ft]
Ho = KR * Ho ‘
tanh (k*d) =
feet. [pause] When divided by the wavelength, L, ---
Ho = [ft] ‘
H/Ho = ‘
α = 9.319 o
H = Ho * (H/Ho ) ‘ ‘
KR =0.9368
H = [ft]

44. Find: αb ? = 0.05353 = Lo=1,151.645 [ft] actual wave d height 0.0460 LHa Hb
L= [ft] = L L
d=20 [ft] Ha
Lo=1, [ft] L
tanh (k*d) =
the quotient, H a over L, equals, --
Ho = [ft] ‘
H/Ho = ‘
α = 9.319 o
H = Ho * (H/Ho ) ‘ ‘
KR =0.9368
H = [ft]

45. Find: αb ? = 0.05353 = Lo=1,151.645 [ft] = 0.0634 actual wave d height
0.0460 L Ha Hb
L= [ft] = L L
d=20 [ft] Ha
Lo=1, [ft] = L
tanh (k*d) =
[pause] Since the actual height of the wave H a, ---
Ho = [ft] ‘
H/Ho = ‘
α = 9.319 o
H = Ho * (H/Ho ) ‘ ‘
KR =0.9368
H = [ft]

46. Find: αb = 0.05353 > Lo=1,151.645 [ft] = 0.0634 actual wave d
height
0.0460 L Ha Hb
L= [ft]
> L L
d=20 [ft] Ha
Lo=1, [ft] = L
tanh (k*d) =
is greater than the breaking height of the wave, then, this wave does not exist, because, it broke, at a depth, ---
Ho = [ft] ‘
H/Ho = ‘
α = 9.319 o
H = Ho * (H/Ho ) ‘ ‘
KR =0.9368
H = [ft]

47. Find: αb = 0.05353 > Lo=1,151.645 [ft] = 0.0634 actual wave d
height
0.0460 L Ha Hb
L= [ft]
> L L
d=20 [ft]
d < db Ha
Lo=1, [ft] = L
tanh (k*d) =
deeper than 20 feet. [pause] For our second iteration, we’ll use a depth based on ---
Ho = [ft] ‘
H/Ho = ‘
α = 9.319 o
H = Ho * (H/Ho ) ‘ ‘
KR =0.9368
H = [ft]

48. Find: αb = 0.05353 > Lo=1,151.645 [ft] d 0.0634 0.0460 L Ha Hb
L= [ft]
> L L
d=20 [ft]
d < db
( HA,i / Li )
di+1= di
Lo=1, [ft] *
( Hb,i / Li )
tanh (k*d) =
our previous depth, and variables H a, and, H b. [pause] After plugging in ---
Ho = [ft] ‘
H/Ho = ‘
α = 9.319 o
H = Ho * (H/Ho ) ‘ ‘
KR =0.9368
H = [ft]

49. Find: αb = 0.05353 > Lo=1,151.645 [ft] d 0.0634 0.0460 L Ha Hb
L= [ft]
> L L
d=20 [ft]
( HA,i / Li )
di+1= di
Lo=1, [ft] *
( Hb,i / Li )
tanh (k*d) =
the appropriate variables, our next depth to test will be, ---
Ho = [ft] ‘
H/Ho = ‘
α = 9.319 o
H = Ho * (H/Ho ) ‘ ‘
KR =0.9368
H = [ft]

50. Find: αb = 0.05353 > Lo=1,151.645 [ft] d 0.0634 0.0460 L Ha Hb
L= [ft]
> L L
d=20 [ft]
( HA,i / Li )
d2= d1
Lo=1, [ft] *
( Hb,i / Li )
tanh (k*d) =
27.58 feet. [pause] When using the depth of feet, ---
d2= [ft]
H/Ho = ‘
α = 9.319 o
KR =0.9368

51. Find: αb ? = ? = Lo=1,151.645 [ft] d ? ? L Ha Hb L= ? [ft] L L
d= [ft]
( HA,i / Li )
Lo=1, [ft]
di+1= di *
( Hb,i / Li )
tanh (k*d) = ?
and performing the same calculations as the first iteration, ---
H/Ho = ? ‘
α = ? o
KR =?

52. Find: αb = 0.06333 < Lo=1,151.645 [ft] d 0.0508 0.0537 L Ha Hb
L= [ft]
< L L
d= [ft]
( HA,i / Li )
Lo=1, [ft]
di+1= di *
( Hb,i / Li )
tanh (k*d) =
we compute the values shown here. At this depth, we notice the actual height of the wave, ---
H/Ho = ‘
α = o
KR =0.9391

53. Find: αb = 0.06333 < Lo=1,151.645 [ft] d 0.0508 0.0537 L Ha Hb
L= [ft]
< L L
d= [ft]
( HA,i / Li )
Lo=1, [ft]
di+1= di *
( Hb,i / Li )
tanh (k*d) =
H a, is less than the breaking height of the wave, H b. Therefore, our subsequent depth, ---
H/Ho = ‘
α = o
KR =0.9391

54. Find: αb = 0.06333 < Lo=1,151.645 [ft] d 0.0508 0.0537 L Ha Hb
L= [ft]
< L L
d= [ft]
( HA,i / Li )
Lo=1, [ft]
d3= d2 *
( Hb,i / Li )
tanh (k*d) =
d 3, will be shallower than feet. For our next depth value we’ll try ---
H/Ho = ‘
α = o
KR =0.9391

55. Find: αb = 0.06333 < Lo=1,151.645 [ft] d 0.0508 0.0537 L Ha Hb
L= [ft]
< L L
d= [ft]
( HA,i / Li )
Lo=1, [ft]
d3= d2 *
( Hb,i / Li )
tanh (k*d) =
26.09 feet. [pause] After running through a third iteration, ---
d3= [ft]
H/Ho = ‘
α = o
KR =0.9391

56. Find: αb ? = ? = Lo=1,151.645 [ft] d ? ? L Ha Hb L= ? [ft] L L
d= [ft]
( HA,i / Li )
Lo=1, [ft]
di+1= di *
( Hb,i / Li )
tanh (k*d) = ?
for this new depth value, we again calculate out all the values, ----
H/Ho = ? ‘
α = ? o
KR =?

57. Find: αb = 0.06150 < Lo=1,151.645 [ft] d 0.05280 0.05338 L Ha Hb
L= [ft]
< L L
d= [ft]
( HA,i / Li )
Lo=1, [ft]
di+1= di *
( Hb,i / Li )
tanh (k*d) =
as shown. [pause] As the actual wave height approaches the breaking wave height, ---
H/Ho = ‘
α = o
KR =0.9387

58. Find: αb = 0.06150 < Lo=1,151.645 [ft] d 0.05280 0.05338 L Ha Hb
L= [ft]
< L L
d= [ft]
( HA,i / Li )
Lo=1, [ft]
di+1= di *
( Hb,i / Li )
tanh (k*d) =
the actual approach angle converges on the approach angle for the breaking wave, alpha b. [pause] The approach angle for the first 3 iterations are, ---
H/Ho = ‘
α = o
KR =0.9387

59. Find: αb = 0.06150 < Lo=1,151.645 [ft] d 0.05280 0.05338 L Ha Hb
L= [ft]
< L L
d= [ft]
( HA,i / Li )
Lo=1, [ft]
di+1= di *
( Hb,i / Li )
tanh (k*d) =
9.319 degrees, degrees, and degrees. Where degrees ---
α1 = 9.319 o
H/Ho = ‘
α2 = o
α = o
α3 = o
KR =0.9387

60. Find: αb = 0.06150 < Lo=1,151.645 [ft] d 0.05280 0.05338 L Ha Hb
L= [ft]
< L L
d= [ft]
( HA,i / Li )
Lo=1, [ft]
di+1= di *
( Hb,i / Li )
tanh (k*d) =
is the most accurate approximation for alpha b, because this value in our algorithm converges to alpha b. [pause]
α1 = 9.319 o
H/Ho = ‘
α2 = o
α = o
α3 = o
KR =0.9387

61. Find: αb = 0.06150 Lo=1,151.645 [ft] d A) 8.9 B) 10.5 L C) 12.1
L= [ft] o
d= [ft]
Lo=1, [ft]
tanh (k*d) =
When reviewing the possible solutions, ---
α1 = 9.319 o
H/Ho = ‘
α2 = o
α = o
α3 = o
KR =0.9387

62. Find: αb = 0.06150 Lo=1,151.645 [ft] d A) 8.9 B) 10.5 L C) 12.1
L= [ft] o
d= [ft]
Lo=1, [ft]
answerB
tanh (k*d) =
the answer is B. [fin]
α1 = 9.319 o
H/Ho = ‘
α2 = o
α = o
α3 = o
KR =0.9387