1. Find: KR,b wave direction αo plan contours view beach db = 15 [ft]
Find the refraction coefficient at the breaking depth, K r b. [pause] In this problem, ---
T = 14 [s] o
αo = 60

2. Find: KR,b wave direction αo plan contours view beach db = 15 [ft]
we are provided with the breakign depth, d b, as well as the period of the wave, ---
T = 14 [s] o
αo = 60

3. Find: KR,b wave direction αo plan contours view beach db = 15 [ft]
T, and the deepwater approach angle, ---
T = 14 [s] o
αo = 60

4. Find: KR,b wave direction αo plan contours view beach db = 15 [ft]
alpha knot. Alpha knot is the angle formed between the crestline of the wave, ---
T = 14 [s] o
αo = 60

5. Find: KR,b wave direction αo plan contours view beach db = 15 [ft]
and the underwater contours of the soil, for deepwater conditions. As the wave encounters these contours, ---
T = 14 [s] o
αo = 60

6. Find: KR,b wave direction plan contours view beach db = 15 [ft]
the wave will begin to turn, ---
T = 14 [s] o
αo = 60

7. Find: KR,b wave direction plan contours view beach db = 15 [ft]
in the direction of the beach, ---
T = 14 [s] o
αo = 60

8. Find: KR,b wave direction plan contours view beach db = 15 [ft]
until the wave breaks, ---
T = 14 [s] o
αo = 60

9. Find: KR,b wave direction plan contours view beach db = 15 [ft]
at a depth of d b. [pause] The sketch shown here, ---
T = 14 [s] o
αo = 60

10. Find: KR,b wave direction plan contours view beach db = 15 [ft]
is in plan view, and overly simplified. [pause] The refreaction coefficient, ---
T = 14 [s] o
αo = 60

11. Find: KR,b wave direction db = 15 [ft] contours T = 14 [s] beach
cos (αo)
KR =
cos (α)
K r, equals, the square root of the cosine of the ----
refraction
coefficient

12. Find: KR,b wave direction db = 15 [ft] contours T = 14 [s] beach
deepwater
cos (αo)
KR =
approach angle
cos (α)
deepwater approach angle, alpha knot, divided by, the cosine of the ---
refraction
coefficient

13. Find: KR,b wave direction db = 15 [ft] contours T = 14 [s] beach
deepwater
cos (αo)
KR =
approach angle
cos (α)
a generic approach angle, alpha. [pause] Since we are asked to find the refraction index ---
approach angle
(generic)
refraction
coefficient

14. Find: KR,b wave direction db = 15 [ft] contours T = 14 [s] beach
deepwater
cos (αo)
KR =
approach angle
cos (α)
for the wave when it breaks, we’ll add a subscript “b”, ----
approach angle
(generic)
refraction
coefficient

15. Find: KR,b wave direction db = 15 [ft] contours T = 14 [s] beach
deepwater
cos (αo)
KR,b =
approach angle
cos (αb)
to the refraction index and to the generic appoach angle, alpha. [pause] Now, to solve for K b r, we need to determine ---
approach angle
when wave breaks
refraction
coefficient
at breaking depth

16. Find: KR,b wave direction db = 15 [ft] contours T = 14 [s] beach
deepwater
cos (αo)
KR,b =
approach angle
cos (αb)
the values of variables alpha knot, and, alpha b. The problem provides us with the deepwater approach angle, ---
approach angle
when wave breaks
refraction
coefficient
at breaking depth

17. Find: KR,b wave direction db = 15 [ft] contours T = 14 [s] beach
deepwater
cos (αo)
KR,b =
approach angle
cos (αb)
alpha knot, which is, 60 degrees.[pause] To determine alpha b, we’ll write out ---
approach angle
when wave breaks
refraction
coefficient
at breaking depth

18. Find: KR,b = wave direction db = 15 [ft] contours T = 14 [s] beach
deepwater
cos (αo)
KR,b =
approach angle
cos (αb)
the coastal engineering version of the law of sines, and solve for alpha b, which turns out to be a function of ---
approach angle
when wave breaks
sin (αb)
sin (αo) = Lb Lo

19. Find: KR,b = wave direction db = 15 [ft] contours T = 14 [s] beach
deepwater
cos (αo)
KR,b =
approach angle
cos (αb)
alpha knot, the deepwater wavelength and the wavelength at the breaking depth. As before, ----
approach angle
when wave breaks
sin (αb)
sin (αo) Lb =
αb = sin-1
sin(αo) * Lb Lo Lo

20. Find: KR,b = wave direction db = 15 [ft] contours T = 14 [s] beach
deepwater
cos (αo)
KR,b =
approach angle
cos (αb)
we know alpha knot equals 60 degrees, so we’re left to find ---
approach angle
when wave breaks o 60
sin (αb)
sin (αo) Lb =
αb = sin-1
sin(αo) * Lb Lo Lo

21. Find: KR,b = wave direction db = 15 [ft] contours T = 14 [s] beach
deepwater
cos (αo)
KR,b =
approach angle
cos (αb)
L b, and L knot. L knot is the deepwater wavelength, ----
approach angle
when wave breaks o 60
sin (αb)
sin (αo) Lb =
αb = sin-1
sin(αo) * Lb Lo Lo

22. Find: KR,b = wave direction db = 15 [ft] contours T = 14 [s] beach
g * T2
deepwater
Lo =
2 * π
cos (αo)
KR,b =
approach angle
cos (αb)
and is a function of the period of the wave, T. After plugging in ---
approach angle
when wave breaks o 60
sin (αb)
sin (αo) Lb =
αb = sin-1
sin(αo) * Lb Lo Lo

23. Find: KR,b = wave direction db = 15 [ft] contours T = 14 [s] beach
g * T2
Lo =
g = [ft/s2]
2 * π
cos (αo)
KR,b =
cos (αb)
14 seconds for T, we compute L knot equals, ---
approach angle
when wave breaks o 60
sin (αb)
sin (αo) Lb =
αb = sin-1
sin(αo) * Lb Lo Lo

24. Find: KR,b = wave direction db = 15 [ft] contours T = 14 [s] beach
g * T2
Lo =
g = [ft/s2]
2 * π
cos (αo)
KR,b =
cos (αb)
Lo =1, [ft]
1, feet. [pause] The wavelength at the breaking depth, L b, equals, --- o 60
sin (αb)
sin (αo) Lb =
αb = sin-1
sin(αo) * Lb Lo Lo

25. Find: KR,b wave direction db = 15 [ft] contours T = 14 [s] beach
g * T2
Lo = db
g = [ft/s2]
2 * π
Lb=
(db/Lb)
Lo =1, [ft]
the breaking depth divided by d over L, at the breaking depth. [pause] From the problem statement, --- o 60 Lb
αb = sin-1
sin(αo) * Lo

26. Find: KR,b wave direction db = 15 [ft] contours T = 14 [s] beach
g * T2
Lo = db
g = [ft/s2]
2 * π
Lb=
(db/Lb)
Lo =1, [ft]
we know the breaking depth, equals, 15 feet. The term d b over L b can be determined --- o 60 Lb
αb = sin-1
sin(αo) * Lo

27. Find: KR,b wave direction db = 15 [ft] contours T = 14 [s] beach
g * T2 db
Lo = db
2 * π
Lb= Lo
(db/Lb)
Lo =1, [ft]
by knowing the value of d b over L knot, which equals, --- o 60 Lb
αb = sin-1
sin(αo) * Lo

28. Find: KR,b = 0.01495 wave direction db = 15 [ft] contours T = 14 [s]
beach o
αo = 60
g * T2 db
Lo = db
2 * π =
Lb= Lo
(db/Lb)
Lo =1, [ft]
[pause] Next, we’ll enter the wave tables between --- o 60 Lb
αb = sin-1
sin(αo) * Lo

29. Find: KR,b = 0.01495 d/Lo d/L 0.014 0.04791 db = 15 [ft] 0.015 0.04964
T = 14 [s] o
αo = 60 db db =
Lb= Lo
(db/Lb)
Lo =1, [ft]
0.014 and 0.015, and interpolate, on a line, between the cooresponding values of d over L, --- o 60 Lb
αb = sin-1
sin(αo) * Lo

30. Find: KR,b
d/Lo
d/L
0.014
db = 15 [ft]
0.015
T = 14 [s]
db/Lb = 0.05 * o
αo = 60
+0.95 * db db =
Lb= Lo
(db/Lb)
Lo =1, [ft]
which is the same as d b over L b, since we’re dealing with the breaking condition, ---- o 60 Lb
αb = sin-1
sin(αo) * Lo

31. Find: KR,b +0.95 * 0.04964 d/Lo d/L 0.014 0.04791 db = 15 [ft] 0.015
T = 14 [s]
db/Lb = 0.05 * o
αo = 60
+0.95 * db
db/Lb=
Lb=
(db/Lb)
Lo =1, [ft]
and it equals, [pause] This makes, the wavelength at breaking, equal to, --- o 60 Lb
αb = sin-1
sin(αo) * Lo

32. Find: KR,b +0.95 * 0.04964 d/Lo d/L 0.014 0.04791 db = 15 [ft] 0.015
T = 14 [s]
db/Lb = 0.05 * o
αo = 60
+0.95 * db
db/Lb=
Lb=
(db/Lb)
Lo =1, [ft]
feet. [pause] Now we can solve for the approach angle at breaking, alpha b, --
Lb= [ft] o 60 Lb
αb = sin-1
sin(αo) * Lo

33. Find: KR,b +0.95 * 0.04964 d/Lo d/L 0.014 0.04791 db = 15 [ft] 0.015
T = 14 [s]
db/Lb = 0.05 * o
αo = 60
+0.95 * db
db/Lb=
Lb=
(db/Lb)
Lo =1, [ft]
and after plugging in the appropriate variables, alpha b, equals, ---
Lb= [ft] o 60 Lb
αb = sin-1
sin(αo) * Lo

34. Find: KR,b +0.95 * 0.04964 d/Lo d/L 0.014 0.04791 db = 15 [ft] 0.015
T = 14 [s]
db/Lb = 0.05 * o
αo = 60
+0.95 * db
db/Lb=
Lb=
(db/Lb)
Lo =1, [ft]
degrees. [pause] At this point, we can return to our original equation for ---
Lb= [ft] o 60 Lb
αb = sin-1
αb = o
sin(αo) * Lo

35. Find: KR,b +0.95 * 0.04964 d/Lo d/L 0.014 0.04791 db = 15 [ft] 0.015
T = 14 [s]
db/Lb = 0.05 * o
αo = 60
+0.95 *
db/Lb=
cos (αo)
KR,b =
cos (αb)
Lo =1, [ft]
the refreaction coefficient at the breaking depth. After plugging in the appraoch angles, --- o 60 Lb
αb = sin-1
αb = o
sin(αo) * Lo

36. Find: KR,b +0.95 * 0.04964 d/Lo d/L 0.014 0.04791 db = 15 [ft] 0.015
T = 14 [s]
db/Lb = 0.05 * o
αo = 60
+0.95 *
db/Lb=
cos (αo)
KR,b =
cos (αb)
Lo =1, [ft]
associated with a deepwater wave and the breaking wave condition, --- o 60 Lb
αb = sin-1
αb = o
sin(αo) * Lo

37. Find: KR,b +0.95 * 0.04964 d/Lo d/L 0.014 0.04791 db = 15 [ft] 0.015
T = 14 [s]
db/Lb = 0.05 * o
αo = 60
+0.95 *
db/Lb=
cos (αo)
KR,b =
cos (αb)
Lo =1, [ft]
K r b, equals, [pause] o
KR,b =0.7197 60 Lb
αb = sin-1
αb = o
sin(αo) * Lo

38. Find: KR,b +0.95 * 0.04964 d/Lo d/L 0.014 0.04791 db = 15 [ft] 0.015
T = 14 [s]
db/Lb = 0.05 * o
αo = 60
+0.95 *
cos (αo)
KR,b =
A) 0.32
B) 0.52
C) 0.72
D) 0.92
cos (αb)
When reviewing the possible solutions, ---
KR,b =0.7197
αb = o

39. Find: KR,b +0.95 * 0.04964 d/Lo d/L 0.014 0.04791 db = 15 [ft] 0.015
T = 14 [s]
db/Lb = 0.05 * o
αo = 60
+0.95 *
cos (αo)
KR,b =
A) 0.32
B) 0.52
C) 0.72
D) 0.92
cos (αb)
the correct answer is C. [fin]
KR,b =0.7197
answerC
αb = o