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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Find: KR,b = wave direction db = 15 [ft] contours T = 14 [s] beach g * T2 Lo = g = 32.16 [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

Find: KR,b = wave direction db = 15 [ft] contours T = 14 [s] beach g * T2 Lo = g = 32.16 [ft/s2] 2 * π cos (αo) KR,b = cos (αb) Lo =1,003.211 [ft] 1,003.211 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

Find: KR,b wave direction db = 15 [ft] contours T = 14 [s] beach g * T2 Lo = db g = 32.16 [ft/s2] 2 * π Lb= (db/Lb) Lo =1,003.211 [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

Find: KR,b wave direction db = 15 [ft] contours T = 14 [s] beach g * T2 Lo = db g = 32.16 [ft/s2] 2 * π Lb= (db/Lb) Lo =1,003.211 [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

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 =1,003.211 [ft] by knowing the value of d b over L knot, which equals, --- o 60 Lb αb = sin-1 sin(αo) * Lo

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 = 0.01495 Lb= Lo (db/Lb) Lo =1,003.211 [ft] 0.01495. [pause] Next, we’ll enter the wave tables between --- o 60 Lb αb = sin-1 sin(αo) * Lo

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 = 0.01495 Lb= Lo (db/Lb) Lo =1,003.211 [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

Find: KR,b d/Lo d/L 0.014 0.04791 db = 15 [ft] 0.015 0.04964 T = 14 [s] db/Lb = 0.05 * 0.04791 o αo = 60 +0.95 * 0.04964 db db = 0.01495 Lb= Lo (db/Lb) Lo =1,003.211 [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

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 * 0.04791 o αo = 60 +0.95 * 0.04964 db db/Lb= 0.04955 Lb= (db/Lb) Lo =1,003.211 [ft] and it equals, 0.04955. [pause] This makes, the wavelength at breaking, equal to, --- o 60 Lb αb = sin-1 sin(αo) * Lo

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 * 0.04791 o αo = 60 +0.95 * 0.04964 db db/Lb= 0.04955 Lb= (db/Lb) Lo =1,003.211 [ft] 302.725 feet. [pause] Now we can solve for the approach angle at breaking, alpha b, -- Lb= 302.725 [ft] o 60 Lb αb = sin-1 sin(αo) * Lo

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 * 0.04791 o αo = 60 +0.95 * 0.04964 db db/Lb= 0.04955 Lb= (db/Lb) Lo =1,003.211 [ft] and after plugging in the appropriate variables, alpha b, equals, --- Lb= 302.725 [ft] o 60 Lb αb = sin-1 sin(αo) * Lo

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 * 0.04791 o αo = 60 +0.95 * 0.04964 db db/Lb= 0.04955 Lb= (db/Lb) Lo =1,003.211 [ft] 15.149 degrees. [pause] At this point, we can return to our original equation for --- Lb= 302.725 [ft] o 60 Lb αb = sin-1 αb = 15.149 o sin(αo) * Lo

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 * 0.04791 o αo = 60 +0.95 * 0.04964 db/Lb= 0.04955 cos (αo) KR,b = cos (αb) Lo =1,003.211 [ft] the refreaction coefficient at the breaking depth. After plugging in the appraoch angles, --- o 60 Lb αb = sin-1 αb = 15.149 o sin(αo) * Lo

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 * 0.04791 o αo = 60 +0.95 * 0.04964 db/Lb= 0.04955 cos (αo) KR,b = cos (αb) Lo =1,003.211 [ft] associated with a deepwater wave and the breaking wave condition, --- o 60 Lb αb = sin-1 αb = 15.149 o sin(αo) * Lo

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 * 0.04791 o αo = 60 +0.95 * 0.04964 db/Lb= 0.04955 cos (αo) KR,b = cos (αb) Lo =1,003.211 [ft] K r b, equals, 0.7197. [pause] o KR,b =0.7197 60 Lb αb = sin-1 αb = 15.149 o sin(αo) * Lo

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 * 0.04791 o αo = 60 +0.95 * 0.04964 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 = 15.149 o

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 * 0.04791 o αo = 60 +0.95 * 0.04964 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 = 15.149 o