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]

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]

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]

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]

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

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

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

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

? 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

? 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

? 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

? Find: db [ft] = Ho’ db T = 13 [s] Ho’ = 10 [ft] breaking d1=13.000 [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

? Find: db [ft] = Ho’ db T = 13 [s] Ho’ = 10 [ft] breaking d1=13.000 [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

? Find: db [ft] = Ho’ db T = 13 [s] Ho’ = 10 [ft] breaking d1=13.000 [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

? Find: db [ft] = Ho’ db T = 13 [s] Ho’ = 10 [ft] breaking d1=13.000 [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

? ? Find: db [ft] = Ho’ db T = 13 [s] Ho’ = 10 [ft] breaking d1=13.000 [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 ?

? Find: db [ft] = Ho’ db T = 13 [s] Ho’ = 10 [ft] g=32.16 [ft/s2] breaking d1=13.000 [ft] height HA Hb L knot equals, 865.013 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 = 865.013[ft]

? Find: db [ft] = =0.01503 Ho’ db T = 13 [s] Ho’ = 10 [ft] g=32.16 [ft/s2] breaking d1=13.000 [ft] height HA Hb equal to, 0.01503. With the d over L knot value, --- = g * T2 d1 L L Lo = =0.01503 2 * π Lo wavelength Lo = 865.013[ft]

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 0.04964 =0.01503 Lo 0.016 0.05132

Find: db [ft] +0.03 * 0.05132 =0.01503 Ho’ db d1/L = 0.97 * 0.04964 d1=13.000 [ft] +0.03 * 0.05132 Ho’ = 10 [ft] to calculate the values of d1 over L, which equals, --- d/Lo d/L d1 0.015 0.04964 =0.01503 Lo 0.016 0.05132

Find: db [ft] +0.03 * 0.05132 =0.01503 Ho’ db d1/L = 0.97 * 0.04964 d1=13.000 [ft] +0.03 * 0.05132 Ho’ = 10 [ft] d1/L = 0.04969 0.04969. [pause] Next, we’ll compute the wavelength, L, --- d/Lo d/L d1 0.015 0.04964 =0.01503 Lo 0.016 0.05132

Find: db [ft] +0.03 * 0.05132 =0.01503 Ho’ db d1/L = 0.97 * 0.04964 d1=13.000 [ft] +0.03 * 0.05132 Ho’ = 10 [ft] d1/L = 0.04969 by dividing 13 feet, by, the value of ---- d1 d1 L= =0.01503 (d1/L) Lo

Find: db [ft] +0.03 * 0.05132 =0.01503 Ho’ db d1/L = 0.97 * 0.04964 d1=13.000 [ft] +0.03 * 0.05132 Ho’ = 10 [ft] d1/L = 0.04969 0.04969, and the wavelength for this first iteration, equals, --- d1 d1 L= =0.01503 (d1/L) Lo

Find: db [ft] +0.03 * 0.05132 = 261.622 [ft] =0.01503 Ho’ db d1/L = 0.97 * 0.04964 d1=13.000 [ft] +0.03 * 0.05132 Ho’ = 10 [ft] d1/L = 0.04969 261.622 feet. [pause] Since this wavelength depends on the depth, we’ll add a subscript, --- d1 d1 L= = 261.622 [ft] =0.01503 (d1/L) Lo

Find: db [ft] +0.03 * 0.05132 = 261.622 [ft] =0.01503 Ho’ db d1/L = 0.97 * 0.04964 d1=13.000 [ft] +0.03 * 0.05132 Ho’ = 10 [ft] d1/L = 0.04969 1, to signify the first iteration. [pause] Next, we’ll return to the tables, and calculate, --- d1 d1 L1= = 261.622 [ft] =0.01503 (d1/L) Lo

Find: db [ft] =0.01503 Ho’ db d1=13.000 [ft] Ho’ = 10 [ft] L1=261.622 [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 =0.01503 Lo 0.016 0.3117 1.288

Find: db [ft] =0.01503 Ho’ db d1=13.000 [ft] shoaling Ho’ = 10 [ft] coefficient L1=261.622 [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 =0.01503 Lo 0.016 0.3117 1.288

Find: db [ft] +0.03 * 0.3117 =0.01503 d1=13.000 [ft] Ho’ = 10 [ft] db L1=261.622 [ft] tanh1(k*d)= 0.97 * 0.3022 +0.03 * 0.3117 tanh k d for a depth of 13 feet, equals, --- d/Lo tanh(k*d) HA/Ho’ d1 0.015 0.3022 1.307 =0.01503 Lo 0.016 0.3117 1.288

Find: db [ft] +0.03 * 0.3117 =0.01503 d1=13.000 [ft] Ho’ = 10 [ft] db L1=261.622 [ft] tanh1(k*d)= 0.97 * 0.3022 +0.03 * 0.3117 tanh1(k*d)= 0.3025 0.3025. [pause] Similarly, we can compute, --- d/Lo tanh(k*d) HA/Ho’ d1 0.015 0.3022 1.307 =0.01503 Lo 0.016 0.3117 1.288

Find: db [ft] +0.03 * 1.288 =0.01503 d1=13.000 [ft] Ho’ = 10 [ft] db L1=261.622 [ft] tanh1(k*d)= 0.3025 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 =0.01503 Lo 0.016 0.3117 1.288

Find: db [ft] = 1.3064 +0.03 * 1.288 =0.01503 d1=13.000 [ft] Ho’ = 10 [ft] db L1=261.622 [ft] tanh1(k*d)= 0.3025 HA,1/Ho’= 0.97 * 1.307 +0.03 * 1.288 = 1.3064 1.3064. [pause] Keep in mind, we want to compare --- d/Lo tanh(k*d) HA/Ho’ d1 0.015 0.3022 1.307 =0.01503 Lo 0.016 0.3117 1.288

Find: db [ft] =0.01503 HA d1=13.000 [ft] L Ho’ = 10 [ft] L1=261.622 [ft] tanh1(k*d)= 0.3025 HA,1/Ho’= 1.3064 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 =0.01503 Lo 0.016 0.3117 1.288

Find: db [ft] =0.01503 HA Hb d1=13.000 [ft] L L Ho’ = 10 [ft] L1=261.622 [ft] tanh1(k*d)= 0.3025 HA,1/Ho’= 1.3064 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 =0.01503 Lo 0.016 0.3117 1.288

Find: db [ft] = =0.01503 HA,1 HA,1 Ho’ d1=13.000 [ft] * L1 Ho’ L1 Ho’ = 10 [ft] L1=261.622 [ft] tanh1(k*d)= 0.3025 HA,1/Ho’= 1.3064 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 =0.01503 Lo 0.016 0.3117 1.288

Find: db [ft] = = 0.04993 =0.01503 HA,1 HA,1 Ho’ d1=13.000 [ft] * L1 Ho’ = 10 [ft] L1=261.622 [ft] HA,1 tanh1(k*d)= 0.3025 = 0.04993 L1 HA,1/Ho’= 1.3064 0.04993. [pause] Next, we can calculate H B 1, --- d/Lo tanh(k*d) HA/Ho’ d1 0.015 0.3022 1.307 =0.01503 Lo 0.016 0.3117 1.288

Find: db [ft] = 0.142 * tanh1 (k*d) = 0.04993 =0.01503 Hb,1 d1=13.000 [ft] = 0.142 * tanh1 (k*d) L1 Ho’ = 10 [ft] L1=261.622 [ft] tanh1(k*d)= 0.3025 HA,1/Ho’= 1.3064 HA,1 divided by L 1, by plugging in tanh 1 K D, we get, --- = 0.04993 L1 d/Lo tanh(k*d) HA/Ho’ d1 0.015 0.3022 1.307 =0.01503 Lo 0.016 0.3117 1.288

Find: db [ft] = 0.142 * tanh1 (k*d) = 0.04296 = 0.04993 =0.01503 Hb,1 d1=13.000 [ft] = 0.142 * tanh1 (k*d) L1 Ho’ = 10 [ft] L1=261.622 [ft] Hb,1 = 0.04296 tanh1(k*d)= 0.3025 L1 HA,1/Ho’= 1.3064 HA,1 0.04296. [pause] Now we can compare H A over L, --- = 0.04993 L1 d/Lo tanh(k*d) HA/Ho’ d1 0.015 0.3022 1.307 =0.01503 Lo 0.016 0.3117 1.288

Find: db [ft] = 0.142 * tanh1 (k*d) = 0.04296 = 0.04993 =0.01503 Hb,1 d1=13.000 [ft] = 0.142 * tanh1 (k*d) L1 Ho’ = 10 [ft] L1=261.622 [ft] Hb,1 = 0.04296 tanh1(k*d)= 0.3025 L1 HA,1/Ho’= 1.3064 HA,1 and H b over L, for this first iteration. [pause] If H A over L is less than, --- = 0.04993 L1 d/Lo tanh(k*d) HA/Ho’ d1 0.015 0.3022 1.307 =0.01503 Lo 0.016 0.3117 1.288

< Find: db [ft] = 0.04993 = 0.04296 HA Hb d1=13.000 [ft] if L L Ho’ = 10 [ft] L1=261.622 [ft] tanh1(k*d)= 0.3025 HA,1/Ho’= 1.3064 HA,1 and H b over L, then wave has not yet broken, --- = 0.04993 L1 tanh(k*d) HA/Ho’ Hb,1 0.3022 1.307 = 0.04296 L1 0.3117 1.288

< Find: db [ft] = 0.04993 = 0.04296 HA Hb d1=13.000 [ft] if L L Ho’ = 10 [ft] then d > db L1=261.622 [ft] tanh1(k*d)= 0.3025 HA,1/Ho’= 1.3064 HA,1 and the breaking depth is less than the depth we used for that iteration. In this case, --- = 0.04993 L1 tanh(k*d) HA/Ho’ Hb,1 0.3022 1.307 = 0.04296 L1 0.3117 1.288

< Find: db [ft] = 0.04993 = 0.04296 HA Hb d1=13.000 [ft] if L L Ho’ = 10 [ft] then d > db L1=261.622 [ft] use smaller d tanh1(k*d)= 0.3025 HA,1/Ho’= 1.3064 HA,1 we should use a smaller depth, d, in the following iteration. [pause] If H A over L is greater than, --- = 0.04993 L1 tanh(k*d) HA/Ho’ Hb,1 0.3022 1.307 = 0.04296 L1 0.3117 1.288

< > Find: db [ft] = 0.04993 = 0.04296 HA Hb d1=13.000 [ft] if L Ho’ = 10 [ft] then d > db L1=261.622 [ft] use smaller d tanh1(k*d)= 0.3025 HA Hb > HA,1/Ho’= 1.3064 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, --- = 0.04993 L1 tanh(k*d) HA/Ho’ Hb,1 0.3022 1.307 = 0.04296 L1 0.3117 1.288

< > Find: db [ft] = 0.04993 = 0.04296 HA Hb d1=13.000 [ft] if L Ho’ = 10 [ft] then d > db L1=261.622 [ft] use smaller d tanh1(k*d)= 0.3025 HA Hb > HA,1/Ho’= 1.3064 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, --- = 0.04993 L1 use larger d Hb,1 = 0.04296 L1

< > = Find: db [ft] = 0.04993 = 0.04296 HA Hb d1=13.000 [ft] if L L Ho’ = 10 [ft] then d > db L1=261.622 [ft] use smaller d tanh1(k*d)= 0.3025 HA Hb > HA,1/Ho’= 1.3064 if L L HA,1 then d < db then the depth, d, equals, --- = 0.04993 L1 use larger d Hb,1 HA Hb = 0.04296 = if L1 L L

< > = Find: db [ft] = 0.04993 = 0.04296 HA Hb d1=13.000 [ft] if L L Ho’ = 10 [ft] then d > db L1=261.622 [ft] use smaller d tanh1(k*d)= 0.3025 HA Hb > HA,1/Ho’= 1.3064 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 --- = 0.04993 L1 use larger d Hb,1 HA Hb = 0.04296 = if then d = db L1 L L

< > = Find: db [ft] = 0.04993 = 0.04296 HA Hb d1=13.000 [ft] if L L Ho’ = 10 [ft] then d > db L1=261.622 [ft] use smaller d tanh1(k*d)= 0.3025 HA Hb > HA,1/Ho’= 1.3064 if L L HA,1 then d < db H A over L greater than H b over L, therefore, ---- = 0.04993 L1 use larger d Hb,1 HA Hb = 0.04296 = if then d = db L1 L L

< > = Find: db [ft] = 0.04993 = 0.04296 HA Hb d1=13.000 [ft] if L L Ho’ = 10 [ft] then d > db L1=261.622 [ft] use smaller d tanh1(k*d)= 0.3025 HA Hb > HA,1/Ho’= 1.3064 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 --- = 0.04993 L1 use larger d Hb,1 HA Hb = 0.04296 = if then d = db L1 L L

> Find: db [ft] = 0.04993 = 0.04296 HA Hb d1=13.000 [ft] if L L Ho’ = 10 [ft] then d < db L1=261.622 [ft] use larger d tanh1(k*d)= 0.3025 HA,1/Ho’= 1.3064 ( 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, --- = 0.04993 L1 Hb,1 = 0.04296 L1

> Find: db [ft] = 0.04993 = 0.04296 HA Hb d1=13.000 [ft] if L L Ho’ = 10 [ft] then d < db L1=261.622 [ft] use larger d tanh1(k*d)= 0.3025 HA,1/Ho’= 1.3064 ( HA,i / Li ) di+1= di * HA,1 ( Hb,i / Li ) we plug in the appropriate variables, and our next depth, d 2, equals, --- = 0.04993 L1 Hb,1 = 0.04296 L1

> Find: db [ft] = 0.04993 = 0.04296 HA Hb d1=13.000 [ft] if L L Ho’ = 10 [ft] then d < db L1=261.622 [ft] use larger d tanh1(k*d)= 0.3025 HA,1/Ho’= 1.3064 ( HA,i / Li ) di+1= di * HA,1 ( Hb,i / Li ) 15.109 feet. [pause] Next, we’ll begin the second iteration, --- = 0.04993 L1 d2= 15.109 [ft] Hb,1 = 0.04296 L1

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

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

Find: db [ft] = ? = ? d2 d2=15.109 [ft] Lo Ho’ = 10 [ft] L2=? Lo = 865.013[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

Find: db [ft] = 0.01747 = ? = ? d2 d2=15.109 [ft] Lo Ho’ = 10 [ft] Lo = 865.013[ft] tanh2(k*d)= ? HA,2/Ho’= ? HA,2 0.01747. [pause] Next we’ll calculate the value for --- = ? L1 Hb,2 = ? L1

Find: db [ft] = 0.01747 =0.01747 d2 d2=15.109 [ft] Lo Ho’ = 10 [ft] Lo = 865.013[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 0.05296 =0.01747 Lo 0.018 0.05455

Find: db [ft] = 0.01747 +0.47 * 0.05455 =0.01747 d2 d2=15.109 [ft] Lo Ho’ = 10 [ft] L2=? Lo = 865.013[ft] tanh2(k*d)= ? d2/L = 0.53 * 0.05296 HA,2/Ho’= ? +0.47 * 0.05455 0.017 and 0.018, for d over L knot, we get d 2 over L, equals, --- d/Lo d/L d2 0.017 0.05296 =0.01747 Lo 0.018 0.05455

Find: db [ft] = 0.01747 +0.47 * 0.05455 =0.01747 d2 d2=15.109 [ft] Lo Ho’ = 10 [ft] L2=? Lo = 865.013[ft] tanh2(k*d)= ? d2/L = 0.53 * 0.05296 HA,2/Ho’= ? +0.47 * 0.05455 d2/L = 0.05370 0.05370. [pause] Again, we’ll add the subscript 2, --- d/Lo d/L d2 0.017 0.05296 =0.01747 Lo 0.018 0.05455

Find: db [ft] = 0.01747 +0.47 * 0.05455 =0.01747 d2 d2=15.109 [ft] Lo Ho’ = 10 [ft] L2=? Lo = 865.013[ft] tanh2(k*d)= ? d2/L = 0.53 * 0.05296 HA,2/Ho’= ? +0.47 * 0.05455 d2/L2 = 0.05370 since we’re on the second iteration. [pause] And to compute L 2, ---- d/Lo d/L d2 0.017 0.05296 =0.01747 Lo 0.018 0.05455

Find: db [ft] = 0.01747 +0.47 * 0.05455 d2 d2=15.109 [ft] Lo Ho’ = 10 [ft] L2=? Lo = 865.013[ft] tanh2(k*d)= ? d2/L = 0.53 * 0.05296 HA,2/Ho’= ? +0.47 * 0.05455 d2/L2 = 0.05370 we divide 15.109 feet by 0.05370, and we get, --- d2 L2= (d2/L2)

Find: db [ft] = 0.01747 +0.47 * 0.05455 = 281.359 [ft] d2 d2=15.109 [ft] = 0.01747 Lo Ho’ = 10 [ft] L2= 281.359 [ft] Lo = 865.013[ft] tanh2(k*d)= ? d2/L = 0.53 * 0.05296 HA,2/Ho’= ? +0.47 * 0.05455 d2/L2 = 0.05370 281.359 feet. [pause] Next we’ll compute the hyperbolic tangent of k d, and, --- d2 L2= = 281.359 [ft] (d2/L2)

Find: db [ft] = 0.01747 +0.47 * 0.05455 =0.01747 d2 d2=15.109 [ft] Lo Ho’ = 10 [ft] L2= 281.359 [ft] Lo = 865.013[ft] tanh2(k*d)= ? d2/L = 0.53 * 0.05296 HA,2/Ho’= ? +0.47 * 0.05455 d2/L2 = 0.05370 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 =0.01747 Lo 0.018 0.3298 1.255

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= 281.359 [ft] tanh2(k*d)= 0.53 * 0.3209 +0.47* 0.3298 interpolation again. [pause] And tanh 2 k d equals, --- d/Lo tanh(k*d) HA/Ho’ d2 0.017 0.3209 1.271 =0.01747 Lo 0.018 0.3298 1.255

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= 281.359 [ft] tanh2(k*d)= 0.53 * 0.3209 +0.47* 0.3298 tanh2(k*d)= 0.3251 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 =0.01747 Lo 0.018 0.3298 1.255

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= 281.359 [ft] HA,2/Ho’ = 1.2635 tanh2(k*d)= 0.53 * 0.3209 +0.47* 0.3298 tanh2(k*d)= 0.3251 1.2635. [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 =0.01747 Lo 0.018 0.3298 1.255

Find: db [ft] = =0.01747 HA,2 HA,2 Ho’ d2=15.109 [ft] * L2 Ho’ L2 Ho’ = 10 [ft] L2= 281.359 [ft] tanh2(k*d)= 0.3251 HA,2/Ho’= 1.2635 and after plugging in the known variables, we get, --- d/Lo tanh(k*d) HA/Ho’ d2 0.017 0.3209 1.271 =0.01747 Lo 0.018 0.3298 1.255

Find: db [ft] = = 0.04491 =0.01747 HA,2 HA,2 Ho’ d2=15.109 [ft] * L2 Ho’ = 10 [ft] L2= 281.359 [ft] HA,2 tanh2(k*d)= 0.3251 = 0.04491 L2 HA,2/Ho’= 1.2635 0.04491. [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 =0.01747 Lo 0.018 0.3298 1.255

Find: db [ft] = 0.142 * tanh2 (k*d) = 0.04491 =0.01747 Hb,2 d2=15.109 [ft] = 0.142 * tanh2 (k*d) L2 Ho’ = 10 [ft] L2= 281.359 [ft] tanh2(k*d)= 0.3251 HA,2/Ho’= 1.2635 HA,2 we’ll plug in the hyperbolic tangent of k d, and compute, --- = 0.04491 L2 d/Lo tanh(k*d) HA/Ho’ d2 0.017 0.3209 1.271 =0.01747 Lo 0.018 0.3298 1.255

Find: db [ft] = 0.142 * tanh2 (k*d) = 0.04616 = 0.04491 =0.01747 Hb,2 d2=15.109 [ft] = 0.142 * tanh2 (k*d) L2 Ho’ = 10 [ft] L2= 281.359 [ft] Hb,2 = 0.04616 tanh2(k*d)= 0.3251 L2 HA,2/Ho’= 1.2635 HA,2 0.04616. [pause] As before, we’ll compare --- = 0.04491 L2 d/Lo tanh(k*d) HA/Ho’ d2 0.017 0.3209 1.271 =0.01747 Lo 0.018 0.3298 1.255

Find: db [ft] = 0.142 * tanh2 (k*d) = 0.04616 = 0.04491 =0.01747 Hb,2 d2=15.109 [ft] = 0.142 * tanh2 (k*d) L2 Ho’ = 10 [ft] L2= 281.359 [ft] Hb,2 = 0.04616 tanh2(k*d)= 0.3251 L2 HA,2/Ho’= 1.2635 HA,2 the computed height of the wave and the breaking height of the wave, --- = 0.04491 L2 d/Lo tanh(k*d) HA/Ho’ d2 0.017 0.3209 1.271 =0.01747 Lo 0.018 0.3298 1.255

< > Find: db [ft] = 0.04491 = 0.04616 HA Hb d2=15.109 [ft] if L Ho’ = 10 [ft] then d > db L2= 281.359 [ft] use smaller d tanh2(k*d)= 0.3251 HA Hb > HA,2/Ho’= 1.2635 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, --- = 0.04491 L2 use larger d Hb,2 = 0.04616 L2

< > Find: db [ft] = 0.04491 = 0.04616 HA Hb d2=15.109 [ft] if L Ho’ = 10 [ft] then d > db L2= 281.359 [ft] use smaller d tanh2(k*d)= 0.3251 HA Hb > HA,2/Ho’= 1.2635 if L L HA,2 then d < db and we should test a smaller depth d. For the next iteration, --- = 0.04491 L2 use larger d Hb,2 = 0.04616 L2

< Find: db [ft] = 0.04491 = 0.04616 HA Hb d2=15.109 [ft] if L L Ho’ = 10 [ft] then d > db L2= 281.359 [ft] use smaller d tanh2(k*d)= 0.3251 HA,2/Ho’= 1.2635 ( HA,i / Li ) di+1= di * ( Hb,i / Li ) HA,2 we’ll compute the next depth to evaluate, d 3, which equals, --- = 0.04491 L2 Hb,2 = 0.04616 L2

< Find: db [ft] = 0.04491 = 0.04616 HA Hb d2=15.109 [ft] if L L Ho’ = 10 [ft] then d > db L2= 281.359 [ft] use smaller d tanh2(k*d)= 0.3251 HA,2/Ho’= 1.2635 ( HA,i / Li ) di+1= di * ( Hb,i / Li ) HA,2 14.700 feet. [pause] If we went through another entire iteration, our fourth depth, d 4, --- = 0.04491 L2 d3= 14.700 [ft] Hb,2 = 0.04616 L2

> Find: db [ft] = 0.04578 = 0.04555 HA Hb d3=14.700 [ft] if L L Ho’ = 10 [ft] then d < db L3= 277.673 [ft] use larger d tanh3(k*d)= 0.3208 HA,3/Ho’= 1.2712 ( HA,i / Li ) di+1= di * ( Hb,i / Li ) HA,3 would equal 14.774 feet. [pause] Next, if we look at our successive approximations --- = 0.04578 L3 d4= 14.774 [ft] Hb,3 = 0.04555 L3

> Find: db [ft] HA Hb if L L then d < db d1=13.000 [ft] use larger d d2= 15.100 [ft] d3=14.700 [ft] ( HA,i / Li ) di+1= di * d4= 14.774 [ft] ( Hb,i / Li ) for the breaking depth, d b, we notice the depth is converging on a value close to ---

> Find: db [ft] HA Hb if L L then d < db d1=13.000 [ft] use larger d d2= 15.100 [ft] d3=14.700 [ft] ( HA,i / Li ) di+1= di * d4= 14.774 [ft] ( Hb,i / Li ) 14.7 something feet. [pause] 14.7 [ft] < db < 14.8 [ft]

> Find: db [ft] HA Hb if L L then d < db d1=13.000 [ft] use larger d d2= 15.100 [ft] d3=14.700 [ft] A) 7.75 B) 11.25 C) 14.75 D) 18.25 d4= 14.774 [ft] When reviewing the possible solutions, --- 14.7 [ft] < db < 14.8 [ft]

> Find: db [ft] HA Hb if L L then d < db d1=13.000 [ft] use larger d d2= 15.100 [ft] d3=14.700 [ft] A) 7.75 B) 11.25 C) 14.75 D) 18.25 d4= 14.774 [ft] the answer is C. [fin] 14.7 [ft] < db < 14.8 [ft] answerC