Find: co [ft/s] ua = 20 [mi/hr] d = 2 [hr] duration F = 15 [mi] d = 65 [ft] depth plan F view A) 5.9 B) 8.6 C) 14.3 D) 24.0 Find the deepwater wave celerity, c knot, in feet per second. [pause] In this problem, we are given that ---- wind direction

Find: co [ft/s] ua = 20 [mi/hr] d = 2 [hr] duration F = 15 [mi] d = 65 [ft] depth plan F view A) 5.9 B) 8.6 C) 14.3 D) 24.0 the wind velocity in the x direction is 20 miles per hour, the length --- wind direction

Find: co [ft/s] ua = 20 [mi/hr] d = 2 [hr] duration F = 15 [mi] d = 65 [ft] depth plan F view A) 5.9 B) 8.6 C) 14.3 D) 24.0 of the fetch is 15 miles long, and the storm event lasts --- wind direction

Find: co [ft/s] ua = 20 [mi/hr] d = 2 [hr] duration F = 15 [mi] d = 65 [ft] depth plan F view A) 5.9 B) 8.6 C) 14.3 D) 24.0 for a duration of 2 hours. [pause] We are also provided --- wind direction

Find: co [ft/s] ua = 20 [mi/hr] d = 2 [hr] duration F = 15 [mi] d = 65 [ft] depth plan F view A) 5.9 B) 8.6 C) 14.3 D) 24.0 the depth of water, d. [pause] The deepwater wave celerity, c knot, equals, --- wind direction

Find: co [ft/s] ua = 20 [mi/hr] d = 2 [hr] duration F = 15 [mi] d = 65 [ft] depth Lo deepwater co = wavelength To the deepwater wavelength, L knot, divided by the deepwater wave period, T knot. [pause] The deepwater wavelength, L knot, --- deepwater period

Find: co [ft/s] ua = 20 [mi/hr] d = 2 [hr] duration F = 15 [mi] d = 65 [ft] depth Lo deepwater g * T2 co = Lo= wavelength 2 * π To equals g times T squared, divided by 2 PI. [pause] The variable T refers to the deepwater period, --- deepwater period

Find: co [ft/s] ua = 20 [mi/hr] d = 2 [hr] duration F = 15 [mi] d = 65 [ft] depth Lo deepwater g * T2 co = Lo= o wavelength 2 * π To T knot. Therefore, we can substitute in this value of L knot, --- deepwater period

Find: co [ft/s] ua = 20 [mi/hr] d = 2 [hr] duration F = 15 [mi] d = 65 [ft] depth Lo deepwater g * T2 co = Lo= o wavelength 2 * π To into our equation for the deepwater wave celerity, c knot, --- deepwater period

Find: co [ft/s] ua = 20 [mi/hr] d = 2 [hr] duration F = 15 [mi] d = 65 [ft] depth Lo deepwater g * T2 co = Lo= o wavelength 2 * π To g * To * T and after simplifying this equation, by cancelling out a T knot term, --- co = o 2 * π * To

Find: co [ft/s] ua = 20 [mi/hr] d = 2 [hr] duration F = 15 [mi] d = 65 [ft] depth Lo deepwater g * T2 co = Lo= o wavelength 2 * π To g * To * T the deepwater wave celerity, c knot, equals, --- co = o 2 * π * To

Find: co [ft/s] ua = 20 [mi/hr] d = 2 [hr] duration F = 15 [mi] d = 65 [ft] depth Lo deepwater g * T2 co = Lo= o wavelength 2 * π To g * To * T g, T knot, over 2 PI. [pause] Variable g, represents, ---- co = o 2 * π * To g * To co = 2 * π

Find: co [ft/s] ua = 20 [mi/hr] d = 2 [hr] duration F = 15 [mi] d = 65 [ft] depth Lo deepwater g * T2 co = Lo= o wavelength 2 * π To gravitational accelration the constant of gravitational acceleration and equals 32.16, --- g * To co = 2 * π

Find: co [ft/s] ua = 20 [mi/hr] d = 2 [hr] duration F = 15 [mi] d = 65 [ft] depth Lo deepwater g * T2 co = Lo= o wavelength 2 * π To gravitational accelration feet per second squared. [pause] We can determine the deepwater wave period, --- 32.16 [ft/s2] g * To co = 2 * π

Find: co [ft/s] ua = 20 [mi/hr] d = 2 [hr] duration F = 15 [mi] d = 65 [ft] depth Lo deepwater g * T2 co = Lo= o wavelength 2 * π To gravitational accelration T knot, by knowing the storm duration, wind speed, --- 32.16 [ft/s2] deepwater g * To period co = 2 * π

Find: co [ft/s] ua = 20 [mi/hr] d = 2 [hr] duration F = 15 [mi] d = 65 [ft] depth Lo deepwater g * T2 co = Lo= o wavelength 2 * π To gravitational accelration and fetch length of the storm. Of the three possible types of storms, --- 32.16 [ft/s2] deepwater g * To period co = 2 * π

Find: co [ft/s] ua = 20 [mi/hr] storm types: fetch-limited? d = 2 [hr] duration-limited? F = 15 [mi] fully developed? 32.16 [ft/s2] g * To we’ll first check to see if the storm is limited by it’s fetch length. If it is not limited by fetch length, --- co = 2 * π

Find: co [ft/s] ua = 20 [mi/hr] storm types: fetch-limited? d = 2 [hr] duration-limited? F = 15 [mi] fully developed? 32.16 [ft/s2] g * To we’ll check to see if the storm limited by its duration. If the storm is not limted by fetch or duration, --- co = 2 * π

Find: co [ft/s] ua = 20 [mi/hr] storm types: fetch-limited? d = 2 [hr] duration-limited? F = 15 [mi] fully developed? 32.16 [ft/s2] g * To it is said to be, a fully developed storm, or, a fully arisen sea. To check for a fetch-limited storm, --- co = 2 * π

Find: co [ft/s] ua = 20 [mi/hr] storm types: fetch-limited? d = 2 [hr] duration-limited? F = 15 [mi] fully developed? 32.16 [ft/s2] g * To we’ll bust out the deepwater wave prediction nomogram for English units, and find the intersection for the lines --- ua co = 2 * π nomogram F

Find: co [ft/s] ua = 20 [mi/hr] storm types: fetch-limited? d = 2 [hr] duration-limited? F = 15 [mi] fully developed? 32.16 [ft/s2] g * To corresponding to a wind speed of, 20 miles per hour wind speeds, and a storm duration of 2 hours. [pause] Tracing downward from this intersection, --- ua co = 2 * π nomogram F

Find: co [ft/s] ua = 20 [mi/hr] storm types: fetch-limited? d = 2 [hr] duration-limited? F = 15 [mi] fully developed? 32.16 [ft/s2] g * To we determine the storm requires a fetch at least 6.4 miles long to fully develop. Since the storm fetch is --- ua co = 2 * π nomogram 6.4 [mi] F

Find: co [ft/s] ua = 20 [mi/hr] storm types: fetch-limited? d = 2 [hr] duration-limited? F = 15 [mi] fully developed? 32.16 [ft/s2] g * To 15 miles long, this storm is not fetch-limited, --- ua co = 2 * π nomogram 6.4 [mi] F

Find: co [ft/s] ua = 20 [mi/hr] storm types: fetch-limited? d = 2 [hr] duration-limited? F = 15 [mi] fully developed? 32.16 [ft/s2] g * To [pause] Next we’ll check to see if the storm is duration limited, --- ua co = 2 * π nomogram 6.4 [mi] F

Find: co [ft/s] ua = 20 [mi/hr] storm types: fetch-limited? d = 2 [hr] duration-limited? F = 15 [mi] fully developed? 32.16 [ft/s2] g * To by identifying the intersection of the lines cooresponding to --- ua co = 2 * π nomogram F

Find: co [ft/s] ua = 20 [mi/hr] storm types: fetch-limited? d = 2 [hr] duration-limited? F = 15 [mi] fully developed? 32.16 [ft/s2] g * To a 20 mile per hour windspeed, and a 15 mile long fetch. This point corresponds to --- ua co = 2 * π F

Find: co [ft/s] ua = 20 [mi/hr] storm types: fetch-limited? d = 2 [hr] duration-limited? F = 15 [mi] fully developed? 32.16 [ft/s2] g * To a duration of 3.6 hours. Since the storm only lasts, ---- ua co = 2 * π d(u ,F) = 3.6 [hr] a F

Find: co [ft/s] ua = 20 [mi/hr] storm types: fetch-limited? d = 2 [hr] duration-limited? F = 15 [mi] fully developed? 32.16 [ft/s2] g * To 2 hours long, this is a duration-limited storm, --- ua co = 2 * π d(u ,F) = 3.6 [hr] a F

Find: co [ft/s] ua = 20 [mi/hr] storm types: fetch-limited? d = 2 [hr] duration-limited! F = 15 [mi] fully developed? 32.16 [ft/s2] g * To and we must use a duration of 2 hours when determining the maximum wave characterisitics. When using --- ua co = 2 * π d(u ,F) = 3.6 [hr] a F

Find: co [ft/s] ua = 20 [mi/hr] storm types: fetch-limited? d = 2 [hr] duration-limited! F = 15 [mi] fully developed? 32.16 [ft/s2] g * To a 20 mile per hour wind speed, and 2 hour storm duration, --- ua co = 2 * π F

Find: co [ft/s] storm types: ua = 20 [mi/hr] duration-limited! d = 2 [hr] T(u ,d) = 2.8 [s] F = 15 [mi] a 32.16 [ft/s2] g * To the wave period equals, 2.8 seconds. [pause] Plugging this value in, --- ua co = 2 * π F

Find: co [ft/s] storm types: ua = 20 [mi/hr] duration-limited! d = 2 [hr] T(u ,d) = 2.8 [s] F = 15 [mi] a 32.16 [ft/s2] g * To for T knot, the deepwater wave celerity, equals, --- ua co = 2 * π F

Find: co [ft/s] storm types: ua = 20 [mi/hr] duration-limited! d = 2 [hr] T(u ,d) = 2.8 [s] F = 15 [mi] a 32.16 [ft/s2] g * To 14.33 feet per second. [pause] Lastly, we need to confirm this is a deepwater wave, --- ua co = 2 * π co = 14.33 [ft/s] F

Find: co [ft/s] ≥ 0.5 ua = 20 [mi/hr] deepwater wave if: d d = 2 [hr] L F = 15 [mi] T(u ,d) = 2.8 [s] a 32.16 [ft/s2] by checking the value of d over L, is greater than, 0.5. [pause] In this expression, variable d, --- g * To co = 2 * π co = 14.33 [ft/s]

Find: co [ft/s] ≥ 0.5 ua = 20 [mi/hr] deepwater wave if: d d = 2 [hr] L depth F = 15 [mi] T(u ,d) = 2.8 [s] a 32.16 [ft/s2] refers to the depth of water, which was provided in the problem statement, as, --- g * To co = 2 * π co = 14.33 [ft/s]

Find: co [ft/s] ≥ 0.5 ua = 20 [mi/hr] deepwater wave if: d d = 2 [hr] L depth F = 15 [mi] d = 65 [ft] depth T(u ,d) = 2.8 [s] a 32.16 [ft/s2] 65 feet. [pause] L knot, equals, g times T squared, --- g * To co = 2 * π co = 14.33 [ft/s]

Find: co [ft/s] ≥ 0.5 ua = 20 [mi/hr] deepwater wave if: d d = 2 [hr] L depth F = 15 [mi] d = 65 [ft] depth T(u ,d) = 2.8 [s] a g * T2 32.16 [ft/s2] Lo= divided by 2 PI. Here, we are assuming a deepwater wave condition. After plugging in --- 2 * π g * To co = 2 * π co = 14.33 [ft/s]

Find: co [ft/s] ≥ 0.5 ua = 20 [mi/hr] deepwater wave if: d d = 2 [hr] L depth F = 15 [mi] d = 65 [ft] depth T(u ,d) = 2.8 [s] a g * T2 32.16 [ft/s2] Lo= g and T, the deepwater wavelength equals, --- 2 * π g * To co = 2 * π co = 14.33 [ft/s]

Find: co [ft/s] ≥ 0.5 = 42.128 [ft] ua = 20 [mi/hr] deepwater wave if: d = 2 [hr] ≥ 0.5 water L depth F = 15 [mi] d = 65 [ft] depth T(u ,d) = 2.8 [s] a g * T2 32.16 [ft/s2] Lo= = 42.128 [ft] 42.128 feet. [pause] And 65 feet, divided by, --- 2 * π g * To co = 2 * π co = 14.33 [ft/s]

Find: co [ft/s] ≥ 0.5 = 42.128 [ft] ua = 20 [mi/hr] deepwater wave if: d = 2 [hr] ≥ 0.5 water L depth F = 15 [mi] d = 65 [ft] depth T(u ,d) = 2.8 [s] a g * T2 32.16 [ft/s2] Lo= = 42.128 [ft] 42.128 feet, is greatter than, 0.5. This confirms that our the value we previously used for T knot, and, --- 2 * π g * To co = 2 * π co = 14.33 [ft/s]

Find: co [ft/s] ≥ 0.5 = 42.128 [ft] ua = 20 [mi/hr] deepwater wave if: d = 2 [hr] ≥ 0.5 water L depth F = 15 [mi] d = 65 [ft] depth T(u ,d) = 2.8 [s] a g * T2 32.16 [ft/s2] Lo= = 42.128 [ft] c knot, are both legitamite, and c knot still equals, --- 2 * π g * To co = 2 * π co = 14.33 [ft/s]

Find: co [ft/s] ≥ 0.5 = 42.128 [ft] ua = 20 [mi/hr] deepwater wave if: d = 2 [hr] ≥ 0.5 water L depth F = 15 [mi] d = 65 [ft] depth T(u ,d) = 2.8 [s] a g * T2 32.16 [ft/s2] Lo= = 42.128 [ft] 14.33 feet per second. [pause] 2 * π g * To co = 2 * π co = 14.33 [ft/s]

Find: co [ft/s] ≥ 0.5 ua = 20 [mi/hr] deepwater wave if: d d = 2 [hr] L depth F = 15 [mi] d = 65 [ft] depth T(u ,d) = 2.8 [s] Lo= 42.128 [ft] a A) 5.9 B) 8.6 C) 14.3 D) 24.0 32.16 [ft/s2] When reviewing the possible solutions, --- g * To co = 2 * π co = 14.33 [ft/s]

Find: co [ft/s] ≥ 0.5 ua = 20 [mi/hr] deepwater wave if: d d = 2 [hr] L depth F = 15 [mi] d = 65 [ft] depth T(u ,d) = 2.8 [s] Lo= 42.128 [ft] a A) 5.9 B) 8.6 C) 14.3 D) 24.0 32.16 [ft/s2] the answer is C. [fin] g * To answerC co = 2 * π co = 14.33 [ft/s]