1. 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

2. 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

3. 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

4. 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

5. 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

6. 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

7. 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

8. 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

9. 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

10. 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

11. 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

12. 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

13. 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 , ---
g * To
co =
2 * π

14. 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

15. 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

16. 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

17. 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

18. 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

19. 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

20. 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

21. 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

22. 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

23. 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

24. 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

25. 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

26. 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

27. 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

28. 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

29. 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

30. 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

31. 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

32. 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

33. 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 = [ft/s] F

34. 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 = [ft/s]

35. 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 = [ft/s]

36. 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 = [ft/s]

37. 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 = [ft/s]

38. 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 = [ft/s]

39. 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=
= [ft]
feet. [pause] And 65 feet, divided by, ---
2 * π
g * To
co =
2 * π
co = [ft/s]

40. 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=
= [ft]
feet, is greatter than, This confirms that our the value we previously used for T knot, and, ---
2 * π
g * To
co =
2 * π
co = [ft/s]

41. 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=
= [ft]
c knot, are both legitamite, and c knot still equals, ---
2 * π
g * To
co =
2 * π
co = [ft/s]

42. 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=
= [ft]
14.33 feet per second. [pause]
2 * π
g * To
co =
2 * π
co = [ft/s]

43. 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= [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 = [ft/s]

44. 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= [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 = [ft/s]