1. Find: To [s] ua = 40 [mi/hr] ua F = 30 [mi] d = 15 [hr] F nomogram
plan F
view
A) 5.9
B) 11.8
C) 16.1
D) 22.4
Find the deepwater wave period, T knot, in seconds. [pause] In this problem, ---
wind
direction

2. Find: To [s] ua = 40 [mi/hr] ua F = 30 [mi] d = 15 [hr] F nomogram
plan F
view
A) 5.9
B) 11.8
C) 16.1
D) 22.4
we are provided the wind velocity in the x direction, u a, as well as, ---
wind
direction

3. Find: To [s] ua = 40 [mi/hr] ua F = 30 [mi] d = 15 [hr] F nomogram
plan F
view
A) 5.9
B) 11.8
C) 16.1
D) 22.4
the length of the fetch, F, and, the duration of time, ---
wind
direction

4. Find: To [s] ua = 40 [mi/hr] ua F = 30 [mi] d = 15 [hr] F nomogram
plan F
view
A) 5.9
B) 11.8
C) 16.1
D) 22.4
d, that the wind is acting upon the given fetch lenght. For this problem we will also need to utilize ---
wind
direction

5. Find: To [s] ua = 40 [mi/hr] ua F = 30 [mi] d = 15 [hr] F nomogram
plan F
view
A) 5.9
B) 11.8
C) 16.1
D) 22.4
a nomograph to determine the period, T. To solve for the period, we first need to determine the nature of the storm, ---
wind
direction

6. Find: To [s] storm types: ua = 40 [mi/hr] fetch-limited F = 30 [mi]
d = 15 [hr]
plan F
view
A) 5.9
B) 11.8
C) 16.1
D) 22.4
whether it is a fetch limited storm, or it is a ---
wind
direction

7. Find: To [s] storm types: ua = 40 [mi/hr] fetch-limited
duration-limited
F = 30 [mi]
d = 15 [hr]
plan F
view
A) 5.9
B) 11.8
C) 16.1
D) 22.4
duration-limited storm, or if it is a fully developed, ---
wind
direction

8. Find: To [s] storm types: ua = 40 [mi/hr] fetch-limited
duration-limited
F = 30 [mi]
fully developed
d = 15 [hr]
plan F
view
A) 5.9
B) 11.8
C) 16.1
D) 22.4
storm. [pause] We’ll first check if the storm storm is fetch-limited, ---
wind
direction

9. Find: To [s] storm types: ua = 40 [mi/hr] fetch-limited
duration-limited
F = 30 [mi]
fully developed
d = 15 [hr]
plan F
view
A) 5.9
B) 11.8
C) 16.1
D) 22.4
by analyzing the wind velocity, u a, and the duration of time, d. We’ll look at the nomograph, ---
wind
direction

10. Find: To [s] storm types: ua = 40 [mi/hr] fetch-limited
duration-limited
F = 30 [mi]
fully developed
d = 15 [hr]
nomograph
for deepwater waves, in English units, and find the intersection of 40 miles per hour, --- ua F

11. Find: To [s] storm types: ua = 40 [mi/hr] fetch-limited
duration-limited
F = 30 [mi]
fully developed
d = 15 [hr]
nomograph
windspeed, and 15 hour duration. Then we’ll trace down, to the fetch length axis, --- ua F

12. Find: To [s] storm types: ua = 40 [mi/hr] fetch-limited
duration-limited
F = 30 [mi]
fully developed
d = 15 [hr]
nomograph
F(u ,d) = 180 [mi] a
and the fetch length, equals, 180 miles. [pause] This means that there needs to be at least 180 miles of fetch length for this storm to be fully developed, --- ua F
180

13. Find: To [s] storm types: ua = 40 [mi/hr] fetch-limited
duration-limited
F = 30 [mi]
fully developed
d = 15 [hr]
nomograph
F(u ,d) = 180 [mi] a
but since there is only 30 miles of fetch length, the storm is considered fetch-limited. [pause] Since we know the storm is fetch limited, we know the duration of time ---
F(u ,d) > F ua a
fetch-limited F
180

14. Find: To [s] storm types: ua = 40 [mi/hr] fetch-limited
duration-limited
F = 30 [mi]
fully developed
d = 15 [hr]
nomograph
F(u ,d) = 180 [mi] a
it will take the waves to reach their maximum height will be less than, ---
F(u ,d) > F ua a
fetch-limited
d(u ,F) < d F
180 a

15. Find: To [s] storm types: ua = 40 [mi/hr] fetch-limited
duration-limited
F = 30 [mi]
fully developed
d = 15 [hr]
nomograph
F(u ,d) = 180 [mi] a
the 15 hour duration of the storm. [pause] Therefore, we will use the wind speed of 40 miles per hour, ---
F(u ,d) > F ua a
fetch-limited
d(u ,F) < d F
180 a

16. Find: To [s] storm types: ua = 40 [mi/hr] fetch-limited
duration-limited
F = 30 [mi]
fully developed
d = 15 [hr]
nomograph
F(u ,d) = 180 [mi] a
and fetch length of 30 miles to determine the characteristics of this fetch-limited deepwater wave. After plotting these values ---
F(u ,d) > F ua a
fetch-limited
d(u ,F) < d F a

17. Find: To [s] storm types: ua = 40 [mi/hr] fetch-limited
duration-limited
F = 30 [mi]
fully developed
d = 15 [hr]
nomograph
F(u ,d) = 180 [mi] a
on the nomograph, the intersecting point corresponds to a duration or approximately, ---
F(u ,d) > F ua a
fetch-limited
d(u ,F) < d F a

18. Find: To [s] storm types: ua = 40 [mi/hr] fetch-limited
duration-limited
F = 30 [mi]
fully developed
d = 15 [hr]
nomograph
4.6 hours, which is less than 15 hours, as we suspected. The nomograph also provides data for the wave height, in feet, --- ua
d(u ,F) = 4.6 [hr] a
d(u ,F) < d F a

19. Find: To [s] storm types: ua = 40 [mi/hr] fetch-limited
duration-limited
F = 30 [mi]
fully developed
d = 15 [hr]
nomograph
Ho =
To =
and the wave period, in seconds. The wave height for this storm equals, --- ua
d(u ,F) = 4.6 [hr] a
d(u ,F) < d F a

20. Find: To [s] storm types: ua = 40 [mi/hr] fetch-limited
duration-limited
F = 30 [mi]
fully developed
d = 15 [hr]
nomograph
Ho = 6.6 [ft]
To =
6.6 feet, and, the deepwater period, equals, --- ua
d(u ,F) = 4.6 [hr] a
d(u ,F) < d F a

21. Find: To [s] storm types: ua = 40 [mi/hr] fetch-limited
duration-limited
F = 30 [mi]
fully developed
d = 15 [hr]
nomograph
Ho = 6.6 [ft]
To = 5.9 [s]
5.9 seconds, which is the value we want to determine. [pause] ua
d(u ,F) = 4.6 [hr] a
d(u ,F) < d F a

22. Find: To [s] A) 5.9 ua = 40 [mi/hr] B) 11.8 C) 16.1 F = 30 [mi]
d = 15 [hr]
Ho = 6.6 [ft]
To = 5.9 [s]
When reviewing the possible solutions, --- ua
d(u ,F) = 4.6 [hr] a
d(u ,F) < d F a

23. Find: To [s] A) 5.9 ua = 40 [mi/hr] B) 11.8 C) 16.1 F = 30 [mi]
d = 15 [hr]
Ho = 6.6 [ft]
To = 5.9 [s]
the correct answer is A. [fin] ua
answerA F