# Finding meridional part for a given latitude

Mariners, I am stumped. While solving a mercator sailing problem using Bowditch II, how do you get a meridional part for a departure latitude? (Lets use the latitude out of the book: 32 degrees, 14.'7 minutes.) What table out of Bowditch should be used, and how do you enter the table, retrieve the correct figure, and thus come to the correct answer for M? The Bowditch explanation is not clear to me how to enter the table step by step get M as 2033.3 for latitude 32 degrees, 14.'7 minutes. Thank you in advance.

[QUOTE=U.S. Marine (ret);79818]Mariners, I am stumped. While solving a mercator sailing problem using Bowditch II, how do you get a meridional part for a departure latitude? (Lets use the latitude out of the book: 32 degrees, 14.'7 minutes.) What table out of Bowditch should be used, and how do you enter the table, retrieve the correct figure, and thus come to the correct answer for M? The Bowditch explanation is not clear to me how to enter the table step by step get M as 2033.3 for latitude 32 degrees, 14.'7 minutes. Thank you in advance.[/QUOTE]

Use table 5. for Lat 32-14, m= 2032.5 and for Lat 32-15, m = 2033.7. Interpolate for 14.7 and you get 2033.3.

K.C.

Kennebec Captain, thank you for clearing that up for me. It makes nearly perfect sense. I do have two more questions after that in which maybe you could give me a slight rudder correction. Upon entering the table for Lat 32-14 it appears that m = 2032.6 and for Lat 32-15 m = 2033.8. (Both being 1/10th off of your reply) Am I missing a rounding step or am I just entering the table incorrectly? I am squared away with the interpolation portion. I appreciate the knowledge.

My Bowditch table 6 shows the same results that you have, U.S. Marine (ret)

I used the 1975 edition of Bowditch. If you look at the “Explanation of Tables” you can see that assumptions are made for the ellipticity and flattening of the earth. The 1975 edition uses the Clarke spheroid of 1866. The 0.1 difference you’re getting probably reflects an updating of the model used for the shape of the earth.

K.C.

In the 2002 Bowditch meridional parts is Table 6 and uses WGS ellipsoid of 1972.

What matters is the difference between the distances. Over short distances the results will be the same. . You’re subtracting one distance to the equator from another to get the distance between two points.

Point being no reason to get bogged down in why the tables are slightly different.

K.C.

The coast guard questions are based off of the 1982 bowditch volume II. Use that to study since that is what you have in the exam room, not the modern edition.

[QUOTE=Capt. Phoenix;80038]The coast guard questions are based off of the 1982 bowditch volume II. Use that to study since that is what you have in the exam room, not the modern edition.[/QUOTE]

You’re right that it is probably smart to use the 1982 edition.

The important thing is to use the same table (Datum) as is used in the example problem if you are trying to learn step by step.

But at the risk of beating a dead horse for solving the problem it doesn’t make a difference in this case.

It likely is not important to someone just trying to pass the test but; both the 1975 editions and the 2002 (and presumably the 1982 ed) use the same example problem.

The problems is:

L1 32-14.7
L2 36-58.7

In 1975
M1= 2033.3
M2= 2377.0

which, subtracting gives a difference of m= 343.7

In 2002, using the same positions but, using a different datum:
M1=2033.4
M2=2377.1

Subtracting m=343.7.

M1 (upper case) is the distance from the L1 to the equator and M2 is the distance from L2 to the equator, but to solve the problem you need the distance from L1 to L2 (lower case m) which is the same (to within 0.1) in both cases.

What you are trying to find is the length of the line from L1 to L2 in the same units (minutes of DLO) as is used in the rest of the triangle you are trying to solve.

By way of analogy it’s like trying to find the distance between two points on a highway by using the mile markers. If point A1 is at mile marker 100 and the exit (point A2) is at mile marker 105 then it’s 5 miles to the exit. If someone adjusts the datum of the mile markers and A1 is now at marker 100.1 and A2 is 105.1, it’s still 5 miles between the two (within 0.1 miles).

K.C.

You are correct, the problem can be done with either book. But he should be used to the lack of information in the old volume two. It’s basically only the tables, the new editions have volume I and II combined so there is more information than would be accessible in the exam room.