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Author
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Topic: 6.6 problems 5 and 6
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louis christensen
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posted June 26, 2005 12:37 PM
In problem 5 the first sunline the longitude is 36.04.0 but it changed to 35.53.2 in the A-lon. Same also for problem 6 on the third sunline the longitude is 62.54.8. But in the A-lon. it is 63.09.7 . I read in 5-8 The degrees part of assumed londitude. For the east longitudes choose a A-lon within 30' of your DR-lon But I guess I dont really understand this. Is this what is going on in the two problems that I am having such a hard time with? I was hoping that you could explan this a little better so I might understand it.
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David Burch
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posted June 26, 2005 01:02 PM
In these running fix problems in section 6.6, we are given a DR position at a DR time, along with our course and speed. For example in #5, the time is 0815 WT and the position is 29° 46.7' S, 36° 25.9' E, with C = 295 T and S= 12 kts.
We are then told the first sunline took place at 1000 WT. Our first job then is to compute or plot what our new DR position is at the time of the sight (1000) ie, after traveling 10h00m - 08h 15m at 12 kt in direction 295 T from the given position. the answer to that was 29° 37.8' S, 36° 04.0' E. See answers on page A-9 in the text.
That is where your first number comes from. There are various ways to solve for this: you can compute the time run = 1h 45m and then the distance run = 21.0 nmi and then plot from the first postion to get the second, you can compute it with a caluclator using the forumlas we provide. See Numerical DR.
The value of the assumed longitude, on the other hand, is determied by both this new DR Lon as well as the GHA of the sun at the time of the sight. The GHA of the sun at sight time was given in the problem itself as GHA = 299° 06.8'. Our job now in selecting the assumed longitude is to choose the nearest longitude to our DR longitude (36° 04.0' E) that we can add to the GHA and end up with minutes part = 60.0'.
In other words, the minutes of our assumed longitude must be 60.0-6.8 = 53.2'.
The only part left is to choose the degrees part of the a-Lon, which will be either 35°, 36°, or 37°, which ever will lead to an assumed longitude closest to the DR lon. The answer in this case is 35 since 35° 53.2' is closer than either of the other choicesm, and is well within 30' of the DR. Note the closest one will always be within 30' of the DR and only one will meet that standard.
Please recall that we have a special section in the In-depth topics of Chapter 11 that is devoted sepcifically to t his task of choosing the assumed longitude. It is 11.19 on page 11-61.
Note that this is an East Longitude problem. In West Longitudes we subtract the a-Lon from the GHA to get LHA, and therefore choose the minutes of a-Lon such that the answer has 0.0'.
From: Starpath, Seattle, WA
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louis christensen
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posted June 26, 2005 03:21 PM
Thank You... I understand it now. the best part i did not think i would get a answer that fast. That is what makes this course so great... Thanks again
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