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Author
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Topic: Problem 1-17
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James Muniz
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posted January 23, 2005 12:30 PM
On problem 1-17 I get 48:16.3 N, 122:58.6 W using the great circle calculator. and using a regular calculator I get 48:16.27 N, 122:58.6 W. The answer shows 48:16.27 N, 122:58.7 W. Is that a miss-print, if not how did we get .1 W difference.
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David Burch
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posted January 23, 2005 01:22 PM
First I should stress that these differences are small in light of this being a plotting problem. That is, 16.30 vs 16.27 in lat = 0.03 x 6000 ft = 60 yards, and 58.6 vs 58.7 in longitude is 0.1' x 6000 x cos 48 = 133 yards.
So we are getting errors of about 100 yards out of 3.5 miles = 100/(3.5x2000) = 1.4%
[If the above reckoning is not clear, please post a question and i can go into more detail in a different topic. It is fair to assume it is a plotting problem because the numerical position was not given for the starting point.]
With that all said, we go back to your question. If you are using a calculator for the solution, then you must have had a digital lat lon for the buoy. What value did you have for that? From there we can look into the differences.... and i must also add that in the world of hi precision GPS (away from this particular problem) these differences are indeed significant and it is fair to raise the question.
But to pursue this we need to know what the input was and what formula you used for the "regular calculator" solution.
Looking ahead, the difference between great circle and Mercator computation for this short distance should not be as large as 0.1' of lon—at least that is my guess till we work it out.
From: Starpath, Seattle, WA
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James Muniz
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posted January 24, 2005 06:17 PM
I was using the position 48:19.8 N, 122:58.6 W from the Light List for Buoy RA. 3.5 n.mi.=3.5'
19.8'-3.5'=16.3'thus 48:16.3 N As for the Longitude I cannot figure that out. If we travel South (180) should the Longitude still be 122:58.6
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