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» Online Classroom   »   » Public Discussion of Inland and Coastal Navigation   » 2-9

   
Author Topic: 2-9
M. C. Rowley


 - posted August 09, 2005 08:21 PM      Profile for M. C. Rowley           Edit/Delete Post 
Would you please discuss what the "acceptable" levels of accuracy (and precision) are for this type of navigation problem? Captioned below are my two (2) solutions compared to the "book" solution:

Part A

Manual: 48 09.20N, 122 53.17W
E-chart: 48 09.15N, 122 53.31W
Book: 48 09.20N, 122 53.35W

Part B

Manual: 2.2nm
E-chart: 2.1nm
Book: 2.3nm

Part C

Manual: 49 fathoms (based on contouring)
E-chart: n/a
Book: 47 fathoms

Part D

Manual: 5.00nm, 071M
E-chart: 5.10nm, 071M
Book: 5.15nm, 074M

The latitutde calculation is basically the same except for the E-chart calculation which is only 0.05nm. My longitutde calculation, however, is slightly more off. I plotted my triangle of error again and basically came up with the same longitude. I do believe, however, that your bearing to Pt. Wilson Buoy 6 is off. If it were indeed 074M then the bearing to the tank at Protection Island would be 203M instead of 206M. Anyways, am I on the right track here and are these errors acceptable?

Thanks........

Mark Rowley
Santa Fe, New Mexico

David Burch


 - posted August 10, 2005 09:23 PM      Profile for David Burch           Edit/Delete Post 
First, we do have an article online that addresses the general subject you raise How close should the answers be. This is referenced from the Tech support section on the coastal nav course.

As for this problem in particular, this is a fix from two compass bearings, so it is rather easy to analyze. The procedure would be to assume some error or uncertainty in each of the bearings, and plot the lines again.

In this problem, you have one bearing line that is 206M. For now, assume this is accurate to ± 2°. Then plot two lines, one at 204M and the other at 208M. Then do the same for the other bearing and you will end up with a trapezoid which reflects your overall uncertainty. It will be come obvious that the farther the target the more sensitive your fix is to correct readings.

We have practice problems on this somewhere in the course. Also we have the small triangle rule that is presented somewhere in the course that can be applied to some extent. if you have an error of 6° in a course heading, then you go off course one tenth of your distance run. Please look up that rule and its discussion.

Put another way, if the target is 1 mile away and your bearing is wrong by 6°, the intersection could be off by 0.1 miles, or if off by 3°, it would be 0.05 miles and so on. this is not a strict analysis, but it gives you a ballpark estimate.

My first guess is your answers are plenty close enough, but the above analysis is the only way to tell for sure.

Also, in principle, you should get more accurate answers from the electronic solutions to problems like this since you can zoom way in and do all the plotting very precisely.

In some cases, the solutions can also be computed, but this is not the subject at hand here.

From: Starpath, Seattle, WA


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