|
|
|
|
|||||
|
||||||
From: Jim Thompson (no email)
Date: Wed Feb 02 2005 - 17:30:54 EST
> -----Original Message-----
> From Bill
> I would like to look at table 18, "Distance of an Object by
> Two Bearings."
>
> My premises:
> Bearing is the direction (cardinal angle) between the observer's north and
> the object, true or magnetic. That being clockwise, 0 to 360 or 0 to
> 359.xxx.
>
> Relative bearing is the angle between the bow end of the
> lubberline and the
> object, clockwise, 0 to 359,xxxd (or 0 to 360d, depending on your religion
> ;-)
>
> Bearing on the Bow is the angle between the bow end of the lubberline and
> the object, clockwise or counterclockwise, from 0-180d.
>
> The 1995 Bowditch, "Explanation of Navigation Tables" states for table 18:
> "To determine the distance of an object as a vessel on a steady course
> passes it, observe the difference between the course and two
> bearings of the
> object....Enter the table with the two differences..."
>
> My example, by the book:
> Course 2d true, object and shoreline to port
> First bearing 295d true, difference 293 d
> Second bearing, 245d true, difference 243d
>
> Now I go to the table. Whoops! No values greater than 160d for entering
> arguments. I can understand that we want an reasonable
> difference in angles
> between two LOP's to obtain an useable running fix; but darn,
> here we have a
> system that only works on the starboard side of the boat!
Bill,
The Glossary in Bowditch 2002 defines "Relative Bearing" this way: "Bearing
relative to heading of a vessel, expressed as the angular difference between
the heading and the direction. It is usually measured from 000? at the
heading clockwise through 360?, but is sometimes measured from 0? at the
heading either clockwise or counterclockwise through 180?, when it is
designated right or left."
I worked your question this way:
Course = 002 T.
Bearing 1 = 295 T, so
RB = 293, or 067 to port.
Bearing 2 = 245 T, so
RB = 243, or 117 to port.
Entering the tables with 067 as "Difference between the course and first
bearing", and 117 as "Difference between the course and first bearing". The
bearings are odd numbers, so we need to interpolate. I get 1.21 for the
first column, and 1.08 for the second column, after interpolation.
If Speed = 8 knots and Time = 30 min, then Distance travelled = 4 NM.
So Distance off the object at the second bearing =
4 NM x 1.21 = 4.8 NM.
And Distance off the object when abeam =
4 NM x 1.08 = 4.3 NM.
I did a rough sketch which suggests this solution is about right. I have not
had time to sit down and draw a proper plot of the solution to check the
result, but plan to do so.
Bowditch 2002 is as sparse on further explanation as your 1995 version. You
might be interested in a short paper on the trigonometry of a similar
situation, posted by a colleague:
http://www.blueheroncps.pe.ca/Courses/SpecialPairs.PDF
|