SV Skalliwag #141
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Distance off problem
This problem appeared on Cruising Compass:
You see a light through a thin layer of fog and take a bearing of 300T at 1210. You are sailing a course of 240T and your speed is 6.0 knots. At 1240 the bearing to that same light is now 347T. The current tables tell you that you've had 1.2 knots of drift with a set of 135T during that time. What is your distance off the light at the second bearing?
I am not sure exactly if I know how to solve this. I got 7 NM. Is that right? How do you solve it?
You see a light through a thin layer of fog and take a bearing of 300T at 1210. You are sailing a course of 240T and your speed is 6.0 knots. At 1240 the bearing to that same light is now 347T. The current tables tell you that you've had 1.2 knots of drift with a set of 135T during that time. What is your distance off the light at the second bearing?
I am not sure exactly if I know how to solve this. I got 7 NM. Is that right? How do you solve it?
Last edited by brokesailor; 03012014 at 01:44 PM.
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Re: Distance off problem
Quote:
Originally Posted by brokesailor
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This problem appeared on Cruising Compass:
You see a light through a thin layer of fog and take a bearing of 300T at 1210. You are sailing a course of 240T and your speed is 6.0 knots. At 1240 the bearing to that same light is now 347T. The current tables tell you that you've had 1.2 knots of drift with a set of 135T during that time. What is your distance off the light at the second bearing?
I am not sure exactly if I know how to solve this. I got 7 NM. Is that right? How do you solve it?
You see a light through a thin layer of fog and take a bearing of 300T at 1210. You are sailing a course of 240T and your speed is 6.0 knots. At 1240 the bearing to that same light is now 347T. The current tables tell you that you've had 1.2 knots of drift with a set of 135T during that time. What is your distance off the light at the second bearing?
I am not sure exactly if I know how to solve this. I got 7 NM. Is that right? How do you solve it?
When I get up in the bridge on watch I'll try to get something..
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Re: Distance off problem
The two bearings to the light cross at the light forming an angle between them (47 degrees). You lay on the first line at 1210 and the second at 1240.
Solve for your motion (speed and direction) between those two times using your indicated speed and course as one side and the set and drift as a second side of a triangle. The third side is your DR speed and course. You can solve it using either a plotting sheet or the law of cosines (since you know two sides of the triangle and the included angle between the two sides (75 degrees  the supplement to 240 minus 135 (105 degrees)).
Once you have solved for the DR speed and course you calculate the distance made good along the DR course during the half hour between the two bearings. Then lay that distance out on the plotting sheet or chart in the direction of the DR you determined  to fit between the two bearing lines to the light.
Measure the distance from the intersection of the second bearing with your DR to the cross bearings at the light and you have your distance from the light.
Note: My included angle was "off" because it didn't account for the set and drift...corrected in a post below.
Solve for your motion (speed and direction) between those two times using your indicated speed and course as one side and the set and drift as a second side of a triangle. The third side is your DR speed and course. You can solve it using either a plotting sheet or the law of cosines (since you know two sides of the triangle and the included angle between the two sides (75 degrees  the supplement to 240 minus 135 (105 degrees)).
Once you have solved for the DR speed and course you calculate the distance made good along the DR course during the half hour between the two bearings. Then lay that distance out on the plotting sheet or chart in the direction of the DR you determined  to fit between the two bearing lines to the light.
Measure the distance from the intersection of the second bearing with your DR to the cross bearings at the light and you have your distance from the light.
Note: My included angle was "off" because it didn't account for the set and drift...corrected in a post below.
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Last edited by fryewe; 03012014 at 06:57 PM.


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Re: Distance off problem
You can obtain the solution graphically by plotting the initial position and Bearing to the target and them plotting an EP taking into account the set and drift over the subsequent 30 minutes; and, the yachts heading and speed, which would give you a CMG of 199.59ºT and distance traveled of 3.21 miles and from that point lay down a line of 347º and measure the distance to the Intersect from the EP; or, by using the law of sines simply calculate the distance from the EP to the intercept which is 4.32 miles
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Re: Distance off problem
"Two trains, heading north & South, leave"
"A 30 foot flag pole casts a 10 foot shadow"
Just use your GPS, much easier.
All that chart stuff can make your head hurt.
Paul T
"A 30 foot flag pole casts a 10 foot shadow"
Just use your GPS, much easier.
All that chart stuff can make your head hurt.
Paul T
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Re: Distance off problem
The course and speed made good using the law of cosines is about 228 at 5.8 knots. The distance traveled in the half hour will thus be 5800 yards along a 228 vector.
Graphically insert this vector between the lines of bearing to the light and where it intersects the second LOB is the position at the time of the LOB and the distance to the light when the LOB was taken can be measured.
Or you can use the values determined using the law of cosines above to solve a triangle to determine the distance...the angle at the light between LOBs is 47 degrees...the angle between the first LOB and the DR vector (300 minus 228) = 72 degrees and the angle between the second LOB and the DR vector (180 minus (47+72)) = 61 degrees. The DR distance vector is 5800 yards so using the law of sines the distance to the light is (D/sin 60)=(5800/sin 47) so D = 5800x0.866/0.73 or 6880 yards.
SVL  our solutions differ and I don't know why for sure. I didn't do this graphically because I don't have a plotting sheet handy. One thing stands out in your solution...the speed made good has to be less than 6 knots because the set is 105 degrees off the bow to port...setting the boat to the south and retarding progress. However the drift is only 1.2 knots and even if it was broad on the beam it would only set the boat to the south by an angle whose tangent is 1.2/6 or about 11 degrees...or to about 229 degrees CMG.
Graphically insert this vector between the lines of bearing to the light and where it intersects the second LOB is the position at the time of the LOB and the distance to the light when the LOB was taken can be measured.
Or you can use the values determined using the law of cosines above to solve a triangle to determine the distance...the angle at the light between LOBs is 47 degrees...the angle between the first LOB and the DR vector (300 minus 228) = 72 degrees and the angle between the second LOB and the DR vector (180 minus (47+72)) = 61 degrees. The DR distance vector is 5800 yards so using the law of sines the distance to the light is (D/sin 60)=(5800/sin 47) so D = 5800x0.866/0.73 or 6880 yards.
SVL  our solutions differ and I don't know why for sure. I didn't do this graphically because I don't have a plotting sheet handy. One thing stands out in your solution...the speed made good has to be less than 6 knots because the set is 105 degrees off the bow to port...setting the boat to the south and retarding progress. However the drift is only 1.2 knots and even if it was broad on the beam it would only set the boat to the south by an angle whose tangent is 1.2/6 or about 11 degrees...or to about 229 degrees CMG.
The web makes everywhere the same place.  Fred Reed, from "Fred on Everything"
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Re: Distance off problem
Quote:
Originally Posted by dabnis
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"Two trains, heading north & South, leave"
"A 30 foot flag pole casts a 10 foot shadow"
Just use your GPS, much easier.
All that chart stuff can make your head hurt.
Paul T
"A 30 foot flag pole casts a 10 foot shadow"
Just use your GPS, much easier.
All that chart stuff can make your head hurt.
Paul T
A little trig can save your life. If you learn to approximate values and if all you need are ballpark numbers to keep you safe you can do the math in your head.
Or you can rely on GPS and the fickleness of thunderstorms never working their magic on your elex...your infallible maintenance...the absolute surety of never having an electrical fire...never having a bug in your computerized plotter...and other serendipity.
Life is full of choices. Some clean their guns when they have a minute to spare...others practice the law of cosines in their head.
The web makes everywhere the same place.  Fred Reed, from "Fred on Everything"
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Re: Distance off problem
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Originally Posted by fryewe
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To paraphrase an old saw..."for the want of an electron a plotter was lost...for the want of a plotter a position was lost...for the want of a position a distance to shoal was lost...for the want of a distance to shoal a boat was lost...for the want of a boat a life was lost."
A little trig can save your life. If you learn to approximate values and if all you need are ballpark numbers to keep you safe you can do the math in your head.
Or you can rely on GPS and the fickleness of thunderstorms never working their magic on your elex...your infallible maintenance...the absolute surety of never having an electrical fire...never having a bug in your computerized plotter...and other serendipity.
Life is full of choices. Some clean their guns when they have a minute to spare...others practice the law of cosines in their head.
A little trig can save your life. If you learn to approximate values and if all you need are ballpark numbers to keep you safe you can do the math in your head.
Or you can rely on GPS and the fickleness of thunderstorms never working their magic on your elex...your infallible maintenance...the absolute surety of never having an electrical fire...never having a bug in your computerized plotter...and other serendipity.
Life is full of choices. Some clean their guns when they have a minute to spare...others practice the law of cosines in their head.
Our boat, below, had a compass & a radio direction finder, that was it. 1964 was a bit before recreational GPS. We did a lot of "chart" work. We fished in British Columbia for 25 years in rental boats. Brought my own GPS in the latter years but always fell back on my charts.
Paul T
SV Skalliwag #141
Re: Distance off problem
Thanks to all. My methodology was correct but still got the wrong answer. I'll have to redo it on a plotting sheet.
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Re: Distance off problem
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Originally Posted by svHyLyte
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You can obtain the solution graphically by plotting the initial position and Bearing to the target and them plotting an EP taking into account the set and drift over the subsequent 30 minutes; and, the yachts heading and speed, which would give you a CMG of 199.59ºT and distance traveled of 3.21 miles and from that point lay down a line of 347º and measure the distance to the Intersect from the EP; or, by using the law of sines simply calculate the distance from the EP to the intercept which is 4.32 miles
Recalibrating, from the initial observation point (OP1), the bearing to the mark is given as 340º. From the second observation point (OP2), the bearing is given as 347º such that the included angle between the two bearings is 7º. One must then determine the position of OP2 relative to OP1 taking into account set, drift, course and speed.
The set is given as 135º (or 45º below the horizontal) at a drift of 1.2 knots. Over the 30 minute interval between observations, the EP would move a distance of .6 mileseast .4243 miles and south .4243 miles (in each case .6 x sin 45º).
The adjusted EP to OP2 would than be taken into account based upon the yacht's heading of 240º at 6 knots or a distance of 3 miles. From the adjusted EP the yacht would move south 1.5 miles (3 x sin 30º) and west 2.598 miles (3 x cos 30º). Relative to OP1, OP2 would be positioned 1.9243 miles south (being 1.5 miles plus .4243 miles) and west 2.1738 miles (being 2.598.4243 miles east). By the square root of the sum of the squares, the distance traveled would have been 2.9031 miles on a heading of 228.49º (270º  arcsin (1.9243/2.1738)). From OP2, the bearing to the mark is given as 347º (see above).
By the law of sines, the lengths of the sides of a triangle are proportional to the sins of the opposing angles. The objective is to find the distance from OP2 to the mark. The included angle facing the unknown side is 111.5153º (being 340º  228.49º). The included angle between OP1 and OP2 at the mark is 7º (being 347º340º) and the opposing side is 2.9031 miles long. And, 2.9031/sin 7º = L/sin 111.5153º. Solving for L one arrives at 22.1619 miles (being 2.9031 x sin 111.5153º/sin 7º). Rather a different answer than my original henscratch, eh? (Intuitively, one can tell the mark is rather a long way off given the minor change in the bearing7ºover 30 minutes running at 6 knots.)
FWIW...
"It is not so much for its beauty that the sea makes a claim upon men's hearts, as for that subtle something, that quality of air, that emanation from the waves, that so wonderfully renews a weary spirit."
Last edited by svHyLyte; 03032014 at 08:45 AM. Reason: correct typo; add addendum

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