Correction for Leeway - Coastal Navigation Question - SailNet Community

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Go Back   SailNet Community > Skills and Seamanship > Seamanship & Navigation
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Seamanship & Navigation Forum devoted to seamanship and navigation topics, including paper and electronic charting tools.


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Old 07-30-2011
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Correction for Leeway - Coastal Navigation Question

Hello All,

Hope you can answer and explain the answer to this question:

Let's say my point of departure is at point A and destination is to point B.

The track (Course to Make Good) from A to B is 145 degrees magnetic.

The winds are west at 22 knots when I depart point A at 1530. Let's say the leeway is 4 degrees and the boat speed is 6 knots. What course, corrected for leeway, should I steer from point A to point B?

Thanks so much.
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Old 07-30-2011
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Though you say the leeway is 4 degrees, it is much easier to identify the leeway as a vector (direction and speed). This vector can be expressed on the chart as a line showing how far the leeway would take you over an arbitrary lenth of time (say two hours); therefore, if the 22 knot wind and what ever associated current were to cause your vessel to move in a direction of 90 degrees at 1 knot, then you can plot a 2 nautical mile line from point A at 90 degrees to represent where you would end up by the "leeway". You can also plot a line for twelve nautical miles from A to B at 145 degrees to represent your intended path for the arbitrary 2 hours. If you were to determine the course direction from the end of the 2 mile line to the end of the 12 mile line; then, this would be the course to use that would correct for the leeway. I'm sure someone could explain this more clearly,- Take care and joy, Aythya crew
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Old 07-30-2011
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I also want to add that distance from point A to point B is 2.9 NM.
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Old 07-30-2011
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What period of time and distance are you using to measure the amount of leeway?
How did you arrive at that number? Is it leeway..or current?

If it's 4 degrees of leeway over the 2.9 miles... the short answer is you steer 141 degrees.

If it's 4 degrees per mile..it's another story..and calculation
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Tempest, the assumption in the problem is 4 degrees of leeway.

If the problem identifies the leeway as vector (direction and speed), then it
would be easier.

My answer would also be 141 degrees magnetic. But the workbook answer is 115 degrees magnetic.

By the way, the variance is 14 degrees EAST.
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Old 07-30-2011
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Enough with the math! Vectors, smectors!

Put the coordinates of Point B in the GPS, note the bearing thereto, use that heading to start and after a few minutes note the COG on the GPS. Adjust your heading so that the COG equals the bearing to Point B.

And dontí tell me about losing power and GPS failure and all that. If that happens youíre just SOL.

PS -- it helps to plot a line from A to B on a real chart so you can be sure there are no hard things lurking along the rhumb line.

But seriously, in small boats itís hard enough to steer within 10 degrees of the desired heading. So pre-planning a few degrees of windage is an interesting exercise, but it shouldnít replace knowing where you are and where youíre going in real time. Thatís what makes GPS so great!
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Most interesting exercise - but somewhat overkill these days.


Use the Smartphone or Ipad app.
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Is there any mention in the problem regarding Deviation?

Whose workbook? I have Tursi's and Pyzel's on my shelf and the CG's , someone elses?

If it's 2.9 miles and the drift is 4 degrees per mile..then 2.9 x 4 = 11.6 degrees of correction add east ..11.6 + 14 = 25.6 or 26 degrees.

145 - 26 = 119 degrees mag. So I'm thinking there's deviation somewhere?

The drift can't be 4 degrees per hour, because it only takes you 29 minutes to get there..so you'd have to halve it to 2...

Something's missing..or I need to put down the rum..
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Quote:
Originally Posted by kwindancer View Post
Hello All,

Hope you can answer and explain the answer to this question:

Let's say my point of departure is at point A and destination is to point B.

The track (Course to Make Good) from A to B is 145 degrees magnetic.

The winds are west at 22 knots when I depart point A at 1530. Let's say the leeway is 4 degrees and the boat speed is 6 knots. What course, corrected for leeway, should I steer from point A to point B?

Thanks so much.
I'd avoid describing the wind like that. It's ambiguous. Is it from the west, or to the west? Better to use the term westerly, or "from the west".

I just did ASA Coastal Navigation and the leeway is normally added at the end. So you want to go 145 degrees and the leeway is 4. If the wind is westERLY, from the west, you should steer 149. From the east, 141.

There is no leeway per mile or any such calculation. Leeway is assumed to be a constant difference between the direction the boat is pointed in, and where it's actually going. So it is added or subtracted from the course to steer depending on the wind direction.

If this is an ASA question it will assume use of the deviation table for your compass.

You want to go 145 mag and want to know the compass reading.

Remember it like this :

TVMDC AW - True Variation Magnetic Deviation Compass, Add West. I remember it like this : Timid Virgins Make Dull Company, Add Whiskey

Desired course = 145 Mag. Find the compass deviation from the table. For example, 10W. Add west, 145 +10 = 155C. Add leeway, 159C.

Interestingly the question throws in some irrelevant data - the boat speed, departure time are both non-sequiturs.

The only way you can get to the 115 in the answers is with the wind from the east, and / or a hugely inaccurate compass!
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"If it's 2.9 miles and the drift is 4 degrees per mile..then 2.9 x 4 = 11.6 degrees of correction add east ..11.6 + 14 = 25.6 or 26 degrees."

I've never heard leeway being done that way. It's a constant error.

Get lost much?
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