Calculating a position in advance, in open water

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BlueSkyNick

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How can you work out where you are going to be in the future, assuming you sail your intended course, without looking at charts (paper or electronic) ?

For example, suppose you leave the Med and are at 36N, 10W exactly. You intend to steer 315degsT for 300 miles. Ignoring drift and leeway, you can calculate (using pythag or trig) your latitude to be 210miles. ie 210minutes so end position is 39degs 30mins N.

You will also be 210miles further west, but what will your longitude be ?

EDIT - changed to 36N from 37 - typo!
 
Bloody hell Nick, Spherical Trigonometry at this time of night!
Your method is presumably an approximation over relatively short distances. So you need to work out the length of a minute of longitude at your average lattitude (37.75 degrees). The radius of the Earth from its axis at this lattitude will be Rxcosine(37.75 degrees). Where R is the radius at the equator, say 3800 miles. So radius at 37.75 degrees lattitude is 3005 miles so one minute is 0.87 miles. Your Westing will be about 210 miles or 240 minutes taking you to 14 degrees West.
N.B. This is an engineer's solution not a navigator's. /forums/images/graemlins/confused.gif
 
[ QUOTE ]


Otherwise, you could just plot this on a plain piece of graph paper, surely.

[/ QUOTE ]dunno how, the squares woud all be the same size, unlike lat and long on the planet.

DJE seems pretty close - but I need to re-read it a few times !
 
Aha! A question on 'The Sailings' at long last - and at close on midnight....

DJE has the best answer for you; failing that, beg, borrow or steal a copy of 'Ocean Passages for the World - HMSO' and follow the exposition in there..... or in Bowditch


/forums/images/graemlins/grin.gif

Edit: Blogger me! It's way past midnight, and I am become like a pumpkin....

/forums/images/graemlins/wink.gif
 
A good half of Nories Nautical Tables is taken up with the Traverse Tables. Surely everyone uses them /forums/images/graemlins/wink.gif

Of course there are bound to be a few gadget freaks who have blind faith in the workings of GPS, poor misguided fools.
 
Nick,

Cold light of day time. I took the radius of the earth at the equator as 3800 miles giving a circumference of about 24000 miles. This of course is land miles. In nautical miles it is about 360 x 60 or 21600 nm. (Or it would be if the Earth was a sphere; in fact it bulges a bit at the equator but I don't know how much.)

Anyway if we stick to the assumption that the Earth is a sphere the calculation is actually simpler than I thought last night. One minute of lattitude is one nautical mile everywhere. One minute of longitude is one nautical mile multiplied by the cosine of your angle of lattitude. So at 37.75 degrees North one minute of longitude is 0.79 nm. Your 210 miles of westing therefore equates to 266 minutes taking you to 14 degrees 26 minutes West.

Edit: Arithmetic corrected!
 
Quite easily, I haven't felt the need to understand how the internal combustion engine works to drive a car, ditto GPS, ditto radar, ditto most things. What does one employ staff for? /forums/images/graemlins/smile.gif
 
What Ho Brylcream! (best Bertie Wooster impression)......

How's it going? Killed any Huns lately? /forums/images/graemlins/grin.gif

Just thought I would bring this thread down to a level I could understand! /forums/images/graemlins/grin.gif
 
Smiffy

I don't have enough hair to keep the brylcreem on with!

All going fine, got bumped around a bit in the night here tho', lots of washing, roof tiles and cats flying around!

How are you, you old fart /forums/images/graemlins/grin.gif
 
This is the much-misspelled Plane Sailing, the simplest of nav calculations, where you assume the earth to be flat (over anything up to 600 miles).

Dep = dlong x cos lat.

Use Traverse Tables, as mentioned, to do the whole thing, or a calculator with trig functions. Work out the dLat and Departure using School-boy Plane Trig on a simple calculator, or with tables if you prefer. Then using the mean latitude in the formula above to calculate the Difference of Longitude from the Departure (the distance sailed east or west calculated originally)
Apply the dlat and dlong to your start position and heh-ho, you’re a real navigator.

Is this stuff not taught by the RYA?
 
Hi This is really a very simple sum or plot which can be carried out in many ways. The most simple way is by using what is known as The Sailings and it's most simple form is tabulated in "Nories Tables" but I am going to assume you do not have these and your sum is roughly right however I make it -212.13 [we know it's West because the answer is negative] and we call the answer to this sum "Departure" which is a measure of longitude. We also need to know the Difference in Latitude which is called "D'Lat" so we work out the D'Lat which ='s Cos Course * Distance = 212.13 [we know it's North because the answer is positive] so now we can add the D'lat to the DR/Plot position and we then have the latitude that we are to arrive at. Add together the commencing DR lat and the worked up lat and then divide them by two and we call this the Mid Lat.
Now we multiply the Cosine of the Mid lat with the departure and the result is the Difference in Longitude or D'Long which when added to the original DR Longitude provides our new DR longitude.
The Mid lat becomes a little more complicated over larger distances i.e. the usual rule of thumb is greater than 600' [N.miles] but this is unusual for most yachtsman and I will leave it out of this debate. Obviously if the N/S result is negative then the answer is Southerly and if the West/East answer is positive then we know that the result is Easterly. Hope that helps.
Cheers,
Kristoria
 
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