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I guess I should have used a better sample question. I used Peter's formula
on a different example
38 deg 45 min N
122 deg 08 min W
traveling 50 Nm
course 045 ( for the change in Lat & long)
I was able to get the correct lat @
39 deg 20.4 min
but could not get the Long to agree
Computer 121 22.7
I think I was getting 121 3.8?

After digesting this responce I will try it again.
Guy

----- Original Message -----
From: "George Huxtable" 
To: 
Sent: Saturday, June 24, 2006 2:17 PM
Subject: Re: I knew where we were, but where are we now?


> Guy Schwartz asked-
>
> This has probably been discussed here before, but hopefully someone
> will help me out. I'm trying to find the formula for caculating your
> DR after traveling from a know position.
> For instance:
> Last Fix
> 40 deg 07 min N
> 44 deg 37 min W
> Course 090
> distance traveled 56.88 Nm
> New DR?
> It would be a big help to me if someone that knows the formula could
> incorparate the above example so I could follow along.
> Thank you,
> Guy
> ===========================
>
> Pierre Boucher provided a helpful response, though his statement -
> "To find DLo (difference of longitude in minutes of longitude)
> distance travel / cos(mid latitude)"
> omits reference to course, and is true only in the special
> circumstances of Guy's Easterly-course example
> Not sure about that from Yanni Nikopoulos. But I think theres a bit
> more to add. We will assume that for our purposes the Earth is a
> sphere, which is usually good enough.
>
> If you are steering a constant course (i.e. a rhumb line), then it's
> always exactly true that your change in latitude in arc-minutes,
> measured positive Northwards, is distance (in miles) x cos course,
> where course is an angle measured clockwise from North, 0 to 360
> degrees. It will be a positive change, if your course has a Northerly
> component, and a negative change if it's at all Southerly. Add that
> change to your initial latitude, allowing for sign, and you have the
> latitude of your destination. That part is simple, then, however long
> your journey is.
>
> Calculating change in longitude is a bit more complicated, because the
> length of a degree of longitude varies, over the Earth's surface, as
> the latitude changes. I always like to consider longitudes as positive
> Westward (though several authorities differ) in order to keep
> longitudes in line with hour-angles, which increase Westward. In which
> case, when travelling small distances (of less than 60 miles or so)
> it's nearly true to say that the change in longitude in arc-minutes,
> measured positive Westward, is distance (in miles) x (- sin course) /
> (cos lat). The problem with that simple approach is that the latitude
> is changing slightly, from one end of the leg to the other, so what
> value of lat do you use in that formula to calculate change of long?
> If there's hardly any change in lat, from one end of the leg to the
> other, either because the leg is so short, or because the course in
> nearly due East or due West, then it just doesn't matter; you can take
> the lat, in that expression, to be the latitude of your starting
> point. Doing the calculation that way is often referred to as "plane
> sailing", because it's treating the Earth locally as if it was a plane
> surface rather than as part of a sphere. In those circumstances, this
> is a good enough approximation.
>
> The example that Guy has quoted is not really an informative one, in
> that with a course of exactly 90 degrees (due East) it's trivially
> simple. The latitude stays the same throughout, at 40 deg 07' N.
> Because it does, our plane-sailing approximation becomes exact, at
> 56.88 x (- sin 90) / cos 40 deg 07', or - 74.4 arc-minutes, so the
> initial long of 44 deg 37' W must be reduced by 74.4 minutes to 43 deg
> 22.6' W. Just as Pierre Boucher showed.
>
> For a longer travel, which involves a change in latitude, you can make
> a much better approximation, quite simply, in the following way.
> First, calculate the change in latitude as before, to get the exact
> latitude at the end of the leg. Now simply average those two values
> for latitude, at the start of the leg and at the end of it. Now use
> this average latitude, or "middle-latitude" in that formula for change
> in long, which becomes-
> Westing in minutes = distance x (- sin course) / cos (mid-lat).
> Calculating in this way was known as "middle-latitude sailing", for
> obvious reasons, and was used in medium-length sea voyages, from one
> headland to another. Even that, however, is only an approximation, and
> for long ocean passages, particularly those that cross the Equator, it
> can be seriously in error.
>
> However, there is a way to calculate a rhumb-line destination
> precisely, for any journey A to B, in which a constant course is
> maintained, no matter how long it might be. This is known as "Mercator
> sailing". First, knowing the latitude of A, work out the exact
> latitude change just as before, so now we have the two latitudes A and
> B. Next, convert each of those latitudes into a value that's known as
> "Meridional Parts", or MP. Tables to do this were available in the
> epitomes that were available to old mariners. But you can convert a
> lat to its equivalent in MP yourself dead easily with a calculator.
> Just halve the lat in degrees (keeping its sign carefully, positive
> North, negative South). Then add 45 degrees (remembering that sign),
> the resulting angle being always positive, between 0 and 90, and find
> the tan of that angle. Now find the log (the ordinary log, to the base
> 10) of that result, and multiply it by 7819. It sounds a handful, but
> a calculator will do it with no trouble at all. The end result, which
> you should be able to check out , is that a lat of zero, on the
> equator, results in a MP of zero; for a lat in the Northern hemisphere
> of +45 deg, it's MP is 2992.9; and for a Southern lat of -45 deg,
> it's -2992.9.
>
> (Note that these values are for a spherical Earth, and to achieve an
> even more precise result, it's possible tinker slightly with the
> conversion of lat to MP to get an even better fit to the Earth's true
> ellipsoidal figure. On that basis are meridional-parts tables, and
> Mercator charts made. We will ignore that tinkering here.)
>
> Now take the difference, by subtracting the initial MP (of A; call it
> MPa) from the final MP (of B; call it MPb), taking account of their
> signs. The result will be positive if the course has any Northing, and
> negative if it has any Southing.
>
> And now, finally,
> Westerly change in long (in minutes)= (MPb - MPa) / (- tan course)
>
> Note that this works precisely under even the most extreme conditions,
> even when travelling between points on opposite sides of the Equator.
> And even if you specify such a daft long distance, and a course
> suitably close to East or West, that the travellers rhumb-line course
> takes him several times round the world before he gets there, it will
> still give the right answer!
>
> Note that for very long legs of a passage, Mercator sailing becomes
> somewhat academic. Great-circle sailing, in which the course is
> regularly adjusted, is almost always significantly shorter, so in
> practice long legs of Mercator sailing are seldom adopted.
>
> I suspect that parts of this answer provide more detail than Guy
> expects or requires,
>
> George.
>
> contact George Huxtable at george@huxtable.u-net.com
> or at +44 1865 820222 (from UK, 01865 820222)
> or at 1 Sandy Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK.
>
>
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