nautical measurements

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DeepAndy

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My GPS gives out a position in metric to 3 decimal points eg. 51 01.123 N.

The last part .123 what part of a fraction are they, i know it's part of 1/60th

So is it tens hundreds & thousands.

BloODY brain has gone dead.

What i think it is , is 1 = 600 sqaure feet 2 = 60 sqaure feet 3 = 6 sqaure feet ...

cheers for help

Andy


<hr width=100% size=1><A target="_blank" HREF=http://www.taurusdiving.co.uk>http://www.taurusdiving.co.uk</A>
 
The formula is: Degrees (minutes/60) (seconds/3600)

Example - 30° 55' 32.5" = 30.92593°

5 decimal places is equal to approximately 3.6 feet.

Forget squares, measure distance from centre of where you are, and possible error

Reasonably good explanation with worked examples <A target="_blank" HREF=http://life.csu.edu.au/geo/dms.html>here</A>

<hr width=100% size=1>There is no such thing as "fun for the whole family."
 
Cheers for that Brendan.

But lets me put it another way , bit more long winded.

Surveying some wrecks and stuff on the Goodwin Sands for a project, i get a hit and for example the Lats say .

Hit A 51 01.123 N
Hit B 51 01.210 N
Hit c 51 01.220 N

What i'm trying to work out is whats the distance between .123 & 210

The end process i'm looking for is that i can make a spreadsheet and converting right down to METERS.

I'm a bit puzzled as i know years ago i did this for an exam somewhere, and my mind has gone a complete blank.

Now , if i split up 51 01.123 N i get this

51 01 . 1 = 1/60 2 = 2/600 3 = 3/6000th of a mile ,,,



still stumped

Andy

<hr width=100% size=1><A target="_blank" HREF=http://www.taurusdiving.co.uk>http://www.taurusdiving.co.uk</A>
 
Sorry, can't work that one out. Sure that some mathematician can provide answer

<hr width=100% size=1>There is no such thing as "fun for the whole family."
 
First of all, 51 01.123 is not a position per se but a latitude only and needs a longitude to go with it to define a precise position.
The definition of a nautical mile is 1minute of arc of Latitude which varies slightly betwween the equator and the poles but is generally accepted as being 6080 feet or 1,853 metres.

A Latitude of 51degrees 01.123minutes North could therefore be interpreted as being 1.123 nautical miles north of Latitude 51degrees north.

The decimal fraction of a minute of latitude therefore becomes a straightforward decimal fraction of a nautical mile eg:
0.123 x 6080 = 747.84 feet .....or..... 0.123 x 1853 = 230.1 metres.
0.210 x 6080 = 1276.80 feet ....or......0.210 x 1853 = 389.13 metres.

HOWEVER, you cant treat decimal parts of a minute of longitude in the same way because a minute of arc of longitude is only equivalent to a nautical mile at the equator which is why, when we lift a distance from the chart with the dividers, we measure it against the Latitude scale

So, I am a little confused as to exactly what you are trying to do here........../forums/images/icons/laugh.gif


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Well you asked ...

Angles are in radians, distances in nm ... I dont want to post ALL the code, so if any functions arent self explanatory do give me a call.

What version of Excel are you using - I migh JUST have a dll that you could call from excel to do the hard work for you!

Function RangeBearing(fromPos, ToPos : LatLon) : RangeAndBearing;
var
dLon_W, dLon_E, d_Phi,d_q : double;
RetRes : RangeAndBearing;
A1,A2,TanA2 : Double;
begin
dLon_W := ModRealTwoPi(FromPos.Longitude-ToPos.Longitude);
dLon_E := ModRealTwoPi(ToPos.Longitude-FromPos.Longitude);
A1 := ToPos.Latitude/2+PiBy4;
A2 := FromPos.Latitude/2+PiBy4;
Tana2 := Tan(A2);
if tanA2 > 0 then
d_Phi := tan(A1)/tanA2
else
d_phi := 0;
if d_phi <> 0 then
begin
d_Phi := ln(d_Phi);
if abs(Topos.Latitude-FromPos.Latitude)< 0.000000001 then
d_q := cos(fromPos.Latitude)
else
d_q := (toPos.Latitude-fromPos.Latitude)/d_Phi;
if dLon_W < dLon_E then
begin
retres.Bearing := ModRealTwoPi(arcTan2(-dlon_W,d_Phi));
retres.Range := sqrt(sqr(d_q)*sqr(dLon_W)+sqr(topos.Latitude-fromPos.Latitude));
end else
begin
retres.Bearing := ModRealTwoPi(arcTan2(dlon_E,d_Phi));
retres.Range := sqrt(sqr(d_q)*sqr(dLon_E)+sqr(topos.Latitude-fromPos.Latitude));
end;
RetRes.range := RetRes.Range*RadiansToNM;
end else
begin
RetRes.Range := 0;
RetRes.Bearing := 0;
end;
result := RetRes;
End;


<hr width=100% size=1>
 
Thanks boatyone.

You quote "So, I am a little confused as to exactly what you are trying to do here "

It's now confusing me.

I have my own dive charter business, and have borrowed a MAGNETOMETER for a little while to find some wrecks on the Goodwin Sands.

I have had some strong hits, which are likely to be cannons, so as i have the positions eg 51 01.123 N & 51 01.220 N , I was trying to work out the distance in meters bewteen the 2,, using Microsoft excel to work out the answers for me.

Then i can use graph paper and try and draw a map.

At present , i don't want to put divers down , bang in a peg, bewteen the 2 points and run out a tape measure.


Andy.

<hr width=100% size=1><A target="_blank" HREF=http://www.taurusdiving.co.uk>http://www.taurusdiving.co.uk</A>
 
Not sure about your diving skills but your navigation seems some what lacking. Sorry if I've got wrong end of stick but B1 has said it all. Wheres your latitude figure??? Anyone else got a starting figure for the Goodwin Sands???

<hr width=100% size=1> <font color=blue>No one can force me to come here.<font color=red> I'm a volunteer!!.<font color=blue>

Haydn
 
hayden

The position is A EXAMPLE.......... Like i really want to give anyone the EXACT position of a newly found pile of CANNONS on the Goodwin sands at the minute.

Part of my CV , states RYA Yachtmaster Offshore ( Motor ) commercial endorsed
To many dive quail's to list, but lets say i have the experience to back up the tickets ;-).

Andy

<hr width=100% size=1><A target="_blank" HREF=http://www.taurusdiving.co.uk>http://www.taurusdiving.co.uk</A>
 
Convert back to degrees, minutes, seconds and use Traverse Tables ?

On 2nd thoughts with the seconds you might have to plot it on graph paper but that shouldn't take long.

Stephen

<hr width=100% size=1>
 
For those who are intrested.

first hit was in position 51 10.431 N 001 30.430 E to get there, a draft of less than ONE meter will help, plus having a mate with the towfish in a £ 25 kids rubber dinghy bought from the local garage ;-).


ANdy


<hr width=100% size=1><A target="_blank" HREF=http://www.taurusdiving.co.uk>http://www.taurusdiving.co.uk</A>
 
Only makes sense if all the points are at the exact same longitude, then you can work it out directly, with one degree equaling 6080 feet. If they're at different longitudes, you'll need to do some trigonometry. Or just plot the buggers on a big chart and measure it off with dividers.

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Or it could be a line of small cannon /forums/images/icons/smile.gif I fail to see what the problem is. I think I can understand what he is trying to do, but its simple, 1 nauticle mile is 2026yds off the top of my head, So .500 is roughly half mile, .250 is quarter mile and so on. The easiest way to measure the distance between the sonar hits is to put both in as waypoints press go to, and read off the difference (in yards not metres)

<hr width=100% size=1>
 
Can I offer a different solution?

Can you plumb both co-ords into yout GPS or chartplotting software as waypoints and let the electo-wossit do the sums for you? From recollections my thingamees go down to yards/metres between two points.

<hr width=100% size=1>
 
Looks like I do have that DLL, callable from Excel if it would make life easier. I've just added functions to return range or bearing betweem two positions.

<hr width=100% size=1>
 
Andy - go to :-

http://www.marquis-soft.com

and download their free prog. "Graphpaper printer".

Open this, go to change / statistics / Mercator
enter 51 in the latitude bit and generally fiddle with sizes etc and then print your own (blank ) Mercator chart onto which you can plot your positions at whatever scale you please.

I have just done this and it is surprisingly accurate.

Not knowing your lats and longs for both points I cannot tell you the answer!!!

Stephen


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