What distance is lat 0.001 lon 0.001?

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Major Catastrophe

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A work colleague has just sent me this mail.

<span style="color:blue"> Could you possibly answer the following question or direct me to somewhere that can.

If I know the latitude and longitude of a place i.e. lat:52.563 lon:-2.142, what area "on the ground" would be covered by a square with corners at lat:52.562 lon:-2.141 and lat:52.564 lon:-2.143. i.e. What distance is lat:0.001 lon:0.001
</span>
 
distance across=.002 cos (52.46degrees) NM=0.0012185 NM
distance up=.002 NM
Area=2.43709e-6 NMxNM
1 NM=1852 m
About 8.36 metre squared.

What units - hectares, acres, barns ...?

Did I understand the question right?

I must be bored!
 
Your question is unanawerable. First the earth is a globe so the area bounded by any pair of lat and long lines is not rectangular - especially towards the Poles. As you approach the Poles the shape enclosed approaches a triangle. That's why there are parallels of latitude but not of longitude.
Second the distance from one parallel of lattitude to another varies with the latitude. So you can give an approximate answer specific to a latitude but not a general answer for all latitudes. I will now sit back and wait to be shot down in flames by the rest of the forum.
 
Correct but ...

He has given a location and geographic data ...... so the area is defined.

Trouble is the area is so small that the error will be massively out of proportion to the actual area to calculate !!

If I had my Nav calculator handy - I would do the calc. (May try MNAVD .... )
 
I'm assuming your measurements are in decimalised degrees, not in degrees and minutes which is the way long and lat are usually measured.

You want the area of a shape that is 0.002 degrees across both in lat and long. At latitude 52.563N, a degree of longitude is 60 nm and a degree of latitude is 36.5 nm.

So your shape is an approximate rectangle with sides 0.120nm and 0.073nm respectively, i.e. an area of 0.00875 sq nm. As a nautical mile is 1852 meters, this is 30,000 square meters.

I think!
 
Assuming + is N / W - is S / E ?? (Not that it really matters - just that my nav calc likes N S E W !!)

Also are the figures minutes.decimals ............ or degrees.decimals ?

eg. 52.563 .... 00deg 52'.563 ............ or 52.563deg
 
You are quite right about his general question - his "square" would not be a square and would have no area at the poles and have the area .002NMx.002NM at the equator. However, he also gave a specific lat and long - for which I might have given the right answer, if I understood him right! The cos bit makes it work.
 
He gave the lat and long as an i.e. So an approximate answer can be provided at that location ignoring the fact that even there the area is not a rectangle but the general answer - for all locations - does not exist
 
About your colleague...

If he is quoting Lat. and Long. to 1/1000ths of degree, should we assume that he is working in land miles rather than natical miles, which are a bit bigger. If he was at sea he would possibly be more likely to quote Lat. and Long. in Degrees/Minutes/Seconds OR Degrees/Minutes/Decimal Fractions of a minute.
 
Re: About your colleague...

Thanks for all the answers. As he didn't specify exactly how he was presenting his coordinate I have simply replied to him that due to the longitude being compressed at the poles it depends where he is on Earth.
 
Re: About your colleague...

I understand his problem a bit more. He is trying to build a service for a customer so that withiin any postcode he can automatically calculate the distance from one point in the postcode to another point by using a formular based upon the know distance of Lat Long minutes and seconds.

I assume due to Longitude meeting at the poles that this would be a variable based upon where in the UK he was.

Is that correct?

I did suggest he used grid references instead as they are based n a Mercator map.

His last mail to me read:

<span style="color:blue">I am not concerned with ultimate accuracy and I am only using co-ordinates within the UK, not for the whole world with extremes of lat/long.

Computers, surprisingly, don't understand 52.56.3N 2.14.2W

These co-ordinates have to be converted into normal decimal numbers, that computers do understand and can work with
so "computer speak" 52.563 is actually "boat speak" 52.33.47N and -2.142 is actually 2.8.31W
</span>
 
Re: About your colleague...

OK:
Your two positions are N1 N, E1 E and N2 N, E2 E.
where N1 ,N2, E1, E2 are all in degrees and decimals of a degree. If the position is W, then it's negative.
b=(N1-N2)
c=(E1-E2)* cos(N1)
(make sure your calculator is using degrees for the cos.)
The distance on the surface of the earth is then:
Radius*arccos(cos(b)*cos(c))*pi/180
where the radius of the earth is 6370 km

... or thereabouts.
 
Re: About your colleague... ok ...

Looks like he's using MS Autoroute ... as that is how it shows lat / long in the data box.
 
Re: About your colleague...

Seen this done at one of the places I was contracting at - has he looked at the shapes of the postcode boundaries and the licencing costs - some of the shapes are very complex and the licencing costs of post codes and boundary data is very high.

I might have all the calcs somewhere for doing exactly what he wants to do so I'll see what I can dig out.
 
Re: About your colleague...

No one seems to have picked up Richardabeaties mistake regarding Latitude.
In Latitude the distance between degrees is 60 Nm regardles of whether you are at the poles or equator. It is Longitude whose degree spacing varies from equator to nil at the poles. olewill
 
Re: About your colleague...

No, he was right, the distance between parallels of latitude varies, being greater at the equator. You were partially right in saying that the distance between parallels of latitude is always 60NM but the length of a NM varies depending on where you measure it.

For that reason, when measuring distances on a chart covering a large area, you should read off the latitude scale at the same latitude as the feature you are trying to measure.

Having said that, the differences would not be significant for this application.
 
Latitude is 1 minute of arc ...

Not a physical measure such as metre etc.

But distortion is based more on chart projection than physical earth ...
 
Re: About your colleague...

[ QUOTE ]
Having said that, the differences would not be significant for this application.

[/ QUOTE ]That's right, hence my earlier answer, which should have been sufficiently accurate for most practical purposes.

But as we all guessed, the original question was wrongly stated. So - garbage in, garbage out.
 
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