Fire Girl firegirl at amorous.com
Sat Aug 9 03:03:50 BST 2008

Rogier (and David too :)  This is quite exciting.  Based on this, I think
I can formulate something.  Thank you for making me understand.

have a great day !! ~~

----- Original Message -----
From: "Rogier Wolff"
To: "Fire Girl"
cord
Date: Fri, 8 Aug 2008 21:41:52 +0200

On Fri, Aug 08, 2008 at 11:02:15AM -0500, Fire Girl wrote:
> I think I am following you! :)
>
> Here is the math I have so far....
>
> Calculation: 5 miles

I'm assuming radius, not diameter. So we'll get a ~10mile square.

> of: 45.5 latitude
> and: 25.5 longitude

Note that Milo thinks this is inaccurate. He's right. But these
"quick" calculations are very efficient. If for example, you want to
select all OSM objects within 5 miles distance from a certain
coordinate, you damn well better do THIS math first, and only then
use
proper geo-valid formulas to select the real ones.

The reason is that if you put all objects in the database through the
"calculate-distance-to-reference-point" formula, this will take ages,
but selecting on a 10 mile min longitude / max longitude and min
latitude max latitude first will be fast: There should be an index
on those elements of the database.

Now for some applications, the "quick-n-dirty" method is already good
enough.

So.... here we go.

kmpermile = 1.609;
circumferencekm = 40000;
circumferencemile = circumferencekm / kmpermile;

latitude = 45.5; // North
longitude = 25.5; // east!

delta = radius / (circumferencemile / 360);

latmin = latitude - delta;
latmax = latitude + delta;

longmin = longitude - delta / cos (latitude);
longmax = longitude + delta / cos (latitude);

which according to my calculator comes to:

latmin = 45.42759500000000000000
latmax = 45.57240500000000000000
longmin = 25.39669848650172993913
longmax = 25.60330151349827006087

Apparently the cosine changes by about 1 in 776 on this 5 mile
stretch
of earth. So this square is only about 9.987 miles wide at the north
end, and 10.013 miles wide a the south end.

Similar differences can be expected to crop up because the earth is
not a perfect sphere.

Roger.

>
> -------------------------------
>
> so, for the latitude, (i.e. add and subtract 2.5/69ths of a degree
of
> latitude)
>
> With 5 miles assumed, and using that forumla, the distance from
45.5
> would be:
>
> 45.505797 (on the plus side)
> 45.494203 (on the minus side)
>
> Right? :)
>
>
> but on the Longitude side, (i.e. 2.5/f) .... f = 69 * sin(latitude)
>
> .... i am not so clear what Latitude number I should be using there
and
> what that might look like. How do I look at least so far in my
> calculations? :) ----- Original Message -----
> From: "David Earl"
> To: "Fire Girl"
> cord
> Date: Fri, 08 Aug 2008 16:02:10 +0100
>
>
> On 08/08/2008 14:30, Fire Girl wrote:
> > I am working with OSM data, and would like to be able to spec out
> > 5 mile bounding boxes from certain GPS points.
> >
> > After research into this problem, I am to understand that each
> > degree of latitude is approximately 69 miles (111 kilometers)
> > apart with a slight variance (68.703 - 69.407 miles) between the
> > equator and the poles, and that each degree of longitude is
> > widest at the equator @ 69.172 miles (111.321 kilometers) and
> > gradually shrinks to zero at the poles. : ) :)
> >
> > So what does this mean? If I want to take a input point, like
lets
> say,
> >
> > 167.9 lat
> > -29.1 lon
> >
> > or
> >
> > -63.1
> > 18.1
> >
> > Can someone say with authority, what the 'calculus' would be to
> > definitivly construct a NSWE bounding box with a 5 mile radius
> > around those points?.... that would be basically close enough or
> > accurate? :)
>
>
> If you want accuracy, then you are asking the wrong question,
> because the "bounding box" will be a curved section of a surface of
> the earth (or, rather, the approximation to it defined by the
> ellipsoid for the datum you're using for the lat/lon coordinates -
> unless you want to start taking altitude and terrain into account),
> not a flat straight-edged box. So if you want to talk in terms of
> flat bounding boxes, you have to start asking "in what projection?"
> etc.
>
> If all you want is the lat/lon of the corners an approximate box
> parallel to lines of latitude and longitude approx 5 miles across,
> then
> you could indeed just add and subtract 2.5/69ths of a degree of
> latitude and 2.5/f of a degree of longitude, where f is the approx
> length of one degree at that latitude, i.e. f = 69 * sin(latitude)
> miles (with the reasonable approximation that the earth is a
> sphere, so the trigonometry is trivial).
>
> David
>
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>

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> http://lists.openstreetmap.org/listinfo/dev

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