# [Proj] Re: Distance measured in Mercator projection

Shannon Scott sscott at locationinc.com
Wed Jul 16 06:53:25 EDT 2008

```Daan,
What I have now seems to be working, but I would like to see the more
complicated formula.
I will use it if I can get it into the code I have.
Thank you.
Shannon

strebe at aol.com wrote:
>
> Glad this works for you. Don't hesitate to ask for the more
> complicated formula if you need greater accuracy — the calculation you
> have will be off by a percent or two because it assumes a spherical earth.
>
> Regards,
> -- daan Strebe
>
>
> -----Original Message-----
> From: Shannon Scott <sscott at locationinc.com>
> To: PROJ.4 and general Projections Discussions <proj at lists.maptools.org>
> Sent: Thu, 10 Jul 2008 4:02 am
> Subject: Re: [Proj] Re: Distance measured in Mercator projection
>
> 1m / cos (phi) was the formula I was looking for.
>
> We have an application that maps data based on a search radius ( ie 5
> miles around Chicago ).
> The maps show the search radius. We had been using UTM for the maps,
> so I was able to plot the search radius with this piece of code:
> <snip>
> // convert miles to meters
> bufdist = (miles * 1609.344);
>
> // create circle
> for (double i=0;i<360;i+=5) {
> deltax = (Math.cos(i) * bufdist);
> deltay = (Math.sin(i) * bufdist);
> pntx = (centerPntx + deltax);
> pnty = (centerPnty + deltay); < br> // plot point
> }
> </snip>
>
> The maps have been converted to work as a google map overlay, so we
> need to use the Mercator projection.
> I have modified the way the search radius is calculated:
> <snip>
> // convert miles to meters
> bufdist = (miles * (Math.abs(1609.344 * (1/Math.cos(latrad)))));
> </snip>
>
> and everything looks good.
> Thanks again.
> Shannon
>
> Ed McNierney wrote:
> > Shannon -
> >
> > First, let me apologize for trying to respond (twice) when I was too
> > rushed to do so properly.
> >
> > Since you say you’re not very familiar with map projections, it
> might > be helpful for you to describe a little bit more about what
> you’re > trying to do. Daan’s and Chris’s comments are both correct
> but could > be confusing to a novice; the scale factor at a single
> point is the > same in all directions for a Mercator projection, but
> strictly > speaking only at that point. If you have another point in a
> different > place, the scale factor at that point will be different
> from the first > unless they’re at exactly the same latitude. So if
> you’re measuring > the distance between two points at a non-trivial
> distance apart, a > 500-mile line will, for example, change length on
> a Mercator map > depending on=2 0its angle with the Equator. What I
> was trying to say in > my first, poorly-worded reply, is that the set
> of points that are all > 500 miles from a center point on a Mercator
> map will NOT form a circle > on that map.
> >
> > If your question really is, “How can I measure the great circle >
> distance between two points given their Mercator coordinates” then >
> that’s rather tricky. And it will depend on how far apart those two >
> points are, and how accurate you need your answer to be.
> >
> > You will find incredibly talented professional expertise on this
> list, > but it’s hard for people to answer the question you didn’t
> ask. If you > can describe – in general terms, rather than specific
> ones – what > you’re trying to do and what questions you’re trying to
> answer, that > will make it easier. If there are constraints on your
> solution (e.g. > “I have to use Mercator so please don’t suggest an
> alternative > projection that would make the math easier”) let us know
> as well. Thanks.
> >
> > - Ed
> >
> > Ed McNierney
> > 205 Indian Hill Road
> > Groton, MA 01450
> > ed at mcnierney.com <mailto:ed at mcnierney.com>
> > +1 (978) 761-0049
> >
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