[Proj] Re: Distance measured in Mercator projection

[Shannon Scott]

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
>
> Thank you all for your comments. 
> 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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