Calculate displacement

[Haavard Holm]

Hello,

  I am writing an application were I need to calculate the distance
between two points.
The two points are given in latitude and longitude coordinates (dd.mm.ss).

How do you calculate this distance ? Is it neccessary in your
application to take into
account that the radius of earth is not constant ?

Best regards

Håvard Holm

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[hamish_b]

Haavard Holm wrote:
>  I am writing an application were I need to calculate the
> distance between two points.
> The two points are given in latitude and longitude
> coordinates (dd.mm.ss).
>
> How do you calculate this distance ?

What you want is a formula to calculate Great-Circle distances, aka
the Geodesic distance.

There is a program called "geod" which comes with PROJ.4 (often in a
linux package called "proj") which will do this for you, if you use
geo applications on Linux it is probably already installed.

https://trac.osgeo.org/proj/

> Is it neccessary in your application to take into account that the
> radius of earth is not constant ?

Yes. Probably the best program out there for this sort of thing is the
more modern version of geod called "geodesic" (same author). You can
find it at
  http://home.comcast.net/~gevenden56/geodesy/project/

The +ellps=<ellipse> command line option tells it which ellipsoid model
of the Earth to use.


More help can be found on the PROJ.4 mailing list.


regards,
Hamish



     

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[R. G. Newbury]

Haavard Holm wrote:

>
>
> Hello,
>
>  I am writing an application were I need to calculate the distance
> between two points.
> The two points are given in latitude and longitude coordinates (dd.mm.ss).
>
> How do you calculate this distance ? Is it neccessary in your
> application to take into
> account that the radius of earth is not constant ?

To determine the rhumb course and distance between {Lat1,Long1} and
{Lat2,Long2) determine X where

X = (Lat2 - Lat1)/ 60
   = difference in latitude in nautical miles
   = the adjacent side of the triangle
then determine Y where

Y = 60 (Long1 - Long2) cos( (Lat1+Lat2)/2)
   = Departure in nautical miles
   =  the opposite side of the triangle

then course to go is Y/X = atan( theta)
   (since tan = opposite over adjacent)
and X = r cos(theta) and Y = r sin(theta)

and distance r = X / cos( theta)   or r = Y / sin (theta)

Note, that if you are doing this from a map or chart, any
degrees,minutes,seconds readings (dd.mm.ss) will have to be transformed
to dd.ddddddd form before doing the trigonometric calculations.
(Good scientific calculators, such as Hewlett-Packard and TI have these
conversions built in....usually as HMS->HR or similar naming)

Also note that you have to be careful with your naming scheme to get the
correct quadrant for the atan coversion angle.

And if you are writing a program in 'C' or 'C++', the math library
(libm) only works with radians, not degrees, so you have to do:

HMS -> HR ( add, subtract, etc) -> radians -> sin or cos etc. *and then
back*!

If you want to learn more (and there is LOTS more to learn) download a
copy of 'The American Practical Navigator' commonly called 'Bowditch'
after the original author (in 1802). You need to read Chapter 24 and
section 2411 particularly. The Bi-Centennial Edition is available as a
37M zip file, expandable to a 42M pdf file from


http://www.nga.mil/portal/site/maritime/?epi_menuItemID=c56aa099e2bff9525b2a7fbd3227a759&epi_menuID=35ad5b8aabcefa1a0fc133443927a759&epi_baseMenuID=e106a3b5e50edce1fec24fd73927a759

Yes, that is all one URL!


There ARE java programs available through the web if you just want to
wnat the course and range between a couple of spots.

Of course, most GPS units make it effectively impossible to determine
the bearing and range between arbitrary positions, although you CAN do
it with some units if you can ping the places as waypoints. But if you
are not there (yet) you cannot.

Also, no standalone GPS can tell you where any arbitrary place is, in
lat-long, given a range and bearing from 'here'....


Geoff




--
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          I may wish to offend you again in the future.

          Tux says: "Be regular. Eat cron flakes."
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[R. G. Newbury]

Haavard Holm wrote:

>
>
> Hello,
>
>  I am writing an application were I need to calculate the distance
> between two points.
> The two points are given in latitude and longitude coordinates (dd.mm.ss).
>
> How do you calculate this distance ? Is it neccessary in your
> application to take into
> account that the radius of earth is not constant ?
>

Addendum. The curvature of the earth is taken care of in the formula I
gave, in the (Lat1 + Lat2)/2 calculation.

This is mid-latitude calculation. This is not "precise" compared to
great circle calculations, but the error is less than about one-tenth of
a mile, at ranges less than 600 miles, and latitudes below 60
degrees...and do you really care that it is 399.4025 miles between 2
places, as compared to '400' if you are flying, sailing, or driving?

Great circle calculations involve a LOT more number crunching.
Geoff

>


--
         Please let me know if anything I say offends you.
          I may wish to offend you again in the future.

          Tux says: "Be regular. Eat cron flakes."
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[hamish_b]

> Haavard Holm wrote:
> >  I am writing an application were I need to calculate the distance
> > between two points.
> > The two points are given in latitude and longitude
> coordinates (dd.mm.ss).
> >
> > How do you calculate this distance ? Is it neccessary in your
> > application to take into account that the radius of earth is not
> > constant ?

R. G. wrote:
> Addendum. The curvature of the earth is taken care of in
> the formula I gave, in the (Lat1 + Lat2)/2 calculation.

curvature, yes, but non-constant radius of Earth, no.

> This is mid-latitude calculation. This is not "precise"
> compared to great circle calculations, but the error is less
> than about one-tenth of a mile,

note that in Bowditch's time sun-shots at noon with a sextant were used
to determine latitude, so simplifications were not such a big deal. A
fairly good guess at the ellipsoid shape of the Earth long predates
modern day technology however.

The accuracy needed will of course depend on the application. But as
you never know how a program you write will be (mis)used, IMO it
is better to err on the side of pedantically correct general solution
than to expect the user to respect the assumptions required to allow
simplifications.


> Great circle calculations involve a LOT more number crunching.

actually for a sphere it is not so bad, see the formula in the attached
image.

Great circle calculations using a realistic ellipsoid shape instead of
a perfect sphere do take a bit more number crunching, but then that's
the exact thing which computer programs are good at, and the code is
already written and available for public use.


For the record, GpsDrive's great circle calculations use the WGS84
ellipsoid for the eccentricity of the Earth.


regards,
Hamish



     

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