# [Proj] Area of a spherical polygon

Noel Zinn (cc) ndzinn at comcast.net
Tue Sep 21 16:20:18 EST 2010

```pi / 2pi, what you really want is tau.  Follow these links!

http://www.math.utah.edu/~palais/pi.pdf

http://tauday.com/

Noel

Noel Zinn, Principal, Hydrometronics LLC
+1-832-539-1472 (office), +1-281-221-0051 (cell)
noel.zinn at hydrometronics.com (email)
http://www.hydrometronics.com (website)

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From: "Karney, Charles" <ckarney at Sarnoff.com>
Sent: Tuesday, September 21, 2010 3:45 PM
To: "Mikael Rittri" <Mikael.Rittri at carmenta.com>; "PROJ.4 and general
ProjectionsDiscussions" <proj at lists.maptools.org>
Subject: Re: [Proj] Area of a spherical polygon

>> From: Mikael Rittri [Mikael.Rittri at carmenta.com]
>> Sent: Tuesday, September 21, 2010 05:06
>>
>> Hello Charles,
>> you wrote:
>>
>> > * the formula for A applies only if the polygon does not include a pole
>> >   (if it does, then add +/- pi to the result);
>>
>> Don't you mean +/- 2pi ?
>
>
>> > * I would be interested to know if this formula for S12 is already
>> >   known.
>>
>> There is an article
>>
>> Robert D. Miller, Computing the area of a spherical polygon, Graphic
>> Gems IV, chapter II.4, pages 132 - 137.  It is on books.google.com,
>> but without pages 135 and 136:
>>
>> Your formula for S12 does not appear (in any obvious form), but
>> Miller's method reminds me a little of yours. Instead of summing the
>> areas between each segment and the equator, Miller sums the areas
>> between each segment and a pole.  I don't know how well-conditioned
>> his method is for small polygons.
>
> Thanks for the pointer.  I have located the code that implements this
> algorithm at
>
> http://people.sc.fsu.edu/~jburkardt/cpp_src/geometry/geometry.html
>
> routine, sphere_polygon_area_3d, in geometry.C.  This uses L'Huilier's
> theorem
>
>
> for the area of a spherical triangle in terms of its three sides.  If
> the edge is specified by the latitudes and longitudes of its endpoints,
> then this entails computing the length of the edge and substituting the
> result in a somewhat elaborate formula.
>
> In contrast, I started with the formula for the area of a spherical
> triangle in terms of two sides and the included angle (on p 71 of
> Todhunter).  If one vertex is the pole then the sides are the two
> co-latitudes, the included angle is the longitude difference, and the
> formula can be used directly without the need to calculate an
> intermediate result.  My (meager) contribution is to recast the formula
> to give the area between the edge and the equator.
>
> My expectation is that this will yield more accurate results than
> Miller's method.  I will benchmark both approaches to confirm or refute
> this.
>
> By the way, Miller's paper contains one geographic howler: In the
> section "Geographic Particulars" he says that ellipsoidal effects can be
> included for small polygons by taking the effective radius to be
>
>    r(phi) = a/sqrt(1 - e^2 * sin(phi)^2)
>
> Surely, this should be
>
>    r(phi) = b/(1 - e^2 * sin(phi)^2)
>
> --
> Charles Karney <ckarney at sarnoff.com>
> Sarnoff Corporation, Princeton, NJ 08543-5300
>
> Tel: +1 609 734 2312
> Fax: +1 609 734 2662
>
>
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