If you wish constructive answers, you need to provide more information:
Purpose of calculation
Accuracy required
"diameter" of polygon. i.e. what is the maximum of distance between
any two points on the polygon.
How is your math? Calculus? Trig?
If you intend to program the solution, what language?

I need to calculate the area in a rescue alpine region.

Hi,
please, someone can tell me how can I to calculate the area of a
polygon on the earth's surface where I know the coordinates (lat/lon)
of its points?

The area is equal to the spherical excess,
which can be calculated directly from the coordinates.

For small polygons use plane trig,

Please, can you teach me or indicate an url where I can learn to
calculate the spherical excees from (lon, lat) coordinates?

Hi,
please, someone can tell me how can I to calculate the area of a
polygon on the earth's surface where I know the coordinates (lat/lon)
of its points?

The area is equal to the spherical excess,
which can be calculated directly from the coordinates.

For small polygons use plane trig,

Please, can you teach me or indicate an url where I can learn to
calculate the spherical excees from (lon, lat) coordinates?
Thanks

Hi,
please, someone can tell me how can I to calculate the area of a
polygon on the earth's surface where I know the coordinates (lat/lon)
of its points?

The area is equal to the spherical excess,
which can be calculated directly from the coordinates.

For small polygons use plane trig,

Please, can you teach me or indicate an url where I can learn to
calculate the spherical excees from (lon, lat) coordinates?

I need to calculate the area in a rescue alpine region.
I suppose that the area will be few km2 and the maximum distance
between adjacent points will be some hundreds meters, therefore I can
suppose the region planar.
I know trig and I've read some articles on internet about this problem
(spherycal trigonometry, spherycal excess, etc.) but I don't find a
complete solution for my problem
For programming the language used is vb.net (or C#).
Thanks.

Choose one of your points as origin and use the formulae from

Choose one of your points as origin and use the formulae fromhttp://williams.best.vwh.net/avform.html#flatto give you distance north
and east for each point. You can then use the formula I posted earlier
in this thread for the area of a polygon in the X-Y plane.

L¢1000 õ Let L = 1000 meters
LL¢45 100 õ Let LL be lat = 45, lon = 100 as reference position
«LL1¢45 100 SodanoD 1000 0 õ LL1 is a point 1000m north of LL
45.00899832 100
«LL2¢45 100 SodanoD 1000 90 õ LL1 is a point 1000m east of LL
44.9999993 100.0126828
LL Sodano LL1 õ distance on the WGS84 ellipsoid betwen LL and LL1
1000.000036 360
LL Sodano LL2 õ distance on the WGS84 ellipsoid betwen LL and LL2
1000.000036 90

If you can tolerate 0.036mm error in a length of 1000m, the flat earth
approximation should do the job for you. :-)

As to the bulge, look up the formula for the area in a figure bounded by
an arc and its chord. For a 1000m chord and a 6000Km radius (roughly
that of Earth), this is (in sq.m)
6E6 ArcArea1 1000
13.88888884
for a 500m chord, this is
6E6 ArcArea1 500
1.736111164
As you can see, this decreases as the cube of the chord length
÷/6E6 ArcArea1 1000 500
7.999999725
A "few km2" is ~3E6m2 so neglecting the above bulge would give an error
of only a few parts per million. Also, note that the area calculated
above is the area bounded by the arc of the Earth's surface and the
chord through the Earth. This is always greater that the area of the
projection of that figure onto the plane of interest so the error
estimate is an upper bound.

I would happily give you my code but it is in APL2. Do you have access
to APL2?

... but I have still a questio about an affermation that I don't
understand.
Why you say " A "few km2" is ~3E6m2"? From which consideration derives
"3E6m2"?

If your polygon is small enough to allow you to ignore the Earth's
curvature, you can use the following formula. You need to convert
lat/lon to x-y coordinates. If the polygon lies within one UTM region,
you can just convert lat/lon to UTM and use the UTM coordinates.

area = sum_{i=0}^{n-1} ((x_i + x_{i+1}) (y_{i+1} - y_i)) / 2
where point 0 = point N.

Is there a general formula in case of the area is between two UTM
regions?


No. If you need to work between two regions then use lat/lon degree

Choose one of your points as origin and use the formulae fromhttp://williams.best.vwh.net/avform.html#flattogive you distance north
and east for each point. You can then use the formula I posted earlier
in this thread for the area of a polygon in the X-Y plane.

L¢1000 õ Let L = 1000 meters
LL¢45 100 õ Let LL be lat = 45, lon = 100 as reference position
«LL1¢45 100 SodanoD 1000 0 õ LL1 is a point 1000m north of LL
45.00899832 100
«LL2¢45 100 SodanoD 1000 90 õ LL1 is a point 1000m east of LL
44.9999993 100.0126828
LL Sodano LL1 õ distance on the WGS84 ellipsoid betwen LL and LL1
1000.000036 360
LL Sodano LL2 õ distance on the WGS84 ellipsoid betwen LL and LL2
1000.000036 90

If you can tolerate 0.036mm error in a length of 1000m, the flat earth
approximation should do the job for you. :-)

As to the bulge, look up the formula for the area in a figure bounded by
an arc and its chord. For a 1000m chord and a 6000Km radius (roughly
that of Earth), this is (in sq.m)
6E6 ArcArea1 1000
13.88888884
for a 500m chord, this is
6E6 ArcArea1 500
1.736111164
As you can see, this decreases as the cube of the chord length
÷/6E6 ArcArea1 1000 500
7.999999725
A "few km2" is ~3E6m2 so neglecting the above bulge would give an error
of only a few parts per million. Also, note that the area calculated
above is the area bounded by the arc of the Earth's surface and the
chord through the Earth. This is always greater that the area of the
projection of that figure onto the plane of interest so the error
estimate is an upper bound.

I would happily give you my code but it is in APL2. Do you have access
to APL2?

Unfortunately I don't have access to APL2 and I don't know it.
However thank you very much for the explanation. It's clear and
precise!
... but I have still a questio about an affermation that I don't
understand.
Why you say " A "few km2" is ~3E6m2"? From which consideration derives
"3E6m2"?
Thanks,
Luca

Calculate the area of the plane polygon (by Method 1 in
http://www.efg2.com/Lab/Graphics/PolygonArea.htm )

ed williams wrote:
Calculate the area of the plane polygon (by Method 1 in
http://www.efg2.com/Lab/Graphics/PolygonArea.htm)

The formula given there is
area = ( sum_{i=0...n-1} (x[i] y[i+1])-(x[i+1] y[i]) )/2
an alternate which, IIRC, is less subject to round off errors is
area = ( sum_{i=0...n-1} (x[i]+x[i+1]) (y[i+1] - y[i]) )/2
which is readily expressed in APL2 as
A{<-}0.5{times}+/(2+/x){times}{neg}2-/y
where x is the x-coordinates and y, the corresponding y-coordinates.

Ted

On Mar 2, 1:49 pm, Ted Edwards <Ted_Espaml...@telus.net> wrote:
area = ( sum_{i=0...n-1} (x[i] y[i+1])-(x[i+1] y[i]) )/2
area = ( sum_{i=0...n-1} (x[i]+x[i+1]) (y[i+1] - y[i]) )/2

You can reduce the number of arithmetic operations and have rounding
benefits by using:
sum_{i=1..n} x(i) (y(i+1)-y(i-1))/2
where we have further extended the array so that (x(n+1),y(n
+1))=(x(1),y(1)) as well as (x(n),y(n))=(x(0),y(0))

ed williams wrote:
On Mar 2, 1:49 pm, Ted Edwards <Ted_Espaml...@telus.net> wrote:
area = ( sum_{i=0...n-1} (x[i] y[i+1])-(x[i+1] y[i]) )/2
area = ( sum_{i=0...n-1} (x[i]+x[i+1]) (y[i+1] - y[i]) )/2

You can reduce the number of arithmetic operations and have rounding
benefits by using:
sum_{i=1..n} x(i) (y(i+1)-y(i-1))/2
where we have further extended the array so that (x(n+1),y(n
+1))=(x(1),y(1)) as well as (x(n),y(n))=(x(0),y(0))

In what follows z is a 182 point GPS track created by combining the
tracks from several walks in the nearby hills. In order to have one
continuous track, I combined them such that there are some legs that go
back over themselves. The result is a very "messy" polygon.
A1 and A2 are the two formulae from my previous post (repeated above)
and A3 is the formula you just gave. Areas are in square meters. Eps
is machine epsilon. A couple of examples simple enough to work by hand
showed zero difference in the results.

z{<-}meadow
A1{<-}0.5{times}+/({times}/0 1{rot_}z)-{times}/1 0{rot_}z
A2{<-}0.5{times}+/(2+/z[;1]){times}{neg}2-/z[;2]
A3{<-}.5{times}+/z[;1]{times}{first}-/1 {neg}1{rot}{each}{enclose}z[;2]
(2-/A1 A2 A3 A1){div}A3 {rem} relative differences
1.914861534E-15 -5.106297424E-16 -1.404231792E-15
{rem} A2-A1 A3-A2 A1-A3
A3
455967.65
Eps
2.220446049E-16
It seems the closest agreement is between A3 and A2 and, as you point
out, A3 involves the least computation.

Ted

For lurkers, the area computed by this formula is signed. The area
is positive if its interior lies to the left of the point sequence
and vice-versa. With a figure-8 loop, you'd get the difference of the
area of the two sub-loops.

ed williams wrote:
For lurkers, the area computed by this formula is signed. The area
is positive if its interior lies to the left of the point sequence
and vice-versa. With a figure-8 loop, you'd get the difference of the
area of the two sub-loops.

Also very useful if you have something with holes. e.g. a sheet metal
layout for which you wish to calculate weight. Get area by going left
around the outline and right around the holes. Also useful for area of
a lake excluding islands.

Ted