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Posted by Poster Matt on February 1, 2008, 7:28 am
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rgc wrote:
>
>>> So if I wish to calculate the radius of the Earth at the latitude of
>>> N56:29:35 (56.492956), what formula can I use?
>> 6372 * cos(56.492956/180*pi) = 3518 km.
>>
>> As you move toward one of the poles, this will finally become 0.
>>
>> Or do you have Earth flattening in mind? If not, the answer can also be that
>> the radius doens't change.
>
> Actually, your approach calculates the radius of a circle of latitude.
> The radius of curvature of the Earth is a bit more complicated.
>
> This issue is discussed in rather excruciating detail in the FAQ 5.1
> article I mentioned in my previous post. (Look for the section labeled
> 5.1a which is just short of half way through the article.) For
> convenience, and because that post could easily get separated from this
> one, here it is again:
>
> http://www.usenet-replayer.com/faq/comp.infosystems.gis.html
Thanks for the pointer Bob - in both posts.
Cheers.
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Posted by Jacquelin Hardy on February 6, 2008, 3:51 pm
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Uffe,
eventhough your question might look clear, it is in fact very vague. If
you are a seaman, you are looking for a seaman answer.
If your are a geodesist, you are looking for a geodesist answer.
If you are a mathematician, you are looking for a mathematical answer.
Here is a possible answer
The geoid:
The hypothetical surface of the Earth that coincides everywhere with
mean sea level and is perpendicular, at every point, to the direction of
gravity. The geoid is used as a reference surface for astronomical
measurements and for the accurate measurement of elevations on the
Earth's surface.
From that point of view, there is no formula to calculate the radius of
Earth, just readings from surveys.
The reference ellipsoid:
In geodesy, a reference ellipsoid is a mathematically-defined surface
that approximates the geoid, the truer figure of the Earth, or other
planetary body. Because of their relative simplicity, reference
ellipsoids are used as a preferred surface on which geodetic network
computations are performed and point coordinates such as latitude,
longitude, and elevation are defined.
If you take WGS84 as a reference ellipsoid, the Earth is flattened by
1/298.257
Then you can calculate the radius for each latitude.
Seamen use the great circle for long planning routes. They assume,
without large errors, that the Earth is spherical.
As you see, a question can rise many more questions
Jacquelin Hardy
a seaman that likes maths.
Uffe Kousgaard a écrit :
>> My question is slightly different. The mean radius of the Earth is approx.
>> 6,372 km, but what formula is used to calculate the radius of the Earth
>> for any given latitude?
>
> R * cos(latitude)
>
> As I understand your question.
>
>
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Posted by Uffe Kousgaard on February 6, 2008, 4:59 pm
Please log in for more thread options > Uffe,
>
> eventhough your question might look clear
Matt asked the question.
Regards
Uffe
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Posted by Paul Cooper on February 7, 2008, 4:42 am
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>> Uffe,
>>
>> eventhough your question might look clear
>
>Matt asked the question.
>
>Regards
>Uffe
>
And I think Uffe knows the answers!
Jacquelin is quite correct. I haven't bothered getting involved in
this discussion because as Jacquelin says, the answer is "it depends
on your application". The original poster was being confused by
attempting to compute distances in a projected coordinate system,
which is a necessarily inaccurate way of carrying out the measurement
(I know, it IS accurate if you are doing distance from the pole of
projection of an equidistant projection, but typically that isn't the
case, and wasn't the case for the OP). In fact I suspect that the
simple cosine rule formula was quite accurate enough for his or her
needs; the difference between spherical geometry, ellipsoidal geometry
and surveyed distance on a geoid is only likely to bother a surveyor
looking for closure in a survey. Over long distances, the difference
is less than a kilometre - and that's for distances between Antarctica
and Europe.
A practical solution: if you want a good geodetic distance calcluator,
then use geod which is bundled with proj. That is a) robust b) written
by an expert in the field of geodesy (Gerry Evenden), c) accurate and
d) free! As far as I'm concerned, this is the gold standard for great
circle calculations on the ellipsoid; it avoids all the
computationally nasty things that can happen if you roll your own! And
as Jacquelin has pointed out, you can't compute distance on the geoid
analytically - but I think the difference would be less than a
millimetre! It is worth noting that the maximum vertical distance
between sphere and ellipsoid is about 11km. The maximum vertical
distance between ellipsoid and geoid is +85 metres to -106 metres);
the ellipsoid is a fantastically accurate depiction of the shape of
the Earth.
Paul
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Posted by rgc on January 31, 2008, 8:10 pm
Please log in for more thread options A later version of the referenced FAQ 5.1 may be found at
http://www.usenet-replayer.com/faq/comp.infosystems.gis.html
>
> The thread 'distance between lon and lat', at the end of last year, was
> interesting. Those with an interest in the subject might want to look at
> the Haversine Formula on the page linked below. I can't remember if this
> was mentioned in the thread or not, the thread is so long it'll take ages
> to check, and my useless newsgroup software won't let me search!
>
> Haversine Formula:
> http://www.movable-type.co.uk/scripts/gis-faq-5.1.html
>
>
> My question is slightly different. The mean radius of the Earth is approx.
> 6,372 km, but what formula is used to calculate the radius of the Earth
> for any given latitude?
>
> I wish my trig was up to working this out but I'm just a humble computer
> scientist with no degree in math.
>
> Thank you all for any help in this.
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