Formula for Earth radius at any given latitude.

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Poster Matt

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Jan 31, 2008, 1:54:17 PM1/31/08
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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.

Uffe Kousgaard

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Jan 31, 2008, 4:34:09 PM1/31/08
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"Poster Matt" <postermatt@no_spam_for_me.org> wrote in message
news:tPooj.326$cJ...@text.news.blueyonder.co.uk...

>
> 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.


Poster Matt

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Jan 31, 2008, 5:17:34 PM1/31/08
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Thanks, but I think I explained my question badly; I'll re-phrase it.

For any specific latitude, what is the formula for calculating the radius
of the Earth at that latitude.

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?

Thanks.

Uffe Kousgaard

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Jan 31, 2008, 5:35:52 PM1/31/08
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"Poster Matt" <postermatt@no_spam_for_me.org> wrote in message
news:2Oroj.493$cJ...@text.news.blueyonder.co.uk...

>
> 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.


rgc

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Jan 31, 2008, 8:10:25 PM1/31/08
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A later version of the referenced FAQ 5.1 may be found at

http://www.usenet-replayer.com/faq/comp.infosystems.gis.html


In article <tPooj.326$cJ...@text.news.blueyonder.co.uk>,

rgc

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Jan 31, 2008, 8:18:41 PM1/31/08
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In article <47a24d4b$0$2112$edfa...@dtext02.news.tele.dk>,
"Uffe Kousgaard" <o...@no.no> wrote:

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

- Bob

Poster Matt

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Feb 1, 2008, 7:28:50 AM2/1/08
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Thanks for the pointer Bob - in both posts.

Cheers.

Jacquelin Hardy

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Feb 6, 2008, 3:51:35 PM2/6/08
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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 :

Uffe Kousgaard

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Feb 6, 2008, 4:59:47 PM2/6/08
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"Jacquelin Hardy" <jach...@videotron.ca> wrote in message
news:d5pqj.15792$ld2....@wagner.videotron.net...

> Uffe,
>
> eventhough your question might look clear

Matt asked the question.

Regards
Uffe


Paul Cooper

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Feb 7, 2008, 4:42:24 AM2/7/08
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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

chuckage

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Feb 8, 2011, 7:13:50 PM2/8/11
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chuckage had written this in response to
http://www.psjournal.com/gis/Re-Formula-for-Earth-radius-at-any-given-latitude-3623-.htm
:

Uffe Kousgaard wrote:

-------------------------------------


-----------------------------------------------------------------------------
function Radius(Latitude : Degrees;
Spheroid : Spheroid_Type) return Meters is
e2 : long_float renames Models(Spheroid).Eccentricity_Squared;
b : Meters renames Models(Spheroid).Semi_Minor_Axis;
f : long_float renames Models(Spheroid).Flattening;
boa : long_float := 1.0 - f;
Phi : Radians := Deg_to_Rad(Latitude);
Psi : Radians := Geocentric_Latitude(Phi, boa);
cl : long_float;
r : Meters;
begin
cl := cos(Psi);
r := b/sqrt(1.0 - e2*cl**2);
return r;
end Radius;

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sweetl...@hotmail.fr

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Oct 10, 2012, 6:39:40 PM10/10/12
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omg thank you so much i have been looking every where on the internet for this ANSWER i needed it for some exercises that i do (I'm in high school)
THANK YOU THANK YOU THANK YOU :)

atob...@gmail.com

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Nov 10, 2012, 12:28:39 AM11/10/12
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The question would require a more complex answer, but if we really need to simplify things, here is an example:

Ex. With a given mean radius of the Earth as 3,440 Nautical Miles, calculate the radius of a parallel of latitude at 46°24'00" North?

Solution:
3,440 NM X Cosine 46° 24' 00" North = Radius

I hope you will get your information from this. If you are studying the science of navigation, try to consult the American Practical Navigator for more concise information.

RCA of Bacolod City
John B. Lacson Colleges Foundation
Bachelor of Science in Marine Transportation
STCW 95 Marine Deck Officer
Founder of Negros Occidental Seafares Association

chris....@baysideflorist.com.au

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Jun 3, 2013, 3:09:49 AM6/3/13
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According to the post at http://gis.stackexchange.com/questions/20200/how-do-you-compute-the-earths-radius-at-a-given-geodetic-latitude
the formula you are looking for is:
Rt= SQRT((((a^2cos(t))^2)+((b^2sin(t))^2))/( ((acos(t))^2)+((b(sin(t))^2)))

Where
Rt = radius of earth at latitude t
a = semi major radius of earth = 6,378,137 meters
b= semi minor radius of earth = 6,356,752.31420 meters




thom...@wildwood.org

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Apr 21, 2014, 1:02:01 PM4/21/14
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hi. im figuring this out now

deem...@gmail.com

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Jun 9, 2015, 4:19:02 PM6/9/15
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I am trying to figure out what city is in the center of the world.

I guess its find out the circumference and find the radius and apply that to a longitude and latitude somehow?

But I reiterate I am not smart.

Samuel W. Flint

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Jun 9, 2015, 9:18:50 PM6/9/15
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deem...@gmail.com writes:

> I am trying to figure out what city is in the center of the world.

It doesn't really work like that. The earth is a sphere(oid), it isn't
flat, which would be required to have a city at the center.

> I guess its find out the circumference and find the radius and apply
> that to a longitude and latitude somehow?
>
> But I reiterate I am not smart.

No, just misinformed/miseducated. And by the way, it's spelled "laymen".

HTH,

Sam

--
Samuel W. Flint
4096R/266596F4
(9477 D23E 389E 40C5 2F10 DE19 68E5 318E 2665 96F4)
(λs.(s s) λs.(s s))

Ian Clifton

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Jun 10, 2015, 6:20:20 AM6/10/15
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swf...@flintfam.org (Samuel W. Flint) writes:

> deem...@gmail.com writes:
>
>> I am trying to figure out what city is in the center of the world.
>
> It doesn't really work like that. The earth is a sphere(oid), it isn't
> flat, which would be required to have a city at the center.
>
>> I guess its find out the circumference and find the radius and apply
>> that to a longitude and latitude somehow?

I guess Accra must be the city closest to long=lat=0°, but the question
is so ill‐posed as to be almost unanswerable.
--
Ian ◎

Ja...@artisticfootprint.com

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Dec 2, 2015, 3:42:35 PM12/2/15
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can you explain the "/180*pi" part? I see it obviously works, but what's the reasoning for it?

Ian Clifton

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Dec 3, 2015, 7:20:32 AM12/3/15
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Ja...@artisticfootprint.com writes:

> can you explain the "/180*pi" part? I see it obviously works, but
> what's the reasoning for it?

I’m not sure of the context of ths question, but surely it’s just
converting degrees to radians?
--
Ian ◎

w.van....@gmail.com

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Feb 18, 2016, 9:04:55 PM2/18/16
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Is the latitude the geodetic latitude or the geocentric latitude?

glauber...@gmail.com

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Jan 12, 2018, 5:49:55 AM1/12/18
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Hello. You could use this online tool.

https://rechneronline.de/earth-radius/

It has also the formula for any latitude, so you can create a function.

If you get the poles and equator radius (6,356.752 km and 6,378.137 km) from this website, you can see you get the same value as you can see in this page https://en.wikipedia.org/wiki/Earth_radius#Global_average_radii.

So I think the formula is correct!
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