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

to

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.

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?

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.

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.

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.

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>,

Jan 31, 2008, 8:18:41 PM1/31/08

to

In article <47a24d4b$0$2112$edfa...@dtext02.news.tele.dk>,

"Uffe Kousgaard" <o...@no.no> wrote:

"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

Feb 1, 2008, 7:28:50 AM2/1/08

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Thanks for the pointer Bob - in both posts.

Cheers.

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 :

Feb 6, 2008, 4:59:47 PM2/6/08

to

"Jacquelin Hardy" <jach...@videotron.ca> wrote in message

news:d5pqj.15792$ld2....@wagner.videotron.net...

> Uffe,

>

> eventhough your question might look clear

news:d5pqj.15792$ld2....@wagner.videotron.net...

> Uffe,

>

> eventhough your question might look clear

Matt asked the question.

Regards

Uffe

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

to

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

Feb 8, 2011, 7:13:50 PM2/8/11

to

chuckage had written this in response to

http://www.psjournal.com/gis/Re-Formula-for-Earth-radius-at-any-given-latitude-3623-.htm

:

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;

##-----------------------------------------------##

Delivered via http://www.psjournal.com/

GPS/GIS Community

Web and RSS access to your favorite newsgroup -

comp.infosystems.gis - 3593 messages and counting!

##-----------------------------------------------##

Oct 10, 2012, 6:39:40 PM10/10/12

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THANK YOU THANK YOU THANK YOU :)

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

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

Jun 3, 2013, 3:09:49 AM6/3/13

to

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

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

Apr 21, 2014, 1:02:01 PM4/21/14

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Jun 9, 2015, 4:19:02 PM6/9/15

to

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.

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.

Jun 9, 2015, 9:18:50 PM6/9/15

to

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))

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.

HTH,

Sam

--

Samuel W. Flint

4096R/266596F4

(9477 D23E 389E 40C5 2F10 DE19 68E5 318E 2665 96F4)

(λs.(s s) λs.(s s))

Jun 10, 2015, 6:20:20 AM6/10/15

to

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
> 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?

is so ill‐posed as to be almost unanswerable.

--

Ian ◎

Dec 2, 2015, 3:42:35 PM12/2/15

to

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

Dec 3, 2015, 7:20:32 AM12/3/15

to

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
> can you explain the "/180*pi" part? I see it obviously works, but

> what's the reasoning for it?

converting degrees to radians?

--

Ian ◎

Feb 18, 2016, 9:04:55 PM2/18/16

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Jan 12, 2018, 5:49:55 AM1/12/18

to

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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