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I'm trying to calculate the distance between the WGS84 meridian at 51°28′40.1″N 0°0′5.3″W and the actual prime meridian at 51°28′40.1″N 0°0′0″W in feet. Wiki claims the two points are 5.3'' from each other. My calculation below results in 334 feet and this differs from Wiki's answer of 336.21 feet. I looked at Google maps and used the measure feature, after which I measured 190 feet between both coordinates and this is pictured.

Does anyone know how to correctly calculate this and why the differences between Google, Wiki and my calculation?

$1^\circ = 60′ = 60\,\text{nm} = 69\,\text{mi}$

$1′ = 60″ = 1\,\text{nm} = 1.15\,\text{mi}$

$1″ = 60″′ = 1/60\,\text{nm} = 101.2\,\text{feet}$

$5.3'' \cdot 101.2\,\text{feet} * cos(51.47°) = 334\,\text{feet}$

notaorb
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2 Answers2

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You have used the simple rule $$1' = 1\ \text{nm}.$$ This rule is correct for distances in north-south direction. But for distances in west-east direction it is not so simple.

You need to consider the fact that neighboring meridians don't have a constant distance. (Just look at a globe to understand this.) Instead, the meridians get closer and closer to each other, when walking away from the equator towards the north or south pole. At the equator you have $$1' = 1\ \text{nm}.$$ But going away from the equator the correct rule for west-east distances becomes $$1' = \cos(\phi)\cdot 1\ \text{nm}$$ where $\phi$ is the geographical latitude. In your case you have $\phi=51.47°$, and hence $\cos(\phi)=0.6228$.

Thomas Fritsch
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  • WGS84 uses an ellipsoid, but over that small distance, the difference between the geodesic length on a sphere vs the ellipsoid is about half a metre. According to https://geographiclib.sourceforge.io/scripts/geod-calc.html The distance is 102.282 m ~= 336'6.85" (that calculation ignores altitude). – PM 2Ring Aug 24 '21 at 19:24
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    @PM2Ring Well, ok. For the sake of simplicity, I have treated the Earth as a sphere instead of an ellipsoid. – Thomas Fritsch Aug 24 '21 at 19:35
  • Sorry but what does $\text{nm}$ stand for here? Obviously not nanometer... – Gert Aug 24 '21 at 20:10
  • Perhaps the ellipsoid geometry needs to be taken into account? It seems 192' (Google), 290' (above), and 336' (Wiki) have large differences. – notaorb Aug 24 '21 at 20:15
  • @Gert nautical mile – notaorb Aug 24 '21 at 20:16
  • Ah. Thank you, friend. – Gert Aug 24 '21 at 20:18
  • @Notaorb Google Earth uses the EGM96 geoid (or maybe 2008), but that differs from WGS84 by 147 feet at Greenwich. – JEB Aug 25 '21 at 00:08
  • @JEB, I get 334' using poster's advice, so that is off by 142'. I looked up EGM96 and read Wiki's Geoid entry, where the Geoid is the shape of the Earth's surface if it were covered by water under Earth's gravity. How would this result in -142' for Google? Is Greenwich part of an oblate ellipsoid that such that the radius to center is less, resulting in shorter length and Google's number? – notaorb Aug 28 '21 at 19:35
  • I'm talking about the height about the ellipsoid vs the elevation above MSL. Most maps (& modern GPS) report elevation, but this calculation requires height. My answer is 100% the correct answer, I know this because google (and DOD, and the CIA, NRO, etc..) got their map from NGA (formerly NIMA, formerly DMA), who got the map from NASA, who got the map from JPL, who got it from section 334 who executed SRTM shuttle mission, and I got a NASA achievement award for calibrating the whole damn thing, and I used the software I wrote for that on this problem. QED. – JEB Aug 28 '21 at 20:20
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The correct answer is 335.573739 feet.

The easiest way to do this is convert lat, lon, height to ECEF and compute the cartesian distance. I just posted the formula here (Actual astronomic latitude (direction of gravity) does not match calculations taking into account centrifugal force of earth).

Of course, that is "as the neutrino flies" and doesn't follow the surface of the Earth. At this distance, the difference is micron tiny.

A more accurate method is to place a locally tangent sphere on one of the points and then project the other onto the sphere, you then get an arc length between them. It gives the same answer to within 1 micron. For instance, if you place the locally tangent sphere at the 1st point, and put the second point on the sphere, you find that it sits at a bearing of 89.9994209 (slightly north of due East), and at a distance of 102.281619 meters, and 0.001028 meters north of great circle pointing due east, with an elevation of 0.9 nanometers (sphere's aren't ellipsoids, but if you use locally Cartesian coordinates, the elevation is -0.8 millimeters, so the sphere is a much better approximation).

An iterative solutions called Vincenty's formula (https://en.wikipedia.org/wiki/Vincenty%27s_formulae) also works.

The 336.5' number is wrong...probably uses the spherical solution.

btw: I used a height for Greenwich of 84.7 meters.

JEB
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