Re: Metric Time (was Re: Why not 13 months? (Was La Systeme Metrique))
Peter Lamb (prl@cbr.dit.csiro.au)
Fri, 20 Oct 1995 03:15:46 GMT
Whittet@shore.net (Whittet) writes:
>In article <45hm7q$5r5@lucas.emi.com>, phaedrus@ says...
>>By your figures, the circumference of the Earth at the equator is
>>24902.72727 miles (BTW, is that statute miles or nautical miles?
>Statute miles, the nautical mile is a completely different animal
> I
>>guess it's statute, since there are 21600 nautical miles around the
>>equator.). That works out to be 131 486 399.9856 feet.
>>
>>I checked, and it works out even IF that pattern of .7272 repeats.
>>Which leads me to suspect that the number above was computed from a
>>figure for feet/equator
Yes, phaedrus is correct; if you take 365.24 days for the year, and
take Whittet's formula that the circumference of the earth in inches
is half the number of seconds in a century, as your definition of the
circumference of the earth, then this does come out at
..
24902.72mi (ie. '72' repeating).
>The figure comes from averaging several sources.
>Funk & Wagnalls encyclopedia gives 24902.4 mi
>The Information Please Almanac gives 24902.4 page 343
>Eighth Edition Surveying Moffitt & Bouchard cites the
>Clarke Spheroid of 1866 = 24901.417 (Northern Hemisphere)
>semimajor axis 6,378,206.4 m
>GRS 80 Spheroid 6,378, 137.0 m
>The Universal Transverse Mercator System also includes
>Clarke 1880 (Africa)
>International (South Pacific)
>Everest (India)
>Bessel (China)
>New Spheroid pending (Australia)
As far as I could tell, all of these values lead to circumferences of
the earth which are less than 24902.72727mi, so I don't know how
this value could be the "average".
Circumference
in miles
Funk & Wagnalls 24902.4
The Information Please Almanac 24902.4
Clarke Spheroid of 1866 (+) 24901.417
GRS 80 Spheroid semi-major axis (6 378 137.0m) 24901.46
Clarke 1880(*) (6 378 249.14m) 24901.90
International(*) (6 378 388m) 24902.44
Everest(*) (6 377 277m) 24898.10
Bessel(*) (6 377 398m) 24898.58
New Spheroid pending (Australia) (**) (6 378 160m) 24901.55
Mean 24901.13
+ I get the circumference of this as being 24901.73mi, but I have
used the value cited (the 6 378 206.4m semi-major axis quoted
is correct for Clarke 1866). Perhaps a different approximation of
pi was used. I used 3.1415926.
* These are calculated from a circle of radius equal to
the semi-major axes of the ellipsoids of the quoted names,
taken from Table E-1 and page 2-4 of "Datums, ellipsoids, grids
and grid reference systems," (US) Defence Mapping Agency, DMA TM 8358.1
** I'm not sure which spheriod is intended here. The semi-major axis
of the current Australian Geodetic Datum (proclaimed April 1966) has
been used ("The Australian Geodetic Datum: technical manual", Nat.
Mapping Council of Aust, Special publication 10, 1986).
>>Where did you get your 24902.72727mi figure.
>See above. It actually was just the best fit between all the values
>I could find.
All of the sources cited give a smaller value than 24902.72727mi.
--
Peter Lamb <peter.lamb@dit.csiro.au>
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