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 semimajor 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 semimajor 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 semimajor axes of the ellipsoids of the quoted names,
taken from Table E1 and page 24 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 semimajor 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>
