Re: Metric Time (was Re: Why not 13 months? (Was La Systeme Metrique))
Whittet (Whittet@shore.net)
22 Oct 1995 07:16:42 GMT
In article <DGq92D.H4F@cbr.dit.csiro.au>, prl@cbr.dit.csiro.au says...
>
>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.
I used 355/113 = 3.1415929+
>
>* 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.
Perhaps, but those are just the first ones I grabbed; as I pointed out
I have been collecting these for some time. There are several calculations
which give values for the earths circumference larger than 24902.72727 feet.
An interesting clue is the stadium which was supposed to be both 600 Greek
feet and 1 600th of a degree of the Earths circumference.
There are three I could find without looking too hard
By my calculations there are 608.733333...feet in 1 600th degree
The Greek stadium of Periclean Athens was 607 feet.
24831.81818182
The Roman Stadium of 606 feet was close
24790.90909091 miles
Another Stadium of 622 feet was also used
this gives a value for the earths circumference of 25445.45454545 miles
imagine the Greeks aproximating PI by using inscribed and circumscribed
circles and see what that gives you.
In the year 1500 1 furlong equaled 625 feet, but it was increased to 660 feet
possibly because a chain is 66 feet and the furlong at 660 feet is forty rods
originally however, the furlong seems associated with the stadium.
Consider also that the perimeter of the Great Pyramid is 1760 cubits, the
number of yards in a mile. The run of that pyramid is 220 cubits and its
side is 440 cubits. We still sprint the 220 and run the 440.
It is interesting that the meter seems to be based on the Belgic Germanic
system where the measures are in triple decimal intervals
a BG foot = 13.22 inches
a BG yard = 39.66 inches (1 meter = 39.37 inches)
a BG fathom = 79.32 inches
a BG chain = 793.2 inches
a BG furlong = 7,932 inches
a BG mile = 79,320 inches
the link is 7.92 inches, aproximately 1/10th fathom
the diameter of the earth is about 7926.783609887 miles
Making the obvious conjecture...
If I use a diameter of 7932 miles I get 24919.11504425 miles
for the circumference of the earth at the equator
24919.11 + 24901.13 x 9/10 = 24902.928
Consider also that in converting from Meters to feet my surveying text
gives 1 US inch = 2.540005+ cm whereas a British inch = 2.54 cm
The point is that there were values I included in my original
calculation which do give a higher average, and probably some
I listed which were not included due to their probably being
repetitions of the same original source.
>
>
>Peter Lamb <peter.lamb@dit.csiro.au>
Steve
