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Re: Strange Maths (was Re: Why not 13 months?)
Whittet (Whittet@shore.net)
26 Jul 1995 17:41:37 GMT
In article <3v24ii$1t8@manuel.anu.edu.au>, christy@rschp2.anu.edu.au says...
>
>doug@netcom.com (Doug Merritt) writes:
>>
>>Take the pyramid height to be 100 units and the wheel to be 1/2 unit
>>in diamater. Then its circumference is pi/4. Now measure out 100
>>rotations of the circle from the center of the pyramid, giving
>>a triangle with base y = (100*pi/4) and height 100. The hypotenuse
>>is then
>> r = sqrt( (100*pi/4)^2 + 100^2)
>> r = sqrt( 6168.5 + 10000 )
>> r = 127.155
>>The angle formed by the hypotenuse and the base is
>> sin theta = y/r
>> sin theta = 100/127.155
>> sin theta = 0.7864
>> theta = 51.85 degrees
>> theta = 51 degrees 51 minutes
>>
>>Ta-da. Simple, and easily within reach of Egyptian technology. All
>>they have to do is measure 100 units for the height and 100 rotations
>>of the wheel for the base. (Obviously one gets the same answer no
>>matter whether this number is 100 or 55 or 723, as long as it's
>>the same for the height and base.)
>> Doug
>>--
>
>And now for an amusing scale-invariant coincidence (?) involving this angle:
>
>The right-angled triangle calculated by Doug above (midpoint of base edge,
>centre of base and apex of pyramid as vertices) shows an inclination
>of the side edges of arctan (4/pi) = 51 deg 51.24' . Fine. As Doug shows,
>it would have been easy for the Egyptians to measure out the base and
>height of such a pyramid, and there is an obvious motivation: *symbolic*
>squaring of the circle (or showing that they could use wheels to generate
>a square with the same area as a given circle).
>
>However:
>
>This angle is closely approximated by arctan (sqrt(phi)), where phi is
>the well-known Golden Ratio, (sqrt(5)+1)/2 = 1.6180339..
>
>This angle is 51 deg 49.64'. The tangents , and hence pyramid heights,
>would differ by only 0.096%. The important feature of this angle is
>that the semi-base of the pyramid and the height (the orthogonal sides
>of the triangle) and the sloping face length (hypotenuse) are in the ratio
>1:sqrt(phi):phi. This is the *only* right-angled triangle for which the
>three side lengths are in a geometric ratio. Phi, of course, also figures
>prominently in the ratios of various inter-apex distances in the pentagon,
>dodecahedron & icosahedron, and is the limiting ratio between successive
>terms of the Fibonacci series 1,1,2,3,5,8,13,21,34,55,89... The Ancient
>Greeks were fascinated by it .
>
>AFAIR Phi was used in architecture by the Greeks in classical times. I
>am not aware of any evidence supporting similar mathematical sophistication
>for the Egyptians, but the coincidence of these two angles would have
>doubtless been exciting if they *had* been into phi.
The Egyptians were into Phi, and your observations have been noted by others.
If the half base of the pyramid is taken as 1 and its apothem as phi, its height
is the square root of phi. If we square a height of 280 cubits we get 78,400
sq cubits, while the area of a triangle with a base of 440 cubits and the
height of the slant side as 356 cubits (220 x 1.6181...) is 78,320 sq cubits.
Also because Pi/2 = 2/square root of phi to a close aproximation, Pi can be usefully
taken as 4/square root of phi (3.144...)a difference of less than half a meter
in the height of the pyramid.
The Kings chamber in the Great Pyramid is also built according to the Phi proportion.
>
>
>Just an idea...
>
> Andy C
>
Steve
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