From: Mike Lorrey (mlorrey@datamann.com)
Date: Sat Mar 09 2002 - 13:56:17 MST
spike66 wrote:
>
> Damien Broderick wrote:
>
> >> Since arctan x = x - x**3/3 + x**5/5 - x**7/7 - ...
> >>
> >> pi/4 = 1 - 1/3 + 1/5 - 1/7 ...
> >>
> > In our world. But if 2 pi radians is defined = 360 degrees, what
> > happens
> > when a circle has more or fewer than 360 degrees, due to space
> > curvature? I have the feeling that by invoking the arctan function
> > you're being, as it were, circular here.
> > Damien Broderick
> > [not any kind of mathematician]
> >
> I disagree sir. In any universe with any curvature, a circle would
> still
> have 360 degrees, for the circle defines the degree, as well as the
> radian.
> The expansion given would work even in a different space curvature,
> and they would derive the same value for pi.
I'm not exactly sure this would occur, but I think that in a steep
gravity gradient, the radius will NOT relate to the circumference by the
same value of pi that we use. See below:
circle from top:
__----------__
/ \
/ \
/ \
| |
| C |
| *------------+R
| |
| |
\ /
\ /
\ /
--____________--
Would appear to be a normal circle, with a normal pi.
Then assume that point C is a gravitational well, such that, looked at
from the point of the gravity gradient:
---------------------------+R < circle, edge on
\ /
\ /
\ /
\ /
\ /
\ /
\ /
\ /
\ /
\ /
C*
Where radius CR does not relate to circumference R by the same ratio of
pi that we use ....
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