Pi-Like Ratios for Circular Lakes
Pi-Like Ratios for Circular Lakes
Everybody knows ==π for a circle. But for a circular lake on the Earth, the ratios are not constant. In the case of a lake, let be the length of the shoreline, its area, and the distance along the surface from the center to the shore. Let the radius of the sphere be .
C
2r
A
2
r
C
A
r
R
The radius of the Earth is 3963.192 miles. A circular lake with lake radius miles has =3. Perhaps whoever was thinking of defining to be 3 had this lake in mind.
r=2075.12
C
2r
π
For a circular lake on a sphere with radius and lake radius , , and .
R
r
C(r,R)=2πRsin(r/R)
A(,R)=C(r,R)dr=2π(1-cos(/R))
r
0
r
0
∫
0
2
R
r
0
The ratios and are candidates for "" for a circular lake. All four of their limits are as or (as the lake flattens and becomes more like a disk).
C(r,R)/(2r)
A(r,R)
2
r
π
π
r0
R∞
The slider ranges from 1 to 3963.192, from the radius of a unit sphere to the radius of the Earth in miles.
R
Set and move to be between 0 and . The circular lake ranges from a point to all of the sphere except a point.
R
r
πR
When , the circular lake is the upper hemisphere and =2. When , both ratios are a little less than 3.
r=
πR
2
C
2r
r=
πR
4
Move , the imitator, to determine values of and )for which =p and =p, for from 1.4 to 3. Then set r equal to or to see what these lakes look like.
p
π
r
(
r
C
r
A
C
2r
A
2
r
p
r
C
r
A