# (Segmented) Spherical Area Formula (2) ~ Revision of Sphere Area Formulas ~

Following up this post: https://gwebanget.home.blog/2019/08/12/segmented-spherical-area-formula I came to a conclusion that SPHERE AREA EQUALS (ALSO) 4 PHEW. So, the school’s theorem seems to be right, in the sense of the mechanism that spherical area equals 4 circle areas, or  “4 C/D” (within 2 dimensions). To me, they only have to change the magnitude of pi=3.14159 to be phew=3.15157.

By series calculation under phew constant, in the range of 60° to 90° or height range = (1-cos60) to (1-cos90) or 0.5 to 1, for n=10, the area equals 3.170486, which exceeds 50% of previous hemisphere’s area AKA 3.1619 of 6.3238

Since h=0.5 AKA 2×60°=120° from the center of the sphere, gives division of the hemisphere area into two identical areas, i.e. 3.17157 + 3.17157, that means, old proposition i.e. 3.1619 of 6.3238 becomes incorrect. It’s a bit smaller than even series calculation, which results the magnitude 3.170486.

Another finding is at n=4 (still at the same case) which gives 3.16855 of surface area. It also exceeds the old figure for h=0.5 to 1 i.e. 3.1619.

This 3.16855 magnitude, equals with the sphere area for h=0 to 0.5 but the n equals 8. It’s double of n=4 as applied at h=0.5 to 1.

Interesting, they’re identical. But I still need further research to figure out anything related to thia matter.

Let’s go to the formulas based on the hemisphere’s position:

“open” means the hemisphere is facing up. With respect to the height the radius is variable.

“closed” means the hemisphere is facing down. With respect to the height, the radius is fixed.

For “open” hemisphere position, Segmented Spherical = (h/r) × Tahu.r² = Tahu.h.r

problem:

r=100 m

h=50 m

Area = 6.34314 × 50 × 100 = 31,715.7 m²

Next :

For “closed” hemisphere position:

Segmented Spherical Area = ( 1+(h/r)² ) × Phew.r² = Phew.h² + Phew.r² = Phew.(h²+r²)

The CLUE of the formula of spherical area is SLANTED LINE.

SLANTED LINE equals scrt (h²+r²) which functions as the growing radius from the initial radius in circular area or the base of the hemisphere.

Spherical Area = Phew.(slanted area)² = Phew.[scrt(h²+r²)]² = Phew.(h²+r²)

Problem:

r=100 m

h=50 m

Area= 3.17157 × [(50m)²+ (100m)²] = 39,644.625 m²

Note:

Tahu is Phew version for Tau. It’s a constant for circumference which equals 2 Phew.

Phew = 3.17157

Tahu = 6.34314