Eclipse on November 23, 2003. The umbra goes from Australia to South Pole and at last on Chili. On globe model, the sun cannot possibly go that way, i.e. crossing the south pole until on “the other side of the earth” (Chili).
Only South Polar Centered Flat Earth AKA PHEW FE map can possibly shape such umbra path.

Pi’s teritory is circular area (2D).

Phew’s territory is circumference (1D).

Let’s examine 1D stuff in which Phew rules.

 

Scale of Radius = 1×100 Cm = 100 Cm AKA 1 Meter

Base Line = 13.0806 Cm

Pi’s arc = 13.0899 Cm

Phew’s Arc = 13.2149 Cm

Neck = 0.2141 Cm

Even worse for Pi:

The difference of Pi’s arc  vs Base Line =

13.0899 Cm – 13.0806 Cm = 0.0093 Cm AKA 0.093 Milimeter.

It even does not touch 1/10 Millimeter.

The question is: If 7.5° Pi’s arc was correct, -after getting bent or pressed with 0.0093 Milimeter displacement- it would be able to lift a neck of 0.2141 Cm as required by a perfect 7.5° of arc.

Can the difference 0.093 Milimeter ‘produce’ a neck of 0.2141 Cm??

No one can imagine it would happen.

What about Phew?

Its arc equals 13.2149 Cm and the difference with the base line equals 13.2149 Cm – 13.0806 Cm =  0.13426 Cm.

Does it make sense that the difference 0.13426 Cm (between Phew’s 7.5 Cm arc and the Base Line) can create a neck of 0.2141 Cm?

Why not? 🙂

This effort requires a RESET of Phew’s curved grids to be like AE FE map’s grids.

There is gonna be absolutely new latitudes and longitudes for any places on earth.

Because at so many cases the existing globe’s longitude and latitude grids are not compatible to figure out a specific direction from certain position. Grids don’t reflect the exact lining up between a place, ‘north pole’ (North Star) and south pole. Equinox realities suggest that solar noon does not precisely occur at 12.00 AM. Solar noon is delayed about 15 minutes.

This reality indicates that longitudes have been given to the wrong places.

In short, grids should be defined, or got in order first before figuring out directions from any place with high precision and efficient.

After everything is served, the steps for figuring out a place’s direction, is as follows (which is a piece of cake) :

1. Define the two radiuses. 1st radius is between South Pole & the departure place, 2nd radius is between South Pole & the destination place.

For places on northern hemiplane:

(nR = Radius for northern hemiplane places. sR = Radius for southern hemiplane places).

nR = (90+lat)/90

For places on southern hemiplane:

sR = (90-lat)/90

 

2. Define the angle.

Angle = Longitude1 – Longitude2

 

3. Draw a line between the two positions.

 

4. Put protactor’s center point at the departure place while its 0° indicator leads to the north direction, and then define the angle between both places.

Note:

For eastwards journey:

Azimuthal Direction = 0° + angle of both places.

For westwards journey:

Azimuthal Direction = 360° – angle of both places.

 

 

 

Capture  from a sunrise footageAfter getting zoomed in…

it’s clear, that’s not ‘Sunrise’, that’s ‘Sunrives’ (Sun Arrives). The sun comes from above the horizon (precisely above the cloud), not from below the horizon. That’s why even in earliest moment of sun appearing, the underneath part of the sun is already visible.

If it were sunrise, the underneath part of the sun wouldn’t be visible at such moment.

It’s the old propisition about Spherical Area and its sub areas.

Things bothering my eyes are, especially, the two blue areas, the two yelow areas and the two green areas. (Basically all of them are incorrect).

They’re not identical twins.

Phew’s 3.17157 size is a bit bigger, while Pi’s 3.14159 size fits the circlular reality.

Phew area old assumption is incorrect. While Pi area/sub areas really apply in this diagram. It is proposional by sight, and accurate by (series) calculation.

However, for 1D, Phew evidence is extremely clear. With the right tools, i.e. a paper tape & a cylinder, it’s extremely hard to meet Pi. The measurement results support Phew more. 2×Pi is a bit shorter for an actual circumference.

Not guarantee, even not the case that C : A = 2 : 1. (Regardless both dimensions)

Curved horizon on the north

 

You can compare with this ‘flat’ south horizon

Pi has 0.285398 which equals some area under (1 – Cos45)

Phew has 0.292893 which equals perfectly (1 – Cos45)

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=3.17157.

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 this 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

New finding for phew: the total length for 90° triangle base line + 90° arc = √2 + 1.5857864 = exact “3”. And for all 360° equals “12”

PS.:

The exact value of 1.5857864 equals 2 – (√2-1)