Is the Earth a Globe or Flat?
Here is a postulate for determining the flatness or sphericalness of the Earth by determining the circumferences of 5 circles:
1) AB = Arctic Barrier circle circumference
2) A = Arctic circle circumference
3) E = Equator circle circumference = 24,901miles
4) AA = Antarctica circle circumference
5) AAB = Antarctica Barrier circle circumference
D = Diameter of Earth axis = 18,047/2(3.14159) = 18,047/6.28318 = 2,872.27168 miles
The equation of a circle passing through three points (x1,y1) (x2,y2) (x3,y3)
Let (j,k) be the coordinates for the center of the circle, with r as its radius.
Then the equation of the circle is:
(x-j)^2 + (y-k)^2 = r^2
(x1-j)^2 + (y1-k)^2 = r^2
(x2-j)^2 + (y2-k)^2 = r^2
(x3-j)^2 + (y3-k)^2 = r^2
The equation of the circle is:
r = squareroot[(x1-j)^2+(y1-k)^2]
|x1^2+y1^2 y1 1|
|x2^2+y2^2 y2 1|
|x3^2+y3^2 y3 1|
j = -----------------------------
|x1 y1 1|
2*|x2 y2 1|
|x3 y3 1|
|x1 x1^2+y1^2 y1 1|
|x2 x2^2+y2^2 y2 1|
|x3 x3^2+y3^2 y3 1|
k = ---------------------------------
|x1 y1 1|
2*|x2 y2 1|
|x3 y3 1|
(j-x1)^2 + (k-y1)^2 = r^2
(j-x2)^2 + (k-y2)^2 = r^2
(j-x3)^2 + (k-y3)^2 = r^2
(j-x2)^2 - (j-x1)^2 + (k-y2)^2 - (k-y1)^2 = 0
(j-x3)^2 - (j-x1)^2 + (k-y3)^2 - (k-y1)^2 = 0
j^2-2(x2)*j+(x2)(x2) - j^2-2(x1)*j+(x1)(x1) + k^2-2(y2)*k+(y2)(y2) - k^2-2(y1)*k+(y1)(y1) = 0
j^2-2(x3)*j+(x3)(x3) - j^2-2(x1)*j+(x1)(x1) + k^2-2(y3)*k+(y3)(y3) - k^2-2(y1)*k+(y1)(y1) = 0
Solve for j and k, then use equation for circle to find r
r = squareroot[(x1-j)^2+(y1-k)^2]
ABa=same latitude point north of Asia (Aa)
ABb=same latitude point north of North America (Ab)
ABc=same latitude point north of Europe (Ac)
Aa=northernmost point on Asia
Ab=northernmost point on North America
Ac=northernmost point on Europe
AAd=southernmost point on Australia
AAe=southernmost point on South America
AAf=southernmost point on Africa
AABd=same longitude point south of Australia (AAd)
AABe=same longitude point south of South America (AAe)
AABf=same longitude point south of Africa (AAf)
ABa: (lat1, long1) = (x1,y1)
ABb: (lat2, long2) = (x2,y2)
ABc: (lat3, long3) = (x3,y3)
Aa: (lat4, long1) = (x4,y1)
Ab: (lat5, long2) = (x5,y2)
Ac: (lat6, long3) = (x6,y3)
AAa: (lat7, long4) = (x7,y4)
AAb: (lat8, long5) = (x8,y5)
AAc: (lat9, long6) = (x9,y6)
AABa: (lat10, long4) = (x10,y4)
AABb: (lat11, long5) = (x11,y5)
AABc: (lat12, long6) = (x12,y6)
Correcting for errors caused by Cartesian system presumptions: 1) curvature of earth is negligible so earth can be presumed “flat” 2) also presumes the grid formed by latitude and longitude is a square. These presumptions are negligible for small distances, however they are compounded over greater distances. Therefore the following offsets for latitude and longitude can be used to compensate for the differences in the earths circumference at the equator and the diameter of the earth at the north/south pole axis:
Constant ratio latitude offset = earth’s circumference/360degrees/60minutes
24,901miles/360/60=1.15282 miles
Constant ratio longitude offset = earth’s circumference/360degrees/60minutes
18,047miles/360/60=0.83553 miles
ABa: (lat1 * 1.15282 miles, long1 * 0.83553 miles) = (x1 * 1.15282 miles,y1 * 0.83553 miles)
ABb: (lat2 * 1.15282 miles, long2 * 0.83553 miles) = (x2 * 1.15282 miles,y2 * 0.83553 miles)
ABc: (lat3 * 1.15282 miles, long3 * 0.83553 miles) = (x3 * 1.15282 miles,y3 * 0.83553 miles)
Aa: (lat4 * 1.15282 miles, long1 * 0.83553 miles) = (x4 * 1.15282 miles,y1 * 0.83553 miles)
Ab: (lat5 * 1.15282 miles, long2 * 0.83553 miles) = (x5 * 1.15282 miles,y2 * 0.83553 miles)
Ac: (lat6 * 1.15282 miles, long3 * 0.83553 miles) = (x6 * 1.15282 miles,y3 * 0.83553 miles)
AAa: (lat7 * 1.15282 miles, long4 * 0.83553 miles) = (x7 * 1.15282 miles,y4 * 0.83553 miles)
AAb: (lat8 * 1.15282 miles, long5 * 0.83553 miles) = (x8 * 1.15282 miles,y5 * 0.83553 miles)
AAc: (lat9 * 1.15282 miles, long6 * 0.83553 miles) = (x9 * 1.15282 miles,y6 * 0.83553 miles)
AABa: (lat10 * 1.15282 miles, long4 * 0.83553 miles) = (x10 * 1.15282 miles ,y4 * 0.83553 miles)
AABb: (lat11 * 1.15282 miles, long5 * 0.83553 miles) = (x11 * 1.15282 miles,y5 * 0.83553 miles)
AABc: (lat12 * 1.15282 miles, long6 * 0.83553 miles) = (x12 * 1.15282 miles,y6 * 0.83553 miles)
= (miles,miles)
AB<A
Therefore
If AA<AAB, then the earth is more flat
or
If AA>AAB, then the earth is more spherical
or
If AA or AAB > 24,901 miles, then the earth is more flat
or
If AA or AAB < 24,901 miles, then the Earth is more spherical
Unfortunately, the last math class I took was Calculus in high school and a physics class at UCLA three years ago, so I need assistance creating a formula to determine distance of a curve based off the circle formed by three points on a sphere or flat surface. However, I only began thinking about a scientific way to determine whether the Earth was more round or flat last night. This is what I’ve come up with so far from this morning.
An airplane flying between any of the points (AAd to AAe or AAe to AAf or AAf to AAd,) on a path on the same plane measuring the nautical miles in between, could then be plugged into the formula to determine the length of the sides of the triangle, based on the distance between two adjacent points on the southern tips of Australia, South America and Africa, to determine the coordinates for the circle’s center which runs along the axis of the North and South poles.
If the world is a globe then a race around Antarctica is possible, however if the world is flat, then the race will be impossibly long. I’ll give it more thought on another day. Any assistance or ideas would be appreciated. Thanks.
John Kee, copyright 2018
