Abstract
Derived formulas, drew diagrams and ran calculations for spherical earth claim
Two individuals on planet earth at the same elevation standing 6 miles apart with a body of water between them should have a visual barrier of 6 feet of water between them
A 6 foot tall barrier of water is the same as average height of European Caucasian humans rounded to the nearest foot
If people approximately 6 feet tall standing at the same elevation (relative to sea level) can see each other's feet with binoculars while standing 6 miles apart then this would refute the claimed radius of the earth if it is a sphere. It would not refute that the earth is a sphere because it could just be a much bigger sphere instead of flat for example which might indicate hidden land.
Intro
My calculatiions could be wrong
If my calculations are correct the ideal distance to test the alleged shape and size of the earth is two people standing 6 miles apart from each other at the same elevation
A way to try to stay the same elevation might be to look at each other across a body of water with their feet standing near where the water meets the land at the coast of the same body of water
The reason for that distance is that at a path of 6 miles between them there should be a visual barrier of water between them that is 6 feet tall
6 feet tall is the average height of Caucasian humans of European ancestry rounded to the nearest foot
If the earth is a sphere of the radius claimed they should be able to see more of their bodies as they get closer to each other than 6 miles and less of their bodies (at a certain point none of their bodies) as they stand farther away than 6 miles
I checked for the alleged radius of the earth at the equator at the highest mountain on the equator where the radius is supposed to be the largest and the alleged radius between the north and south poles where the earth's radius is supposed to be the smallest and in all cases two individuals standing 6 miles apart as you would walk the curve would have a visual barrier approxiamately 6 feet tall between them (between 5.989 feet and 6.016 feet) this is approximately the standing height of a person rounded to the nearest foot
The following is an explanation for how to calculate the height of a visual barrier of water between two people standing on a spherical planet followed by diagrams that go with the explanation for how to get the formula then followed by calculations using the radius of the earth according to sources listed later below, screenshots of calculations and diagrams were created at websites listed in links below in the last section, along with various websites by other people related to the same topic for additional reading, the information in some of these websites do not give the same results as one another and it is important to understand the difference between a visual barrier due to curvature and a visual drop due to curvature I am hoping by viewing these other websites along with my own explanation of how to calculate a visual barrier this will become more clear to the reader.
Method of Formulation
If two points A and B are selected on a outer surface of a sphere with less than one fourth of the circumference of a circle with an equal radius to the sphere apart from each other as traveled by the shortest arc path along the surface of the sphere the following can be calculated using a diagram of a two dimensional circle instead of a more complicated diagram of a three dimensional sphere because only two orthogonal dimensions are needed to represent the shortest path of travel in Cartesian coordinates.
Let us call the direction of the shortest straight line between the two points (A and B) horizontal and the other direction perpendicular to it vertical.
If one were to try to imagine eyes being at the one point and looking at the other point (at a distance of zero units from the surface) the eyes could not see the other point because there would be an opaque barrier between the points and the vertical height and horizontal width of the barrier could be calculated.
Let R be the radius of the sphere and P be the shortest arc distance between the two points (A and B) that one could travel walking on the surface of the sphere to get from one point to the other. Let H be the shortest straight line distance between the two points (A and B) which would be the horizontal width of the visual barrier at it's thickest in the horizontal direction or the horizontal width of the visual barriers base. Let V be the height of the opaque barrier at it's highest in the vertical direction which would be the distance measured between a point called W halfway between the two points (A and B) to a point called T which shall be located at the top of the sphere halfway between the two points horizontally but not vertically. Let C be a point at the center of the sphere.
Let the angles measured be in radians. Let us call the angle ACB the name Theta. The angle ACW has an equal value to the angle WCB both being one half of Theta. Two symmetric triangles with right angles are formed one contains points A, C and W and the other containing points W, C and B. Using those two triangles the following can be calculated.
P = length of curved Path between points A and B
V = height of Visual barrier = length of line segment WT
Length of WT + Length of CW = V + Length of CW = R
Length of CW = R * cos(Theta/2)
V = R - Length of CW
V = R-R*cos(Theta/2) = R*[1-cos(Theta/2)] = R*[1-cos(P/2R)]
H = width of base of visual barrier
H = the straight Horizontal distance between points A and B
H = length of line segment AB = length of AW + length of WB
length of AW = length of WB = R * sin(theta/2)
H = 2R*sin(Theta/2) = 2R*sin(P/2R)
R = the Radius of the planet
Theta is an angle in radians
Theta = P/R
Theta/2 = P/2R
Theta/2 = inverse sine of H/2R
The straight line length from A to T is equal to the straight line length from T to B is equal to the square root of V squared plus (H/2) squared.
Now imagine an individual whose eye height from the bottom of their feet to their eyes was a height of Z was standing on point A and a duplicate of that individual was standing on point B. If they were both oriented as line segments that were purely in the vertical direction with no horizontal components in order to see each other's eyes Z would need to be greater than V. But if the line segment of the person standing on point A was parallel to the line segment AC and the line segment of the person standing on point B was parallel to the line segment BC.
Then in order for the two people to see each other's eyes
Z = Standing eye height
Z*cos(Theta/2) > V
Z*cos(P/2R) > R*[1-cos(P/2R)]
And
Z > V / cos(Theta/2)
Z > R*[1-cos(P/2R)] / cos(P/2R)
And the horizontal distance between their eyes length N would be
N = H+2Z*sin(Theta/2) = [2R*sin(P/2R)]+[2Z*sin(P/2R)]
Z and N are not yet included in diagrams at the present time
Diagrams for formulation made with Geogebra and IMGflip
Calculations done with geogebra and duckduckgo
Links to geogebra produced diagrams modified in IMGflip
http://web.archive.org/web/20211014154142/https://imgflip.com/i/5qespv
http://web.archive.org/web/20211014154142/https://i.imgflip.com/5qespv.jpg
http://web.archive.org/web/20211014155226/https://imgflip.com/i/5qeulb
http://web.archive.org/web/20211014155208/http://i.imgflip.com/5qeulb.jpg
http://web.archive.org/web/20211014160801/https://imgflip.com/i/5qex46
http://web.archive.org/web/20211014160802/https://i.imgflip.com/5qex46.jpg
Terms of copyright
Copyright Carl Janssen 2020, 2021
Material Above is copyrighted some of the material below is from outside sources and not intellectual property of Carl Janssen
Permission to duplicate copyrighted portion in entirety with attribution providing the link to the original article
http://web.archive.org/web/*/http://teachingthenarrowway.blogspot.com/2020/02/visual-border-height-on-perfect.html
No permission is granted to duplicate a portion without duplicating the entire article
Sources of data for radius, and calculations for Radius and double radius
http://web.archive.org/web/20211008193450/space.com/17638-how-big-is-earth.html
http://web.archive.org/web/*/https://nssdc.gsfc.nasa.gov/planetary/factsheet/earthfact.html
Double Equator Radius
7926 miles
https://duckduckgo.com/?q=%3D3963*2&ia=calculator
the highest point above Earth’s center is the peak of Ecuador’s Mount Chimborazo, located just one degree south of the Equator where Earth’s bulge is greatest. The summit of Chimborazo is 20,564 feet above sea level
https://oceanservice.noaa.gov/facts/highestpoint.html
https://web.archive.org/web/20211013024510/https://oceanservice.noaa.gov/facts/highestpoint.html
Mount Chimbazo summit approx distance in miles from center of earth
Chimbazo Summit radius = Chimbazo Elevation + Equator Radius
3963 miles + 20564 feet / 5280 feet per mile
3966.89469697 miles
https://duckduckgo.com/?q=%3D3963%2B(20564%2F5280)&ia=calculator
Double Chimbazo Summit Radius
7933.78939394 miles
https://duckduckgo.com/?q=%3D(3963%2B(20564%2F5280))*2&ia=calculator
Earth's polar radius is 3,950 miles
http://web.archive.org/web/20211008193450/space.com/17638-how-big-is-earth.html
Double Earth's polar radius
7900 miles
https://duckduckgo.com/?q=%3D3950*2&ia=calculator
Additional Reading Recommendations
https://duckduckgo.com/?q=8+inches+per+mile+squared&ia=web
https://duckduckgo.com/?q=earth+curvature+formula+in+miles&ia=web
https://duckduckgo.com/?q=earth%27s+curvature+per+mile&ia=web
https://duckduckgo.com/?q=horizon+drop+per+mile&ia=web
https://earthcurvature.com/
http://web.archive.org/web/20211007043857/https://earthcurvature.com/
http://web.archive.org/web/20200219191233/https://flatearth.ws/8-inches-per-mile
Online Tools used in calculations, graphics and diagram creation
https://geogebra.org/graphing
https://www.math10.com/en/geometry/geogebra/fullscreen.html
https://imgflip.com/memegenerator
copied and pasted angle symbol charecter from here
http://web.archive.org/web/20210307090638/wumbo.net/symbol/angle/