Wednesday, November 23, 2016

How to calculate curvature on a globe model

Many flat earthers calculate the curvature of the globe based on a radius of 6371 km. Logic then states that every km of distance there will be a curvature or drop of the earth of 7,8 cm. For the following km this is the same but then we also have to take into account the angular rotation over the first km. The second km actually results in a 31 cm drop in reference to our position. At 5 km this is 1,96 m and at 10 km this is 7,8 m. So you see there is quadratic acceleration of that drop or curvature. This method (R^2+b^2)^(1/2)-R is used by many flat earthers to point out that a far away place cannot be seen on a globe as the curvature of the earth would prevent this. Many times they have a point but there are instances that they are wrong as they don't take into account your line of sight. How does this work and how should it be used to prove a flat earth?

Used theory

The most flat earthers use a diagram showing the curvature of the earth versus the distance. That is based on triangle-calculation of the earth. If the globe earth has a radius of R then the curvature drop over distance b is calculated as: 
x=(R^2+b^2)^(1/2)-R in which R=6371 km. In this case x can be your eye-height watching to the horizon with your line of sight, but x can also mean a building or ship behind the horizon. This means that for proving that an object has disappeared completely behind the horizon of the curved globe then you will need to use your line of sight to calculate it. In that case it is necessary to deduct the distance from your location to the horizon from the distance from you location to the specific object. Based on that distance you must calculate the curvature-drop from the horizon on. If you don't do it like that than it is impossible to prove or disprove a flat earth or a globe.

Example calculation curvature with line of sight

From Washington's Rock (400 ft elevation (120 m high) you can see the skylines of both New York and Philadelphia, total distance 120 miles (192 km /2 = 96 km) and the skylines should be hidden behind 800 fl (240 m) of curvature (1). In this case the line of sight result in a distance to the visible horizon on a globe model of (6371120^2-6371000^2)^(1/2) = 39,1 km. This number has to be deducted from the total distance aka 96-39,1=56,9 km. With this distance you will need to calculate the curvature drop after the horizon. x=(6371^2+56,9^2)^(1/2)-6371 = 254 m aka 846 ft. Based on the source (2) actually 17 buildings would be visible of the New York skyline as these buildings are higher than 254 m. But on the photo in the book of Eric Dubay you can actually see lots of buildings, and actually see complete buildings from the ground up. This suggests based on the calculations with line of sight that there is something wrong with this calculation or that there is something wrong with the globe model. If one of these statements is true than it certainly would suggest that it is not a globe with a radius of 6371 km and would support a flat earth. The forgoing method of calculation (based on a globe) in my opinion is the correct one. When proving or disproving you should remember to deduct the distance until the horizon first and based on that calculate the curvature drop after the horizon.

Sources
(1) 200 Proofs Earth is not a spinning ball, Eric Dubay 
(2) https://en.wikipedia.org/wiki/List_of_tallest_buildings_in_New_York_City

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