Rabinoz's calculations for the circumference of the earth (The circumference can be calculated from (distance from Vaupes) * 360°/(angle difference of sun from Vaupes)) and based entirely on what I claim are fictitious globe numbers. Also I see it could be possible the sun rises and falls in elevation on a flat earth. So no proof here either.
So, you don't believe the distances we get from Google Earth or the sun elevations from
Sun Earth Tools?
Well, lets's just forget that and uses figures all contained in your Wiki. Can't be fairer than that, can we?
I will use the same locations, so the latitudes will be the same, but even that doesn't matter.
Finding your Latitude and Longitude
Latitude
To locate your latitude on the Flat Earth, it's important to know the following fact: The degrees of the Earth's Latitude are based upon the angle of the sun in the sky at noon equinox.
That's why 0˚ N/S sits on the equator where the sun is directly overhead, and why 90˚ N/S sits at the poles where the sun is at a right angle to the observer. At 45˚ North or South from the equator, the sun will sit at an angle 45˚ in the sky. The angle of the sun past zenith is our latitude.
Knowing that as you recede North or South from the equator at equinox, the sun will descend at a pace of one degree per 69.5 miles, we can even derive our distance from the equator based upon the position of the sun in the sky.
Here is the basic method, often called "Voliva's Method".
Around 1899,
Thomas Winship, author of Zetetic Cosmogony, provides a calculation demonstrating that the sun can be computed to be relatively close to the earth's surface
if one assumes that the earth is flat:
On March 21-22 the sun is directly overhead at the equator and appears 45 degrees above the horizon at 45 degrees north and south latitude. As the angle of sun above the earth at the equator is 90 degrees while it is 45 degrees at 45 degrees north or south latitude, it follows that the angle at the sun between the vertical from the horizon and the line from the observers at 45 degrees north and south must also be 45 degrees. The result is two right angled triangles with legs of equal length. The distance between the equator and the points at 45 degrees north or south is approximately 3,000 miles. Ergo, the sun would be an equal distance above the equator.
This is illustrated in this diagram from
Modern Mechanics - Oct, 1931:
Voliva's Flat Earth Sun Distance. This is also shown in the Wiki under
Distance to the Sun under the section
Sun's Distance - Modern Mechanics.
But this calculates the height from only
ONE location, Latitude 45°.
In would seem that we would get
a more accurate result by taking measurements from a
number of different latitudes and comparing the results.
So this time, I will present the sun elevations from six locations, close to longitude 70°W as in
So you think the sun is about 5,000 km high? « on: August 23, 2016, 08:22:33 PM »And I will calculate the sun's elevation as just 90˚ at the equator (Lat = 0˚), decreases by 1˚ for we move North of South of the equator. The distance from the equator will be calculated as
(degrees of lat) x 69.5 miles as in the Wiki.
Also, I will make it more like the "Voliva Method" and just take all distances from the equator, not Vaupes.
The following table gives the data for each location. As for Voliva the date and time are for the overhead sun at an equinox, here
UTC 20/Mar/2016 16:48 - the time the sun was overhead at long 70˚ West.
Location
|
Latitude
|
Sun Elev
| Dist from
Equator |
Flat Sun Ht
| Calc
Circum |
Kimmirut, Canada | 62.847° | 27.15° | 4,368 miles | 2,240 miles | 25,020 miles |
Bingham, Maine, USA | 45.059° | 44.94° | 3,132 miles | 3,125 miles | 25,020 miles |
Santo Domingo | 18.486° | 71.51° | 1,285 miles | 3,843 miles | 25,020 miles |
Equator | 0° | 90° | 0 miles | - - - - | - - - - |
Chupa District, Peru | -15.109° | 74.89° | 1,050 miles | 3,889 miles | 25,020 miles |
Los Tamariscos, Argentina | -45.033° | 44.97° | 3,130 miles | 3,126 miles | 25,020 miles |
Punta Arenas, Chile | -53.164° | 36.63° | 3,695 miles | 2,747miles | 25,020 miles |
Once we have the elevation angle of the sun from the site the height of the sun can be calculated from:
h = d x tan(ElevAng) .
Also, if the earth is taken as a Globe the calculated circumference is found from
(distance from equator) x 360°/(latitude difference from equator).
This time, I must stress, the sun elevations and distances are calculated exactly as in "the Wiki".
So, please no excuses like blaming Google earth or "Globalist" figures - they are Flat Earth distances from the Society's Wiki!
Using this method to find the height of the sun on the Flat earth gives measurements from 2,240 miles (for Kimmirut to the Equator) to 3,889 miles (for Chupa District to the Equator) depending on the spacing of the measurement sites.
Also for at locations close to 45° from the equator we get the usual Flat Earth sun height of "a bit over 3,000 miles, vis 3,125 miles and 3,126 miles.
In other words, claiming that the Flat Earth sun is at about 3,000 miles altitude has no foundation whatever.
It is very telling when we use the same data to calculate the earth's circumference, as Erosthanes did, we get an absolutely consistent 25,020 miles circumference.
Explain that!This consistency is simply because we used "the Wiki's" figure of exactly 69.5 miles/degree, whereas the actual sun elevation figures from SunEarthTools would have little errors.
Certainly these figures would indicate that the earth is a globe with a distant sun.
<< typo, swapped let for lat >>