Tuesday, November 29, 2016

Distance to the visible horizon on a flat earth versus globe

If we live on a globe then the curvature of the earth goes with some distance quadruple downwards showing us a distinct curve. But if it is a flat earth or a plane then what you can see has a linear relationship. How can you calculate the distance to the visible horizon on a globe model and a flat earth? 

Globe model

We have been told that the earth is a globe with a radius of 6371 km. This means that from our point of view the edge of water or the horizon actually will block the waters that lie behind this line. But for the argument if it is a globe model then how do you calculate the distance to the horizon? So if you are on an outlook of 7,5 m then your horizon would be due to the curvature of the earth at (6371007,5^2-6371000^2)^(1/2) =  9,7 km.

Flat earth model

If you look at ongoing water such as the sea on a plane or flat earth there appears to be a mirror behind the visible horizon. This is caused due to perspective as all the lines of sight converge near the horizon. The visible horizon is created due to to much information or colours behind the horizon. It results in a mirroring effect. From our point of view also are line of sight is mirrored at the horizon. This is very typical as when you follow ships over the horizon, they seem to be disappearing from down to up. But they disappear above the visible horizon. Based on the theory of line of sight and a mirror after the visible horizon and the distance to the Sun you can find a relationship like this for the distance to the flat earth horizon: dh = distance horizon = ht/(tan(tan-1(6,35/15650))) = ht/0,40575 with in that ht = height of eye in m and dh in km. So if you are at an outlook-point of 7,5 m then your horizon would be at dh = 7,5/0,40575 = 18,5 km. Based on this theorised model you can see much further on a flat earth then you can do on a globe.

PS: I am not saying that the forgoing formula for the distance calculation of the visible horizon on a flat earth is completely correct, but it does give a direction on how much it might be.

What is the distance to the sun at the horizon?

Our Sun combined with our line of sight and the mirroring effect actually debunks the globe model. You cannot see it with your own eyes unless you have the right enhancement combined with some Sun-protection but then you might see it. Photo's do prove it actually. The following photo is from Rijkswaterstaat (the Dutch government civil-engineering department) and shows the Sun mirrored. Of-course to see this you will have to enhance the location of the Sun at the horizon greatly.

Shape of the mirrored object

All lines of sight and lines of perspective converge at the horizon. Depending on our height we cannot see further then the visible horizon. But behind the visible horizon the lines of perspective do continu. As there is to much information aka colours behind the visible horizon there will be mirroring. Normally you will see the sky mirrored on that plane. From our point of view this mirror extends endlessly but will appear as a very small area above the visible horizon. In this mirror ships are visible for a short time, until they disappear from the bottom up due to our "mirrored" line of sight. The most interesting thing about the Sun is that it actually shows how far the mirror extends above the visible horizon. The Sun is mirrored but due to our mirrored line of sight. How this works, you can read in "calculating how long ships are visible cruising over the horizon". As the Sun is mirrored over a great distance and is cut by our line of sight, at the visible horizon it appears smaller. At the moment that the Sun sets in perspective, then it crosses our mirrored line of sight.

Dimensions of Sun and mirrored Sun

When enhancing the Sun at the horizon you can see the mirrored shape just above the visible horizon. It is a very small area but seems to go outward a little bit as it approches our visible horizon. When you measure this it has a ratio of 5,3 at the visible horizon where as the Sun measures 6,8. As in another article on calculating the distance to the Sun, moon and planets I stated that the Sun on a flat earth model would be at a distance of 6207 km with a radius of 28,8 km. This is a diameter of 57,6 km. This would mean that the mirrored Sun at our visible horizon appears to be 57,6/6,8*5,3 = 44,9 km. So over the length of the mirror our mirrored line of sight reduces the size by 57,6-44,9=12,7 km. If you look at one sight of the Sun for triangular calculation it means a reduction of 6,35 km over the distance. 

Distance to the Sun

Looking at the photo it appears to be taken just above the waterlevel. So assuming it was taken at eyelevel it was taken at approximately 1,75 m. Assuming also that the visible horizon is at about 5600 m then the distance to the Sun over our line of sight would be 6,35/(tan(tan-1(2/5600))) = 17780 km. I know this method has several vaiables and therefor the answer can be various also. But the point of this method of calculation is that it shows the order of magnitude on how far the Sun would be based on our line of sight. With an average speed of the Sun of 2617 km/hour (spring/autumn) in six hours the sun would travel approximately 15702 km (summer = 1950*8=15600 km). So the forgoing method of calculation actually is a good approximation on how far the Sun is at when crossing the horizon based on our line of sight.

How far is the visible horizon?

So if the Sun sets it is at an approximate distance of 15650 km. Based on a eye-height of 2 m how far would actually the visible horizon be? This would result in the following distance to the visible horizon: 2/(tan(tan-1(6,35/15650)) = 4,93 km. This can also be simplified saying distance to the horizon dh=ht/0,40575 with ht is heigt eye in m and dh in km.

PS: this kind of mirroring is only posible on a flat earth model as the plane continu's and continu's as on a globe it would be blocked by the curvature of the earth.

Flat earth or hollow

I do realise that there is one problem with the forgoing theory. There are many many video's in which you actually can see very far over a watersurface using a strong telescope. For instance if you set up a telescope at 0,6 m above waterlevel then based on that on a globe you would see the visible horizon at 2,76 km and on a flat earth model at 1,47 km. Both answer actually don't fit with what we see. There are many video's in which you can see much further than that. If both of these answers are wrong and we can see much further, then this would suggest a hollow earth model with us already on the inside. In that model the sun would also be at the centre of the universe and revolving in 24 hour cirlcles. At the horizon you still would have a small band working as stated as a mirror, after which our line of sight is mirrored even more steep.

Photo: Rijkswaterstaat / Harry van Reeken, https://beeldbank.rws.nl

Monday, November 28, 2016

How big is the circumference of Antarctica on a flat earth?

On a flat plane the globe actually has been pushed flat with the continent Antarctica cracked open as being the inside out. This means that at the far edges of our know world there is a big rim or ring of ice. So based on the know dimensions of a globe earth how far away from the noth-pole lies Antarctica and what is the circumference?

Known dimensions of Antarctica

Based on source (1) Antarctica has a surface-area of exactly 14.000.000 km^2. This number is very striking in my opinion, as it looks made-up or the dimensions have been guessed or it is disinformation. Based on this figure we can estimate the distance of the average radius of the continent on a globe model and then we know what the average latitude is. If we know that we can translate this information into a flat earth model.

Calculating the radius and latitude

Based on the "known" area we can calculate the radius of that area, taking the south-pole as the centre of Antarctica. It results in [14.000.000/pi]^(1/2) = 2111 km. This average distance from the south-pole would be located at sin-1(2111/6371) = 19,35 degrees from the south-pole and that would mean it lies at a latitude of 70,65 degrees south. This then actually is 90+70,65 = 160,65 degrees from the north-pole.

Calculating the circumference of Antarctica

On a flat earth the radius up to 160,65 degrees from the north-pole would result in a radius of 2*pi*6371*160,65/360 = 17863 km. So in this case the rim of ice is at this distance from the north-pole on a flat plane. Based on this radius the circumference of Antarctica (inside is outside) would actually be 17863*2*pi = 112.236 km.

Calculating the circumference up to the south-pole

Given the forgoing method of calculation, on a flat earth the south-pole (180 degrees from the norht-pole), the radius would be 2*pi*6371*0,5 = 20.015 km and the circumference 20.015*2*pi = 125.758 km. This maybe coincides with the distance to the firmament or dome, or maybe there is even more land beyound the told south-pole (as there might not be 360 degrees of earth (north-south) but maybe 400 degrees). This of-course is guessing.

Read further suggestion:
Black dot in the Sun at Antarctica, is it a Sun simulator?
Sailing around the world at 24 knots on a flat earth is possible

(1) https://nl.wikipedia.org/wiki/Antarctica

On a flat earth the sun has different speeds during a year

Every year we experience spring, summer, autumn and winter, caused by the position of the Sun. On a globe model the sun appears to be travelling from 23 degree latitude north to 23 degree latitude south. But how would this work on a flat earth model?

How can the speed vary?

On a flat plane the earth is the centre of the universe with the Sun revolving at a constant 24 hours per day. In this case the easiest way is to consider that the north-pole is actually a pole and at its highest point a beam is holding the sun in position. As the pole rotates in exactly 24 hours then the beam must be able to lower the sun as the seasons change. As the constant is 24 hours, then the distance increases with the lowering of the Sun resulting in a greater speed. So at 23 degrees north the Sun actually would go the slowest and at 23 degrees south the fastest.

How fast does the Sun go?

To calculate how fast the sun would go, we need to determine the radius of the Sun's path at 23 degree north, south and at the equator. This I will do based on the know radius of the globe aka 6371 km. As the earth is pushed into a flat earth then the distance from the north-pole to the equator would be 6371*2*pi/360*90 = 10007 km resulting in a circumference of 10007*2*pi = 62831 km giving a speed at the equator of the sun of 62831/24 = 2617 km/hour. At 23 degree north this means a radius of 7450 km, a circumference of 46810 km and a speed of 1950 km/hour. At 23 degree south this means a radius of 12565 km, a circumference of 78948 km and a speed of 3290 km/hour. So if the earth is a flat plane than the Sun must have different speeds!

PS: the following impression isn't how the world looks like, it is a simple way to show how the speed could change with a change of position based on a continuous 24 hour time-frame.

Thursday, November 24, 2016

Tracking birds with 5 radars from 200 m and up suggests flat earth model

To know where birds are the KNMI has developed a method using 5 radars in the Nederlands and Belgium, so that they scan from 200 m till 1,6 km (first layer, divided in layers of 100 m). Second layer is from 1,6 km and up. This is actually a very useful tool, so that birds can be avoided by planes. So how does this registration of birds go on a globe or a flat earth as the radar beam can only go in a straight , and what kind of conclusion can we make out of these facts?

5 Radars used

For the Nederlands they used Radar ID 6234 Den Helder and Radar ID 6260 De Bilt and for Belgium Radar ID 6410 Jabbeke, Radar ID 6451 Zaventem and Radar ID 6477 Wideumont (1).

Principles of a radar on a globe

A big problem for radars on a globe is that they can only register what is above their line of sight. For-instance if the radar is approximately 50 m high, their line of sight crosses the horizon at (6371050^2-6371000^2)^91/2) = 25,2 km. From that point on the radar cannot see what is below their line of sight, behind the horizon. In that case for more information they will need to use other radars as well. A high grid of radars would be required.

Radar on a flat earth

If a radar is located on a flat earth, it would work the best. Beams from the radar are send out and come back with the required information on weather, planes and in this case birds. As the radar beam goes in a straight line and comes back it can give much information from ground-level to several km high. A low grid of radars is needed.

Radar Den Helder and bird tracking near Eemshaven

For the Nederlands and Belgium they have full coverage of the 2 lands and can track all the birds from 200 m and up (2). If we for-instance look at Eemshaven, it is located at 148,4 km (3) from the nearest radar being Den Helder. With a horizon distance for the approximately 50 m high radar of 25,2 km, the earth still would over 148,4-25,2 = 123,2 km have a curvature drop of: (6371^2+123,2^2)^(1/2)-6371 = 1,19 km. As the radar beam cannot turn and see over the curvature, how come you can see bird-movement from 200-1600 m in source (4) whilst at Eemshaven you can only see birds above 1,19 km height. This would be on a globe an impossibility as radar beams go in a straight line, so you cannot see over the horizon. Unless it isn't a globe model, but a flat earth model. You judge for yourself, but this surely suggests a flat plane.

(1) https://github.com/enram/case-study/blob/master/data/radars/radars.geojson
(2) https://github.com/enram/case-study/tree/master/data/bird-migration-altitude-profiles
(3) http://www.afstand-berekenen.com/zoeken?from=den+helder&to=Eemshaven%2C+Netherlands
(4) http://enram.github.io/bird-migration-flow-visualization/viz/2/nl-be/

Are tides related to the moon or is it a north-south relationship

Everyone knows that the tides are caused by the gravitational pull of the moon. It is also told that if the moon is on one side of the globe then on the other-side it also will be high tide. This is a very interesting relation and should be tested if it is true. Everywhere the moon is at 90 degrees the water will be pulled and there is a high tide. As water is very flexible it should rise at the right moment the moon passes from the east to the west. High tide should then occur when the moon passes the meridian or passes at the 90 degree point for a certain location. Lets put this to the test.

10 Locations and relation moon and tides

Based on source (1) I came with the local time for the high-tide and based on source (2) for the moon-time passing the meridian. The following data is for 25 November 2016:
  • Reykjavik Iceland: moon = 10:25, high-tide = 10:46, checks out;
  • Skopun, Faroe Islands: moon = 9:22, high-tide = 11:17, tide is 2 hours slow;
  • Den Helder Netherlands: moon = 9:34, high-tide = 9:42, checks out;
  • Ireland: moon = 9:20, high-tide = 15:40, high-tide is 6 hours before and after, moon coincides with lowtide;
  • Saint-Gilles-Croix-de-Vie, France: moon = 9:56, high-tide = 7:41, 2 hours to quick;
  • Hammonds Plains, Canada: moon = 8:57, high-tide = 3:32, 5 hours to quick;
  • Ponta Delgada, Portugal: moon = 9:40, high-tide = 6:34, 3 hours to quick;
  • La Orotava, Spain: moon = 10:29, high-tide = 5:30, 5 hours to quick;
  • Bahamas: moon = 9:07, high-tide = 4:10, 5 hours to quick;
  • Barbados: moon = 9:00, high-tide = 12:05, 3 hours to late.
If you do a statistical regression analysis on this data you will find that the relationship moon-time and high-tide-time is unprovable, as apparently the difference between moon-time and high-tide-time simply is to big.

Relation to latitude and longitude

Some time ago I did a statistical regression analysis of the tides comparing the positions of these and other locations on the earth. I did this in order to know if there is a relationship between the tides and latitude and/ or longitude. This analysis was based on time (GMT data of 8 November 2016) for good comparison. In this analysis we prove the following: H0 = there is no relationship, H1 = there is a relationship between the tides and latitude or longitude. It gave the following results: 
  • high tide vs Latitude F-test = 0,173 T-test = 0,589
  • high tide vs Longitude F-test = 0,016 < 0,05 test fails
  • low tide vs Latitude F-test = 0,468 T-test = 0,967
  • low tide vs Longitude F-test = 0,282 T-test = 0,751.
Striking is the fact that the Longitude high-tide F-test failed, actually suggesting that there is no east-west based relation with the tides. For the Latitude all test pass and give good results also. This actually can support the theory that tides have a north-south relation and given the previous results for the moon and tides no east-west relation. So if this actually is the case, then the next question would be what is causing this?

PS: the forgoing statistical regression analysis has been based on 15 locations, which is the minimal needed for this kind of analysis. The reliability would improve if this can be done for over 50 or 100 locations worldwide.

Additional proof on north-south tides

If you go to the source (3) (added on 1 December 2016) and select Ocean, Currents and Currents projected on AE (flat earth) you will get the following picture. Very very striking on this is the circular outward movement of the tides, with the north-pole as a centre aka the cause of the tides going outward. This is the same effect as when you trow a stone into water. This absolutely suggests that tides are caused by something at the north-pole. The picture confirms the forgoing found theory that tides are north-south based and not moon related aka east-west based. Based on the tides a flat earth model makes much much more sense.

(1) https://www.worldtides.info/
(2) https://www.timeanddate.com
(3) https://earth.nullschool.net/#current/ocean/surface/currents/azimuthal_equidistant added 1-12-2016

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.

(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

Distance to Sun, Moon and planets on a flat earth

If the earth is flat and a non-moving plane, then all the rest revolves around our planet. If that is the case then all distances concerning the universe, the Sun, Moon and planets are completely wrong. It would actually mean that we are in the centre of our universe, making us people very important. So what is triangular based calculation and how can it calculate the distance to the Sun, Moon and planets.

Triangular time-based distance calculation

For all the planets the Sun and Moon the rise and set time are all very well known. In the used source (1) you can check for yourself the rise and set-times for the planets and source (2) for the Sun and Moon. If you see the sun from two places on the earth under a certain angle with a certain distance between these places, you can calculate the height hc to the object. For instance for the Sun:

Data on Bangkok (Thailand)

  • rise-time Bangkok = 6:18 = 6,30;
  • set-time Bangkok = 17:47 = 17,783;
  • time = 14,00;
  • length of daylight = 17,783-6,30 = 11,483;
  • halve daylight time till 90 degree = 11,483/2 = 5,741;
  • how much time is passed = 14,00-6,30 = 7,70;
  • time passed 90 degree = 7,70-5,741 = 1,959;
  • time still to go till set = 5,741-1,959 = 3,782;
  • angle (time based) = tan-1(3,782/1,959) = 62,63 degree

Data on Sanaa (Yemen)
  • rise-time Sanaa = 6:05 = 6,083;
  • set-time Sanaa = 17:30 = 17,50;
  • time = 10,00;
  • length of daylight = 17,50-6,083 = 11,417;
  • halve daylight time till 90 degree = 11,417/2 = 5,709;
  • how much time is passed = 10,00-6,083 = 3,917;
  • time till 90 degree = 5,709-3,917 = 1,792;
  • angle (time based) = tan-1(3,917/1,791) = 65,42 degree.

If a and b are those angles at the same time in Bangkok and Sanaa with a distance between them of 6053 km (C) then all other dimensions can be calculated as follows:

  • c = angle at the sun = 180-62,63-65,42 = 51,95 degree;
  • B = 6053 / sin(51,95) * sin(62,63) = 6826 km;
  • A = 6053 / sin(51,95) * sin(65,42) = 6990 km;
  • hc = distance to the sun = 6990 * sin(62,63) = 6207 km;
  • hc = 6826 * sin(65,42) = 6207 km.

Distance Moon and planets

Based on this type of calculations, I repeated it for the Moon and planets. With this method I have the following surprising results:

  • Mercury = 3937 km;
  • Moon = 4851 km;
  • Venus = 5239 km;
  • Saturn = 5714 km;
  • Mars = 6010 km;
  • Jupiter = 6014 km;
  • Sun = 6207 km;
  • Uranus = 6737 km;
  • Neptune = 6982 km.

If we live on a flat earth then these object can turn independent of each-other as there is distance between them. These distances can also explain how the moon can go in-front of the sun (as it is on a lower plane) resulting in eclipses on a flat earth.

Van Allen Belt

Coincidentally if these distances are correct, it would place them almost completely in the First Van Allen Radiation Belt (3) which is located form 1000-6000 km. Maybe there is something wrong with forgoing calculation and should it actually be within the Van Allen Belt, or maybe this Belt actually is larger. Only the order of magnitude of the distance to these object seems to be correct and coincides with the Belt.

Size Sun and Moon

If the radius of the Sun and Moon actually is dependend on the height, then based on the known supposedly fake radius then the corrected radius would now be:
  • radius Sun = 695700*6207/149.600.000 = 28,8 km;
  • radius Moon = 1737*4851/384.400 = 21,9 km. 
As 2*tan-1(28,8/6207) = 0,53 degree and 2*tan-1(21,9/4851) = 0,51 degree it is visible for the naked eye and would appear/ look like the same size.

(1) http://theskylive.com/planets used data of 15 november 2016 for the planets
(2) https://www.timeanddate.com used data of 1 november 2016 for Sun and Moon
(3) https://en.wikipedia.org/wiki/Van_Allen_radiation_belt#Inner_belt

Calculating how long ships are visible cruising over the horizon

An interesting problem which I was faced with is how can I explain boats going over the horizon and disappearing on a flat earth model. I did think on this quite long and did not see anyone give a proper calculated answer. You only see movies of the ships disappearing without the correct theory or real answer behind it. In this case I did find it and have found a way to calculate the relationship between eye-height, distance to visible horizon, cruising speed of the ship and how long you can still see it over the horizon till it disappears in the horizon-mirror.

What is the horizon-mirror?

You cannot see further than the horizon. This is 100% depended on the height of your eye. You line of sight is most important in this. Everyone knows that the higher you are the further you can see. The horizon always levels up at your line of sight (this actually is an important argument for a plane). The horizon-mirror occurs due to the fact that all the lines of sight converge in such a way, that there is to much information. This means to much colour at a very slope angle (think of 0.02 degree or something like that). It results in to much info which results in a mirror. After the visible horizon mirroring occurs which normally results in a mirrored sky. Directly after the horizon the mirroring-effect starts.

Theory on mirroring after horizon

You can argue in two ways. Or your line of sight goes down and meets the horizon, or the plane line of sight goes up. From your point of view this actually is not of importance as the angle between these is very little. But having said that, almost none realises that our line of sight is mirrored from the point of the visible horizon. This means that the mirrored line of sight is going up from the horizon. Object which are located behind that point on a plane, will be cut of from the below up. This type of mirroring is everywhere visible. On the beach, desert or at sea. It doesn't show what is below the mirrored line of sight, but mirrors what is above. This principle is very important to understand the type of calculations in this (see impression below article).

How much can you see of buildings behind horizon

The mirroring-effect results in only one thing. Everything is cut from below and the rest is mirrored below the mirrored line of sight. This principle you can observe in many video's trying to prove or disprove a globe or flat earth. But no-one takes in concern our line of sight. This is most important as it explains almost everything on this subject. What is cut of from the expected picture is in direct relation to the angle of your line of sight. That angle is calculated with tan-1(h/d) with h=height and d=distance. If your eye-height is at 2 m and the horizon is at 5,6 km (this still is somewhat guessing) then the angle would be tan-1(2/5600) = 0,02 degree. From the horizon the mirrored line of sight goes up at an angle of 0,02 degree. Based on this you can calculate how much of buildings (measured from sealevel) have disappeared in the mirror. If the building is at 1 km behind the visible horizon than actually 1*tan(tan-1(2/5600))*1000 = 0,35 m would be invisible. At 10 km that would be 3,5 m. If you look at a globe than that would give 0,078 m (1 km) and 7,8 m (10 km). PS these calculations start from the point of the horizon. It suggests that on a globe the drop will be much quicker in relation to the flat earth.

Calculations on disappearing-time ships over horizon

When looking at a video on ships disappearing over the horizon this relation can be used to calculate how long it will take for the ship to disappear for real over the horizon on a flat earth model. If a ship is cruising at 30 km/hour and in the video the ship is visible 27 min, then how high is that ship up to the top of the shipmast: (30/60*27)*tan(tan-1(2/(56000)))*1000 = 4,8 m. You can turn this around also: How long would the ship be visible if it has a topmast of 7 m then 7/[(30/60)*tan(tan-1(2/5600)]/1000 = 39.2 min. What would be the speed if it is 6 m high and travels 40 min: [6/(tan(tan-1(2/5600))*60/40]/1000 = 25,2 km/hour In this case there are several variables and can only be checked if you know the dimensions of the ship, your eye-height, the cruising speed and the real distance to the visible horizon. But given the video and these results it shows that the relation is most-likely correct and therefor can be used as additional proof for a flat earth.

Comparison to a globe-model

If the forgoing actually would be on a globe earth, then ofcourse the  figures would be different. From the point of the visible horizon the boat gradually would sink behind the wall of water. Every km it would go down (6371^2+1^2)^(1/2)-6371 = 7,8 cm. For the following km this is the same but from our point of view the curvature of the first km needs to be accounted also. So at 2 km the boat would already have disappeared 31 cm. At 10 km this already would be 7,8 m which means that the ship already would have disappeared completely. In the video the author says that the ships is cruising at considerable speed so as stated the ship would probably have a speed of 30 km/hour (about 16 knots) or more. In the mentioned 27 min it would have travelled 13,5 km at which the boat would have disappeared behind 14,3 m of water. This is absolutely not the case as shown in the video, as the ship clearly disappears above the visible line of horizon. Next to that the video shows that the boat disappears on eye-level or our line of view in the so-called mirror. These two facts and the forgoing analysis support the flat earth model and don't support the globe-model.

Tuesday, November 22, 2016

Excellent video on boats vanishing over horizon

The following video is an excellent explanation on how ships sink below the horizon. It is based on mirroring. The further a ship is over the horizon the higher the mirror-line gets. This certainly is suggesting a continuation of a plane and not a globe. The author of this video forgets to mention that from our point of view the line of sight actually cuts of the underside of the boat. The further the more it cuts of... Over this mirrored line of sight everything that is above that line of sight will be mirrored over that mirrored line of sight. This works as shown in the following impression.

Flat earth or globe, introduction

Many people around the world (including myself) realise that there is something wrong with this world. Everything is changing and a big question is, is everything true about what we have learned at school and is the earth really a globe? When I myself came across this subject I only was interested so that I could simply disprove the flat earthers. But after studying the material further and further I came to realise that many things concerning a globe cannot be proven and other things can be proven. This is also the case for a flat earth. So what is it, what are we living on? Is it a round plane, or is it a proven fact that we live on a globe. At this point in time I have come to believe that we have been misled from the start of our kindergarten-career and that the globe is actually some kind of flat plane. What kind of shape it really is, for now that is the quest to be answered.

Goals of this blog

I have the intent to prove and disprove arguments which will confirm a flat earth or deny a globe. In this there are many discussion-points as everyone can imagine, but the biggest method of interpretation for me is calculation. Based on the supposedly radius of the earth, it is possible to check if some far away locations should be possible or not. Having said that it can be used to prove both sides on the quest. I in that case have found several arguments of other flat earthers which argument wrongly and therefore actually undermining their argument. Thus one of the point of this blog will also be to show them how to properly prove or disprove their statements.

Why have we been lied to?

Another important question is: if we do live on an flat earth like model, then why have we been lied to and why have we been unluminated (contradiction of illuminated). Many of these kinds of questions are hard to answer and often must require stamina and good insight. Einstein did say on the subject of the movement of the earth: "Since then I have come to believe that the motion of the earth cannot be detected by any optical experiment, though the Earth is revolving around the Sun" (1). Aka Einstein tells us that the rotation of the earth and twirling movement through the universe cannot be seen or measured and assumes that the proclaimed movement must be simply true. This statement is simply unbelievable. His point can be turned around saying: "there is no movement and thus the Sun must be revolving around the (plane) earth". You tell me, but the quest in this is to prove it without relying on fake Nasa immages.

Anyone is free to argument on this blog if only in a rational way, without insulting others or each-other. It is not allowed to post links to other sited, unless asked for. Interesting links I will post myself.

(1) http://photontheory.com/Einstein/Einstein02.html

746 and Sun enters Cancer

Somehow the system wants to show the number 746 aka 21-06-2018. It is the day the Sun enters Cancer.  777 is 22-07-2018 aka is my birthday....