Sunday, February 19, 2017

Explanation on big chuncks of ice falling from the sky


From time to time you see unusual news like chuncks ice fall from the sky. The sky is clear but from out of nothing a big chunck of ice falls down. In many cases they will explain this phenomenon due to ice-buildup on the wings of plains. But in many cases the chuncks are enormous and might be the cause of crater-holes, and therefore are not caused by plains. Think about a strange phenomenon like gigantic blue-like icebergs made of 100% pure clean frozen water. So where does that come from. These most-likely also fall from the sky.

Firmament and ice-buildup

On a flat earth as described in the bible there is a firmament over the motionless plane. This glass like bowl is considered to be at a 100 km height above the earth. At that altitude it can get extremely cold and so moist and maybe condensation would freeze instantly as it would come into contact with the extreme cold glass-layer. This ice would then steadily grow in size up to the point it lets go. Small rock-sized chunks fall all over the world, but from time to time these can be much bigger.

Round craters

As the firmament is at 100 km height, the speed that the ice-chuncks would have, would be very great. Lets say about 200-300 km/hour. If a huge chunck would come down at great speed, it has an immense power. It might result in craters of several hundreds of meters in diameter. From the moment the ice lets loose the chunck falls in a straight line down (as the flat earth isn't rotating). This would mean that the chunck would crash into the earth at 90 degree angle, always resulting in an almost round crater. This is why craters all over the world seem to be round, and cannot be explained by meteors crashing down, at all kinds of angles on a rapidly rotating earth.

Wednesday, January 18, 2017

Thoughts on gravity

Gravity = the acceleration-speed

If above is above and below is below then what is above will drop down and it will increase in speed according to the acceleration-speed. On a flat plane gravity doesn't exist unless it means the acceleration-speed. On a globe we revolve at an incredible speed but the gravity on the equator and poles is the same.. So if we wouldn't get crushed at the poles then we would fly away at the equator. We revolve at a speed at around 1000 miles or 1600 km per hour at the equator but at the poles at 0 km per hour. Many things in this short text doesn't add up and would suggest a flat non-motioning plane.

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.

Sources:
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.

Sources:
(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.

Sources:
(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/