HAPPY HOAX ANNIVERSARY!!! (Rockets can't fly in a vacuum)

But Enceladus does not have a density of 1000kg/m^3, in fact it is mostly hollow, that is why your calculations are useless.

In order to calculate the thrust, with the new figures (at least 33% cavities), we need the pressure (as I have said from the start).

If it is mostly hollow, why did it eject ice from it's surface as such.

wew lad, did I miss a few days.  Okay, quick recap...

cikljamas still doesn't understand Newton's second law and refuses to acknowledge how the exhaust is acting on the rocket.  Then cikljamas claims that rockets working in a vacuum is equivalent to creating something out of nothing, a la the creation of the universe.  So presumably, when cikljamas gets around to seeing videos of rockets working in a vacuum, it's also evidence that the universe was created out of nothing.  That'll be awkward.

Then sandokhan fires up his Hollow Enceladus Theory, and claims that Enceladus is in fact a tiny rocket-sized ball shooting water vapor like rocket exhaust, and the fact that it doesn't move like a rocket is proof that rockets don't work in a vacuum.  Seems to me like it's actually proof that Enceladus is not a tiny rocket-sized ball, but to each his own.  Surprisingly, sandokhan also doesn't understand Newton's second law, or at least I am surprised at this.  Coming from a man who loves to wade in piles of unitless numbers, I supposed I just assumed he'd want to calculate f=m*a on everything.  Guess I was wrong.

Three questions I still have:
1) Do flatearthers believe f = m a ..?
2) How did we acquire those images of Enceladus, if not from a space probe that used a rocket?
3) If Jupiter et al are tiny little balls just a little further beyond the clouds, why don't they fall to earth?

I have many other questions, but I gotta learn to stay focused.

There were no Voyager or Cassini missions.

In FET, Jupiter has the same diameter as that of the Sun/Moon/Black Sun/Shadow Moon, some 636 meters.

Saturn, then, has some 200 meters in diameter.

Enceladus, some 20 meters in diameter at most.

That is, Enceladus is as large as a modern rocket.

Really?

Geocentrism was the standard model throughout the Western world during antiquity, the Middle Ages and the Renaissance. The earliest known attempt at a mathematical model came from Eudoxus of Cnidus in the 4th century BCE (his works are lost, but his solar system is described by Aristotle).

*A* According to Aristarchus (3rd B.C.) diameter of the sun was about 70 000 km (5,5 times wider than the diameter of the earth)
*B* Ptolemy (2nd A.D.) was in agreement with Aristarchus regarding the distance to the sun (7 000 000 km) and consequently regarding the diameter of the sun, as well.
*C* Tycho Brahe (1546-1601) THE GREATEST ASTRONOMER OF ALL TIME was still in agreement with Ptolemy and Aristarchus on this question.

So, according to Ptolemy and Tycho Brahe the sizes of the spheres were calculated in the following way. The Moon's distance from Earth was thought to vary between 33 Earth radii (ER) and 64 ER. The Sun‟s distance was calculated to be between 1160 ER and 1260 ER.

THE DISTANCE TO THE MOON :
33*6400 = 211 200 km
64*6400 = 409 600 km

THE DISTANCE TO THE SUN :
1160*6400 = 7 424 000 km
1260*6400 = 8 064 000 km



THE DISTANCE TO THE SATURN :
between 72 000 000 and 80 000 000 km

The distances between the equator and the north pole cannot be the same for a globe and a flat earth. Just think of half a ball. To draw the distance between the top and the bottom of the half ball, you would have to draw along the curve of the ball. Now, if you flatten the half ball, you will get the same distance between the center of the circle and the outside edge. However, the size of the bottom of the half ball has to expand out to a larger circle. So, either your distance around the equator is wrong or your distance between the equator and the north pole is wrong or your flat earth model is wrong. Take your pick.

Since the known distance from the north/south pole to the equator is 10,000 km (5400 nm)...and since the known circumference of the earth is 40,000 km (21600 nm)...then it follows that the earth cannot be a flat disk since the radius of a disk of 40,000 km is 6369 km...not 10,000 km...So, the only way around this simple argument is providing any evidence to the contrary regarding the circumference of the equator or the distance from the poles to the equator or both. There are 90 degrees of distance from the equator to the North Pole. Each degree has 60 minutes, each minute = 1 nautical mile, therefore 60 x 90 = 5,400 nautical miles = *10 000 km.* Btw, what would be meaning of the word EQUATOR on the flat earth?

Now, if the earth were a flat disc (on which the distance from the NP to the Equator would be 10000 km, as it is the case in our reality), then the circumference of such a disc (at the equator) would be 62800 km, not 40000 km!!! This number (62800 km) is absolutely preposterous (and in every conceivable aspect beyond the wildest imagination of an utter lunatic), so that only a complete idiot would give any attention to such ludicrous fanciful value. So, since the KNOWN distance from the North Pole to the Equator is out of question, and since the KNOWN value for the circumference of the Equator is out of question, also, your flat earth dreams end up right here, once and for all.

So, my question for you is this :

Do you really believe that Aristarchus, Ptolemy and Tycho Brahe were NASA shills???

If you do, then i can sell you the Brooklyn bridge, would you like to buy it?

It is important to note that the geysers from Enceladus have shut down: we are discussing here only the data pertaining to the year 2005.

https://www.space.com/31385-saturn-moon-enceladus-geysers-losing-steam.html

Yet, there is no difference in the orbital parameters of Enceladus, with or without the geysers.

Without the Airy hypothesis, Enceladus becomes an icy hollow shell.

The radio tracking data of the Cassini spacecraft cannot be relied upon when it comes to very sensitive observations (see the references provided).

What if the radius of Enceladus is not 500 km, but a much lower figure?

All of these questions can be answered once we infer that Cassini has a hemispherical resonator gyroscope onboard which uses the Coriolis effect to detect rotation.

It is important to note that the geysers from Enceladus have shut down: we are discussing here only the data pertaining to the year 2005.
https://www.space.com/31385-saturn-moon-enceladus-geysers-losing-steam.html

Its important to note that you are wrong, firstly you have already posted a paper from this year that discusses variations in plumes from individual sources to position in orbit. I cant find any other refference at this moment relating to the plumes stopping.

Secondly as you know, you posted the links, the last Cassini pass through the plume was in 2015.

Yet, there is no difference in the orbital parameters of Enceladus, with or without the geysers.

 

Well as pointed out above the jets as far as we know have not stopped.

But in any case that’s my position not yours?? I have been consistent the geysers, relativley low mass as it is ejected cannot provide sufficent force to move a small moon.

Without the Airy hypothesis, Enceladus becomes an icy hollow shell.

 

Again your paper, your link.

We established that your papar was using various Airy assumptions to hypothesise what interior structures could be present to match the known mass and observable activity of the moon.

The radio tracking data of the Cassini spacecraft cannot be relied upon when it comes to very sensitive observations (see the references provided).
 

Im going to assume you mean the RSS system on Cassini, a passive experiment which was capturing data on radio waves to and from earth and the affect of these waves passing through or near a variety of objects. It was used to gather more information on an acceleration anomoly, as you pointed out.
This was an added bonus of the RSS not its primary function and you havent provided any evidence that the RSS system was in anyway defficent for the rest of its tasks?

What if the radius of Enceladus is not 500 km, but a much lower figure?
 

I raised this a couple of days ago, why don’t you play around with the mass of Enceladus and and come up with a number of sceanarios and the amount of force potentially required to creat observable movement or variation of orbit.

You wittered on about not being able to work out force as you didn’t know the pressure???

But please go ahead, I don’t mind joining in on a bit of a thought experiment, or do you actually believe it is smaller, if so can you provide evidence.

All of these questions can be answered once we infer that Cassini has a hemispherical resonator gyroscope onboard which uses the Coriolis effect to detect rotation.

It has infer away

Since the known distance from the north/south pole to the equator is 10,000 km (5400 nm)...and since the known circumference of the earth is 40,000 km (21600 nm)...then it follows that the earth cannot be a flat disk since the radius of a disk of 40,000 km is 6369 km...not 10,000 km...So, the only way around this simple argument is providing any evidence to the contrary regarding the circumference of the equator or the distance from the poles to the equator or both. There are 90 degrees of distance from the equator to the North Pole. Each degree has 60 minutes, each minute = 1 nautical mile, therefore 60 x 90 = 5,400 nautical miles = *10 000 km.* Btw, what would be meaning of the word EQUATOR on the flat earth?

Now, if the earth were a flat disc (on which the distance from the NP to the Equator would be 10000 km, as it is the case in our reality), then the circumference of such a disc (at the equator) would be 62800 km, not 40000 km!!! This number (62800 km) is absolutely preposterous

The RE have the same problem:

https://www.theflatearthsociety.org/forum/index.php?topic=64953.msg2197506#msg2197506

Can you explain to your readers how four trillion billion liters of water stay glued next to the outer surface of a sphere?

You want to use gravitons or spacetime?

Why is there no curvature whatsoever across lake Ontario (as an example)?

There were no Voyager or Cassini missions.

In FET, Jupiter has the same diameter as that of the Sun/Moon/Black Sun/Shadow Moon, some 636 meters.

Saturn, then, has some 200 meters in diameter.

Enceladus, some 20 meters in diameter at most.

That is, Enceladus is as large as a modern rocket.


Then, if we put these numbers into the thrust equation, with m_rf = 1000kg/s, A = 6.25 m², V = 583m/s, and calculate the mass of an icy hollow discoidal shell with an outer radius of 20 meters and an inner radius of 15 meters, we see that the force developed by the geyser jets is formidable, given the true size of Enceladus.

What I have to do is to prove that Cassini never orbited the solar system, and also answer the question regarding the source of matter for the geyser jets given the true size of Enceladus.

Those are not the figures you quoted in your peer reviewed papers!!!!!

But it is entirely based ON THE AIRY HYPOTHESIS.

Please read.

Page 74 (page 18 of the pdf document)

Lateral variations in ice shell thickness assuming complete Airy compensation of all known topography [up to spherical harmonic degree 8 (Nimmo et al., 2011)].

What was assumed in that document and what was the purpose of the assumption doesn't change the
calculation of the Enceladus' mass based on speed and diameter of it's orbit.

If Enceladus had different mass it would have different orbital parameters around Saturn.
If Saturn had different mass it would have different orbital parameters around the Sun.
And the chain goes on...

Since the known distance from the north/south pole to the equator is 10,000 km (5400 nm)...and since the known circumference of the earth is 40,000 km (21600 nm)...then it follows that the earth cannot be a flat disk since the radius of a disk of 40,000 km is 6369 km...not 10,000 km...So, the only way around this simple argument is providing any evidence to the contrary regarding the circumference of the equator or the distance from the poles to the equator or both. There are 90 degrees of distance from the equator to the North Pole. Each degree has 60 minutes, each minute = 1 nautical mile, therefore 60 x 90 = 5,400 nautical miles = *10 000 km.* Btw, what would be meaning of the word EQUATOR on the flat earth?

Now, if the earth were a flat disc (on which the distance from the NP to the Equator would be 10000 km, as it is the case in our reality), then the circumference of such a disc (at the equator) would be 62800 km, not 40000 km!!! This number (62800 km) is absolutely preposterous



The RE have the same problem:

https://www.theflatearthsociety.org/forum/index.php?topic=64953.msg2197506#msg2197506

Can you explain to your readers how four trillion billion liters of water stay glued next to the outer surface of a sphere?

You want to use gravitons or spacetime?

Why is there no curvature whatsoever across lake Ontario (as an example)?

The force in KN of gravity on earth far exceeded the centrifugal force created by its rotation, in answer to your water problem, also observation I can see its not.

Curvature, there is curvature over lake Ontario as there is over the the entire globe. Not the lake specifically but any plane or geodetic survey is sufficent to measure the curvature of the earth. Can you explain why you think there isnt a curve?

Also should this not be a new thread, even compared to your previous leaping around, we are getting further and further away from the thread title.

Shh shh, guys, I want to see two flateathers argue about gravity around other planets.

Curvature, there is curvature over lake Ontario as there is over the the entire globe.



https://www.flickr.com/photos/chris_baird/14067034302

Taken from a viewing stand at Beamer Memorial Conservation Area, Grimsby

DISTANCE 55 KM ; CURVATURE OF 59 METERS


Beamer's Falls #071114
River Forty Mile Creek
Class Ramp
Size Medium
Height: 45
Crest: 20
The Niagara Peninsula Conservation Authority acquired Beamer Memorial Conservation Area in 1964, to protect and preserve the Niagara Escarpment and the Forty-Mile Creek valley system. The site is home to a variety of Carolinian plants and wildlife.

http://www.gowaterfalling.com/waterfalls/beamer.shtml


Therefore, from 45 meters in altitude, we should see a huge 59 meter curvature right in front of us, and a visual obstacle of some 65 meters.


Here is the other photograph from Beamer Falls:




http://www.flickr.com/photos/suckamc/53037827/#

Again, no curvature whatsoever across a distance of 55 km, no 59 m midpoint visual obstacle.


Ms. Kerry Ann Lecky-Hepburn took these photographs some years ago: the RE called her, and were told they were taken at an altitude of 170 m in Grimsby.



No ascending slope, no midpoint visual obstacle of 59 meters, no curvature whatsoever.

From the very same spot, Ms. Lecky-Hepburn used a reflector telescope for this zoom:



No curvature whatsoever across a distance of 55 km.


Another photograph signed Mrs. Lecky-Hepburn:



http://www.flickr.com/photos/planetrick/487755017/#

http://www.flickr.com/photos/planetrick/487726854/#in/photostream

No curvature whatsoever, from Hamilton to Lakeshore West Blvd: no visual obstacle, just a perfectly flat surface of the water all the way to the other shoreline.




http://www.flickr.com/photos/tundrabluephotography/312939439/#

No 59 meter curvature whatsoever, a perfectly flat surface of the water.



Let us go to lake Michigan now.

 
Grand Haven Daily Tribune   April 3, 1925

Captain Wm. J. Preston and his U. S. Coast Guard crew at Grand Haven harbor witnessed a strange natural phenomenon last night, when they saw clearly the lights of both Milwaukee and Racine, shining across the lake.  As far as known this is the first time that such a freak condition has prevailed here.

 The phenomena was first noticed at shortly after seven o’clock last night, when the lookout called the keeper’s attention to what seemed to be a light flaring out on the lake.  Captain Preston examined the light, and was of the impression that some ship out in the lake was “torching” for assistance.

Launch Power Boat

   He ordered the big power boat launched and with the crew started on a cruise into the lake to locate, if possible, the cause of the light.  The power boat was headed due west and after running a distance of six or seven miles the light became clearer, but seemed to be but little nearer.  The crew kept on going, however, and at a distance of about ten and twelve miles out, a beautiful panorama of light unfolded before the eyes of the coast guards.

 Captain Preston decided that the flare came from the government lighthouse at Windy Point at Racine.  Being familiar with the Racine lights the keeper was able to identify several of the short lights at Racine, Wis.

Saw Milwaukee Also

   A little further north another set of lights were plainly visible.  Captain Preston knowing the Milwaukee lights well, easily distinguished them and identified them as the Milwaukee lights.  The lights along Juneau Park water front, the illumination of the buildings near the park and the Northwestern Railway station were clearly visible from the Coast Guard boat.  So clearly did the lights stand out that it seemed as though the boat was within a few miles of Milwaukee harbor. 

   Convinced that the phenomenon was a mirage, or a condition due to some peculiarity of the atmosphere, the keeper ordered the boat back to the station.  The lights remained visible for the greater part of the run, and the flare of the Windy Point light house could be seen after the crew reached the station here.


DISTANCE GRAND HAVEN TO MILWAUKEE: OVER 80 MILES (128 KM).

http://www.coastwatch.msu.edu/images/twomichigans2a.gif


Windy Point Lighthouse:

http://upload.wikimedia.org/wikipedia/commons/thumb/5/5f/Wind_Point_Lighthouse_071104_edit2.jpg/800px-Wind_Point_Lighthouse_071104_edit2.jpg

The lighthouse stands 108 feet (33 m) tall

THE CURVATURE FOR 128 KM IS 321 METERS.

Using the well known formula for the visual obstacle, let us calculate its value:

h = 3 meters BD = 1163 METERS

h = 5 meters BD = 1129 METERS

h = 10 meters BD = 1068 METERS

h = 20 meters BD = 984 METERS

h = 50 meters BD = 827.6 METERS

h = 100 meters BD = 667.6 METERS


No terrestrial refraction formula/looming formula can account for this extraordinary proof that the surface across lake Michigan is flat.



Moreover, as we have seen, the light from Windy Point was continuously observed, during the approach, and during the return to the station:

The power boat was headed due west and after running a distance of six or seven miles the light became clearer, but seemed to be but little nearer.  The crew kept on going, however, and at a distance of about ten and twelve miles out, a beautiful panorama of light unfolded before the eyes of the coast guards.

The keeper ordered the boat back to the station.  The lights remained visible for the greater part of the run, and the flare of the Windy Point light house could be seen after the crew reached the station here.


More information on lake Michigan here:


http://theflatearthsociety.org/forum/index.php?topic=30499.msg1591587#msg1591587


Another photograph signed Ms. Kerry-Ann Lecky Hepburn, no curvature whatsoever across a distance of 55 km (Grimsby-Toronto), the boat is not part of either an ascending slope or a descending slope:



Two photographs taken from the Niagara escarpment: the boats are not part of either an ascending slope or a descending slope, no curvature of 59 meters whatsoever all the way to the other shoreline:




Port Credit - Toronto, 14.5 km, 4 meters curvature, absolutely nonexistent, there isn't one centimeter/one inch of curvature over this distance:










Let us increase the distance to 33.6 km, zero curvature (supposed to be 22 meters), Oakville - Toronto:



We now go to Etobicoke, some 6 miles from Toronto, no 1,8 meter curvature, no ascending slope:





There is no curvature whatsoever across lake Ontario.

There is no curvature whatsoever across lake Ontario.

If there was curvature, what would you expect it to look like?

OK. For the people arguing if rockets wont work in a vacuum, is there any thing an astronaut can do to 'move' in space?

Lets say you are all by yourself not gravitationally bound to anything nearby, so you are completely motionless. The only thing you have on you is your oxygen tank. If one were to pierce it (causing a sudden escape of all the oxygen very quickly) would the astronauts position move at all?

The simple answer is, no. But then again in the situation we are forced to accept about so called space, there would be no direction. No up, down or left to right or right to left..... Nothing.
Basically suspended animation...assuming anything could be in that which obviously....or should be to most....they can't.

Pierce a tank in low pressure and it tries to equalise with the low pressure to make that pressure marginally higher. This would be fine inside a small chamber but in so called space as we are told is a vacuum, the tank would simply lose its gas in super short order.

It's like the opposite of snapping off the sealed nib at the base of a glass thermos flask and listening as the atmosphere under pressure fills that lower pressure inside the thermos in a fraction of a second. It's because there is little resistance other than the walls of the thermos itself.

A  ruptured tank in an extreme low pressure environment would do no work to initiate a spin or movement because it's not releasing its gas against any opposite reaction of gas or resistance of any kind. It simply expands into the vastness of the extreme low pressure.

It would certainly react if it was in a tiny caravan sized evacuation chamber because it would hit resistance of its own gas as it builds up to equalise, so it would likely move a little...but not much, depending on tank size, obviously.

Curvature, there is curvature over lake Ontario as there is over the the entire globe.

There is no curvature whatsoever across lake Ontario.

Lets take all the photographs out of it, millions of debunkings already there.

First thing you need to do is get rid of this concept of the bulge, its a common misconception. There is no bulge because the flat plane you are expecting the bulge to rise from does not exist.

Assuming you are not going to trust GPS I would use a theodolite.

Set up two stations using an assumed datum or tie into the local or geodetic grid depending on the size or nature of the survey.

Theodolite allows you to measure both the horizontal and vertical at the same time.

The measurement on the horizontal plane determines the horizontal distance between the two stations. However, the true horizontal distance is actually curved like the Earth’s surface. So when setting up the instuments, even if they are very close together,  the direction of gravity is different at each point, which means the vertical axes are not parallel to each other.

Vertical distances are measured along the vertical axis to determine the difference in elevation between points. Vertical angles are measured in the vertical plane either above or below the horizontal plane of the theodolite. Zenith angles, are used as a reference for measuring vertical angles, they are defined as 0° directly overhead and 90° at the horizontal plane. There is always some variation in the hundredths or thousandths fraction of seconds, so measurements are taken, back sites and forsites to each station, IE moving the theodolite to each station and measuring multiple times. I usually work to standards where 6-10 measurements of the zenith angles are required. These very small errors are eliminated using a calibrated GPS Surveying Station.

The slope distance is the shortest distance from the instrument to the target, station to station. This distance is the hypotenuse of the horizontal and vertical distances.

If the stations were a km apart the angular differences would show a curvature of 80mm

Quote from: sandokhan on September 04, 2019, 08:21:44 AM

There is no curvature whatsoever across lake Ontario.

If there was curvature, what would you expect it to look like?

Something along the lines of inserting the observer’s height and object distance into the curvature calculator .... add some percentage for refraction and this should basically give you a fairly accurate amount of what should be hidden in the far distance.

But of course during the day atmospheric conditions may obscure distant objects, but never should anyone see the full amount of distant buildings (as displayed over lake Ontario) .
No ‘superiour mirage’ or other exotic phenomena can reason away what thousends of people know for a fact.
‘One can see much further under ideal conditions than what should be possible in the current globe model with a curvature drop of appr. 8 inches per mile squared.

Quote from: markjo on September 04, 2019, 09:04:58 AM

Quote from: sandokhan on September 04, 2019, 08:21:44 AM

There is no curvature whatsoever across lake Ontario.

If there was curvature, what would you expect it to look like?

Something along the lines of inserting the observer’s height and object distance into the curvature calculator .... add some percentage for refraction and this should basically give you a fairly accurate amount of what should be hidden in the far distance.

But of course during the day atmospheric conditions may obscure distant objects, but never should anyone see the full amount of distant buildings (as displayed over lake Ontario) .
No ‘superiour mirage’ or other exotic phenomena can reason away what thousends of people know for a fact.
‘One can see much further under ideal conditions than what should be possible in the current globe model with a curvature drop of appr. 8 inches per mile squared.

so if you are looking at a photograph from sandokhans post, can you tell me the atmospheric conditions, height of camera, focal length of the lense etc etc.

If there was curvature, what would you expect it to look like?

Exactly what the following precise formulas imply:

CURVATURE

C = R(1 - cos[s/(2R)]) - angle measured in radians


R = 6378,164 km

s = distance



VISUAL OBSTACLE




BD = (R + h)/{[2Rh + h²]^1/2(sin s/R)(1/R) + cos s/R} - R


BD = visual obstacle

h = altitude of observer


No curvature across the strait of Gibraltar, no ascending slope, no midpoint 3.5 meter visual obstacle, a perfectly flat surface of the water all the way to Africa:

http://www.dailymotion.com/video/x42v7ip

38:28 to 38:35




From the same spot, a splendid photograph:



http://www.flickr.com/photos/carlosromero/130948289#

If there was curvature, what would you expect it to look like?

Exactly what the following precise formulas imply:

CURVATURE

C = R(1 - cos[s/(2R)]) - angle measured in radians


R = 6378,164 km

s = distance



VISUAL OBSTACLE




BD = (R + h)/{[2Rh + h²]^1/2(sin s/R)(1/R) + cos s/R} - R


BD = visual obstacle

h = altitude of observer


No curvature across the strait of Gibraltar, no ascending slope, no midpoint 3.5 meter visual obstacle, a perfectly flat surface of the water all the way to Africa:

http://www.dailymotion.com/video/x42v7ip

38:28 to 38:35




From the same spot, a splendid photograph:



http://www.flickr.com/photos/carlosromero/130948289#

You are still looking for a bulge, it only exists in your mind, try and think in 3 dimensions.

You are a scientist stop cut and pasting photographs you have been posting for years, your threads are the only thing that comes up when you google BD = (R + h)/{[2Rh + h2]1/2(sin s/R)(1/R) + cos s/R} - R 

too many variables in photographs

Do it with numbers

Tactical argument switch.

Anyways, the curve is shown in the sign.

The English Channel: 34 km distance from Cap Gris Nez to Dover, a curvature of some 22.4 meters on a round earth.






The original webpages, as they were posted on flickr.com


The photographers located between Cap Blanc Nez and Cap Gris Nez: we will ascend to 30 meters.



And now the photograph itself: no curvature whatsoever, all the way to the other shoreline, the Dover cliffs seen in their entirety (on a round earth, from 30 meters, we could not see anything under 16.5 meters from the other side), the ships are not part of an ascending/descending slope, no midpoint curvature of 22.4 meters:




Another photograph taken right on the beach of Cap Gris Nez: no curvature over a distance of 34 km:





Dover cliffs:

Dover cliffs:

See how in the top picture the land above the cliffs appears to touch the water, but then in the next picture you see it never does.

Classic failure.

Dover cliffs and these pictures were covered long ago.

If you are located right on the Cap Gris Nez, facing Dover, all you'd see is a huge wall of water (22.4 meters in height), the midpoint curvature.

Not to mention the visual obstacle.

Even if you ascend to 45 meters (maximum height on the French side) it still won't help you.

NO curvature whatsoever across a distance of 34 km.

But you seem not to be convinced.

Then, I will increase the distance to 5,200 km.

If you are located right on the Cap Gris Nez, facing Dover, all you'd see is a huge wall of water (22.4 meters in height), the midpoint curvature.

Not to mention the visual obstacle.

Even if you ascend to 45 meters (maximum height on the French side) it still won't help you.

NO curvature whatsoever across a distance of 34 km.

But you seem not to be convinced.

Then, I will increase the distance to 5,200 km.

There is no hill of water, there is no bulge

A photograph with an exposure time of 20 seconds taken at 10.5 p.m., July 1, 1908 by George Embrey of Gloucester.

http://www.phenomena.org.uk/features/page88/page88.html

JULY 1, 1908 LETTER SENT TO THE LONDON TIMES

http://www.nuforc.org/GNTungus.html

“TO THE EDITOR OF THE TIMES.”

“Sir,--I should be interested in hearing whether others of your readers observed the strange light in the sky which was seen here last night by my sister and myself. I do not know when it first appeared; we saw it between 12 o’clock (midnight) and 12:15 a.m.  It was in the northeast and of a bright flame-colour like the light of sunrise or sunset.  The sky, for some distance above the light, which appeared to be on the horizon, was blue as in the daytime, with bands of light cloud of a pinkish colour floating across it at intervals.  Only the brightest stars could be seen in any part of the sky, though it was an almost cloudless night.  It was possible to read large print indoors, and the hands of the clock in my room were quite distinct.  An hour later, at about 1:30 a.m., the room was quite light, as if it had been day; the light in the sky was then more dispersed and was a fainter yellow.  The whole effect was that of a night in Norway at about this time of year.  I am in the habit of watching the sky, and have noticed the amount of light indoors at different hours of the night several times in the last fortnight.  I have never at any time seen anything the least like this in England, and it would be interesting if any one would explain the cause of so unusual a sight.

Yours faithfully,
Katharine Stephen.
Godmanchester, Huntingdon, July 1.”


Let us remember that the first newspaper report about the explosion itself ONLY appeared on July 2, 1908 in the Sibir periodical.



A report from Berlin in the New York Times of July 3 stated: 'Remarkable lights were observed in the northern heavens on Tuesday and Wednesday nights, the bright diffused white and yellow illumination continuing through the night until it disappeared at dawn...'

On July 5, (1908) a New York Times story from Britain was entitled: 'Like Dawn at Midnight.' '...The northern sky at midnight became light blue, as if the dawn were breaking...people believed that a big fire was raging in the north of London...shortly after midnight, it was possible to read large print indoors...it would be interesting if anyone would explain the cause of so unusual a sight.'


The letter sent by Mrs. Katharine Stephen is absolutely genuine as it includes details NOBODY else knew at the time: not only the precise timing of the explosion itself (7:15 - 7:17 local time, 0:15 - 0:17 London time), BUT ALSO THE DURATION OF THE TRAJECTORY OF THE OBJECT, right before the explosion, a fact uncovered decades later only by the painstaking research of Dr. Felix Zigel, an aerodynamics professor at the Moscow Institute of Aviation.


If the light from the Sun could not reach London due to curvature and/or any light reflection phenomena, then certainly NO LIGHT from an explosion which occurred at some 7 km altitude in the atmosphere could have been seen at all, at the same time, on a spherical earth.


It could not have been caused by either a comet, or an asteroid, or a meteorite.

The precise geomagnetic pulses were observed THREE DAYS BEFORE THE EVENT:






TWO OBJECTS CAUSED THE TUNGUKSA EXPLOSION:

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1995026#msg1995026

The initial map of the trajectory:



The final map: two trajectories, whose paths were modified in mid-air, no natural object is capable of such a feat.




https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1995521#msg1995521



It was found that the pattern of ytterbium’s distribution at Tunguska follows the projection of the “southern” TSB’s path on the ground. Similar shapes have been formed at Tunguska for the surface distribution of lanthanum, lead, silver and manganese (Zhuravlev & Demin, 1976). Only these five elements have patterns of distribution in Tunguska soils and peats that follow the projection of the TSB path on the ground, and only ytterbium follows this path strongly enough to be considered as the most likely main ingredient of the TSB substance.

This is an amazing outcome, one should note. This soft silvery-white rareearth metal, discovered in 1878, is now used mainly for improving the hardness of stainless steel, as well as in making high-power lasers. Definitely, if the chief chemical component of the TSB was ytterbium it hardly could have been a natural space body.

If the light from the Sun could not reach London due to curvature and/or any light reflection phenomena, then certainly NO LIGHT from an explosion which occurred at some 7 km altitude in the atmosphere could have been seen at all, at the same time, on a spherical earth.

A photograph with an exposure time of 20 seconds taken at 10.5 p.m., July 1, 1908 by George Embrey of Gloucester.

http://www.phenomena.org.uk/features/page88/page88.html

JULY 1, 1908 LETTER SENT TO THE LONDON TIMES

http://www.nuforc.org/GNTungus.html

“TO THE EDITOR OF THE TIMES.”

“Sir,--I should be interested in hearing whether others of your readers observed the strange light in the sky which was seen here last night by my sister and myself. I do not know when it first appeared; we saw it between 12 o’clock (midnight) and 12:15 a.m.  It was in the northeast and of a bright flame-colour like the light of sunrise or sunset.  The sky, for some distance above the light, which appeared to be on the horizon, was blue as in the daytime, with bands of light cloud of a pinkish colour floating across it at intervals.  Only the brightest stars could be seen in any part of the sky, though it was an almost cloudless night.  It was possible to read large print indoors, and the hands of the clock in my room were quite distinct.  An hour later, at about 1:30 a.m., the room was quite light, as if it had been day; the light in the sky was then more dispersed and was a fainter yellow.  The whole effect was that of a night in Norway at about this time of year.  I am in the habit of watching the sky, and have noticed the amount of light indoors at different hours of the night several times in the last fortnight.  I have never at any time seen anything the least like this in England, and it would be interesting if any one would explain the cause of so unusual a sight.

Yours faithfully,
Katharine Stephen.
Godmanchester, Huntingdon, July 1.”


Let us remember that the first newspaper report about the explosion itself ONLY appeared on July 2, 1908 in the Sibir periodical.



A report from Berlin in the New York Times of July 3 stated: 'Remarkable lights were observed in the northern heavens on Tuesday and Wednesday nights, the bright diffused white and yellow illumination continuing through the night until it disappeared at dawn...'

On July 5, (1908) a New York Times story from Britain was entitled: 'Like Dawn at Midnight.' '...The northern sky at midnight became light blue, as if the dawn were breaking...people believed that a big fire was raging in the north of London...shortly after midnight, it was possible to read large print indoors...it would be interesting if anyone would explain the cause of so unusual a sight.'


The letter sent by Mrs. Katharine Stephen is absolutely genuine as it includes details NOBODY else knew at the time: not only the precise timing of the explosion itself (7:15 - 7:17 local time, 0:15 - 0:17 London time), BUT ALSO THE DURATION OF THE TRAJECTORY OF THE OBJECT, right before the explosion, a fact uncovered decades later only by the painstaking research of Dr. Felix Zigel, an aerodynamics professor at the Moscow Institute of Aviation.


If the light from the Sun could not reach London due to curvature and/or any light reflection phenomena, then certainly NO LIGHT from an explosion which occurred at some 7 km altitude in the atmosphere could have been seen at all, at the same time, on a spherical earth.


It could not have been caused by either a comet, or an asteroid, or a meteorite.

The precise geomagnetic pulses were observed THREE DAYS BEFORE THE EVENT:






TWO OBJECTS CAUSED THE TUNGUKSA EXPLOSION:

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1995026#msg1995026

The initial map of the trajectory:



The final map: two trajectories, whose paths were modified in mid-air, no natural object is capable of such a feat.




https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1995521#msg1995521



It was found that the pattern of ytterbium’s distribution at Tunguska follows the projection of the “southern” TSB’s path on the ground. Similar shapes have been formed at Tunguska for the surface distribution of lanthanum, lead, silver and manganese (Zhuravlev & Demin, 1976). Only these five elements have patterns of distribution in Tunguska soils and peats that follow the projection of the TSB path on the ground, and only ytterbium follows this path strongly enough to be considered as the most likely main ingredient of the TSB substance.

This is an amazing outcome, one should note. This soft silvery-white rareearth metal, discovered in 1878, is now used mainly for improving the hardness of stainless steel, as well as in making high-power lasers. Definitely, if the chief chemical component of the TSB was ytterbium it hardly could have been a natural space body.

If the light from the Sun could not reach London due to curvature and/or any light reflection phenomena, then certainly NO LIGHT from an explosion which occurred at some 7 km altitude in the atmosphere could have been seen at all, at the same time, on a spherical earth.

So lets look at the facts

5200km distance from event to observation

basic curvature calculation puts this at a curvature of 2006.89m

Tunguska meteorite exploded at a height of 28k feet or 8534.4m

So from the observers point of view a fireball explosion of the equivalent magnitude of 185 Hiroshima nuclear bombs would be visible at a heaigt of approximately 6527m

Can you take me through why you think you couldn't see it?

basic curvature calculation puts this at a curvature of 2006.89m

Tunguska meteorite exploded at a height of 28k feet or 8534.4m

You need new batteries for your calculator.

You mean: 2006.89 KILOMETERS

Please read:

Over the next few days night skies in Asia and Europe were aglow, with contemporaneous reports of photographs being successfully taken at midnight in Sweden and Scotland. It has been theorized that this effect was due to light passing through high-altitude ice particles that had formed at extremely low temperatures—a phenomenon that many years later was produced by space shuttles. In the United States, a Smithsonian Astrophysical Observatory program at the Mount Wilson Observatory observed a months-long decrease in atmospheric transparency consistent with an increase in suspended dust particles.

Two questions, which I know will be ignored:

1) Why are you assuming that what was seen was the explosion itself rather than anything else associated with it, such as the incoming object, or fragments or vaporized material from it, the afterglow which lasted for days, etc?

2) What does this have to do with rockets in a vacuum?

basic curvature calculation puts this at a curvature of 2006.89m

Tunguska meteorite exploded at a height of 28k feet or 8534.4m

You need new batteries for your calculator.

You mean: 2006.89 KILOMETERS

Aye hands up on that one.

Assuming the light from the explosion radiated in all directions, it would and indeed was visable.

How does this incident prove FE?

Ps why do you only respond when I have made an error, can I assume you believe my other points true

No incoming object.

Local fragments/vaporized material.

The explosion was seen INSTANTANEOUSLY, same second: read the letter sent to the Times.

It has been theorized that this effect was due to light passing through high-altitude ice particles that had formed at extremely low temperatures—a phenomenon that many years later was produced by space shuttles.

You still don't get it.

If the light from the Sun could not reach London due to curvature and/or any light reflection phenomena, then certainly NO LIGHT from an explosion which occurred at some 7 km altitude in the atmosphere could have been seen at all, at the same time, on a spherical earth.

Let us remember the discussion we had here a long time ago...

Not so.  In the right circumstances refraction can continue to refract light indefinitely, parallel to the earth's surface.  If the earth were flat, the refraction would eventually cause the light to be pushed to the ground, but on a curved surface, the refraction continues to refract the light parallel to the earth's surface and for great distances.

To talk about ice crystals, with an explosion at some 7 km in the atmosphere on one side of the globe, and a very clear view of the initial trajectory/flash of the explosion from the other side of the hypothetical globe, means that you have no explanation for the facts involved here.

According to your explanation, we should have a 24 hour a day constant sunlight...this is what you wrote:

In the right circumstances refraction can continue to refract light indefinitely, parallel to the earth's surface.

Certainly the sun's rays of light (official theory) will be parallel to some portion of the surface at some time in the earth's rotation...that is why I invited you to think.

20 second exposure for the picture shows it wasn’t direct light.

No incoming object.

Local fragments/vaporized material.

The explosion was seen INSTANTANEOUSLY, same second: read the letter sent to the Times.

It has been theorized that this effect was due to light passing through high-altitude ice particles that had formed at extremely low temperatures—a phenomenon that many years later was produced by space shuttles.

You still don't get it.

If the light from the Sun could not reach London due to curvature and/or any light reflection phenomena, then certainly NO LIGHT from an explosion which occurred at some 7 km altitude in the atmosphere could have been seen at all, at the same time, on a spherical earth.

Quote from: sandokhan on September 07, 2013, 01:39:20 AM

Let us remember the discussion we had here a long time ago...

Not so.  In the right circumstances refraction can continue to refract light indefinitely, parallel to the earth's surface.  If the earth were flat, the refraction would eventually cause the light to be pushed to the ground, but on a curved surface, the refraction continues to refract the light parallel to the earth's surface and for great distances.

To talk about ice crystals, with an explosion at some 7 km in the atmosphere on one side of the globe, and a very clear view of the initial trajectory/flash of the explosion from the other side of the hypothetical globe, means that you have no explanation for the facts involved here.

According to your explanation, we should have a 24 hour a day constant sunlight...this is what you wrote:

In the right circumstances refraction can continue to refract light indefinitely, parallel to the earth's surface.

Certainly the sun's rays of light (official theory) will be parallel to some portion of the surface at some time in the earth's rotation...that is why I invited you to think.

8.5km elevation I got that bit right


There are still too many variables to make any definitive statements on it.

You certainly haven't demonstrated that an air blast of that magnitude could not have been seen.

So stepping back a step you were using photographs to demonstrate that distant objects can be seen due to lack of curvature, which in my view you did by stating that there was no bulge.

You ignored any questions or positions that showed you that the bulge would not exist.

So looking at the mathematical formula you invented, why dont you run some numbers through it. Surely that would put this to bed once and for all.