Photographic Evidence of Curvature Beyond the Visible Horizon

Your desperate attempt to link looming with the fact that a 40 meter building was seen from Holland, over a distance of 128 km, shows that you have nothing else at your disposal.

You have been shown, with the most precise proofs, that gravity is not attractive, therefore the water could not stay curved on a surface of a round earth.

Again, my answer will be the same as here http://www.theflatearthsociety.org/forum/index.php?topic=39108.msg1249791#msg1249791 :

YOU HAVE NOT PROVIDED ANY FORMULA TO BACK YOUR WILD CLAIMS!

LOOMING CANNOT CHANGE THE FACTS: there is a 984 meter visual obstacle over a distance of 128 km, from Holland to Racine.

WHY do you keep bullshitting everybody here?

And as stated, the type of superior mirage that occurs mostly in polar regions is not the kind of superior mirage that was seen at Racine.

WHERE IS YOUR PROOF, THE PRECISE FORMULA?


There is no formula whatsoever which will explain the fact that a 40 meter building, can be seen from a 128 km distance.


Just as in the case of attractive gravity, YOU HAVE NO SCIENTIFIC DATA AT YOUR DISPOSAL, AND YOU ARE MAKING UP FACTS AS YOU GO ALONG, THINKING IT WILL WORK. IT WON'T. Not with me.

Please research your bullshit claims very carefully, using any formula you can find; it won't help you, not at all.

You cannot see a 40 meter building, given the 984 meter visual obstacle.


Now, the chart does provide an indication of what could be expected, also, in the looming phenomenon.

You already have at your disposal the terrestrial refraction formula, which does not help at all.


Please understand: there is no looming formula which can help you...not over a distance of 128 km...with a 40 meter building...be my guest, and do the research, before coming overhere with bullshit claims...looming cannot work miracles, and what you need IS A MIRACLE!

To what altitude should you ascend to actually see the County Courthouse, from Holland? EXACTLY! To no less than 871 meters.

From Holland it is absolutely impossible to see anything from Milwaukee or Racine; the curvature measures some 321 meters, and the visual obstacle measures 984 meters; no looming or terrestrial refraction will help you, as we have seen...



The trajectory of the ball lightning was precisely seen from both Irkutsk and the Gobi desert.

The facts again:

The first report of the explosion was in the Irkutsk paper dated July 2, 1908, published two days after the explosion:

...the peasants saw a body shining very brightly (too bright for the naked eye) with a bluish-white light.... The body was in the form of 'a pipe', i.e. cylindrical. The sky was cloudless, except that low down on the horizon, in the direction in which this glowing body was observed, a small dark cloud was noticed. It was hot and dry and when the shining body approached the ground (which was covered with forest at this point) it seemed to be pulverized, and in its place a loud crash, not like thunder, but as if from the fall of large stones or from gunfire was heard. All the buildings shook and at the same time a forked tongue of flames broke through the cloud.

The sky was cloudless; the precise trajectory was observed, and the explosion itself, no refraction/reflection...

The visual obstacle measures 67.5 km. It is the end of the round earth theory.


And now even more proofs about the Tunguska explosion.
http://www.halexandria.org/dward232.htm

Herdsman in the Gobi desert to the south described a fireball streaking across the sky along a flight path (based on a later reconstruction) at about 10o, just slightly east of true north.  Along this direction, the object approached Keshma from the south.  Then the object was observed by others moving very nearly due east toward Preobrazhenka.  This was followed by the object moving slightly north of due west toward Vanavara.  The explosion itself was oval shaped, suggesting a prior motion in the westerly direction.     


The Gobi desert is over 2000 km away from Tunguska...the explosion was seen again clearly...no refraction/reflection...the facts are very clear...       


Let us calculate the precise numbers.

Elevation of the Gobi desert: 900 - 1500 meters; to satisfy the RE requirements, we will ascend to 2000 meters.

Distance to Tunguska, over 2000 km...

Then the visual obstacle will measure: 275 KILOMETERS (two hundred and seventy-five kilometers).


The precise trajectory and explosion seen all the way from the Gobi desert, despite a 275 km visual obstacle (on a spherical earth)...it is over for the round earth theory...


AND YOU HAVE NOT ANSWERED THE FACT THAT THERE IS NO ATTRACTIVE GRAVITY...my last message here...

COMETS DO NOT OBEY AN ATTRACTIVE GRAVITY LAW:

The tails of the comets do not obey the principle of gravitation and are repelled by the sun. “There is beyond question some profound secret and mystery of nature concerned in the phenomenon of their tails” ; enormous sweep which it (the tail) makes round the sun in perihelion, in the manner of a straight and rigid rod, is in defiance of the law of gravitation, nay, even of the recorded laws of motion” Comets’ tails are clearly subject to some strong repulsive force, which drives the matter composing them away from the sun with enormously high velocities” (W.H. Pickering)

Wrong.  The above quote is from John Herschel's 'Outlines of Astronomy' 1849.  Since then, it has been shown that a comet's tail is pushed away from the sun by solar wind.  There is no defiance of gravity.

You have not answered any of the facts I provided for you...with this exception...your last comment.

Solar wind presupposes the existence of a spherically shaped sun...I did prove to you that there is no way the Sun could have attained a spherical shape...please read carefully...

And the solar wind is just a pseudoexplanation, given the solar paradoxes, that is, the neutrinos that cannot be detected, the faint young sun paradox.


Clouds and gases do not obey an attractive gravity law. The barometric pressure paradox shows the same thing.

Since you do not have at your disposal anything else, you will superficially try to divert attention using random quotes from my messages; in the same message I proved to you that the sun could not have attained a spherical shape to start with. Did you also quote that section of my message?

YOU HAVE NOT PROVIDED ANY FORMULA TO BACK YOUR WILD CLAIMS!

If the vertical temperature gradient is +12.9°C per 100 meters (reminder: Positive sign means temperature gets hotter as one goes higher), then horizontal light rays will just follow the curvature of the Earth, and the horizon will appear flat. If the gradient is less the rays are not bent enough and get lost in space. That is the normal situation of a spherical, convex horizon. But if the gradient gets larger, say +18°C per 100 meters, an observer would see the horizon as concave, the right and left ends turned upwards as if one were standing at the bottom of a saucer

http://en.wikipedia.org/wiki/Mirage

In the figure at the left, the heavy arc denotes the surface of the Earth, with center at C, and the lighter arcs AB and A'B' concentric with it represent the sides of a beam of light propagating at constant height above the surface, from AA' to BB'. We can regard AA' and BB' as wavefronts of the (bending) horizontal beam at two different places; the direction of propagation is perpendicular to the wavefront. (Equivalently, we can say that the direction of propagation of a horizontal beam must always be perpendicular to the local vertical; the radial lines CAA' and CBB' radiating from the center of the Earth are the local verticals at A and B respectively.)

Evidently, to make the beam follow the Earth's curvature, its lower edge AB must travel more slowly than its upper edge A'B'. The speed of propagation is in fact just inversely proportional to the refractive index, n. As the distance to be traversed by our circulating beam is proportional to the distance R  from the center of the Earth, we require that 1/n (which is proportional to the speed) be proportional to R ; or, in other words, we need the product nR  to be constant, independent of height above the Earth. (This will make the “optical path length” along AB the same as along A'B'.)

[If you find this argument too superficial to be convincing, you can go read the more rigorous derivations of the nR = constant condition by Biot (1836), by Auer and Standish (2000), or one of the papers cited in my “horizontal-ray paradox” file, such as Bravais (1856) or Thomson (1872). There's also a wonderful mathematical treatment of circulating rays by Kummer (1860); a French translation of it by Verdet (1861) is available on Gallica.]

To have nR  remain constant with height, n must decrease by 1 part in 6.4 × 106 for every meter of height, as R = 6400 km = 6.4 × 106 m. But the refractivity ( n − 1 ) is only about 1/3200 of n ; so the density of the air [which is proportional to ( n − 1 ), not to n] must fall by about 3200/6.4 × 106 m, or 1 part in 2000 for every meter.

Now, the decrease in density due to the decrease in pressure with height (1 part in 8000 per meter) is only 1/4 of this, so we need 3 times as much decrease from the temperature gradient, or 3 parts in 8000 per meter. That means the temperature must increase by 3 parts in 8000 of the 300 K, or about 900/8000 of a degree = 0.11° per meter. So a temperature inversion (i.e., increasing upward instead of the usual decrease) of about 0.11°/m will produce a circulating beam or ray.

As a check, we can do the arithmetic a little differently. Because the refractive index n decreases by 1 part in a million per degree, and we need a decrease in n of 1 part in 6.4 million per meter, we would need about 1/6.4 of a degree per meter if the pressure stayed constant. But, as the pressure alone gives an effect equivalent to 3/80 of a degree per meter, we really need only ( 1/6.4 − 3/80 ) or, expressing the fractions as decimals, about 0.16 − 0.04 = 0.12°/m.

http://mintaka.sdsu.edu/GF/explain/atmos_refr/bending.html
San Diego State University

Further scientific work showing how looming can follow the curve of the earth:

William Latham (1798)
Wegener (1918)
J. de Graaff Hunter (1913)

You have not answered any of the facts I provided for you...with this exception...your last comment.

Solar wind presupposes the existence of a spherically shaped sun...I did prove to you that there is no way the Sun could have attained a spherical shape...please read carefully...

And the solar wind is just a pseudoexplanation, given the solar paradoxes, that is, the neutrinos that cannot be detected, the faint young sun paradox.


Clouds and gases do not obey an attractive gravity law. The barometric pressure paradox shows the same thing.

Since you do not have at your disposal anything else, you will superficially try to divert attention using random quotes from my messages; in the same message I proved to you that the sun could not have attained a spherical shape to start with. Did you also quote that section of my message?

I have a life.  I am responding to your quotes as I get time.  I am inclined not to bother, since every quote of yours that I have analysed so far is based on science from the 1800's and is shown to be flawed.

Regarding your response to the solar wind: Solar wind is composed primarily of electrons and protons.  Protons have mass.

Regarding your claim the sun is not spherical, there is ample evidence that the sun is spherical, most obvious of which is the astronomical observations made by telescopes since Galileo's times.  Sunspot's have been observed rotating across the sun's surface  and changing shape proportionately as they only could on a spherical surface.

The absence of an explanation of why the photosphere is not at the  pressure as it should be due to gravity isn't proof the sun is not spherical; it's simply a fact which yet requires explanation.  Just as your inability to grasp that the failure of ozone to settle out as you expect that it should can be easily explained ie: ozone falls to layers where there is less penatration of ultravoilet light, which allows it to degrade to oxygen which combines to form air (O₂).

Indeed an explanation is offered, which you dismiss casually:

But if this is so, why is the lowest pressure of the solar atmosphere observed over the sunspots where the light pressure is least?

Seems to me, sunspots actually represent deep holes in the sun's surface, and hence may offer less gravitational pull?  Or perhaps the electromagnetic field has something to do with it.  I confess don't know, I would need to do more research in this area, but I can promise you that I will not be accepting your un-referenced cut and pastes without further information.   Show me a source, and put your quotes in quotation marks atleast so I know who is saying this (because I'm not going to take your word for it), then I will come up with a cohesive response.

I'm not playing tit for tat here; if you want me to respond to something, give me your sources, otherwise I might as well be a puppet dancing on the strings you manipulate.

WHY do you keep bullshitting everybody here?

I'm not bullshitting anyone here.  I am quoting current reputable mainstream sources for my information.

You on the other hand are quoting 19th century disproved theories....

 

And now even more proofs about the Tunguska explosion.
http://www.halexandria.org/dward232.htm

Herdsman in the Gobi desert to the south described a fireball streaking across the sky along a flight path (based on a later reconstruction) at about 10o, just slightly east of true north.  Along this direction, the object approached Keshma from the south.  Then the object was observed by others moving very nearly due east toward Preobrazhenka.  This was followed by the object moving slightly north of due west toward Vanavara.  The explosion itself was oval shaped, suggesting a prior motion in the westerly direction.     


The Gobi desert is over 2000 km away from Tunguska...the explosion was seen again clearly...no refraction/reflection...the facts are very clear...       


Let us calculate the precise numbers.

Elevation of the Gobi desert: 900 - 1500 meters; to satisfy the RE requirements, we will ascend to 2000 meters.

Distance to Tunguska, over 2000 km...

Then the visual obstacle will measure: 275 KILOMETERS (two hundred and seventy-five kilometers).


The precise trajectory and explosion seen all the way from the Gobi desert, despite a 275 km visual obstacle (on a spherical earth)...it is over for the round earth theory...

If you head up from the Gobi desert on a heading of around 10 degrees, you'll find you don't intersect the Tunguska area at all.

It is not a good idea to argue with me about gravity...

WTF....?   Who made you the expert?  As in, show me your peer reviewed articles, and your university degrees.  I challenge you; prove to me just how you are such an authority.  Or are you just an ordinary Joe like me with an interest in science and no more reason than anyone else to claim righteousness on the topic of gravity..?

METEORITES DO NOT OBEY AN ATTRACTIVE GRAVITY LAW:

Meteors, after entering the terrestrial atmosphere at about 200 km. above the ground (in the official theory), are violently displaced toward the east. These displacements of the meteors are usually ascribed to winds blowing in the upper atmosphere. The atmospheric pressure at a height of 45 km. is supposed to be but “a small fraction of one millimeter of mercury.” On the other hand, the velocity with which the meteors approach the earth is between 15 and 75 km. per second, on the average about 40 km. per second or over 140,000 km. per hour. If winds of 150 km. per hour velocity were permanently blowing at the height where the meteors become visible, it would not be possible for such winds of rarefied atmosphere to visibly deflect stones falling at the rate of 140,000 km. per hour.
Approaching the earth, the meteorites suddenly slow down and turn aside, and some are even repelled into space. “A few meteors give the appearance of penetrating into our atmosphere and then leaving it, ricocheting as it were.”

btw, I worked out where you are getting this "no gravity" material.  Solely from the work of Immanuel Velikovsky.  Velikovsky's theories have not been supported by any significant scientist or theorist.  He is alone in his beliefs and never showed any proof of any of his claims.  His understanding of the actual science behinds his claims is poor. Furthermore, he did not support a flat earth.  His theories supported a broadly Copernican model that posited magneto-electrical forces as the cause of the effect we ascribe to gravity.

And despite quoting him solely on 'no gravity', you haven't once quoted the man.  Very poor show old chap.

Re: meteors. The faster they hit the upper layer and the more oblique the angle, the harder the impact.  This is why the space shuttle required such serious shielding, and why for re-entry they need to get the angle just right or they will indeed bounce of the earth's upper atmosphere and back out into space. 

The 'daylight fireball' in August 1972 was filmed bouncing of the earth's atmosphere. 

http://www.liveleak.com/view?i=503_1215418331
http://adsabs.harvard.edu/full/1994A%26A...283..287C

It all depends on the angle.  If the angle is too steep though, the meteor will fall towards the earth.  There are no winds blowing in the upper atmosphere, and I have never heard, and can find no reference other than Velikovsky's words, of any theory that meteor's are buffeted by strong winds in the upper atmosphere.  The buffeting they experience is due to friction heating and energising the air around the meteor as it enters, or bounces off the atmosphere. 

Violently displaced towards the east?  The Earth spins from the west to east, hence the sun comes up in the east.  So when a meteor hits the atmosphere, which is travelling east along with the rest of the planet, of course the friction will displace the meteor in this direction.  It's called "wind resistance" aka air pressure.  The same phenomena keeps an airplane aloft.  Not magic.  Not mystery.  Science. 

pitroidtech...your responses, as usual, have very little to do with science...

You have not provided any formula, as I have asked.

We need here a looming formula, which will provide exact answers to your concerns, do I have to do your homework for you?

I have already found such a formula...where is your research on this subject? You posted general facts about thermal inversion, that is very nice of you, but we need the formula itself.


You have not answered to any of the points I raised which do prove that gravity cannot be attractive. The article Cosmos without Gravitation by Velikovsky QUOTES from scientific sources, see the bibliography here:

http://www.varchive.org/ce/cosmos.htm

The fact that a cloud is made up of water droplets which do defy gravitation was commented by Velikovsky, as were the other points he raised, but the scientific facts are beyond dispute, please read carefully. The facts about the atmosphere of the sun, and the sunspots are very well known, as you should know by now.


Ozone is created in the upper atmosphere.  As it falls back to earth it degrades back to oxygen.

DO YOU UNDERSTAND WHAT WE ARE TALKING ABOUT HERE? Certainly you do not...

Ozone-oxygen cycle: the atomic oxygen IMMEDIATELY REACTS WITH other oxygen molecules, to form ozone again.

The overall effect of the ozone-oxygen cycle is to convert penetrating UV radiation into heat, WITHOUT ANY NET LOSS OF OZONE.

Thus, the ozone layer is kept in a stable balance. And, moreover, in the stratosphere, the ozone layer concentrations are about 2 to 8 parts per million, which is much higher than in the lower atmosphere.


Now, we get back to what I told you before.

Ozone, though heavier than oxygen, is absent in the lower layers of the atmosphere, is present in the upper layers, and is not subject to the mixing effect of the wind. The presence of ozone high in the atmosphere suggests that oxygen must be still higher: As oxygen is less dense than ozone, it will tend to rise to even greater heights. Nowhere is it asked why ozone does not descend of its own weight or at least why it is not mixed by the wind with other gases.

With attractive gravity, OZONE WOULD DESCEND IMMEDIATELY AS ITS SPECIFIC WEIGHT IS GREATER THAN THAT OF OXYGEN.


Nitrogen is only 3% lighter than air.  This along with the constant state of motion of the air in the atmosphere (below the stratosphere, air is constantly on the move), means any particle are held in suspension, much like shaking a bottle of oil and water prevents the oil and water from settling out to different layers.

Please give up these childish, unscientific notions...

IF THERE IS ATTRACTIVE GRAVITY, THEN GASES MUST SEPARATE INTO LAYERS, ACCORDING TO THEIR SPECIFIC WEIGHTS.

Then the moment the wind subsides, the nitrogen should stream upward, and the oxygen should drop, preceded by the argon. If winds are caused by a difference in weight between warm and cold air, the difference in weight between heavy gases high in the atmosphere and light gases at the lower levels should create storms, which would subside only after they had carried each gas to its natural place in accordance with its gravity or specific weight. But nothing of the kind happens.

When some aviators expressed the belief that pockets of noxious gas are in the air, the scientists replied:

There are no pockets of noxious gas. No single gas, and no other likely mixture of gases, has, at ordinary temperatures and pressures, the same density as atmospheric air. Therefore, a pocket of foreign gas in that atmosphere would almost certainly either bob up like a balloon, or sink like a stone in water.

Why, then, do not the atmospheric gases separate and stay apart in accordance with the specific gravities?

End of story. Gases do not obey an attractive gravity law.


Water is not held as droplets it is held as water vapour. 

You are dreaming, again...

DEFINITION OF A CLOUD:

A CLOUD IS A VISIBLE MASS OF DROPLETS. The small droplets of water WHICH DO MAKE UP A CLOUD, will have 0.01 mm in diameter.
The tiny particles of water are very densely packed, and may even combine to form larger water molecules, which ARE denser than the surrounding air.

IT IS RIGHT AT THIS POINT, WHERE WE ADDRESS THE ISSUE YOU DODGED: Water, though eight hundred times heavier than air, is held in droplets, by the millions of tons, miles above the ground. Clouds and mist are composed of droplets which defy gravitation. For quite a while, that cloud will hold those droplets of water, DEFYING ATTRACTIVE GRAVITY, that is what are talking about here.

Let us take a look at the weight of some clouds.

Clouds can have a height ranging from 50 meters to over 5 km, and a length ranging from 100 meters to 1000 km; a cumulus cloud, 1 kilometer in diameter, will weigh 5 MILLION TONS, or about the weight of 1 million cars. A cumulonimbus cloud, 5 kilometers in height, and having a diameter of 15 kilometers, will actually weigh 1 BILLION TONS.


Clouds and mist DO defy attractive gravity...


Obviously, you have never thought about these things...you have just accepted everything, just as you accept the round earth nonsense.


ntheleast...do not try such nonsense, here with me...those inhabitants of the Gobi desert saw both the trajectory AND the explosion itself from some 2000 km distance...we have then a 275 km visual obstacle...

I have already given you the precise observation about the trajectory itself (which, by the way, proves it could not have been either a comet or a meteorite):

T.R. LeMaire, a science writer, continues this thought, by suggesting "The Tunguska blast's timing seems too fortuitous for an accident" (LeMaire 1980). He claims that a five-hour delay would make the target of destruction St. Petersburg, adding that a tiny change of course in space would have devastated populated areas of China or India.

Can we assume that the 'pilot' chose a cloudless day with excellent visibility from aloft to assure a safe drop? American Military strategy called for identical weather conditions; for a perfect strike on Hiroshima's industrial heart, the Enola Gay's bombardier was forbidden to release through a cloud cover: he had to see the target below. To maximize blast destruction, minimize radiation perils: the bomb was set to explode at a high altitude rather than against the ground. Similarly, the Siberian missile detonated high in the air, reducing or even eliminating fallout hazard (LeMaire 1980).

LeMaire maintains the "accident-explanation is untenable" because "the flaming object was being expertly navigated" using Lake Baikal as a reference point. Indeed, Lake Baikal is an ideal aerial navigation reference point being 400 miles long and about 35 miles wide. LeMaire's description of the course of the Tunguska object lends credence to the thought of expert navigation:

The body approached from the south, but when about 140 miles from the explosion point, while over Kezhma, it abruptly changed course to the east. Two hundred and fifty miles later, while above Preobrazhenka, it reversed its heading toward the west. It exploded above the taiga at 60º55' N, 101º57' E (LeMaire 1980).


The trajectory changed course several times...that is what the people from Irkutsk and Gobi saw very clearly...and Gobi is a very large desert...that is why I provided the 2000 m altitude for our calculations, even though the highest altitude there is around 1500 m...

And now even more proofs about the Tunguska explosion.
http://www.halexandria.org/dward232.htm

Herdsman in the Gobi desert to the south described a fireball streaking across the sky along a flight path (based on a later reconstruction) at about 10o, just slightly east of true north.  Along this direction, the object approached Keshma from the south.  Then the object was observed by others moving very nearly due east toward Preobrazhenka.  This was followed by the object moving slightly north of due west toward Vanavara.  The explosion itself was oval shaped, suggesting a prior motion in the westerly direction.     


The Gobi desert is over 2000 km away from Tunguska...the explosion was seen again clearly...no refraction/reflection...the facts are very clear...       


Let us calculate the precise numbers.

Elevation of the Gobi desert: 900 - 1500 meters; to satisfy the RE requirements, we will ascend to 2000 meters.

Distance to Tunguska, over 2000 km...

Then the visual obstacle will measure: 275 KILOMETERS (two hundred and seventy-five kilometers).


The precise trajectory and explosion seen all the way from the Gobi desert, despite a 275 km visual obstacle (on a spherical earth)...it is over for the round earth theory...

You need to work on refining your posts. Pages and pages of text don't work in your favor when trying to make a point.

Stuff about how clouds should instantly plummet and the atmosphere should separate into layers.

Not believing in mixtures and aerosols is all fine and good, but I'm not sure how it acts as evidence for a flat Earth, as things should settle just as much in a flat Earth model as in a round Earth model. The fact that clouds do not appear to fall from the sky should be a good indication that your hypothesis that clouds should fall from the sky is wrong.

Herdsman in the Gobi desert to the south described a fireball streaking across the sky along a flight path (based on a later reconstruction) at about 10o, just slightly east of true north.  Along this direction, the object approached Keshma from the south.  Then the object was observed by others moving very nearly due east toward Preobrazhenka.  This was followed by the object moving slightly north of due west toward Vanavara.  The explosion itself was oval shaped, suggesting a prior motion in the westerly direction.     

If you head up from the Gobi desert on a heading of around 10 degrees, you'll find you don't intersect the Tunguska area at all.

Levee, I refuse to debate this with you any further.  Apart from growing tired of your condescending attitude, I find your continual cut and pasting of the same information to be a trifle boring.  You keep telling me to provide formulas, then when I show how the refractive index of an inversion later is calculated, you completely ignore the information and tell me yet again that I haven't shown the formula.  Yet the formula's you quoted explicitly state that they do not allow for temperature inversion!  So you have not shown any calculations yourself.

It's obviously quite pointless having any kind of debate with you.  You believe what you want to believe in the face of any reasonable evidence to the contrary, and ignore the facts when it suits you.

Levee, I refuse to debate this with you any further.  Apart from growing tired of your condescending attitude, I find your continual cut and pasting of the same information to be a trifle boring.  You keep telling me to provide formulas, then when I show how the refractive index of an inversion later is calculated, you completely ignore the information and tell me yet again that I haven't shown the formula.  Yet the formula's you quoted explicitly state that they do not allow for temperature inversion!  So you have not shown any calculations yourself.

It's obviously quite pointless having any kind of debate with you.  You believe what you want to believe in the face of any reasonable evidence to the contrary, and ignore the facts when it suits you.

So, the FE'rs have gotten under you skin.  +1 for the FET.  Next subject, please.

Really, don't get frustrated about the fact that they ask for proof, you give it to them, and then they run in circles telling you that you should believe your own eyes or that your proof is wrong/lie or that it is part of the conspiracy.  This is their M.O.  They will demand proof from you, and then try to debunk every bit of evidence that you present.

Quote from: pitdroidtech on October 03, 2011, 07:02:54 AM

Levee, I refuse to debate this with you any further.  Apart from growing tired of your condescending attitude, I find your continual cut and pasting of the same information to be a trifle boring.  You keep telling me to provide formulas, then when I show how the refractive index of an inversion later is calculated, you completely ignore the information and tell me yet again that I haven't shown the formula.  Yet the formula's you quoted explicitly state that they do not allow for temperature inversion!  So you have not shown any calculations yourself.

It's obviously quite pointless having any kind of debate with you.  You believe what you want to believe in the face of any reasonable evidence to the contrary, and ignore the facts when it suits you.

So, the FE'rs have gotten under you skin.  +1 for the FET.  Next subject, please.

Really, don't get frustrated about the fact that they ask for proof, you give it to them, and then they run in circles telling you that you should believe your own eyes or that your proof is wrong/lie or that it is part of the conspiracy.  This is their M.O.  They will demand proof from you, and then try to debunk every bit of evidence that you present.

No you are wrong.  FET have not got under my skin and it is most assuredly not a win to FET.  Levee is unique even amongst FETists.  It is quite specifically Levee who get's under my skin.  I am quite happy to continue arguing with other FET supporters.  You can see I have had a long running and quite reasonable discussions with Ski for example or even Tom (believe it or not).

Levee on the other hand is the only one who cut and pastes pagefuls of THE SAME MATERIAL again and again without making any effort to respond to legitimate questions.  If I thought Levee and I could have a legitimate to and fro intellectual discussion I would continue to discuss it with him, but he has shown again and again a complete inability to grasp scientific concepts and simply cut and pastes reams of information from sources that he doesn't really understand.  It's pointless arguing with him and I will do so no longer.

If any other FE'er comes along and wants to discuss points with me I will do so gladly.

BTW, this is my thread, I am the OP, and I am fed up with Levee continually pulling the thread off topic.  The Topic is " Photographic Evidence of Curvature Beyond the Visible Horizon".  The emphasis is on "Photographic".  Discussions about meteors, comets, tunguska, gravity etc are not on topic.  Discussion about refractive index of inversion layers certainly are relevant to the topic, given the affect of refraction on visibility of objects on the horizon in photographs, but I have provided a link to the workings regarding refractive index of an inversion layer running parallel to the earth, and Levee has not responded to this.  Infact he has categorically refused to acknowledge it at all.  Therefore I consider his ongoing input to this thread irrelavant and it will be ignored by me. 

I couldn't help it, so I did my own photographic experiment across the surface of a body of water with different magnifications and different elevations.  I still need to do some comparisons between them, determine where the actualy water line should be in the refraction across the surface in the low elevation photos, and research again what the curvature height should be across a distance of one mile. 

I took several pictures, but my 'setup' isn't ideal, so some pictures are better than others.  I'm still sorting them.

I'm thinking about getting another set of the low elevation pictures now that the temperature has dropped over the last couple days to see if there's a difference. 

* 8 inches per mile.  I'm using a large rock as my target, but the opposite bank works too and is a bit further.  I'm going to move over a bit which will put me parrallel to the opposite bank and it's foilage, instead of it being a little closer and angled toward me at my current spot.

I couldn't help it, so I did my own photographic experiment across the surface of a body of water with different magnifications and different elevations.  I still need to do some comparisons between them, determine where the actualy water line should be in the refraction across the surface in the low elevation photos, and research again what the curvature height should be across a distance of one mile. 

I took several pictures, but my 'setup' isn't ideal, so some pictures are better than others.  I'm still sorting them.

I'm thinking about getting another set of the low elevation pictures now that the temperature has dropped over the last couple days to see if there's a difference. 

* 8 inches per mile.  I'm using a large rock as my target, but the opposite bank works too and is a bit further.  I'm going to move over a bit which will put me parrallel to the opposite bank and it's foilage, instead of it being a little closer and angled toward me at my current spot.

That means from a height of 8 inches, the horizon is 1 mile distant. 

Good luck with your experiment.  I look forward to seeing your results.

The relationship between altitude and distance to the horizon on a round earth is not linear.  The horizon will approach 1 quarter of the earth's circumference but never reach it, regardless of altitude.

That means from a height of 8 inches, the horizon is 1 mile distant. 

Good luck with your experiment.  I look forward to seeing your results.

8 inches height equals 1 mile distant, right.  Looked it up really quick earlier as I was getting ready to head out, misread it and posted.  Too much on my mind.

I'm going to get another set and see if I can get a little lower.  The refraction across the surface will be a problem though, and I'm not sure if I'll really be able to tell anything or not because of the scale I'm working with.

Distance- .67 mile to rock, .75 to opposite shore.
elevation- about 3.5 inches (center of scope), 15 inches, 20 feet, and 35 feet (these are rough estimates, I'll use a tape measure next time too)
scope- 15-60x 60mm

The water is almost like glass when there is no wind.  Maybe some faint ripples out in the middle.  It's a gravel pit pond.

I'd head out to the rock in my kayak and use some white tape at water level and mark every 6 inches upwards for about 3 feet.  That might help.

You keep telling me to provide formulas, then when I show how the refractive index of an inversion later is calculated, you completely ignore the information and tell me yet again that I haven't shown the formula.

Your effort is greatly appreciated...but, you only posted a formula for the determination of the ray curvature...and you know very well what we need here, don't you? As usual, I have to do the research for the RE...

HERE IS THE ONLINE CALCULATION OF THE LOOMING/THERMAL INVERSION EFFECTS ON CURVATURE:

http://mintaka.sdsu.edu/GF/explain/atmos_refr/altitudes.html

We need to specify the following: temperature/height at the place of observation, temperature/height of the visual target, distance, and most important of all the STANDARD OR ARBITRARY LAPSE RATE (an arbitrary lapse rate corresponds to taking into account the looming/thermal inversion effects).

How to compute the arbitrary lapse rate is described in the paragraphs preceding the formula, and also here (you missed these things, didn't you?):

http://mintaka.sdsu.edu/GF/explain/atmos_refr/bending.html


Let us give it a try with a distance of 50 km, temperature of 20C, observer at 2 meters, and visual target at 180 meters.

We will get a POZITIVE VALUE for the angle in question, that is, the visual target can be seen/observed.

Let us now change the height of the visual target to 20 meters, something we know that it could not be seen on a round earth. Also, we will use an arbitrary lapse rate.

We will get a NEGATIVE VALUE for the angle, that is, the visual target could not be seen.


For Lake Michigan, we will use multiple sets of numbers, to satisfy the requirements of the RE. Temperatures at Holland of 5, 10, 20C (remember, it was Memorial Day 2003, we could expect some 10-15C there), and at Milwaukee/Racine of 5, 10, 20C; height of the resident at some 20 meters (and I am being very generous here; actually we could take some 5-10 meters, but I will look the other way), height of visual target: first 40 meters, then 183 meters.

We will use, of course, ONLY AN ARBITRARY LAPSE RATE, to take into account the full LOOMING/THERMAL INVERSION/DUCT FORMATION EFFECTS.


NO MATTER HOW THE FORMULA IS USED WE GET A NEGATIVE VALUE/ANSWER: that is, the visual target CANNOT BE SEEN.

This was to be expected of course...as I have clearly described the facts.

The final word and conclusion: there is no curvature over Lake Michigan, across a distance of 128 km.



Not believing in mixtures and aerosols is all fine and good, but I'm not sure how it acts as evidence for a flat Earth, as things should settle just as much in a flat Earth model as in a round Earth model. The fact that clouds do not appear to fall from the sky should be a good indication that your hypothesis that clouds should fall from the sky is wrong.

The fact that clouds do not fall from the sky points clearly to another explanation of gravity.

Again, for you...

DEFINITION OF A CLOUD:

A CLOUD IS A VISIBLE MASS OF DROPLETS. The small droplets of water WHICH DO MAKE UP A CLOUD, will have 0.01 mm in diameter.
The tiny particles of water are very densely packed, and may even combine to form larger water molecules, which ARE denser than the surrounding air.

IT IS RIGHT AT THIS POINT, WHERE WE ADDRESS THE ISSUE YOU DODGED: Water, though eight hundred times heavier than air, is held in droplets, by the millions of tons, miles above the ground. Clouds and mist are composed of droplets which defy gravitation. For quite a while, that cloud will hold those droplets of water, DEFYING ATTRACTIVE GRAVITY, that is what are talking about here.

Let us take a look at the weight of some clouds.

Clouds can have a height ranging from 50 meters to over 5 km, and a length ranging from 100 meters to 1000 km; a cumulus cloud, 1 kilometer in diameter, will weigh 5 MILLION TONS, or about the weight of 1 million cars. A cumulonimbus cloud, 5 kilometers in height, and having a diameter of 15 kilometers, will actually weigh 1 BILLION TONS.


Each droplet of water in the clouds defies attractive gravity, can you understand this?


The gases in the atmosphere DO NOT obey an attractive gravity law. The barometric pressure paradox shows that gravity could not possibly be attractive.


If you head up from the Gobi desert on a heading of around 10 degrees, you'll find you don't intersect the Tunguska area at all.

We would have to know first, FROM WHERE, what place, the observation was actually made. Then, and only then, we could comment on the intersection with Tunguska.

Please read again.

Herdsman in the Gobi desert to the south described a fireball streaking across the sky along a flight path (based on a later reconstruction) at about 10o, just slightly east of true north.

We would have to know the exact location of those hersman to make any comments on the intersection, but...

Along this direction, the object approached Keshma from the south.  Then the object was observed by others moving very nearly due east toward Preobrazhenka.

Therefore, the ball lightning changed path several times before approaching Tunguska, as I have indicated from the very beginning.

This was followed by the object moving slightly north of due west toward Vanavara.  The explosion itself was oval shaped, suggesting a prior motion in the westerly direction.     


The inhabitants of the Gobi desert described clearly not only the erratic path of the object, but also the shape of the explosion.

The facts are very clear even for a stubborn RE supporter; the visual obstacle measures some 275 km, no way the object could have been seen from some 2000 km distance. Please wake up.


Perhaps, Nelgrande, you would like then to watch these videos:

(photograph pasted to look like the Earth in space, fake Apollo 11 mission)

(Apollo 16, astronauts suspended from cables)

You like to be treated like a chump, by Nasa?

And here is how the Moon became spherical:

http://web.archive.org/web/20090126060440/http://geocities.com/apolloreality/


You will not believe a very clear description of the fact that the explosion from Tunguska was seen both from the Gobi desert and from Irkutsk, but you will believe that the fake Apollo missions (please watch the videos to convince yourself) did actually take place...

For Lake Michigan, we will use multiple sets of numbers, to satisfy the requirements of the RE. Temperatures at Holland of 5, 10, 20C (remember, it was Memorial Day 2003, we could expect some 10-15C there), and at Milwaukee/Racine of 5, 10, 20C; height of the resident at some 20 meters (and I am being very generous here; actually we could take some 5-10 meters, but I will look the other way), height of visual target: first 40 meters, then 183 meters.

We will use, of course, ONLY AN ARBITRARY LAPSE RATE, to take into account the full LOOMING/THERMAL INVERSION/DUCT FORMATION EFFECTS.

NO MATTER HOW THE FORMULA IS USED WE GET A NEGATIVE VALUE/ANSWER: that is, the visual target CANNOT BE SEEN.

This was to be expected of course...as I have clearly described the facts.

The final word and conclusion: there is no curvature over Lake Michigan, across a distance of 128 km.

I'll humour you.  Let's look at the equations:

Please enter the temperature and height at the Observer:
Temp. :   5  °C       Height :     2    meters
and the temperature and height at the Target:
Temp. :    23  °C     Height :     183     meters

This gives a laps rate of -99.4 degrees C/km  (which is actually very close to the .11 degrees/m that the very bending page you linked to claims is the level at which a refracted ray travels parallel to the earth)

A positive lapse rate indicates the rate at which temperature falls with increasing altitude.  Therefore a NEGATIVE lapse rate indicates the rate at which temperature RISES with altitude, ie: an inversion.

So the lapse rate of -99.4 and a distance of 128km gives us a POSITIVE apparent altitude of the target:  0.015 degrees of arc

As the results of the calculation show:

These values place the target 1.84 minutes of arc, or 0.031 degrees, above the apparent sea horizon. The target is visible.

You can see that at a low height of the observer of 2m, and a large temperature differential, can result in the 183m target being visible at a distance of 128km.

Now the temperature differential is large, but keeping in mind that 1. this occurance has only been seen 12 times in 30 years, and 2. the temperature difference over a height differential of 181m is entirely plausible for a thermal inversion, the ability to see  the horizon at this distance is not proof of a flat earth.

Let's look at something lower to the ground and the temperature differential that would be needed over height (ie the strength of the inversion)

Please enter the temperature and height at the Observer:
Temp. :   6.5  °C       Height :     2    meters
and the temperature and height at the Target:
Temp. :    10.8  °C     Height :     40     meters

= -113.16 lapse rate @ 128km
=   0.51 minutes of arc  =  0.008 degrees of arc

That's a difference of 4.3 degrees over 38 metres height, not unreasonable.

BTW, we don't use an arbitrary lapse value; we use the lapse value that the formula calculates from the temperature and height differentials.  Perhaps this is where you are making your mistake?  Please show how you calculated the lapse value.

Or perhaps the mistake you have made is to assume that the temperature is constant all the way from the ground to the top of the building. If this was the case then the formula yields negative values for arc, but this only occurs if there is no inversion in effect. When the temperature for both heights is the same, the lapse rate is zero.

Here is the calculation for a ray that curves at the same rate as the surface of the earth:

To have nR  remain constant with height, n must decrease by 1 part in 6.4 × 106 for every meter of height, as R = 6400 km = 6.4 × 106 m. But the refractivity ( n − 1 ) is only about 1/3200 of n ; so the density of the air [which is proportional to ( n − 1 ), not to n] must fall by about 3200/6.4 × 106 m, or 1 part in 2000 for every meter.

Now, the decrease in density due to the decrease in pressure with height (1 part in 8000 per meter) is only 1/4 of this, so we need 3 times as much decrease from the temperature gradient, or 3 parts in 8000 per meter. That means the temperature must increase by 3 parts in 8000 of the 300 K, or about 900/8000 of a degree = 0.11° per meter. So a temperature inversion (i.e., increasing upward instead of the usual decrease) of about 0.11°/m will produce a circulating beam or ray.

Let's plug .11 degrees per metre (ie 110 degrees per km) into the formula, with height 2m of observer and height 183m of target:

If the observer temperature at 2m is 5 degrees, then the target temperature at 183m is 25 degrees. Again, not common, but not unheard of for an inversion; completely within bounds. 

If we plug 110 degree C per 1km and respective heights of 2m and 40m, we get a temperature differential of 4.18 degrees. 

If the warm air layer is being heated by the sun, and the cold air layer is being kept cold by water currents for example, then the difference in temperature between the warm air layer and the cold air layer can become extreme enough to cause the observed results.

Normally, air temperature decreases at a rate of 3.5°F for every 1000 feet (or roughly 6.4°C for every kilometer) you climb into the atmosphere. When this normal cycle is present, it is considered an unstable air mass and air constantly flows between the warm and cool areas. As such the air is better able to mix and spread around pollutants.
During an inversion episode, temperatures increase with increasing altitude. The warm inversion layer then acts as a cap and stops atmospheric mixing. This is why inversion layers are called stable air masses.

Temperature inversions are a result of other weather conditions in an area. They occur most often when a warm, less dense air mass moves over a dense, cold air mass. This can happen for example, when the air near the ground rapidly loses its heat on a clear night. In this situation, the ground becomes cooled quickly while the air above it retains the heat the ground was holding during the day. Additionally, temperature inversions occur in some coastal areas because upwelling of cold water can decrease surface air temperature and the cold air mass stays under warmer ones.

http://geography.about.com/od/climate/a/inversionlayer.htm

The fact that clouds do not fall from the sky points clearly to another explanation of gravity.

Intelligent falling, perhaps?

You're not really providing anything to explain why things can't mix. Why should gases fight Brownian motion and separate themselves into layers? Where's the maths to suggest the tiny differences in mass are strong enough to overcome all the other forces fighting to mix the gases up and keep the clouds in the sky?

We would have to know first, FROM WHERE, what place, the observation was actually made. Then, and only then, we could comment on the intersection with Tunguska.

It doesn't matter where in the desert the observed it from. Even following the most westerly parts of the desert up on a 10 degree heading will just miss the point where the Tunguska event occurred.

To be fair, it seems silly to assume that this event is the same one that occurred over Tunguska. We would have to know first, FROM WHERE, what place, the observation was actually made. Then, and only then, we could comment on the intersection with Tunguska.

Therefore, the ball lightning changed path several times before approaching Tunguska, as I have indicated from the very beginning.

This seems to be even more evidence that the object seen in the desert was different to the one that exploded over Tunguska. Especially as your sources suggest that the object ended up 'slightly north of due west', nowhere near the direction of Tunguska.

@Levee, I'd love to answer your moon landing conspiracy videos, but as I said before, I'm not going to engage you on issues that are not relevant to the thread topic.  Go and post your moon stuff in the thread for it, and I will answer you there.

In the meantime, I request that you please refrain from posting off topic posts in this thread.

I have some good photographs of clouds appearing to touch the horizon at sea as photographed from an altitude of approximately 100 feet above sea level. I will post them later. They demonstrate the same effect as seen by the OP's photos.

I have some good photographs of clouds appearing to touch the horizon at sea as photographed from an altitude of approximately 100 feet above sea level. I will post them later. They demonstrate the same effect as seen by the OP's photos.

Great 

I will get some more photos uploaded myself soon. 

You cannot have 5C in Holland, and some 23C in Milwaukee, it would be a 18C difference over a distance of 128 km. Please check the archives of the local newspapers, and the fact that the residents did mention it was WARM DAY (Memorial Day).

With some 10C in Holland, you will get a negative result.


And you will get a negative result also for 6.5 (Holland), and 10.8 (Racine), no way the County Courthouse could be seen from 128 km distance.

Therefore, the formulas are pretty clear, and will tell you that those buildings could not and would not be seen from such a great distance.

It seems strange that you are satisfied with using an 18C difference, if you just try 10C instead, the result will be negative.

IN FACT IT WAS SOME 60 DEGREES F IN MILWAUKEE, 15C, ON MEMORIAL DAY, AND NOT 23C:

http://www.jsonline.com/historicarchive/search/?searcgVt-word&searchText=2003 memorial day

And the lapse rate will be much lower...and the result, of course, negative.

I did not make any mistakes at all; I used the right formula, you used the wrong temperatures for that day.


n, you have not explained (and actually dodged the question) of the following fact: droplets of water contained in a cloud defy attractive gravity, again, can you understand this? Gases do not stay separated according to their specific weights in the atmosphere, as they should, given the existence of attractive gravity.


But it does matter where it was observed from.

Herdsman in the Gobi desert to the south described a fireball streaking across the sky along a flight path (based on a later reconstruction) at about 10o, just slightly east of true north.

Perhaps, the description is about the flight path itself, as it was seen from the Gobi desert.

The description it pretty clear; your constant denial only adds to the fact that you cannot accept that an explosion was seen from some 2000 km distance, despite a 275 km visual obstacle.

Along this direction, the object approached Keshma from the south.  Then the object was observed by others moving very nearly due east toward Preobrazhenka.

Therefore, the ball lightning changed path several times before approaching Tunguska, as I have indicated from the very beginning.

This was followed by the object moving slightly north of due west toward Vanavara.  The explosion itself was oval shaped, suggesting a prior motion in the westerly direction.     


The inhabitants of the Gobi desert described clearly not only the erratic path of the object, but also the shape of the explosion.

The facts are very clear even for a stubborn RE supporter; the visual obstacle measures some 275 km, no way the object could have been seen from some 2000 km distance. Please wake up.


The herdsmen observed even the erratic path, the explosion itself...no need to add or comment anything else.



If you do not want to accept the facts, please watch those videos again...that is the world you live in, faked Nasa missions...


Here is the report from Irkutsk itself, some 1000 km distance from Tunguska:

The first report of the explosion was in the Irkutsk paper dated July 2, 1908, published two days after the explosion:

...the peasants saw a body shining very brightly (too bright for the naked eye) with a bluish-white light.... The body was in the form of 'a pipe', i.e. cylindrical. The sky was cloudless, except that low down on the horizon, in the direction in which this glowing body was observed, a small dark cloud was noticed. It was hot and dry and when the shining body approached the ground (which was covered with forest at this point) it seemed to be pulverized, and in its place a loud crash, not like thunder, but as if from the fall of large stones or from gunfire was heard. All the buildings shook and at the same time a forked tongue of flames broke through the cloud.

All the inhabitants of the village ran out into the street in panic. The old women wept, everyone thought that the end of the world was approaching.

Pretty clear again, isn't it? No refraction/reflection...with a visual obstacle of 67,7 km, nobody could have seen anything from 1000 km distance.


“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.”

Even the trajectory of the explosion itself was seen, all the way from London...given the 7463 km visual obstacle nobody could have observed anything, from the other side of a globe...

Levee, have you considered that RM might be a better forum for you?

n, you have not explained (and actually dodged the question) of the following fact: droplets of water contained in a cloud defy attractive gravity

So do planes. There's more at play in the Earth's atmosphere than just gravity.

More Gobi desert stuff

Why do you think the thing over the Gobi desert is the same as the one that happened at Tunguska? If I can see the lights from a plane in the sky, does that mean that all the people in the world who can see lights in the sky are looking that the same plane that I am?

Here is the report from Irkutsk itself, some 1000 km distance from Tunguska:

Just because something was published in a paper in Irkutsk doesn't mean it happened in Irkutsk.

Also, a lot of your other reports seem to suggest this thing was moving around all over the place. If it was moving around a lot, what's wrong with the idea that it passed within sight of Irkutsk before exploding over the Tunguska area?

Even the trajectory of the explosion itself was seen, all the way from London

How does anyone observe a trajectory? I can't recall any reports from London of a dotted line in the sky marking the path the object was taking...

You cannot have 5C in Holland, and some 23C in Milwaukee, it would be a 18C difference over a distance of 128 km. Please check the archives of the local newspapers, and the fact that the residents did mention it was WARM DAY (Memorial Day).

You really don't seem to get this do you??

The temparature at ground level could have been more or less the same all the way from Holland to Milwaukee.  It's the temperature differences AT ALTITUDE that causes the inversion.  If it was 13 degrees at both locations at 2m, but 31 degrees at 183m, and remember this is a (relatively) cool layer of air with a warmer layer of air above, heated by the sun - not inconceivable that such a temperature difference could exist.  As to reports of a warm day, what does that mean?  It's a bit open ended.  And the majority of the refraction has already occurred over the cold ocean surface where the temperature differential between air layers is more extreme.

The fact is it is uncommon to see mirages such as this.  A fact that is evident by the research conducted by MARIE HAVENGA http://www.sandhillcity.com/mirage_articles.htm So given that such mirages are uncommon, it's reasonable to propose that a temperature inversion of unusually extreme conditions existed.

No matter which way you choose to twist it Levee, without an exact record of the temperatures at all points between Holland and Milwaukee and at different altitudes on that day at the time of the mirage, it's impossible to maintain that the degree of refraction needed could not have been attained.

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.

Grand Haven Daily Tribune   April 3, 1925

Now I ask you, if the surface of the earth is flat, then would not ANY cool clear day cause Milwaukee to be visible from Grand Haven harbor?  The answer is yes.  But yet, it's only very rarely visible.

The temperature records for Milwaukee are very clear: RANGE BETWEEN HIGH OF 60F TO A LOW OF 47F (8,3 C TO 15,5 C).

The same goes for Holland, Michigan.

With these values in mind we get:

A lapse rate of about -44C per km (way below the values for looming/thermal inversion layers).

I used 8.3C for Holland, and 15.5 for Milwaukee, and 183 meters the height of the target.

The result will be negative, of course.


To dream about impossible differences in temperature, over a distance of 128 km is pure nonsense.

There is no way a building of some 40 meters could be seen with a visual obstacle of 984 meters in front of the observer.

The highest temperature for Milwaukee was, as recorded in the data I provided, some 60F, you cannot use anything else.

As I said before, the case is closed; the only way the County Courthouse from Racine could have been seen, is if the surface of Lake Michigan is flat.

The only way those lights from Racine could have been seen (remember the date of 1925; no 40 meter buildings then in Racine) from Grand Haven, is of course, if the surface of Lake Michigan is completely flat.


On June 30, 7:15 - 7:20 am, there was sunlight all over Siberia. In London, at 0:15, of course, there was darkness, around midnight. We are told, in the official theory, that this happens because of the curvature of the Earth, which blocks the rays of light from the Sun.

Then, how could an explosion which did take place at some 7km in altitude, on a cloudless day, be seen from Irkutsk, the Gobi desert, and from London itself?

We have the glow of the trajectory, on one side of a globe, and the immediate, instant observation on the other side the same globe, from London.

At an instant, at some 7:20, newspapers can be read in London, photographs can be taken in Stockholm...therefore the intensity of the light of the explosion itself was seen, no question about it, as we have the newspaper eyewitness accounts.

The intensity decreased gradually over a period of one hour.

The visual range limit for the observation of the explosion itself is some 400 km. Period.

It was visible from Irkutsk, with a 67.5 km visual obstacle.

It was seen from London, even the trajectory itself, the glow from it, despite a 7643 km visual obstacle.


Regular changes, daily things do not evoke astonishment and do not stir the imagination of peoples.

The herdsmen of Gobi lacked the modern protection against the elements of nature, and they were much more accustomed to disturbances of weather patterns.

No other explosion took place over the sky of Asia at that time, June 30, 7:20, visible from the entire area, please give up the nonsense.

The herdsmen described even the erratic path of the object itself, and the explosion, all this despite a 275 km visual obstacle.


So do planes. There's more at play in the Earth's atmosphere than just gravity. The caliber of this statement puts you right along pitroidtech, with his immortal line: clouds are made up of vapour...no wonder you can be fooled so easily by Nasa...

Then, how could an explosion which did take place at some 7km in altitude, on a cloudless day, be seen from Irkutsk, the Gobi desert, and from London itself?

Simple. It wasn't seen from Irkutsk because the quote you keep pointing out is from a newspaper report talking about a village north of Irkutsk. It wasn't seen from the Gobi desert because the reports from there don't place the object in the direction of Tunguska. It wasn't seen from London because nobody in London reported an explosion. They all saw a long lasting glow.

No other explosion took place over the sky of Asia at that time, June 30, 7:20, visible from the entire area, please give up the nonsense.

How do you know?

So do planes. There's more at play in the Earth's atmosphere than just gravity. The caliber of this statement puts you right along pitroidtech, with his immortal line: clouds are made up of vapour...no wonder you can be fooled so easily by Nasa...

I'm not sure I follow. What does NASA have to do with aerosols and mixtures?

The highest temperature for Milwaukee was, as recorded in the data I provided, some 60F, you cannot use anything else.

The temperature is recorded at ground level, not at the level of the warm layer.  If you don't believe it's possible for a warm 31 degrees C layer of air to exist above a cold air layer of 13 degrees C, then it's not possible to argue with you anymore. 

So do planes. There's more at play in the Earth's atmosphere than just gravity. The caliber of this statement puts you right along pitroidtech, with his immortal line: clouds are made up of vapour...no wonder you can be fooled so easily by Nasa...

Your insults and slings don't become you Levee...

Precipitation forms when cloud droplets (or ice particles) in clouds grow and combine to become so large that their fall speed exceeds the updraft speed in the cloud, and they then fall out of the cloud. If these large water drops or ice particles do not re-evaporate as they fall farther below the cloud, they reach the ground as precipitation.

http://www.weatherquestions.com/What_causes_precipitation.htm 
Rain in the form of droplets are held up by wind resistance, for the same reason a leaf sometimes blows on the wind.  When the updraft ceases, the water droplets fall.

But the droplets don't start as droplets, they start as water vapour.  When the clouds become saturated with water vapour, ie when they reach the dew-point, droplets form and start to fall to the ground.  If their are no updrafts, the droplets continue to fall, gathering more water molecules as they fall forming bigger rain drops.  In a similar way, but in more extreme conditions, snow and hail forms from water vapour in the air.

Clouds also consist of microscopic ice crystals and microscopic water droplets, so small they are easily blown along on the air currents.

You badly need to study up on air pressure and aerodynamics if you want to discuss this stuff sensibly.

We know that clouds are made of water vapor, what we don’t know or at least forget is the important role that condensation plays in making clouds visible. For the most part water vapor is invisible. This is proven by the fact that the air we breathe regularly has some water vapor as part of its composition. However we don’t see it since its apart of the air. Condensation is what makes water vapor visible. Basically high temperatures excite water molecules until they change from a liquid state to a gaseous one. However lower temperatures can cause enough water vapor to condense back into liquid form. This small amount stays as very small droplets that can stay suspended in the air mostly thanks to small dust particles that they attach themselves to. It is pretty much the same way you see small bits of glitter suspended in clear glue. The drops are small enough to stay trapped in the air until condensation reaches a point of no return making rain. One result of this is that light becomes reflected and refracted. This is what makes clouds visible.

http://www.universetoday.com/73198/what-are-clouds-made-of/

First, clouds are made of tiny ice crsytals and water droplets. Water evaporates from lakes, rivers, and streams, and once it gets into the atmosphere, it condenses because of the cooler temperatures, and then clouds are formed.

http://www.severe-weather-fan.com/what-are-clouds-made-of.html
Water vapour is lighter than air, and is able to make it's way up to higher altitudes where it condenses but is held aloft by wind currents until the cloud reaches reaches dew-point.

Now let me ask you a question, since I have answered so many of yours.  Even if you don't like my answers, atleast you may do me the privilege of answering this question: why if there is no gravity, does the rain then fall?  And why does it fall once it reaches the dew-point and not a lot bigger or a lot smaller?