I think i found the final nail in the coffin in this idea for this thread. Running the wattage/distance squared equation for the Sun. It's pretty funny and I'll show the steps.
3.83x10^26th power divided by 300,000^2 = square root the whole lot= 1.957x10^13 divided by 300,000= 65233333.33
So 65,233,333.33 is the number seconds the light travels before it will not illuminate anything.
65233333.33 divided by 3600(60 secs x 60 minutes)=18,120.37 hours of travel time.
18,120.37/ 24= days time=755.02 light days
What the hilarious part is that if you take the light hours number and divide it by the year(365) without the day(24) conversion step: 18,120.37/365= 49.65! I find this hilarious in light of the 50.71 ly number that our big bang boys bandy about. It looks like some boys years ago might have forgot a step, and that's being ignorantly generous.
The number when applied in this full format equates to 2.05 light years for the Sun's Wattage/distance squared. As for Katirai's number of 6.09 ly, that may or may not be wrong but 2.05x3= 6.15 which is awfully close.
I still remember when they were saying it was 33 lightyears.
sjw40364
Re: Large Double Layer... I Mean 'Bone' Found In Space
I'm sorry, but I just don't see those 200 to 400 billion stars in that "dark" ribbon. Matter of fact I don't see any stars except a few in the background. See lots of out of focus foreground objects in front of the ribbon and lots of reflected light, but just can't make out those billions of stars. I do however believe they are looking at one of the plasma filaments making up the arm, just they are looking out at it, not inward.
You see, all those light sources must be between us and the galactic core since supposedly the dust is too thick to see even the galactic core made up of millions of stars, so those individual stars can not be behind the disk or in the disk, or they would certainly be invisible since I cant see millions of stars due to the dust. I like the way the dust is so thick I cant see millions together, but can see individual objects and clouds of plasma just fine. 2+2+?
Vasa
Re: Distances in Astronomy?
That's only because there are so many black holes in the galactic center, the light from thosr hundreds.of millioms of stars can't reach us through the infinitely curved space-time, and gravitational lensing converges the light of these millions of stars into a few points.
The maths don't lie my friend.
GaryN
Re: Distances in Astronomy?
@kalensar
I think i found the final nail in the coffin in this idea for this thread. Running the wattage/distance squared equation for the Sun. It's pretty funny and I'll show the steps.
There is no possible way that reflected light can account for the visibility of distant objects, and nobody has come forward to attempt to prove that it can. So what alternatives explanations are there? I've been reading this fellows PhD thesis, and it may explain light travelling in the vacuum.
Abstract
Beams of light tend to broaden as they propagate in linear medium. However, in a nonlinear medium (in which the index of refraction is proportional to the intensity) self-focusing can compensate for diffraction. In this case, the beam can create an index change similar to the one of an optical fiber and to guide itself in this self-induced fiber. These beams keep their shape and dimensions invariant along propagation. We called this type of beams spatial solitons. The beam that creates the fiber can be composed of more than one mode (e.g. TEM00 + TEM01+...), in this case we called this type of soliton: composite soliton.
Search Tal Carmon PhD thesis to find the pdf
And search this for another pdf about the EM Vacuum.
Nonlinear Structure of the Electromagnetic Vacuum
Lloyd
Re: Distances in Astronomy?
Maximum Reach of Solar Radiation? Well, Kalensar, your calculation above at http://thunderbolts.info/forum/phpBB3/viewtopic.php?f=10&am~ is very interesting, though I haven't yet tried to analyze it for potential errors. But the general idea seems potentially sound.
More from Katirai (http://www.scribd.com/doc/61291192/AstronomyDec28-2008) As usual, if anyone disagrees with any of the statements below, please say so and say why. My silence doesn't mean that I agree with it all, just that I haven't had time yet to study everything carefully. I tend to agree with the general direction of the arguments, but not necessarily with some or many details.
_{Sun Composition} _"The core of the sun must, therefore, be made of radioactive elements. _"This enormous energy, including the heat energy that is generated by massive amounts of radioactive elements that exist inside the sun, plus the hot, turbulent ocean-like liquid gases over the surface that create the waves, vibrations, friction and heat, can explain why the sun is a source of light, heat, and other radiation. _"It also explains why the sun has given energy at a steady rate for millions of years. _"Despite the fact that scientists analysing the atmosphere of the sun have discovered 65 elements and only years later, small traces of hydrogen and helium, theoreticians have suggested that hydrogen and helium atoms make up 99.9% of the volume of the sun. _"They believe that the sun gives light and heat because hydrogen is being converted into helium. _"Theoreticians claim that "The changing of hydrogen into helium in the sun results in the release of the sun's energy in the form of heat and light. _"" If this idea were true, then during the millions of years that the sun has existed, all of its mass and at least most of its atmosphere should have been converted into helium. _"The fact that hydrogen and helium constitute a small percentage of the atmosphere of the sun lead us to the understanding that the idea could be false. _"In contrast, the presence of a very large amount of radioactive atoms inside the sun is a sufficient and plausible explanation for the sun to produce its great energy. _"If one studies the characteristics of radioactive elements, one finds that these elements naturally radiate heat, as well as x-rays and gamma rays.
_{Solar Wind vs. Gravity} _"The force of the magnetic wind, then, must cancel most of the sun's gravitational force. _"Were it not for the solar wind, the earth and all the planets in the solar system would have collapsed into the sun in a relatively short period of time. _"The force of the solar wind, plus the centrifugal force resulting from the orbital motion of the planets, keep the planets from being drawn into the sun. _"Physicists did not bring into account the force of the solar wind when calculating the mass of the sun. _"It is a gross error, because the force is so great that it cancels at least 90 percent of the sun's gravitational force. _"In other words, the actual mass of the sun must be far greater than what astronomers have traditionally calculated.
_{Milky Way Composition} _"If it is understood that the sun is composed of very heavy elements that are radioactive, then a rough calculation will show that the average density of the sun must be at least 3.5 times higher than the earth. _"Since the sun's volume is 1,300,000 times the earth, its mass must then be at least 4,550,000 times the earth. _"To see how large the mass of the sun is, let us assume that the sun is the centre of the Milky Way. _"Then, let us estimate how many objects, such as luminous spheres, planets, and asteroids could be circling around it. _"Let us consider that the sun's mass is equal to the rest of the Milky Way. _"Calculation shows that in the Milky Way, there could be one million planets similar to Mercury, one million planets similar to Mars, one hundred thousand planets like Earth, hundreds of millions of asteroids, each hundreds or even thousands of kilometres wide, fifty planets --- as large as Jupiter, plus clouds, rocks, minerals, and gas filling all the spiral arms belonging to the Milky Way. _"The total mass of all the planets, --- asteroids and clouds combined would still be less than the mass of the sun.
GaryN
Re: Distances in Astronomy?
As usual, if anyone disagrees with any of the statements below, please say so and say why.
_{Sun Composition}...
I don't believe Katirai had ever considered an EU or ES model, but I get the feeling that if he had, he would have embraced it. As my model of the Sun is different from all the other alternatives though, I like to think he would have picked mine.
"They believe that the sun gives light and heat
Not proven. Still waiting for NASA to offer up the evidence of any transverse waves being emitted by the Sun.
kalensar
Re: Distances in Astronomy?
Thanks for the kind remark there Lloyd. Here's another in the same vein that I have not shown. SImply the same equation ran a slightly different way on the calculator. This one will probably make your heads spin guys.
I'm treating this one as simply a divisor. There will be no preliminary square rooting of the whole equation.
1.182098766×10¹²/24= 49,254,115,226.33744856 light days
49,254,115,226.33744856/ 365.25= 134,850,418.141923199 light years.
Pardon the hugely long decimals. This particular run is probably the biggest laugher of all the numbers. Our sun illuminates for 134 million light years, possibly?! The numbers of the Sun are fun. Still nowhere have I gotten anywhere close to 50 light years or even close to that number in specifically light years; no factors of 2.5 or less. either.
Lloyd
Re: Distances in Astronomy?
Kalensar, what does 3.83x10^26 refer to, which you divide by c^2 to get t (time in seconds)? Is it Energy?
Last Quote from Katirai This is from the last part of his book at http://www.scribd.com/doc/61291192/AstronomyDec28-2008. If there are disagreements, please say why. _{Milky Way Object Count} _"In fact, if we carefully study the Milky Way, we find that in its plane, there are no more than several thousand objects that could actually be called planets. _"As demonstrated in the preceding text, the idea that there exists hundreds of billions of stars is an illusion that is based on the idea that every point of light is a star. _"In reality, it is much more likely to be a planet, planetoid, asteroid, or clouds of dust and minerals. _"Are there that many planets or planetoids in the Milky Way? The answer is no. _"If one looks at the sky on a clear night, there appear to be millions of stars visible to the naked eye. _"In reality, the number of visible stars is only 2,800. _"the total number that could be seen all around the earth is about 6000. _"Astronomers claim that there are four hundred billion stars in the Milky Way. _"On the contrary, when we look at many photographs of the Milky Way, we see that all of them show a mass of clouds with only a small number of objects in between the clouds. _{Stellar Visibility Range} _"Astronomers believe that the sun from a distance greater than 50.71 light-years would not be visible to the naked eye. _"The distance of 1,000 light years is so great, that no star could be visible to the naked eye even it were thousands of times brighter and larger than the sun. _"If the suggestion that the 2,000 light year width of the Milky Way were true, then the distribution of stars visible to the naked eye would be the same in all directions. _"We would not be able to see the configuration of the plane of the Milky Way in any direction. _"In other words, we would not see the plane of the Milky Way. _"The fact that our naked eye can see the plane of the Milky Way as a band of misty light so clearly proves that the width of the Milky Way must be much less than 2,000 light years. _"This simple experiment demonstrates that the sun from the distance of 6.09 light-years would not be visible to the naked eye. _"the sun, at a distance of 32 light-years, appears as a star of fifth magnitude. _"In other words, the sun appears 2.512 times brighter than a star that is barely visible to the naked eye. _"This means that the sun from a distance of 50.71 light-years would be barely visible to the naked eye. _{Galactic Rotation Time} _"Pluto takes about 250 years to complete its orbit. _"the most distant objects circling the sun should not be too far away from Pluto. _"A rough calculation would show that these objects are moving slowly and would take many thousands of years to complete their orbit around the sun. _"This is exactly what the Dutch-American astronomer Adrian Van Maanen (1884-1946) discovered regarding the rotation of the Andromeda galaxy. _"He found that Andromeda completes its rotation in less than 1,000 years.
kalensar
Re: Distances in Astronomy?
3.83x10^26 is the wattage of the Sun, Lloyd. 3.83x10^26 Watts! Its a bright light bulb which can be treated exactly like a LED bulb because that is technically what it is under the Electric Star model. I am sorry if I did not make that clear. This was covered by Katirai as well and here is the experiment he mentioned:
"A simple experiment conducted by Laurence A. Marchall, a teacher at Gettysburg College, which directly gives a rough idea about the luminosity of the sun. For this experiment one simply needs a 100-watt light bulb, a wax photometer made of two pieces of wax separated by a piece of tinfoil and a measuring rod. In order to find the luminosity of the sun we can place a 100-watt light bulb and the sun on opposite sides of a wax photometer and vary the distance from the photometer to the 100-watt bulb until the brightness of bulb at the photometer is the same as the brightness of the sun at the photometer. It was found that the bulb at a distance of about 8cm equals the sun in brightness. Since we have the distance of the photometer to both the light bulb and the sun (1.5 x 10^13cm), using the inverse square law we can calculate the luminosity of the sun by the following formula. The formula simply states; the luminosity of the sun divided by the square of its distance to the photometer, is equal to, the luminosity of the bulb divided to the square of its distance to the photometer. L sun / d^2 sun = L bulb / d^2 bulb L sun / (1.5 x 10^13cm) 2 = 100 watt / (8 cm) 2 L sun = 3.5 x 10^26 watts This is an approximate value for the luminosity of the sun. However, a more accurate value obtained by highly accurate photometers is 3.83 x 10^26 watts."
And:
"For the next experiment, a very small (1.7 mm x 3 mm) light bulb was found with wattage (0.0375-watt) determined by the manufacturer(112). It was found that the light of the mini light bulb, outside the city and away from city light(footnote 113), in a dark and clear night, at the distance of 570 meters, appeared so faint that the naked eye could not see it. Knowing the actual luminosity of the light bulb and the distance that it became invisiblThe experiment was carried out in Botswana by Mr. Ghodrati and his team. e, by using the inverse square law of light, we can calculate at what distance a 100-watt light bulb would become equally faint. 0.0375 watts / (570 metes) 2 = 100 watts / d 2 d = 29,434 meters or 29.434 km Since the luminosity of the sun is 3.83 x 10^26 watts, using the inverse square law we can determine at what distance the light of the sun becomes equally faint. 0.0375 watt / (570 meters) = 3.83 x 10^26 watt / d2 meters d = 5.76 x 1016 meters or 5.76 x 10 13 km Since light travels 300,000 km per second then we have: d = 192,015,900 light-seconds = 6.09 light-years."
"Footnote 112: Mini Lamp: 1.5 volts, 25mA, its size; 1.7 mm x 3 mm. Manufactured in China and distributed by Orbyx Electronics, LLC. Footnote 113: The experiment was carried out in Botswana by Mr. Ghodrati and his team."
Lloyd
Re: Distances in Astronomy?
Kalensar, I commented on Katirai's experiment before. I said it seems to me that, since his experiment was performed in Earth's atmosphere, the air may have absorbed some of the light, whereas in space the light might go much further. Do you have any data on how much light is scattered or absorbed in air?
The Earth's atmosphere slightly tinges the color of the Sun toward yellow, but the size is not mitigated or changed. Earth's atmosphere does slightly bend the light, this is admitted and known, while in space Stars and other objects that are not close by, like Mars, will be white in color.
A great example is Saturn. We know the color of this planet is yellow but it shines blue in earth's atmosphere with the naked eye, while in space it appears as white, or through a telescope it's true colors will be shown.
Lloyd
Re: Distances in Astronomy?
Distance of Sun's Visibility I think the maximum distance the Sun is visible would be dependent on the intensity of its radiation. So can you tell us how much the intensity is reduced by Earth's atmosphere? If the atmosphere reduces it much, then Katirai's calculation would be based on a wrong assumption.
Betelgeuse Pavlink posted links to Betelgeuse images on this thread: http://thunderbolts.info/forum/phpBB3/viewtopic.php?f=3&~. Here's the first image. The orange spherical blob in the middle is apparently what we normally see as the star, but here it looks like an atmosphere or cloud around an inner white star. The semicircle to the left of it seems to be the bow of its plasmasphere. The article that was linked in the thread said the star is approaching a wall of dust, but the linear feature appears more likely to be a Birkeland current, since it seems to be helical.
Pavlink said Betelgeuse is actually a triple star system. Maybe the next image shows the approximate locations of all 3. I quoted Katirai recently regarding Betelgeuse. He showed that the conventional estimates of its mass would make the red/orange sphere so thin in density that it would surely be invisible. So I think it's likely that it's much closer than claimed. If it has a plasmasphere, it may be close to the size of Jupiter, as a brown dwarf star. It's said to be 640 ly away and its diameter is 43 to 56 milliarcseconds, which converts to 730 to 1200 solar radii, or 3.4 to 5.6 AU in radius, about the distance minimally to the outer part of the asteroid belt (beyond Mars) or maximally to just past Jupiter's orbit. Let's estimate 4.5 AU. One Google book says for Jupiter, "the plasmasphere has a radius only about eight times the radius of Jupiter, or 8 x 143,000 km, which comes to just over 1 million km. 4.5 AU is about 670 million km. That's a 670 times reduction of the diameter, which would entail a 670 times reduction of the distance. 640 ly divided by 670 is just under one lightyear away. If that's correct, it would be about 12 times farther than what Katirai estimated for the diameter of the Milky Way. In that case Betelgeuse is outside the Milky Way, or I need to reduce the estimate of the Betelgeuse plasmasphere diameter to less than 100,000 km, which may be too small to be possible.
kalensar
Re: Distances in Astronomy?
Those are not Katiriai's numbers for the Sun's wattage, Those are the generally agreed upon numbers the World over for the last few decades or more. What Katirai's personal experiment does is verify that the way the light works, regardless of atmospheric medium because of the short distances involved, is not anywhere close to what the mainline astronomers say it does. His own experiment shows that a light of known wattage works perfectly with the known maths of that substance on our personal scale. He doesn't state it as directly as I do that the Stellar Magnitude system is the primary problem of over-complication within astronomy. We don't use a magnitude system for candles over distances, but if we know the distance at which a candle becomes invisible then we can deduce the energy intensity of that candle.
Betelguese, man I love that object because it represents the prime problems in astronomy. It's plainly visible, variates it's luminosity over time, and is one of the most heavily studied objects out there simply because of it's mystique. It has one of the largest margins of error for the visible night sky objects, a margin of +or- 45 parsecs, and this simply because of it's luminosity variations.
That was excellent information on Jupiter's invisible plasmasphere, thanks for sharing that!
