Two Suns | |
Anonymous Coward User ID: 79399167 United States 09/20/2020 07:05 PM Report Abusive Post Report Copyright Violation | |
Anonymous Coward User ID: 69236336 United States 09/20/2020 07:20 PM Report Abusive Post Report Copyright Violation | About 24 years ago (I remember the time because my boys were 3 and 4 at the time) I had the most vivid apocalyptic, desolate dream---something I had never had. I dreamed I was with my boys and my parents at Disneyworld. Suddenly we walked out of one of the attractions and the park was almost empty, the atmosphere ashen, sky was grey, and I want to say smoky--very few people, but those few around were wandering, disoriented. Visibility was very muted. The false facades of the Disney streets were in collapse, surreal, everything seemed monochromatic. I started to panic and reached out to grab my children's hands and found my father a few steps from us, but could not find my mother. I looked down the Disney fantasy street toward the horizon and the sky was grayish-red, thick and THERE WERE TWO SUNS IN THE SKY, their diffuse light penetrating the smoky, filmy atmosphere. One was slightly higher than the other but they were both fairly low in the sky. One seemed slightly larger. I understand in the dream, reality as we know it, had been devastated, forever altered and it was paralyzing. I knew my mother was gone. All of this was acutely REAL. And then I woke up. I have always known this dream was significant--I HAD NO KNOWLEDGE ABOUT ANYTHING CONCERNING "TWO SUNS" at that time--a brand new concept to me. Never heard anything about such a thing. Now 24 years later, both sons are grown, married and successful, my parents are still living, but very elderly (90s)--my father amazingly well for his advanced age, my mother completely, profoundly "gone" from Alzheimers, also in failing physical health. I wonder about their 'status' being a marker for an event possibly foretold in a symbolic nightmare sent to me 24 years ago. In other words, when she 'leaves' the world, just like I could not find her in the dream, it that an inflection point? I know it sounds eerie, but I've wondered. "Two suns" references have popped up in the last 10-15 years I've noticed. |
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Strate8
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Anonymous plato User ID: 79262705 United States 09/20/2020 09:47 PM Report Abusive Post Report Copyright Violation | an extra-solar body sure would cause a ruckus here on Terra. Quoting: Anonymous plato 79262705 maybe this explains all the fuckery we see going on. there is a chance after all...... that bill gates DOESN"T actually want to genocide everyone with satanic nanobots. an extra Sun, on the other hand, well shyt. starting to add up. "There will be signs in the Sun, the Moon and Stars, and on earth nations will be in dismay, perplexed by the roaring of the sea and waves. People will die of fright in anticipation of what is coming upon the world, for the powers of Heaven will be shaken. Luke 21:25-26 These verses describe some sort of supernatural cosmic upheaval that will affect the motions of the sea, and which will cause those who believe that God is a Delusion, to lose their collective shit. fuck it,, im ready, You? LETS DO THIS |
Big Duke6
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Anonymous Coward User ID: 70985397 United States 09/21/2020 06:21 AM Report Abusive Post Report Copyright Violation | The Nemesis system is here. Has been here over a decade now. Our Sun's binary twin star system that intersects our Solar system on an eliptical orbit. 8 ENORMOUS planets, each with moon(s), the Nemesis star <only seen in IR> and occupied starship Nibiru. NOTHING outside of Earth's atmosphere abides by 3D physics. NOTHING. |
Fraust66
Forum Administrator User ID: 76954750 United Kingdom 09/21/2020 06:28 AM Report Abusive Post Report Copyright Violation | |
anonymous_one
User ID: 78969238 South Africa 09/21/2020 07:04 AM Report Abusive Post Report Copyright Violation | So I am stating that these videos of a second sun or a large planet "manifesting" in near earth orbit or worse earth atmosphere are obviously fake (or at least illusions). Why? First, a planet or sun (star) cannot "appear and disappear", do you ever see the sun/moon/mercury/venus etc phasing in and out of view? Of course not, if there was a planet there or a second sun, it would remain in view all the time, all over the world. Its a bloody great ball of gas or hunk of rock. Second, a planet or (visible) star this close in (between earth and sun) would be visible to people with small telescopes, and the net would be inundated with actual evidence, not dubious fuzzy pictures of a half formed something. *** We know this because we can see Mercury/Venus/Mars/Jupiter etc with our little telescopes at home, hence a planet size object that does not belong would quickly be noticed (and every amateur astronomer would love to see something undiscovered and get his name on it). Thirdly, a sun size object or large (gas giant) passing between earth and sun would have significant effects on inner planet orbits (not just a wobble) as well as earth's tides etc (and these have not been impacted at all, hence we know there is not even a moon size object near by) I am not saying there is nothing out there, official NASA research (although suppressed) has indicated there is a high probability of an unknown object in the solar system. And I agree, effects of something have been seen on outer planets, and changes in Earth itself which may (no actual evidence) be attributed to magnetic/gravitational pull of an incoming body (be it a planet or more likely brown dwarf star, a brown dwarf as this is dark, and not really visible except in infra-red (aka the important of the IR telescope in Antartica) and hence not seen by amateurs but has an extensive magnetic/gravitational field) It may or may not be part of the trigger of the micro-nova event (there is no actual evidence either way). I don't even disagree that if its there it is coming "soon", given world events, I believe the cataclysm or something is imminent. It is not however "close" in that sense, as in like the moon, it has to be "out in the solar system" somewhere, far enough that the impacts have been minor so far. So "coming soon" means it will in near future get close enough to have a tremendous impact on Earth, but given its a star, it will still be very far away in actual distance, or it would rip the planets out of orbit, and we know this does not happen (as Earth is still in Earth orbit after the last cataclysm cycle) *** as opposed to incoming asteroids/meteorites which are very small and dark and not even NASA sees many (or they claim not to which makes one wonder) |
Anonymous Coward User ID: 79339701 Australia 09/21/2020 08:28 AM Report Abusive Post Report Copyright Violation | Moon's doing 180 degree cartwheels. "Field rotation" is the lame explanation offered by nasa shills. Gee I wonder what covid is distracting us all from. Gee I wonder why we're all locked down and unable to move from the coast lines. Gee I wonder why there's so many obvious cartoon villains broadcasting a stupid dr evil plan. Gee I wonder if they really meant it when they said they wanted 90% to die and whether the ocean will be the thing to do it. |
Astromut
Senior Forum Moderator 09/21/2020 09:36 AM Report Abusive Post Report Copyright Violation | Moon's doing 180 degree cartwheels. Quoting: Anonymous Coward 79339701 "Field rotation" is the lame explanation offered by nasa shills. Field rotation is correct. Suck on it, Eric Briggs. Take the center point of the moon in equatorial coordinates and add .25 degrees to declination (approximate radius of the moon). This will give you the equatorial coordinates of both the center point of the moon and the north point of the moon on the equatorial grid. Then calculate the position angle of the moon's bright limb relative to the north equatorial point: x = atan((cos(sun declination)*sin(sun ra - moon ra))/(cos(moon dec)*sin(sun dec) - sin(moon dec)*cos(sun dec)*cos(sun ra - moon ra))) Then to calculate for field rotation, convert the equatorial coordinates of both the center point and the north point of the moon to horizon coordinates. This is given by the following formulae. For azimuth, the formula is: tan(az)=(-sin(Hour angle in degrees +delta)*cos(declination))/(sin(declination)*cos(latitude)-cos(declination)*sin(latitude)*cos(hour angle in degrees+delta)) where delta is: tan(delta)=(p*cos(theta')*sin(hour angle in degrees))/((distance of moon in km/6378.14)*cos(declination)-p*cos(theta')*cos(hour angle in degrees)) For altitude the formula is: sin(altitude)=(sin(declination)*sin(latitude))/(cos(declination)*cos(latitude)*cos(hour angle in degrees)) Now once you have converted both the center point and north point to altitude and azimuth, take the difference of each (altitude and azimuth) between the north and center points to find delta alt and delta az. From there, calculate the angle of the line from the center point to north point relative to the horizon: arctan(delta alt/delta az) Then add the position angle of the moon. Now by doing this over the course of the moon rising to setting you can see that its apparent orientation SHOULD flip up to about 180 degrees from rising to setting. I created a spreadsheet to predict the position and orientation of the moon based on equations from books that are decades old and it all still works and accurately predicts both the moon's position and orientation: [link to www.youtube.com (secure)] |
Astromut
Senior Forum Moderator 09/21/2020 09:41 AM Report Abusive Post Report Copyright Violation | Lots of videos of two suns lately. Many of which are being posted by normies, not suntards. Makes ya wonder... Quoting: Anonymous Coward 24125589 [link to www.instagram.com (secure)] [link to www.instagram.com (secure)] [link to www.instagram.com (secure)] [link to www.instagram.com (secure)] [link to www.instagram.com (secure)] Wow. That third video was compelling. Filter flare. |
Anonymous Coward User ID: 79349473 Slovakia 09/21/2020 09:42 AM Report Abusive Post Report Copyright Violation | Moon's doing 180 degree cartwheels. Quoting: Anonymous Coward 79339701 "Field rotation" is the lame explanation offered by nasa shills. Field rotation is correct. Suck on it, Eric Briggs. Take the center point of the moon in equatorial coordinates and add .25 degrees to declination (approximate radius of the moon). This will give you the equatorial coordinates of both the center point of the moon and the north point of the moon on the equatorial grid. Then calculate the position angle of the moon's bright limb relative to the north equatorial point: x = atan((cos(sun declination)*sin(sun ra - moon ra))/(cos(moon dec)*sin(sun dec) - sin(moon dec)*cos(sun dec)*cos(sun ra - moon ra))) Then to calculate for field rotation, convert the equatorial coordinates of both the center point and the north point of the moon to horizon coordinates. This is given by the following formulae. For azimuth, the formula is: tan(az)=(-sin(Hour angle in degrees +delta)*cos(declination))/(sin(declination)*cos(latitude)-cos(declination)*sin(latitude)*cos(hour angle in degrees+delta)) where delta is: tan(delta)=(p*cos(theta')*sin(hour angle in degrees))/((distance of moon in km/6378.14)*cos(declination)-p*cos(theta')*cos(hour angle in degrees)) For altitude the formula is: sin(altitude)=(sin(declination)*sin(latitude))/(cos(declination)*cos(latitude)*cos(hour angle in degrees)) Now once you have converted both the center point and north point to altitude and azimuth, take the difference of each (altitude and azimuth) between the north and center points to find delta alt and delta az. From there, calculate the angle of the line from the center point to north point relative to the horizon: arctan(delta alt/delta az) Then add the position angle of the moon. Now by doing this over the course of the moon rising to setting you can see that its apparent orientation SHOULD flip up to about 180 degrees from rising to setting. I created a spreadsheet to predict the position and orientation of the moon based on equations from books that are decades old and it all still works and accurately predicts both the moon's position and orientation: [link to www.youtube.com (secure)] Never in my long life have I even seen the moon flip 180 degrees..this is absolutely not normal.. And you can't hide these new planet and Sun's now..they are and will be visible to the whole world by October... A huge blue ultraviolet spectrum Sun and a smaller red sun... Ra and horus... Blue and red kachina... |
Astromut
Senior Forum Moderator 09/21/2020 09:45 AM Report Abusive Post Report Copyright Violation | Moon's doing 180 degree cartwheels. Quoting: Anonymous Coward 79339701 "Field rotation" is the lame explanation offered by nasa shills. Field rotation is correct. Suck on it, Eric Briggs. Take the center point of the moon in equatorial coordinates and add .25 degrees to declination (approximate radius of the moon). This will give you the equatorial coordinates of both the center point of the moon and the north point of the moon on the equatorial grid. Then calculate the position angle of the moon's bright limb relative to the north equatorial point: x = atan((cos(sun declination)*sin(sun ra - moon ra))/(cos(moon dec)*sin(sun dec) - sin(moon dec)*cos(sun dec)*cos(sun ra - moon ra))) Then to calculate for field rotation, convert the equatorial coordinates of both the center point and the north point of the moon to horizon coordinates. This is given by the following formulae. For azimuth, the formula is: tan(az)=(-sin(Hour angle in degrees +delta)*cos(declination))/(sin(declination)*cos(latitude)-cos(declination)*sin(latitude)*cos(hour angle in degrees+delta)) where delta is: tan(delta)=(p*cos(theta')*sin(hour angle in degrees))/((distance of moon in km/6378.14)*cos(declination)-p*cos(theta')*cos(hour angle in degrees)) For altitude the formula is: sin(altitude)=(sin(declination)*sin(latitude))/(cos(declination)*cos(latitude)*cos(hour angle in degrees)) Now once you have converted both the center point and north point to altitude and azimuth, take the difference of each (altitude and azimuth) between the north and center points to find delta alt and delta az. From there, calculate the angle of the line from the center point to north point relative to the horizon: arctan(delta alt/delta az) Then add the position angle of the moon. Now by doing this over the course of the moon rising to setting you can see that its apparent orientation SHOULD flip up to about 180 degrees from rising to setting. I created a spreadsheet to predict the position and orientation of the moon based on equations from books that are decades old and it all still works and accurately predicts both the moon's position and orientation: [link to www.youtube.com (secure)] Never in my long life have I even seen the moon flip 180 degrees..this is absolutely not normal.. It is normal and has been going on longer than you have been alive. You're just not that observant. You cannot refute my calculations. Do you hear me? You have completely failed to refute ANY of my calculations, you will never refute them, I am right, I am smarter than you, I know what I'm talking about, I have backed it up mathematically. Last Edited by Astromut on 09/21/2020 09:46 AM |
Anonymous Coward User ID: 79349473 Slovakia 09/21/2020 09:47 AM Report Abusive Post Report Copyright Violation | Moon's doing 180 degree cartwheels. Quoting: Anonymous Coward 79339701 "Field rotation" is the lame explanation offered by nasa shills. Field rotation is correct. Suck on it, Eric Briggs. Take the center point of the moon in equatorial coordinates and add .25 degrees to declination (approximate radius of the moon). This will give you the equatorial coordinates of both the center point of the moon and the north point of the moon on the equatorial grid. Then calculate the position angle of the moon's bright limb relative to the north equatorial point: x = atan((cos(sun declination)*sin(sun ra - moon ra))/(cos(moon dec)*sin(sun dec) - sin(moon dec)*cos(sun dec)*cos(sun ra - moon ra))) Then to calculate for field rotation, convert the equatorial coordinates of both the center point and the north point of the moon to horizon coordinates. This is given by the following formulae. For azimuth, the formula is: tan(az)=(-sin(Hour angle in degrees +delta)*cos(declination))/(sin(declination)*cos(latitude)-cos(declination)*sin(latitude)*cos(hour angle in degrees+delta)) where delta is: tan(delta)=(p*cos(theta')*sin(hour angle in degrees))/((distance of moon in km/6378.14)*cos(declination)-p*cos(theta')*cos(hour angle in degrees)) For altitude the formula is: sin(altitude)=(sin(declination)*sin(latitude))/(cos(declination)*cos(latitude)*cos(hour angle in degrees)) Now once you have converted both the center point and north point to altitude and azimuth, take the difference of each (altitude and azimuth) between the north and center points to find delta alt and delta az. From there, calculate the angle of the line from the center point to north point relative to the horizon: arctan(delta alt/delta az) Then add the position angle of the moon. Now by doing this over the course of the moon rising to setting you can see that its apparent orientation SHOULD flip up to about 180 degrees from rising to setting. I created a spreadsheet to predict the position and orientation of the moon based on equations from books that are decades old and it all still works and accurately predicts both the moon's position and orientation: [link to www.youtube.com (secure)] Never in my long life have I even seen the moon flip 180 degrees..this is absolutely not normal.. It is normal and has been going on longer than you have been alive. You're just not that observant. You cannot refute my calculations. Do you hear me? You have completely failed to refute ANY of my calculations, you will never refute them, I am right, I am smarter than you, I know what I'm talking about, I have backed it up mathematically. That was an extremely fast reply...hmm... |
Astromut
Senior Forum Moderator 09/21/2020 09:48 AM Report Abusive Post Report Copyright Violation | ... Quoting: Astromut Field rotation is correct. Suck on it, Eric Briggs. Take the center point of the moon in equatorial coordinates and add .25 degrees to declination (approximate radius of the moon). This will give you the equatorial coordinates of both the center point of the moon and the north point of the moon on the equatorial grid. Then calculate the position angle of the moon's bright limb relative to the north equatorial point: x = atan((cos(sun declination)*sin(sun ra - moon ra))/(cos(moon dec)*sin(sun dec) - sin(moon dec)*cos(sun dec)*cos(sun ra - moon ra))) Then to calculate for field rotation, convert the equatorial coordinates of both the center point and the north point of the moon to horizon coordinates. This is given by the following formulae. For azimuth, the formula is: tan(az)=(-sin(Hour angle in degrees +delta)*cos(declination))/(sin(declination)*cos(latitude)-cos(declination)*sin(latitude)*cos(hour angle in degrees+delta)) where delta is: tan(delta)=(p*cos(theta')*sin(hour angle in degrees))/((distance of moon in km/6378.14)*cos(declination)-p*cos(theta')*cos(hour angle in degrees)) For altitude the formula is: sin(altitude)=(sin(declination)*sin(latitude))/(cos(declination)*cos(latitude)*cos(hour angle in degrees)) Now once you have converted both the center point and north point to altitude and azimuth, take the difference of each (altitude and azimuth) between the north and center points to find delta alt and delta az. From there, calculate the angle of the line from the center point to north point relative to the horizon: arctan(delta alt/delta az) Then add the position angle of the moon. Now by doing this over the course of the moon rising to setting you can see that its apparent orientation SHOULD flip up to about 180 degrees from rising to setting. I created a spreadsheet to predict the position and orientation of the moon based on equations from books that are decades old and it all still works and accurately predicts both the moon's position and orientation: [link to www.youtube.com (secure)] Never in my long life have I even seen the moon flip 180 degrees..this is absolutely not normal.. It is normal and has been going on longer than you have been alive. You're just not that observant. You cannot refute my calculations. Do you hear me? You have completely failed to refute ANY of my calculations, you will never refute them, I am right, I am smarter than you, I know what I'm talking about, I have backed it up mathematically. That was an extremely fast reply...hmm... Yes, you think I just did the calculations and made all those videos? Look at the upload dates dumbass, I've been debunking this shit for YEARS! |
Anonymous Coward User ID: 79349473 Slovakia 09/21/2020 09:50 AM Report Abusive Post Report Copyright Violation | ... Quoting: Astromut Field rotation is correct. Suck on it, Eric Briggs. Take the center point of the moon in equatorial coordinates and add .25 degrees to declination (approximate radius of the moon). This will give you the equatorial coordinates of both the center point of the moon and the north point of the moon on the equatorial grid. Then calculate the position angle of the moon's bright limb relative to the north equatorial point: x = atan((cos(sun declination)*sin(sun ra - moon ra))/(cos(moon dec)*sin(sun dec) - sin(moon dec)*cos(sun dec)*cos(sun ra - moon ra))) Then to calculate for field rotation, convert the equatorial coordinates of both the center point and the north point of the moon to horizon coordinates. This is given by the following formulae. For azimuth, the formula is: tan(az)=(-sin(Hour angle in degrees +delta)*cos(declination))/(sin(declination)*cos(latitude)-cos(declination)*sin(latitude)*cos(hour angle in degrees+delta)) where delta is: tan(delta)=(p*cos(theta')*sin(hour angle in degrees))/((distance of moon in km/6378.14)*cos(declination)-p*cos(theta')*cos(hour angle in degrees)) For altitude the formula is: sin(altitude)=(sin(declination)*sin(latitude))/(cos(declination)*cos(latitude)*cos(hour angle in degrees)) Now once you have converted both the center point and north point to altitude and azimuth, take the difference of each (altitude and azimuth) between the north and center points to find delta alt and delta az. From there, calculate the angle of the line from the center point to north point relative to the horizon: arctan(delta alt/delta az) Then add the position angle of the moon. Now by doing this over the course of the moon rising to setting you can see that its apparent orientation SHOULD flip up to about 180 degrees from rising to setting. I created a spreadsheet to predict the position and orientation of the moon based on equations from books that are decades old and it all still works and accurately predicts both the moon's position and orientation: [link to www.youtube.com (secure)] Never in my long life have I even seen the moon flip 180 degrees..this is absolutely not normal.. It is normal and has been going on longer than you have been alive. You're just not that observant. You cannot refute my calculations. Do you hear me? You have completely failed to refute ANY of my calculations, you will never refute them, I am right, I am smarter than you, I know what I'm talking about, I have backed it up mathematically. That was an extremely fast reply...hmm... Which Eric Briggs?... The real one called Angry Catfish Briggs on YouTube...or the paid government/NASA shill called Eric Briggs.. Hmmm... |
Astromut
Senior Forum Moderator 09/21/2020 09:53 AM Report Abusive Post Report Copyright Violation | ... Quoting: Anonymous Coward 79349473 Never in my long life have I even seen the moon flip 180 degrees..this is absolutely not normal.. It is normal and has been going on longer than you have been alive. You're just not that observant. You cannot refute my calculations. Do you hear me? You have completely failed to refute ANY of my calculations, you will never refute them, I am right, I am smarter than you, I know what I'm talking about, I have backed it up mathematically. That was an extremely fast reply...hmm... Which Eric Briggs?... The real one called Angry Catfish Briggs on YouTube...or the paid government/NASA shill called Eric Briggs.. Hmmm... More than one person has the name Eric Briggs you fucking moron. I meant the Eric Briggs who doesn't understand field rotation, not the Eric Briggs who compliments my channel and understands field rotation. |
Anonymous Coward User ID: 79401030 United Kingdom 09/21/2020 10:02 AM Report Abusive Post Report Copyright Violation | |
Astromut
Senior Forum Moderator 09/21/2020 10:12 AM Report Abusive Post Report Copyright Violation | |
Anonymous Coward User ID: 79386029 Canada 09/21/2020 10:15 AM Report Abusive Post Report Copyright Violation | binaries and even trinaries etc., are in fact the standard in our galaxy. our binary is a failed, collapsed, sun that is called a dark star.. these videos may be fake or just show reflection. this second sun would show up like a moon in the day light since it does not emit any light of its own. |
Anonymous Coward User ID: 79401030 United Kingdom 09/21/2020 10:15 AM Report Abusive Post Report Copyright Violation | |
Anonymous Coward User ID: 79349473 Slovakia 09/21/2020 10:17 AM Report Abusive Post Report Copyright Violation | |