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# Can you still see Polaris even if you are in the south pole?

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I haven't been to south pole but can the Polaris still be viewed if the viewer is in the south pole? Or this question makes no sense at all?

Currently Polaris is at a declination of a bit over 89 degrees, which means that no one south of 1 degree south latitude can see Polaris. That's almost all of the Southern hemisphere, let alone the South Pole.

Polaris won't be the North Star forever, thanks to axial precession. In about 13000 years or so, Polaris will have a declination of about 46 degrees or so (twice the 23 degree axial tilt). Polaris will thus be visible in 13000 years or so as a wintertime star to all of Africa, all of Australia, and most of South America, but none of Antarctica.

After millions of years, proper motion may make Polaris visible over Antarctica. But then again, being a yellow supergiant, its unlikely that Polaris will be visible anywhere (without a telescope). It will instead be dead.

While the majority of the celestial sky is visible on both hemispheres, you are not able to see Polaris on the south pole, since Polaris is pointing directly towards the north pole. I know that during winter time, you can definitely just see the plough/big dipper (part of the Ursa Major constellation) as far south as Uluru/Ayers Rock in Australia, but that is not enough to see the northern star. The northern star will generally speaking disappear below the horizon when you are at around the equator. There is no "southern star" to orient yourself on the southern hemisphere, but there are a number of rules of thumb to find the south celestial pole on the southern hemisphere as well, several of them involving the "southern cross", or the Crux constellation.

At the South pole no stars north of the celestial equator will be visible.

The general equations of the declination limit at a given latitude:

declination limit = $latitude - 90$ (for the northern hemisphere, $latitude > 0$)

declination limit = $90 + latitude$ (for the southern hemisphere, $latitude < 0$)

## The Atlantean Conspiracy

NASA and modern astronomy say the Earth is a giant globe spinning 1,000 mph around its central axis, traveling 67,000 mph circles around the Sun, spiraling 500,000 mph around the Milky Way, while the entire galaxy rockets a ridiculous 670,000,000 mph through the Universe, with all of these motions originating from an alleged “Big Bang” cosmogenic explosion 14 billion years ago. That’s a grand total of 670,568,000 mph in several different directions we’re all supposedly speeding along at simultaneously. No one has ever seen, felt, heard, measured or proven such motion, yet the vast majority of people unquestioningly accept that the clearly motionless Earth beneath their feet is actually moving over six hundred million miles per hour!

This is probably my favorite proof of the Flat Earth. How can anyone possibly defend the Heliocentric model when Polaris never moves? It is mind-boggling what people will accept without giving one ounce of effort of looking into it for themselves… or rather, people believe that being told what to think IS doing research.

I was having a discussion with a close friend last night and brought up this point. Up until then he had a healthy skepticism of the evidence. When I showed him this he paused. "I thought you couldn't see Polaris past the equator.." he said, stroking his beard. I showed him that ancient navigators used Polaris even into many degrees South of the equator.
"Hmm, I may have been wrong this whole time. "

It was awesome to see a man I consider closer than a brother slowly awakening to the truth.

I didnt beleve none of this flat earth stuff at first untill iwatched one of your videos I only have one question tho if a train cant go uphill or on a curve then how do the trains in the Himalayan mountains go up and down

out of everything ive researched on the flat eart that is the only thing I still cant get my head around if you know I would appreciate it if you could reply to this thank you

What are your thoughts on the procession of the equinoxes? The mainstream model has earth tilting away from Polaris as it goes through its cycle. Does the flat earth model take procession into account?

Man i am truly greatful for the information you provide we truly are in an age of great awakening would love to go about looking for spiritual awakening as well i think just now my mind is becoming clear and trying to open minds to a different perspective is difficult but as in part to being lied to its very hard to accept when stating things that are very obvious yet go against everything we have been told does put up barriers keep fighting the good fight and i am a strong believer in what your opions will keep trying on My side of the map

When I encountered the Hollow Earth theory in the late 1990's I was already a "conspiracy theorist" aware of the machinations of the wicked Cult of the Illuminati and their 'Long Range Plan.' I had studied Atlantology, Egyptology and Ufology in depth and the Hollow Earth was intriguing, it seemed to provide some answers to nagging questions, like where the 'gods' originated from a Holy Mountain at the center of the Earth (Olympus, Meru, Zion etc.)

Therefore, I deduced that if there was any truth in our globe being hollow, with a world within, our ancient forebears with such advanced knowledge of science and technology (pyramids) would have known this. I dug out the myths and legends. First, I examined Egypt's beliefs and instantly realised it was all focused on one realm, the Underworld. Then I looked at the Maya, also with an Underworld, which was represented by the Ballcourt.

And so it went on, Celts, Maoris, Greeks, Indians. everyone knew about the Underworld. How curious, since supposedly none of these cultures ever met. The Underworld is considered by all academics to be an imaginary place, but that is not how it is described. It is accessible, through caves, lakes, waterfalls and mysterious locations, such as the World's End.

A solid sphere neither has a center or an end, only a hollow one could accommodate an Underworld, below ground. It was said to be reached in the far north a Paradise where there is neither night or day and the 'gods' have immortality. That inevitably lead Hollow Earthers to the North Pole, where we assumed there was an opening, protected by the military and disguised by NASA. There are many anomalies recorded in this region.

The further north that explorers travelled past the ice, they noted warm seas, an abundance of insects, pollen and even foxes. It should not be so. There were also reports of witnessing a small hazy sun, which never rose or set. Hollow Earthers presumed these navigators had entered the polar depression and glimpsed the 'inner sun.'

However, if the Flat Earth is true, it could be our sun which is always visible, above the horizon at the extreme north. Many of the findings of Hollow Earth studies can be translated to the Flat Earth model. It has a center, which is due north, it also has an end. if you encounter the Antarctic ice barrier encompassing the seas and it must have an underside, the Underworld.

I can assure you Hollow Earthers are not controlled opposition, we are genuine and passionate in our beliefs, but our entire hypothesis is based on the Heliocentric globe model which, like everyone else, we had no reason to doubt, conditioned from birth. If it is a lie, it is monstrous and unforgivable and the people of Earth will denounce all authority, forever.

If we are living on a flat plane, surrounded by ocean with a canopy of stars, nothing more than lights reflecting water and a localised Sun and Moon, we can finally regain our sanity, our reality and most importantly, our spirituality and trust in God. The 'Son's of Baal' will resist this realisation with all their power, but that is rapidly diminishing and will evaporate when the truth comes out. Be prepared to have "aliens" and "asteroids" hurled at you!

It cannot stand, if there is no "outer-space" there is no threat, period! Hollywood's mission will be over and NASA's the governments, media, scientists, church and scholars will have much explaining to do, but who would believe anything they ever said again? Astronomers, physicists, mathematicians and sages knew the nature of our world for thousands of years and that we are at the center of this Universe, just like the Bible said.

The Flat Earth appears to be the catalyst for the 'awakening' we have been anticipating. In 1895, William Carpenter wrote:

“And what then? What then? No intelligent man will ask the question and he who may be called an intellectual man will know that the demonstration of the fact that the Earth is not a globe is the grandest snapping of the chains of slavery that ever took place in the world of literature or science. The floodgates of human knowledge are opened afresh and an impetus is given to investigation and discovery where all was stagnation, bewilderment and dreams!"

These wise words are even truer now, than they were then.

Has anyone found a proof of the stars being visible from impossible latitudes. I can't seem to. If you do please send a link. If you can't I consider this a loss for the fet. And glad I didn't tell everyone about it.

the old mariners could judge the position of the pole star from the pointer stars. I have personally done this at 17 degrees south and it is easy. the pole star remains below the horizon, but its position can be judges by the rotation of the pointers easy

## 27 Replies to &ldquoPolestar proves flat earth&rdquo

We know that in the flat earth model, the south pole star gets ignored. The flat earthers acknowledge that the north star exists but they refuse to admit that there is a south pole with a south star and their model leave that completely out. Flat earthers cant explain away the south star.

This may be the first time I’ve heard anyone claim the existence of the “South Star.” People have claimed the existence of the “Southern Cross,. Please provide evidence of this supposed Southern Star, photographic images might help if you can prove their of the southern sky and identify the source.

Thanks for all your effort in this regard.

There is no star in the exact position of Southern Celestial Pole. But there is a Southern Celestial Pole. A constant point in the sky around which all the luminaries in the Southern sky rotate.

They also don’t explain why Polaris can’t be seen from the southern hemisphere.

“They also don’t explain why Polaris can’t be seen from the southern hemisphere.“

It can be seen South of Equator.

Yes, On the Level, from the top of a mountain a few hundred miles South of the Equator…

If you were honest you would admit that Polaris is directly above, in the center of the sky, at the North Pole, and it descends in the sky one degree down for every degree of latitude as you travel towards the equator. So you can tell your latitude by measuring the number of degrees above the horizon Polaris is. It is 90 degrees above the horizon at the North Pole, so latitude is 90 degrees and it is on the horizon at the equator, so latitude is zero.

This is EXACTLY what we would expect to experience if we were traveling around the surface of a globe. But this behavior of Polaris is very very difficult to explain if we were on a flat plane.

Take your dinner plate. Stick a straw in the middle and stick Polaris on the top of that. Move anywhere on your dinner plate and look up. Polaris will always remain up there. It will never come down to the horizon. But try on a balloon, then you will see you are getting the same effect as we actually experience.

This is why in some places I say flat earthers do not look at the sky. Because if you understand what is happening in the sky, then it is very difficult to understand how we could be seeing this manifestation from a flat plane.

Thank you Flat Earth Facts for clarifying your claim that Flat Earthers don’t look at the sky. You state:

“ If you were honest you would admit that Polaris is directly above, in the center of the sky, at the North Pole, and it descends in the sky one degree down for every degree of latitude as you travel towards the equator. So you can tell your latitude by measuring the number of degrees above the horizon Polaris is. It is 90 degrees above the horizon at the North Pole, so latitude is 90 degrees and it is on the horizon at the equator, so latitude is zero.”

If in fact, Earth were a ball as described Polaris should in fact be at the horizon near the equator as you claim. Yet you list the following as a Flat Earth supporting proof:

“ 99) Viewed from a ball-Earth, Polaris, situated directly over the North Pole, should not be visible anywhere in the Southern hemisphere. For Polaris to be seen from the Southern hemisphere of a globular Earth, the observer would have to be somehow looking “through the globe,” and miles of land and sea would have to be transparent. Polaris can be seen, however, up to over 20 degrees South latitude.”

You also list the following:

“103) There are several constellations which can be seen from far greater distances over the face of the Earth than should be possible if the world were a rotating, revolving, wobbling ball. For instance, Ursa Major, very close to Polaris, can be seen from 90 degrees North latitude (the North Pole) all the way down to 30 degrees South latitude. For this to be possible on a ball-Earth the Southern observers would have to be seeing through hundreds or thousands of miles of bulging Earth to the Northern sky.“

For Polaris to be seen 20 to 30 degrees South of the equator it could not have sunk to horizon level at the equator. 20 to 30 degrees South of the equator is far more than a few hundred miles, more like 2500-3750 miles. These proofs have been attested to by mariners, navigators and many others. As Mr. William Carpenter states :

“The astronomers’ theory of a globular Earth necessitates the conclusion that, if we travel south of the equator, to see the North Star is an impossibility. Yet it is well known this star has been seen by navigators when they have been more than 20 degrees south of the equator. This fact, like hundreds of other facts, puts the theory to shame, and gives us a proof that the Earth is not a globe.” -William Carpenter, “100 Proofs the Earth is Not a Globe” (71)“

On the website below bloggers posted videos showing Polaris apparently to be significantly above the equator from Malaysia and Indonesia. The Malay location was apparently North of equator and the Indonesian location approximately at the equator.

It’s not that Flat Earthers don’t look at the sky as your post erroneneously asserts. They obviously do and observe completely different phenomenon than claimed by globists. Your own list of “Flat Earth” proofs show that to be the case.

If you’re honest you’ll admit the claims Made by both camps in this regard conflict. No model can resolve the conflict only facts and repeated observation can.

Yes. You know I am independent. I am not on your side and I am not on the globe earth side. So you are correct. There is a very strong bias towards the globe earth model and there is a very strong tendency to ignore any evidence that may contradict the globe earth model.

So if you look at the sky how do you explain this, that Polaris descends one degree in the sky as you move one degree in latitude towards the equator?

That is exactly the behavior we would expect on a globe, but seems impossible on a flat plane?

As far as observations from points very close to the equator, that is not very relevant. From points very close to the equator, if you are above sea level, you will be able to see Polaris for some time. So if you get on top of a big mountain quite a distance north of the equator, you may be able to see Polaris. But this is the exception, not the rule.

If you are at sea level at the equator you will see Polaris on the horizon. So this behavior is totally consistent with moving about on the surface of a globe and totally inconsistent with moving about on a flat plane. It is for this reason I suggest that flat earthers do not look at the sky.

Flat Earth Facts please note what Eric Dubay observed regarding the globe model’s failure to account for the many Polaris and other star observations from 20-30 degrees and more South of the equator. It was in the late 19th century that Heliocentrists contrived the idea that the Earth tilted 23.5 degrees to attempt to account for these observations. In additional, remember Foucalt even used electromagnets to fake pendulum oscillations supposedly mirroring Earth’s supposed rotation.

“To account for this glaring problem in their model, desperate heliocentrists since the late 19th century have claimed the ball-Earth actually tilts a convenient 23.5 degrees back on its vertical axis. Even this brilliant revision to their theory cannot account for the visibility of many other constellations though. For instance, Ursa Major, very close to Polaris, can be seen from 90 degrees North latitude (the North Pole) all the way down to 30 degrees South latitude. The constellation Vulpecula can be seen from 90 degrees North latitude, all the way to 55 degrees South latitude. Taurus, Pisces and Leo can be seen from 90 degrees North all the way to 65 degrees South. Aquarius and Libra can be seen from 65 degrees North to 90 degrees South! The constellation Virgo is visible from 80 degrees North down to 80 degrees South, and Orion can be seen from 85 degrees North all the way to 75 degrees South latitude! An observer on a ball-Earth, regardless of any tilt or inclination, should not logically be able to see this far.“

The globe/heliocentric theory despite your claims never held up to factual analysis. That’s why so many mutations were Made to make it appear that it someone could. Note also globists have a tough time locating precisely where the Celestial poles are there said to be moving all the time. The globular theory appears unable to cope. As a globist noted:

The great tragedy of science, the slaying of a beautiful theory by an ugly
fact.
— Thomas Henry Huxley

Yes, On the Level, it was a fantastic idea, brilliant, to tilt the spinning globe on that 23.5 degree tilt. Truly a stroke of genius, it works so well. But I am not sure that this will help in seeing Polaris from the south. Because no matter how they tilt it they still have to have Polaris directly above the North Pole. So you still should not be able to see Polaris south of the equator. Unless I am missing something here?

The reason they tilted the spinning globe 23.5 degrees is to explain the different lengths of the days during the year and to explain the 24 hour sun and 24 hour darkness days on North and South poles. It is really brilliant and does really work, in the sense that it predicts very accurately what we experience in the various places on the ‘globe’.

But titling the spinning ball like this is not going to help you see Polaris in the South. I think better not to rely on observations from the 19th century. You need to have a look today, now, and see the situation. I think you will find that Polaris is down at the horizon on the equator at sea level. If not please let me know. You can see it, from the top of a mountain, south of the equator, but not at sea level. So check it out. If you find Polaris 30 degrees above the horizon at sea level on the equator then the globe earthers have a problem, but I don’t think you will find this.

Yes. Of course, I agree, globe earth model has been patched together and adjusted to suit the observations, however the resulting model they have is really a very good predictive scientific model. They have a logical idea they can explain which could work. And the point I am trying to make is the flat earth people do not have a model they agree on, they do not even have a map of the flat earth that could work, and they can’t explain what is happening in the sky. Like this behavior of Polaris, dropping down one degree in the sky as you move towards the equator by one degree of latitude. That is what would happen if you were moving on a ball, its hard to imagine how this could happen if you were moving on a flat plane…

Yes. Foucault’s Pendulum, I think it is real. But just because the pendulum rotates does not mean it is the earth that is rotating. There is a rotation in the system, we all agree on that. Either it is the earth that is rotating or it is the system that is rotating around the earth. So in either case, the rotation of the earth or the rotation of the system around the earth, that could cause the pendulum to rotate.

“But titling the spinning ball like this is not going to help you see Polaris in the South. I think better not to rely on observations from the 19th century. You need to have a look today, now, and see the situation. I think you will find that Polaris is down at the horizon on the equator at sea level. If not please let me know. You can see it, from the top of a mountain, south of the equator, but not at sea level. So check it out. If you find Polaris 30 degrees above the horizon at sea level on the equator then the globe earthers have a problem, but I don’t think you will find this.”

Next time I’m South of the equator I’ll have a look myself. It’s been over a decade since the last opportunity.

Actually it can, to the southern tropic line because of the axis of the earth.

we are living in a time like no other time like today but then again yesterday was like no other day like yesterday Nicky Minaj.

Basically, if you are stretching out the south pole that means you are seeing the high number of north degree before the equator to the south, after the equator, but you are not seeing the southern stars, but north stars that stretch out and counter-rotate that is beyond the horizon but do not go down but counter-rotate, so it is not south stars, but still north star which we can see Southern Cross in the north when should be southern cross or Midpoint Cross. So, but too south you can’t see it because it will go to horizon, but less than the horizon, it is counter-rotate to the north direction, not southern direction. The further to the south the less you will see which you should see that is equal to the north, but not, right.

That is on way to proof flat earth so it is not the same equal of degree for north and south, but higher in north for north pole and little lower in south for north pole, but not opposite for south pole which can’t be seen in north and south, so it is a very low range of south degree to see south stars. The range of south stars is lower than range for the north star.

What a load of derp. Polaris cannot be seen at “over 20 degrees south latitude” The rest of this stuff is just fantasy wish fulfillment . For instance: Vulpecula is a seasonal constellation. Visible in northern hemisphere in the summer, and winter in the southern hemisphere. Funny how you forgot to mention that – and the reason why it is so. Hint: earth is a spheroid and is tilted on its axis. The rest of your astronomy is equally derp ridden. Flat earth fail.

The reason why the pole star is not visible from points in the Southern Hemisphere is clearly because the light from the pole star passes through too much of the atmosphere and is all absorbed. The atmosphere IS NOT perfectly clear. There is a slight opacity to it which increases with distance. Hence you cannot see the PS from, for example Australia.

……well. no. It’s because you don’t have vision that goes on forever. You can only see so far. Image a giant room. A hallway going o. Forever, it’s going to look like a pin point at some point. Same thing, just on a larger scale. The distance convergaces. We have circle vision. You can’t see on forever. Solo the whole flat earth can see on forever thing is stupid

For all the amount of time you have taken to think about this point, it is a pity that you didn’t look for information sources about the main concern: how does Polaris remain aligned with the Earth’s North Pole?

The first assumption is simply wrong: Polaris is currently but has not always been nor it will be forever the Pole Star. Just like Einstein introduced his cosmological constant to allow his theory to match what he thought he was observing (that the Universe didn’t change) here it is assumed that Polaris has always been in the North direction. Well that is not true.

In reality, the rotation axis of the Earth precesses in a cycle that lasts 26000 years, this is the tilt that was asked for above.

As is taught in every school in the World, the Earth is affected by two movements, one of which is the orbit around the Sun. It needs to be stated here, that the whole Solar System does. Every planet except Pluto (is it a planet? yes? no?) follows the same (very similar) path as seen from the Earth. This is because the whole Solar System is moving within the same plane. Motion around the Sun is done in the Ecliptic plane. There is the Sun, then there is this tablecloth around it. The planets are moving in circles within this cloth to a very good approximation.

Picture a marble moving through the tablecloth. At the same time the marble is spinning with its axis pointing towards the room’s ceiling (Polaris).

Some information about the Pole Star including equinox’s precession: https://en.wikipedia.org/wiki/Pole_star
Information about the Ecliptic Plane: https://en.wikipedia.org/wiki/Ecliptic

We know that some stars that can be seen from the Northern Hemisphere cannot be seen from the Southern Hemisphere. There are some stars that can be seen in the Southern Hemisphere that cannot be seen from the Northern Hemisphere. When Magallanes started his voyage around the globe 500 years ago (which was completed by Elcano) he discovered some stars no one (in Europe) had ever seen before. Nebulae that today we name after him (like both Magallanic clouds). Sailors used the stars to guide them and they knew them very well…. in the Northern Hemisphere. The Southern Hemisphere was completely unknown to them and they had a wee bit of issues navigating using those stars. The national flags of several countries (New Zealand, Fiji, Australia…) portray the Southern Cross, which is a constellation that you cannot see from (most places in) the Northern Hemisphere. Likewise, Polaris cannot be seen from the Southern Hemisphere.

What do modern European Astronomy institutions do to be able to watch the stars (and galaxies and blazars, magnetars, quasars, pulsars, clouds, nebulae…) that can only be seen from the Southern Hemisphere? Well, they build the European Southern Observatory in some country in the Southern Hemisphere. The European Southern Observatory bridges a gap for European astronomers. Located in Atacama’s desert in Chile, possibly one of best places in the whole Earth to observe the night sky.

Not only observation is excellent from Atacama, but also, because of its location south of the Equator, it is possible from E.S.O. to have a look at a region of the sky that we normally don’t have access to, from the Northern Hemisphere. The real scientific value of the E.S.O. is that it is one of the very few (and one of the very best) observatories placed in the Southern Hemisphere.

This is because depending on what Hemisphere on the Earth you are placed, you look towards one part of the sky or the other. Which only happens in a globe.

Yes. Good points. It is hard, if not impossible, to explain what we see in the sky presuming the earth was flat. If the earth was flat and there was a polestar above the North Pole that would be visible all over the flat earth. Think of the earth as a dinner plate and the polestar a light suspended above the centre of the plate. You could see it from everywhere.

But on a globe, you would be looking in a different direction into the sky from the Southern Hemisphere than you would be looking from the Northern Hemisphere, thus seeing a different part of the sky, different stars, etc, which is exactly what we experience.

So what we observe in the sky, that is confirming the globe earth model, and contradicting the predictions of the flat earth model.

If the earth were a globe Polaris should not be visible from most if not all the southern hemisphere. In fact, people claim to see it from Australia.

It can also be seen from south america.

Yes. I am in Australia and if the earth was flat I would most certainly be able to see Polaris. Just take a dinner plate and imagine it is the flat earth and put a stick in the middle and put Polaris on the top of that stick and imagine where Australia would be on the flat earth map and imagine you were a little ant and stand there and look up. Can you see Polaris. Of course you can. But I am here in Australia and I can’t see Polaris. It is below the horizon here. Way below the horizon. Of course Australia is big. So if you went very far North, like Darwin, we have even got more North possible than that, you are getting up close near the Equator. So maybe in some cases way up there close to the equator you may see it. But if you could it would be way down near the horizon, not up in the center of the sky as we would expect it to be on a flat earth.

Same story of course. South America. You can’t see it from Argentina for example. But Equator goes through South America around Bogata so you can be in South America but North of the equator. You can see Polaris there, but again near the horizon, not up in the middle of the sky as we would expect on a flat earth.

It would be possible to get this effect on a flat plane but you would have to bring down the stars very very very low…

If you have a light in the middle of your room and walk from one side of the room to the other in the middle of the room it is above your head and on either side it goes down, if your room was big enough it would go down to the horizon.

But for this to work with polaris you are talking of having it thousands of miles up, not light years up….

Anyhow my point is globe earthers have got their story straight. They have a model, which every globe earther accepts, and which is a logical reasonable story that explains what we see happening in the sky. It may or may not be a true story. But it is a logically consistent and believable story. But flat earthers do not have any agreed upon model and can not explain how any of the details work.

It may be possible to construct a valid predictive scientific model based on the premise that the earth is flat. I don’t know. But flat earthers so far have not done this and scientifically inclined people can not take it seriously unless you can present a valid predictive model that explains our observations.

Personally I don’t know. I am not a flat earther and not a globe earther. I see faults in both ideas and suspect the reality is something different.

Polaris cannot be seen at 20* South latitude, or 10*, or 5*. As you travel north or south in the northern hemisphere, the angel of Polaris corresponds closely to the latitude from which you observe it. That could only happen if the Earth is a sphere. As you travel south (in the northern hemisphere) Polaris’ relative position in the sky sinks in the sky correspondent with latitude until it disappears below the horizon at about the equator. As you head south in the northern hemisphere, Polaris doesn’t fade away or get dimmer (which would be the case if the light were “absorbed”) but sinks below the horizon. This would not happen with a “flat earth”.

You can prove this fact for yourself by measuring the angle of Polaris from the horizon with a protractor as you travel north or south in the northern hemisphere. You don’t need to rely on websites or photos from space. Prove it for yourself based on your own observations. It is observable, verifiable, irrefutable proof the earth is a sphere.

And then maybe you’ll start to wonder if believing junk theories and assertions from untrained people is really the best way to live life.

But! If earth was flat then polaris would be visible everywhere instead of up to 20 degrees south

## How to see the Southern Cross from the Northern Hemisphere

View at EarthSky Community Photos. | Kannan A in Woodlands, Singapore, captured this photo of the Southern Cross on March 8, 2021. He wrote: “The Southern Cross constellation seen here in the morning in Singapore looking south. On the left of this cross are the 2 pointer stars, Alpha Centauri (Rigel Kentaurus) and Beta Centauri (Hadar). They point to the Southern Cross.” Thanks, Kannan!

The Southern Cross – also known as Crux – is an iconic constellation for people south of the equator. It’s visible every clear night, and its stars shine brightly enough to be picked out pretty easily even from urban locations.

If you’re in the Northern Hemisphere, you too can see the famous Southern Cross, if you’re far enough south, and know where and when to look.

View at EarthSky Community Photos. | Prateek Pandey in Bhopal, India, caught the Southern Cross while at its highest point around midnight (its midnight culmination) on March 6, 2021. In April and May, the Southern Cross reaches its highest point in the sky earlier in the evening. Thank you, Prateek!

Finding the Southern Cross from the Northern Hemisphere.
At 35 degrees south latitude and all latitudes farther south, you can see the Southern Cross at any hour of the night all year around. In that part of the Southern Hemisphere, the Southern Cross is circumpolar, which means it circles the sky close to the celestial pole and is always above the horizon.

However, for much of the Northern Hemisphere – including most of the United States – the Southern Cross never rises above the horizon, so it can never been seen from our middle and far northern skies.

You can see see all of Crux from the U.S. state of Hawaii. In the contiguous U.S., you need to be in southern Florida or Texas (about 26 degrees north latitude or farther south). Even from the far-southern contiguous U.S., you have a limited viewing window for catching the Southern Cross. It has to be the right season of the year. It has to be the right time of night. And you have to look in the right direction: south!

For the Northern Hemisphere’s tropical and subtropical regions, the month of May is a good time for finding Crux in the evening sky. You can see the Southern Cross at other times of the year, but not at such a convenient time. In middle March, for instance, you have to wait till about 1 a.m. to catch the Southern Cross at its highest point in the sky. In December and January you have to catch Crux before dawn.

No matter the hour or date, the Southern Cross climbs to its highest point in the sky when it’s due south. The Cross is fairly easy to visualize, because it stands upright over the horizon.

Bright stars Alpha and Beta Centauri pointing to Crux, or the Southern Cross, from Stephen Green in Waikoloa, Hawaii, on April 26, 2019. Stephen is at about 20 degrees north latitude. Thank you, Stephen!

How to use the Big Dipper as a guide. Although the Big Dipper is a fixture of the Northern Hemisphere skies, this star formation has a close kinship with the Southern Cross. Both the Big Dipper and the Southern Cross reach their highest point in the sky in unison. Remember spring up and fall down. That’s Northern Hemisphere spring we’re talking about.

The Big Dipper soars highest in the sky on late northern spring evenings. When the Big Dipper is seen above Polaris, the North Star, the Southern Cross is seen standing over the southern horizon in southern Florida and Texas.

For the Southern Hemisphere, by the way, it works the same way – but in reverse. The Big Dipper can actually be seen in the Southern Hemisphere at opportune times from about 26 degrees south latitude and all latitudes farther north. But to spot it, the Big Dipper has to be viewed at the right season of the year and the right hour of the night. When the Southern Cross sails highest up in the Southern Hemisphere sky, the “upside-down” Big Dipper is seen just above the northern horizon at latitudes near the tropic of Capricorn (23.5 degrees south latitude).

Dr Ski in Valencia, Philippines – 7 degrees north latitude – captured this image on April 30, 2019. He wrote: “The Southern Cross transits the meridian (reaches its highest point). The top of the Cross (Gamma Crux) is 24 degrees above my horizon when it culminates. Alpha and Beta Centauri are the bright stars at the lower left, pointing to the Cross.” Thank you, Dr Ski!

Southern Cross in navigation. When European sailors journeyed south of the equator, they found that the North Star had disappeared below the horizon. As they sailed even farther south, the Big Dipper dropped out of sight as well. Unlike the Northern Hemisphere, the Southern Hemisphere has no bright pole star to highlight the celestial pole. Fortunately, the Southern Cross acts as a navigational aid.

There are various ways to find the direction due south using the Southern Cross as a guide. For example, a line drawn from the star Gacrux through the star Acrux points in the general direction of the south celestial pole (the point in the sky directly above the Earth’s south pole).

View larger. | Here is the Southern Cross as seen from Manila – at 14 degrees north latitude – on April 29, 2012. Image via our friend JV Noriega.

Bottom line: The Southern Cross can be seen by northerners, as long as you’re below 26 degrees north and know where and when to look!

## Polaris is bugging me.

Ok, now that I am spending more time learning the night sky and watching the comings and goings of the various heavenly objects, one thing is bugging the heck out of me. I get why Polaris would appear not to move if you resided at or near the northern axis of our turning globe. But we live at 28 degrees here in Tampa. That is much closer to the equator than the north pole. I just can't wrap my head around still being able to see Polaris in the same spot every night as we rotate. It's driving me crazy! (Not that I can actually see it in my suburb, but you know what I mean)

I have even tried to take a ball and put a dot on it to represent where we are and rotate the ball around the poles (using toothpicks to represent) and we move out of view every time with my simulation. I can comprehend that we might still be able to see the North Star, but, in the very same spot all night long? No way!

Are there any videos that simulate this so that I can see how this works? Tell me I am not the only one who is struggling with this.

### #2 RyanSem

I'm a visual learner, so videos like this help me understand it on an internal level better than just reading about it. Here's a timestamped link: https://youtu.be/HsJxGpDmJrQ?t=51

This was shot at 32 degrees lat, which is actually really close to what you're seeing!

Edited by RyanSem, 11 January 2020 - 08:32 PM.

### #3 DLuders

You could download the free planetarium program http://stellarium.org , set your location to Tampa, FL, and zoom in on Polaris. Then, set your location to Anchorage, Alaska and see that there's no real difference.

Ryan's example is excellent.

Edited by scadvice, 11 January 2020 - 08:49 PM.

### #5 bumm

The farther away something is, the less it appears to move when viewed from different positions. Polaris does have a measurable parallax, but it is so far away at 434 light years that for our purposes it might as well be engraved on steel.

### #6 outofsight

It probably has something to do with the Earth's diameter being only about 8,000 miles, so you're only about 4,000 miles away from the polar axis, and that Polaris is pretty close to the axis, but Polaris is about 2,528,000,000,000,000 miles away. And I could be off by a few zeros, either way.

So it probably doesn't appear to move much because it is relatively close to the axis and we are relatively close to the axis, but it is very^n far away. Something like that probably has something to do with its lack of apparent motion.

But you should wait for more (better?) explanations and you'll probably get a better idea. Remember, a lot of this stuff in astronomy is only Relative and Apparent, or lack of Apparent, at first glance it may not be what it actually is, or isn't.

Edited by outofsight, 11 January 2020 - 08:52 PM.

### #8 Brett Waller

The reason you don't see any apparent movement is that the diameter of the Earth is insignificant when compared to the distance of Polaris, so there is effectively no change in the direction of the line-of-sight as the Earth rotates. Even the diameter of the Earth's orbit is effectively insignificant with respect to the distance of even the nearest stars, and the parallax of the nearest stars rarely exceeds one arc second which is totally undetectable for most practical purposes. That is why you don't see any movement as the Earth rotates, no matter where you are on the Earth's surface.

Now, if you are having trouble wrapping your head around the enormity of the distance between the stars, then join the crowd. It is so vast relative to any distances you can visualize as to be difficult to conceptualize.

### #9 TOMDEY

You're in very good company! The ancients puzzled over the same thing, and came to the correct conclusion: The reason for the almost perfect stasis of the NCP relative to the stars is that the stars are so far away, relative to the diameter of the earth. It does indeed execute little bitsy loops, but many many orders of magnitude too small to detect. The annual circle is, of course bigger, but only reveals the few closest stars as detectably parallaxed, over the course of the year. There are other factors with small effects, the finite speed of light causing the images to lean into our direction of travel and the proper motions of heavenly bodies, etc, etc. They are all very small. Tom

### #10 jmillsbss

It probably has something to do with the Earth's diameter being only about 8,000 miles, so you're only about 4,000 miles away from the polar axis, and that Polaris is pretty close to the axis, but Polaris is about 2,528,000,000,000,000 miles away. And I could be off by a few zeros, either way.

So it probably doesn't appear to move much because it is relatively close to the axis and we are relatively close to the axis, but it is very^n far away. Something like that probably has something to do with its lack of apparent motion.

But you should wait for more (better?) explanations and you'll probably get a better idea. Remember, a lot of this stuff in astronomy is only Relative and Apparent, or lack of Apparent, at first glance it may not be what it actually is, or isn't.

### #12 Dalew

Ah, I figured I would get a lot of help on this. Makes sense that the vast vast distance of Polaris could certainly render a few thousand miles insignificant. I'm not sure that gets me totally there but it helps quite a bit. Still on the hunt for a visual that helps me see how it is that my town doesn't rotate around to the opposite side of the earth out of sight of Polaris. Clearly it doesn't, since Polaris is up there every night. But I would still like to see it from a different perspective than my driveway. Yeah. I am kind of like that. I'm sure I drove my parents around the bend.

### #13 Dalew

Oh, and yes, I do have Stellarium and use it regularly. In fact I just pretended I was in Melbourne, Australia the other night. Whole different sky!!

### #14 havasman

"The illusion of perception can be a seductively potent distractor from fact." -Glenn LeDrew, amateur astronomer and forum stalwart

### #15 BFaucett

Why Doesn't Polaris Move?

Polaris is very distant from Earth, and located in a position very near Earth's north celestial pole.

Earth rotates once a day on its axis, an imaginary line that passes through Earth from its north pole to its south pole. If that imaginary line — the axis — is projected into space above the north pole, it points to Earth's north celestial pole. As the Earth spins on its axis, it also “spins around” the north celestial pole. Polaris, located almost exactly at the north celestial pole, the center of spin, stays in the same place, while stars farther away from the north celestial pole can be seen to move in a wider circle around Polaris as viewed from Earth during its daily rotation.

Polaris actually lies just a short distance away from where Earth's axis points. Polaris is located about 1 degree off to the side of the north celestial pole, so Polaris does move a little, tracing a very small arc in the night sky, around which the other visible stars make wider circles.

This picture of the night sky above Hawaii was taken by leaving the camera shutter open for a long time. The picture captures the apparent movement of the stars caused by Earth's rotation on its axis. Polaris is the star in the center of the star field it shows essentially no movement. Earth's axis points almost directly to Polaris, so this star is observed to show the least movement. The other stars appear to trace arcs of movement because of Earth's spin on its axis.

### #16 JG-42N

Try your experiment with the ball with much longer toothpicks. The further above the ball (planet) that you would place Polaris (offset from the center of the rotation a bit since it is not exactly at the pole), the better it represents how the angle (from your horizon) changes from your spot on the ball as it would rotate. You could simulate it with a small circle drawn on a piece of paper and a long line for the axis.

### #17 Sky Muse

Polaris does move, but in a very tight circle so tight that it's perceptible only by observing it over several nights, whilst noting its position each night. Polaris A, the main star, has a small companion-star, Polaris Ab, orbiting it.

Ab's position can be noted, too, night after night, and will disappear behind A even, for a time.

But Polaris never strays far, at all, from the North Celestial Pole(NCP). It's only the star's vertical position in the sky, from horizon to zenith, that changes, and from this latitude and that.

This is the view of Polaris from the north-door of my home.

That's the only way it moves across the sky, up and down, vertically, and as you travel from latitude to latitude.

Edited by Sky Muse, 11 January 2020 - 11:31 PM.

### #18 clearwaterdave

You are doing your experiment the wrong way.,You should be inside the ball looking out.,with the "DOT" being Polaris.,( marked on the top of the ball ).,As the world turns the sky appears to turn overhead.,With the DOT being you.,you are putting yourself up in the sky and YOU are going around the earth.,

Picture the sky being the ball.,revolving around you.,not you revolving around the sky.,good luck.,

### #19 kathyastro

Still on the hunt for a visual that helps me see how it is that my town doesn't rotate around to the opposite side of the earth out of sight of Polaris.

Here you go. That's you at 28N on the left side of the Earth at sunset, and again on the right side of the Earth at sunrise. (Diagram is based on the equinox for simplicity.) In both cases, you are looking North towards Polaris (blue lines). The blue lines are parallel because Polaris is so far away.

### #20 Dalew

Here you go. That's you at 28N on the left side of the Earth at sunset, and again on the right side of the Earth at sunrise. (Diagram is based on the equinox for simplicity.) In both cases, you are looking North towards Polaris (blue lines). The blue lines are parallel because Polaris is so far away.

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That is most helpful, Kathy! I think I get it now. You made my day!

### #21 tommm

Still on the hunt for a visual that helps me see how it is that my town doesn't rotate around to the opposite side of the earth out of sight of Polaris.

Draw a circle about the size of a quarter at the bottom of a piece of paper. Draw a line along a radius of the circle at your latitude, i.e. 28 deg above a horizontal line through the center of the quarter. Now draw a line tangent to the circle at the end of the radius you drew, i.e. perpendicular to the radius, and just touching the circle, and extend it to intersect a vertical line drawn through the center of the circle up to the top of the page. That line defines the limit of how low you can see in the sky looking north. You can see anything above that line.

You can see that the intersection of the two lines, the vertical one and the one tangent to the circle surface at 28 deg, is not all that many circle diameters above the circle. Polaris is many, many circle diameters above that point, and just a short (scaled) distance from the vertical line, so you can always see it no matter whether your spot on the earth is on the same side of the vertical line as Polaris or not.

Extend the tangent line on past the intersection with the vertical line, and then imagine it fixed to the circle. Imagine rotating the circle with the attached line. The part of the line above its intersection with the vertical line sweeps out a cone in space as you rotate the circle. Polaris is well within that cone, so it is always visible to you.

Edit: Just saw kathyastro's diagram. You can see the vertical lines she drew are well inside the cone swept out by the tangent line I described above. Of course they are not quite vertical, but they differ from vertical by only a tiny amount.

## Debunking the Flat Earth

Flat-earthers will often protest that it's impossible for Polaris to appear always above the North Pole considering that the earth is traveling around the sun along an orbital path 940 million kilometers in circumference. This argument is actually irrelevant with respect to the shape of the earth and is merely a (bad) argument for a stationary earth. As you will see illustrated below, even if the earth is stationary, its shape can only be spherical.

Nonetheless, flat earthers simply lack understanding of the geometry of the heliocentric model. The distance the earth travels during its annual orbit is minuscule compared to the distance of Polaris.

Consider this:
- The diameter of earth's orbit is about 300 million km (186 million miles).
- The distance to Polaris is around 3.6 quadrillion km (2.4 quadrillion miles).
- That's a distance ratio of 1 : 12,000,000

To put that into perspective, imagine you are staring straight ahead at a distant mountain located 100 kilometers away. Now take a step eight millimeters to the left. Obviously you are still staring straight at the mountain. If you are the earth and the mountain is Polaris, that 8 mm distance is the equivalent of the change in earth's relative position after six months of orbit.

The exact distance of Polaris is not what's important for this discussion. What matters is that it's far enough away that its rays are parallel, that all its light comes in at the same angle. The result of this position of Polaris in relation to the earth -- its location and its parallel rays -- is that the apparent altitude of Polaris above the horizon is determined solely by the curvature of the earth.

### How Looking Up ↑ Reveals Earth's Curvature

An observer at the North Pole will see Polaris 90° directly overhead. The location of the North Pole is latitude N 90°. At the equator an observer will see Polaris 0° on the horizon. The equator is located at latitude 0°. To a sailor or traveler in the northern hemisphere moving directly south, for every one degree of latitude traveled, Polaris will appear one degree lower in the sky. The altitude of Polaris is therefore always equal to an observer's latitude.

 (click to see full size) The angles at which we observe Polaris from any location in the northern hemisphere match perfectly with the geometry of a spherical earth.

### Polaris Can't Help Mariners Find Their Latitude on a Flat Earth

 Illustration showing altitude angles of Polaris on a flat earth. Distances are to scale. (click to view full size).
Where is Polaris on a flat earth? This illustration shows why the altitude of Polaris cannot match an observer's latitude on a flat earth.

### The Sun Also Reveals the Shape of The Earth

 (click to see full size) Because the earth is a sphere, on the equinox, at solar noon, an observer's latitude is always equal to the sun's altitude subtracted from 90°. latitude = (90° - altitude)

Cool post man, looking forward to more.

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https://youtu.be/VKoeDOSQCVY
25 Flat Earth claims that 'Rocked' my world view 5- 5

Earth is covered with a glass dome(firmament) so when you look up your seeing reflections, Polaris is not a planet. It's the reflection of the black hole sun. That's why the path is a spiral just like the flat earth sun.

Unknown poster on December 29, 2016. You are controlled opposition. That's a strawman argument and you know it. No one says that polaris is the reflection of the sun.

Please let me know if my reasoning is wrong. In northern hemisphere stars rotate CCW CW in southern. Facing north in New York stars rise from my right side: CCW. Facing south, stars rise from my left side: CW. If facing north in both hems, whether on a ball or a plane, stars rise from the same direction. The reason it's said that stars rotate differently in both hemispheres is simply the direction the observer is facing.

That is exactly right, John.

So if the earth is traveling 1000 mph in a cycle and orbiting the sun at 66.600 mph and the sun is moving 450.000 thru space then y are all the constellations the same . almost 7000 years of recorded time and they have not moved 1 bit .

Not just that y do we see mercury and Venus there closer to the sun then us but we see them at night . We should only see then at day time o we cant see star at day time hmmm ? Wake up people don't believe every thing u are told . If there where a flat earth and we could touch the dome then we would ask someone put this here there must be a GOD . but we live where the satin is the master of this world and a master description . he don't want us to believe in GOD.

They have the greeks described ursa major as different to how we see it no. But you wouldnt know that because you dont read.

We aren't living on a planet dude, it's a flat plane. You can draw all the little science diagrams and refer to refraction as much as you like but if you actually use scientific method and do these experiments you'll find the truth. The more people that learn the truth for themselves the better this world would be. Neil De Grasse Tyson is only where he is because he swallows the crap theories written before him and applies his bull crap to them. It's just an echo chamber of lies that the majority of people believe because they aren't smart enough to think for themselves. Take this blog for example it's called 'debunking flat earth', you have set out with an agenda to debunk something rather than go and look for the truth whatever that may be due to the actual recordable findings!

+cheex
You're just one more commenter who can't address or refute the information presented in this article. It's based on observable, empirical, facts. I understand that you don't like it, but you need to demonstrate why it's wrong.

John T. Banewicz. Thank you, this Saul dude is running all the math, but leaving out the one key component which is the direction of the observer.
shields himself by spouting a whole lot of angles and distances, positioning, but intentionally leaves out directional observation very carefully as to not prove Flat Earths existence.

I'd love to hear this guys explanation for the speed of light in relation to our speed hurling through space.

(or how he explains how nothing ever hits Earth even know we've been flying vast lengths at furious speeds for billions of years)

Excellent post! Bookmarked for future reference.

the earth doesn't only travel 300 million kilometers a year. the sun travels as well draging the earth along with it.

what about this argument? Viewed from a ball-Earth, Polaris, situated directly over the North Pole, should not be visible anywhere in the Southern hemisphere. For Polaris to be seen from the Southern hemisphere of a globular Earth, the observer would have to be somehow looking “through the globe,” and miles of land and sea would have to be transparent. Polaris can be seen, however, up to over 20 degrees South latitude.

Where did you get that information? I'm sure you didn't get it by your own observation because it's simply false. Polaris can often be seen slightly south of the equator due to atmospheric refraction, but, as Neil DeGrasse Tyson recently tweeted, "Polaris is gone by 1.5 deg S. Latitude."

so. how would one actually see Polaris if the earth was somehow a fast spinning 1600km/h spheroid speeding around the sun at 66,600km/h, and the earth is now half way around the sun (6 months time), and it is now night time??

You can't see Polaris because of atmosphere, light pollution and its weak light.
This what he said.
Then, he must pick up the refraction to justify it.

You can't see Polaris because of atmosphere, light pollution and its weak light.
This what he said.
Then, he must pick up the refraction to justify it.

You can go outside whatever city you're in, out into a rural area, then you can look up and see the night sky with little or no light pollution. Including Polaris.

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The visible stars are mostly in the same arm of the galaxy as We are, and moving in the same direction as We are. The difference in speed and scale are the factors We need to look at. The difference in speed is much smaller than Our specific speed, and the vast scale makes any changes take lifetimes - many MANY - to observe.

Exactly.. So HOW exactly has the calculations been made if it takes many MANY lifetimes to observe just a teeny tiny fraction of a broad scale. Even IF there are historical records that are reliable and provable that prove a tiny move, it doesn't even come CLOSE to calculating a ridiculous speed that we have never tested or observed.. PLUS..It STILL doesn't prove a ball earth IF the stars were adjusting in the sky either..

Even though Polaris is the North Star today, this has not always been the case. The place in the sky that the Earth's north pole points at changes slowly over time. This movement is called stellar precession. In 3000 BC, a faint star called Thuban in the constellation of Draco was the North Star. Polaris did not become the North Star until about AD 500. It will get closer to straight above the Earth's north pole until sometime in 2102. Then it will move away again. It will be the closest star to the pole until about AD 3000.
Source: https://simple.wikipedia.org/wiki/Polaris_(star)

As God said to Job. "Where wast thou when I laid the foundations of the earth? declare, if thou hast understanding."
Neither was Wikipedia there 3,000 years ago. They are 'suggesting what they think was the North Star', but they do not know. Plus they prove themselves time and again to be untrustworthy and agenda driven. The only way to clear up this question is to compare the allegations with reality. An example of that is the Earth's atmosphere. The behavior of water and air teach us that were this world actually spinning, both the water and the atmosphere would have been tossed off like from clothes in a spinning washing machine - eons ago. Water doesn't bend and the song is right, you can't pin a cloud to the sand.

how about someone just go to Cape Horn and look? . freaking Tyson tweets. seriously, don't be a religious zealot. I've yet to read where you account for the exponential/parabolic diminishing of visibility due to perspectival vanishing. It's something I as an architect deal with on regular occasion, since buildings are never experienced in 'elevation' (like your diagrams show), but rather perspective. Lastly, your 'just look at the geometry' argument is a bit empty considering an inversion of the relative azimuthal/latitudinal viewing angles with a lower, closer sun on a relatively flat surface. open up your brain a little to consider these and calculate them to show falsity or hypothetical necessity to test it's ramifications for coherency with measurable experience.

BOOOOMMM y'all got lit up cuz!!

Explain Polaris NOW. HA AH AH A A OHHHHHHHHHHHHHHHHHHHHHHHHHHHHH

"Saul TraneMarch 3, 2016 at 10:23 PM
Where did you get that information?"
I believe I can answer that. ahem. EVERYWHERE?!
NASA, SCHOOL, PUBLIC LIBRARY, ANY DOCUMENTARY ABOUT SPACE EVER MADE?! SHALL I CONTINUE?! L O L

Correct! Space does not exist! The GLOBE does not exist! You need a proof? Proof is YOU, if you know, how to use your brain. Watch this 11 minute movie until the end. The story lacks narrating- good change for you to think along. You will see how easily we have been fooled. https://youtu.be/F9VesA_InEk

the sun travels 500,000 mph around the Milky Way, while the entire galaxy rockets a ridiculous 670,000,000 mph. so the earth travels across the milky way 4 380 000 000 miles per year. over 2000 years that is 4,380,000,000,000. shouldn't this effect us viewing the stars? just a honest question i want to figure this out.

i am also excited to how you answer the star argument thess are a couple of them. i haven't seen it adressed anywhere.

9) Viewed from a ball-Earth, Polaris, situated directly over the North Pole, should not be visible anywhere in the Southern hemisphere. For Polaris to be seen from the Southern hemisphere of a globular Earth, the observer would have to be somehow looking “through the globe,” and miles of land and sea would have to be transparent. Polaris can be seen, however, up to over 20 degrees South latitude.

100) If Earth were a ball, the Southern Cross and other Southern constellations would all be visible at the same time from every longitude on the same latitude as is the case in the North with Polaris and its surrounding constellations. Ursa Major/Minor and many others can be seen from every Northern meridian simultaneously whereas in the South, constellations like the Southern Cross cannot. This proves the Southern hemisphere is not “turned under” as in the ball-Earth model, but simply stretching further outwards away from the Northern center-point as in the flat Earth model.

103) There are several constellations which can be seen from far greater distances over the face of the Earth than should be possible if the world were a rotating, revolving, wobbling ball. For instance, Ursa Major, very close to Polaris, can be seen from 90 degrees North latitude (the North Pole) all the way down to 30 degrees South latitude. For this to be possible on a ball-Earth the Southern observers would have to be seeing through hundreds or thousands of miles of bulging Earth to the Northern sky.

104) The constellation Vulpecula can be seen from 90 degrees North latitude, all the way to 55 degrees South latitude. Taurus, Pisces and Leo can be seen from 90 degrees North all the way to 65 degrees South. An observer on a ball-Earth, regardless of any tilt or inclination, should not logically be able to see this far.

105) Aquarius and Libra can be seen from 65 degrees North to 90 degrees South! The constellation Virgo is visible from 80 degrees North down to 80 degrees South, and Orion can be seen from 85 degrees North all the way to 75 degrees South latitude! These are all only possible because the “hemispheres” are not spheres at all but concentric circles of latitude extending outwards from the central North Pole with the stars rotating over and around.

None of those spurious arguments, whether you buy them or not, changes the fact that the positions of the celestial objects as we observe them in the sky are impossible on a flat earth.

I totally agree with saladin89, if we're travelling at these speeds, we wouldn't even see th light travveling as we wouldn't be In its light path.. And if we are on a ball, and we see the northern star becuz it's the light still travveling, then we would see It in multiple positions at the same time in the sky, we don't, we see one.. why don't we shoot past new galaxy's, or collide into another one, why does water always find its level. Why do no sattelites have a live stream, why do none of them ever get wiped out by the thousands of metorites every day.. cm'on man, so many questions, we haven't even been through the van Halen belt, a nasa astronaut even said that

No they aren't impossible "Saul" As you've been told time and time again. Being ignorant isn't going to cut it anymore. Those celestial patterns so to speak are a result of magnetism.

Thermal imaging proves the world is flat, it can't be hidden, it's over.

Apparently, you didn't understand the above article. I'm not talking about "celestial patterns so to speak", I'm talking about the angles at which we observe the celestial bodies. They are geometrically impossible on a flat earth. If you can prove otherwise, please do.

Thermal imaging does not prove the world flat. Please provide a thermal image of the sun from a location after the sun has set.

What frame of reference are you measuring these speeds in, specifically the speed of the galaxy?
Other galaxies are moving away from us, resulting in them appearing to be red-shifted.
But the nearest galaxy is 2.5 million light years away, so that speed isn't going to do much.

The sun is orbiting the galactic centre, But the orbit is also quite large and the other stars are orbitting it as well, so it changes angle at a very slow rate.

So yes, all of this will effect the stars, and mean their position is not constant, but at a very slow rate.
The north star has not always been Polaris and Polaris isn't directly above the north pole, it is slightly out.
In the year 3000 Gamma Cephei will become the north star due to these movements.

9 - It isn't. Polaris is not visible significantly south of the equator. The best reference for that was a navy captain referencing a pole star without specifying Polaris. He could have been referring to Sigma Octantis, the south pole star. No one has ever been able to show Polaris from a significantly south position.

100 - No. This depends on how close to the pole they are. If they are above the north pole, like Polaris or Ursa Minor, then they can be seen simultaneously.
If they are not, then it will be dependent on Earth facing that direction.
The same applies to the south.
The Southern Cross is not a south pole star, it merely points the way to the south. The southern cross is only 60 degrees south. That means when it is directly overhead some location, it will be visible on that meridian up to roughly 30 degrees north, and 180 degrees around Earth, it will only be visible below 30 degrees south.
This is quite simple to do. Treat the location as number, from +90 (north) to -90 (south). For the southern cross that means -60. Then add and subtract 90 to that to get the range (and truncate at + or - 90, or you can figure out what it would be on the other side). That means it ranges from -60+90=+30, to -60-90=-150 (so all the way to -90, and on the other side it would extend to -90 (SP) -(-150--90)=-30.

You can also do a range. The southern cross isn't just at -60. It ranges from -55 to -65. For the range, take the north extend and subtract 90 and the southern extent and add 90. This means it will range from (-65+90)=+25, to -55-90=-145, so due south. The only time that gets more complicated is when it goes over the pole, which I wont go into.

It won't appear to everyone in the southern hemisphere at the same time, just like the sun doesn't, even when it is over the tropics.
The other issue is the day light making it hard to see the stars.

103 - This is akin to the southern cross.
Ursa Major is a large constellation which extends from 30 to roughly 65 degrees north.
It is not as close to the North pole/Polaris as this argument pretends.
Its position means the constellation in its entirety would be visible at times up to 25 degrees south, and partially up to roughly 60 degrees south.

The rest is the same.
Vulpecula is located between 20 and 30 degrees north. As such you should be able to see it down to 50-60 degrees south.
Taurus, Pisces and Leo are located between 0 and 30 degrees north, so they would be visible down to between 60 and 90 degrees south.
Aquarius and Libra are the same but in the south.
Virgo is in the middle, ranging from 10 degrees south to 10 degrees north, meaning it should be visible in its entirety between 80 degrees south and 80 degrees north.
Orion ranges from 20 degrees north to 10 degrees south and thus would be visible from 70 degrees south to 80 degrees north.

All of these are consistent with a globe Earth. In every case it is visible for a hemisphere.
None of those are impossible on a round Earth.
But every one is inconsistent with a flat Earth.
If Earth was flat everyone should be able to see every star. Polaris should be visible from Antarctica, but instead it disappears at the equator.

So those points don't show Earth is flat. They show it can't be flat and that observations are consistent with a round Earth.

## Does Mars have a North Star?

As the sun sets over the stark Martian landscape, stars come into view. Image by Mars rover Spirit, 2005, via NASA.

The North Star for Earth is Polaris. Does our next-door neighbor planet, Mars, have the same North Star as Earth? If not, does Mars have a star located more or less above its North Pole?

Let’s talk about what we mean by North Star. Every planet in our solar system spins on its axis. Earth’s spin is what defines the length of our day of approximately 24 hours. If you continue the imaginary line of a planet’s axis out into space – in a northern direction as measured from earthly north – it might point to a star that’s visible to the eye. Or it might not. We call such stars pole stars, or North Stars. On Earth, that northern pole star – less than a degree from the north celestial pole – is the beloved star Polaris. Scouts and hikers know you can use Polaris to find the direction north, when compasses fail.

Meanwhile, Earth’s Southern Hemisphere doesn’t have a comparable South Star. The nearest visible star to the south celestial pole of Earth is about 9 degrees away.

So, does Mars have a North or South Star? The answer is … not in any satisfying way. There’s no bright North Star, and only a modestly-bright South Star, for Mars.

Earth doesn’t orbit exactly upright with respect to our orbit around the sun. Instead, Earth’s tilt is about 23 1/2 degrees. Meanwhile, Mars’ tilt is about 25 degrees. What’s more, the rotational axes of the Earth and Mars don’t tilt in the same direction, as in this illustration. For both reasons – different amount of tilt, and different direction of tilt – Mars doesn’t have the same North Star as Earth.

In the northern sky as seen from Mars, the best candidate for a North Star is located on Mars’ sky dome about half a degree from Mars’ north celestial pole. That’s closer than Polaris is to Earth’s north celestial pole, but, while Polaris is relatively bright (50th brightest of all stars in the night sky), the star near Mars’ north celestial pole is faint.

In fact, this star is barely within the limit of visibility to the eye alone.

Mars’ North Pole points to a spot in the sky that’s about midway between Deneb, the brightest star in the constellation Cygnus the Swan, and Alderamin, the brightest star in the constellation Cepheus the King. Click here to see the position of Mars’ north celestial pole between the constellations Cygnus and Cepheus.

Meanwhile, in the southern sky as seen from Mars, Kappa Velorum is only about three degrees from the Martian south celestial pole. That’s not as close as Polaris is to Earth’s north celestial pole, plus this star is only modestly bright, not nearly as bright as Polaris.

Future Mars colonists aren’t going to have a bright North Star – like our Polaris – to guide them.

On the other hand, if you were standing outside at night on the surface of Mars, you’d see some other cool stuff!

Earth and moon, as seen from Mars by the Curiosity rover in 2014. From Mars, you’d see both the Earth and moon with the eye alone. Read more about this image.

As seen from Mars, you could see Earth’s moon orbiting around Earth once each month. From Earth, we can’t see any other planets’ satellites with the unaided eye, but this amazing sight on Mars would be visible to the eye alone. Both the Earth and the moon would appear starlike.

In general, the Earth as seen from Mars would somewhat mimic our view of Venus as seen from Earth. By that we mean that – like Venus in relationship to Earth – Earth in relationship to Mars is an inner planet. It orbits closer to the sun than Mars. Thus Earth as seen from Mars would be a morning or evening “star” – just as Venus is as seen from our world.

And although both the Earth and moon would appear as stars to the unaided eye, observers on Mars with telescopes would sometimes see them as crescent worlds – just as we do Venus.

So … no North Star for Mars. But Martian stargazers wouldn’t lack for things to see!

This series of images shows the Martian moon Phobos as it crossed in front of the sun, as seen by NASA’s Curiosity Mars rover on Tuesday, March 26, 2019 (Sol 2359). Image via NASA/JPL-Caltech/MSSS. Read more about this image.

Bottom line: Does planet Mars have a North Star akin to Earth’s North Star Polaris?

## Why Polaris is so Fascinating

By watching the motion of Polaris A’s companion star, Polaris Ab, astronomers expect to learn not only the stars’ orbits but also their masses. Measuring the mass of a star is one of the most difficult tasks.

Astronomers want to determine the mass of Polaris accurately because it is the nearest Cepheid variable star a type of star that pulsates radially, varying in both diameter and temperature and producing changes in brightness with a well-defined stable period and amplitude.

Cepheid variable stars can be used to measure the distance to galaxies and the expansion rate of the universe. This is why it is so important to understand their physics and evolution.

The most important part of this process is knowing a Cepheid’s mass. The North Star (Polaris) is a Cepheid variable, but its periodic variations in brightness are steadily decreasing in amplitude. This has been observed to be the case for the last hundred years.

“Around the beginning of the 20th Century, Polaris’ brightness fluctuated every four days by 10%. Only ten years ago this variation had dropped to 2%, leading astronomers to believe this steady decline in the variability of the star was about to end.

That was until recent observations uncovered an increase in variability to 4%. Polaris is an odd star in that it is a Cephid variable with a declining variability, and now astronomers are baffled as to why the brightness fluctuation has been revived.

The NASA’s Hubble Space Telescope has been able to have photographed the close companion of Polaris for the first time.” – Center for Astrophysics.

The researchers plan to continue observing the Polaris system for several years. During that time, the movement of the small companion in its 30-year orbit around Polaris A, should be detectable.

The ultimate goal is the get an accurate mass for Polaris by measuring the motion of the companion, Polaris Ab in its orbit.

Polaris and Comet Lovejoy (NASA APOD) by Rogelio Bernal Andreo.


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# Orders on the Cartesian product of partially ordered sets

The wikipedia article about partially ordered set describes three of the possible partial orders on the Cartesian product of two partially ordered sets, and mentions that similar constructions can be done on the Cartesian product of more than two sets. However, if there are more than two sets, the constructions can be combined with each other in various ways. If we restrict ourselves to the lexicographic order and the product order, how can we describe the various way in which these orders can be combined?

For three sets, we could define $(a_1, a_2, a_3) \leq(b_1,b_2,b_3)$ via:

• $a_1 < b_1$ or ($a_1=b_1$ and ($a_2 < b_2$ or ($a_2=b_2$ and $a_3 \leq b_3$))) (lexicographical order)
• $a_1 \leq b_1$ and $a_2 \leq b_2$ and $a_3 \leq b_3$ (product order)
• $a_1 < b_1$ or ($a_1=b_1$ and $a_2 \leq b_2$ and $a_3 \leq b_3$) (first lexicographical order, then product order)
• ($a_1 \leq b_1$ and $a_2 \leq b_2$ and ($a_1 < b_1$ or $a_2 < b_2$)) or ($a_1 = b_1$ and $a_2 = b_2$ and $a_3 \leq b_3$) (first product order, then lexicographical order)
• $a_1 \leq b_1$ and ($a_2 < b_2$ or ($a_2=b_2$ and $a_3 \leq b_3$)) (first product order, then ...)
• ($a_1 < b_1$ or ($a_1=b_1$ and $a_3 \leq b_3$)) and ($a_2 < b_2$ or ($a_2=b_2$ and $a_3 \leq b_3$)) (first product order, then ...)

I think these are all orders I want to consider for three sets (ignoring permutations), but how can I be sure without a systematic way to describe these orders. Is there a systematic way to describe the orders on the Cartesian product of a finite number of partial ordered sets?

Does the description gets easier, if we look at bounded lattices (or semilattices with identity element) instead of partially ordered sets?

You can systematize the various ways of imposing an order structure on the cartesian product of the underlying sets of the given posets. However, why are you interested in enumerating all possible such orders? Thinking categorically, one is usually motivated by constructing orders with particular properties rather than random mixing of possible definitions. For instance, the product order yields a categorical product in the category $Pos$ of posets. It is more natural to think of induced orders in this way, unless of course there is some particular, application driven, reason to consider a particular induced order.

• I want to systematically describe these orders for posets, because I want to try to characterize those which give me bounded lattices (or semilattices with identity element). The application behind this is to better understand how an efficient data-strucure for a semilattice might look like (especially if I want to use a join-semilattice instead of a tree for representing hierarchies). Perhaps I would be better off thinking about relational representations instead, since this is what databases did when trees were no longer enough. Oct 26, 2012 at 8:12

I found the trick required to construct similar orders as the ones I gave as examples. The crucial step is to look at preorders instead of partial orders, and ignore the Cartesian product as much as possible.

For a preorder relation $\leq$, we can define the strict preorder relation $<$ via

• $a < b$ if and only if $a \leq b$ and not $b \leq a$.

If two preorders $\leq_1$ and $\leq_2$ on the same set $P$ are given, we can define preorder analogs of lexicographical and product orders via

• $a \leq_L b$ if and only if $a <_1 b$ or ($a \leq_1 b$ and $a \leq_2 b)$.
• $a \leq_P b$ if and only if $a \leq_1 b$ or ($a \leq_1 b$ and $a \leq_2 b)$.

We can also define the analog of the product of the strict orders

• $a \leq_S b$ if and only if ($a <_1 b$ and $a <_2 b$) or ($a \leq_1 b$ and $b \leq_1 a$ and $a \leq_2 b$ and $b \leq_2 a$).

The following preorders are canonically defined on the direct product of $n$ partially ordered sets:

• For $1 \leq i \leq n$: $(a_1,\ldots,a_n) \leq_i (b_1,\ldots,b_n)$ if and only if $a_i \leq b_i$.

The partial orders I gave as example simply arise by successively defining new preorders on the direct product via $\leq_L$ and $\leq_P$ starting from the preorders $\leq_i$. This will give rise to a partial order on the direct product if each $\leq_i$ occurs at least once in the constructed preorder.

Does the description gets easier, if we look at bounded lattices (or semilattices with identity element) instead of partially ordered sets?

My guess would be that in this case, each $\leq_i$ must occur exactly once in the constructed preorder and that $\leq_S$ may not be used, but I haven't checked this yet. I guess the first step would be to define a notion of independent preorders, and then check the behavior of $\leq_L$, $\leq_P$ and $\leq_S$ for (independent) preorder analogs of lattices. I guess the result is that $\leq_S$ doesn't give rise to a preorder analog of lattices, $\leq_L$ will only work for bounded lattices (or semilattices with identity element), and $\leq_P$ will work without restrictions (except that the preorders should be independent).

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# §10.74 Methods of Computation

## §10.74(i) Series Expansions

The power-series expansions given in §§10.2 and 10.8, together with the connection formulas of §10.4, can be used to compute the Bessel and Hankel functions when the argument or is sufficiently small in absolute value. In the case of the modified Bessel function see especially Temme (1975).

In other circumstances the power series are prone to slow convergence and heavy numerical cancellation.

If or is large compared with , then the asymptotic expansions of §§10.17(i)10.17(iv) are available. Furthermore, the attainable accuracy can be increased substantially by use of the exponentially-improved expansions given in §10.17(v), even more so by application of the hyperasymptotic expansions to be found in the references in that subsection.

For large positive real values of the uniform asymptotic expansions of §§10.20(i) and 10.20(ii) can be used. Moreover, because of their double asymptotic properties (§10.41(v)) these expansions can also be used for large or , whether or not is large. It should be noted, however, that there is a difficulty in evaluating the coefficients , , , and , from the explicit expressions (10.20.10)–(10.20.13) when is close to 1 owing to severe cancellation. Temme (1997) shows how to overcome this difficulty by use of the Maclaurin expansions for these coefficients or by use of auxiliary functions.

Similar observations apply to the computation of modified Bessel functions, spherical Bessel functions, and Kelvin functions. In the case of the spherical Bessel functions the explicit formulas given in §§10.49(i) and 10.49(ii) are terminating cases of the asymptotic expansions given in §§10.17(i) and 10.40(i) for the Bessel functions and modified Bessel functions. And since there are no error terms they could, in theory, be used for all values of ; however, there may be severe cancellation when is not large compared with .

## §10.74(ii) Differential Equations

A comprehensive and powerful approach is to integrate the differential equations (10.2.1) and (10.25.1) by direct numerical methods. As described in §3.7(ii), to insure stability the integration path must be chosen in such a way that as we proceed along it the wanted solution grows in magnitude at least as fast as all other solutions of the differential equation.

In the interval , needs to be integrated in the forward direction and in the backward direction, with initial values for the former obtained from the power-series expansion (10.2.2) and for the latter from asymptotic expansions (§§10.17(i) and 10.20(i)). In the interval either direction of integration can be used for both functions.

Similarly, to maintain stability in the interval the integration direction has to be forwards in the case of and backwards in the case of , with initial values obtained in an analogous manner to those for and .

For the function , for example, can always be computed in a stable manner in the sector by integrating along rays towards the origin.

Similar considerations apply to the spherical Bessel functions and Kelvin functions.

For further information, including parallel methods for solving the differential equations, see Lozier and Olver (1993).

## §10.74(iii) Integral Representations

For evaluation of the Hankel functions and for complex values of and based on the integral representations (10.9.18) see Remenets (1973).

For applications of generalized Gauss–Laguerre quadrature (§3.5(v)) to the evaluation of the modified Bessel functions for and see Gautschi (2002a). The integral representation used is based on (10.32.8).

For evaluation of from (10.32.14) with and complex, see Mechel (1966).

## §10.74(iv) Recurrence Relations

If values of the Bessel functions , , or the other functions treated in this chapter, are needed for integer-spaced ranges of values of the order , then a simple and powerful procedure is provided by recurrence relations typified by the first of (10.6.1).

Suppose, for example, and . Then and can be generated by either forward or backward recurrence on when , but if then to maintain stability has to be generated by backward recurrence on , and has to be generated by forward recurrence on . In the case of , the need for initial values can be avoided by application of Olver’s algorithm (§3.6(v)) in conjunction with Equation (10.12.4) used as a normalizing condition, or in the case of noninteger orders, (10.23.15).

For further information see Gautschi (1967), Olver and Sookne (1972), Temme (1975), Campbell (1980), and Kerimov and Skorokhodov (1984a).

## §10.74(v) Continued Fractions

For applications of the continued-fraction expansions (10.10.1), (10.10.2), (10.33.1), and (10.33.2) to the computation of Bessel functions and modified Bessel functions see Gargantini and Henrici (1967), Amos (1974), Gautschi and Slavik (1978), Tretter and Walster (1980), Thompson and Barnett (1986), and Cuyt et al. (2008).

## §10.74(vi) Zeros and Associated Values

Newton’s rule (§3.8(i)) or Halley’s rule (§3.8(v)) can be used to compute to arbitrarily high accuracy the real or complex zeros of all the functions treated in this chapter. Necessary values of the first derivatives of the functions are obtained by the use of (10.6.2), for example. Newton’s rule is quadratically convergent and Halley’s rule is cubically convergent. See also Segura (1998, 2001).

Methods for obtaining initial approximations to the zeros include asymptotic expansions (§§10.21(vi)-10.21(ix)), graphical intersection of graphs in (e.g., §10.3(i)) with the -axis, or graphical intersection of complex-variable surfaces (e.g., §10.3(ii)) with the plane .

To ensure that no zeros are overlooked, standard tools are the phase principle and Rouché’s theorem; see §1.10(iv).

### ¶ Real Zeros

See Olver (1960, pp. xvi–xxix), Grad and Zakrajšek (1973), Temme (1979a), Ikebe et al. (1991), Zafiropoulos et al. (1996), Vrahatis et al. (1997a), Ball (2000), and Gil and Segura (2003).

### ¶ Complex Zeros

See Leung and Ghaderpanah (1979), Kerimov and Skorokhodov (1984b, c, 1985a, 1985b), Skorokhodov (1985), Modenov and Filonov (1986), and Vrahatis et al. (1997b).

### ¶ Multiple Zeros

See Kerimov and Skorokhodov (1985c, 1986, 1987, 1988).

## §10.74(vii) Integrals

### ¶ Hankel Transform

See Cornille (1972), Johansen and Sørensen (1979), Gabutti (1979), Gabutti and Minetti (1981), Candel (1981), Wong (1982), Lund (1985), Piessens and Branders (1985), Hansen (1985), Bezvoda et al. (1986), Puoskari (1988), Christensen (1990), Campos (1995), Lucas and Stone (1995), Barakat and Parshall (1996), Sidi (1997), Secada (1999).

### ¶ Fourier–Bessel Expansion

For the computation of the integral (10.23.19) see Piessens and Branders (1983, 1985), Lewanowicz (1991), and Zhileĭkin and Kukarkin (1995).

### ¶ Spherical Bessel Transform

The spherical Bessel transform is the Hankel transform (10.22.76) in the case when is half an odd positive integer.

See Lehman et al. (1981), Puoskari (1988), and Sharafeddin et al. (1992).

### ¶ Kontorovich–Lebedev Transform

See Ehrenmark (1995).

### ¶ Products

For infinite integrals involving products of Bessel functions of the first kind, see Linz and Kropp (1973), Gabutti (1980), Ikonomou et al. (1995), Lucas (1995), and Van Deun and Cools (2008).

## §10.74(viii) Functions of Imaginary Order

For the computation of the functions and defined by (10.45.2) see Temme (1994c) and Gil et al. (2002d, 2003a, 2004b).


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# An object with a mass of 4 kg, temperature of 270 ^oC, and a specific heat of 25 J/(kg*K) is dropped into a container with 36 L  of water at 0^oC . Does the water evaporate? If not, by how much does the water's temperature change?

Aug 30, 2017

Yes, the water evaporates.

#### Explanation:

According to principle of calorimetry,

${Q}_{o} = {Q}_{w}$

${\left(m C \Delta T\right)}_{o} = {\left(m C \Delta T\right)}_{w}$

$4 \times 25 \times \left(270 - {T}_{f}\right) = 36 \times 4.2 \times {T}_{f}$

$\Rightarrow {T}_{f} \approx {107.5}^{o} C$

As ${T}_{f} > {100}^{o} C$ , the water evaporates


text:
Math Help - 2 easy double integrals

1. 2 easy double integrals

I need those 2 integrals solved until tonight , cause tomorrow morning i'm writing a text and i need them.have tried and failed so far !

View image: MSP7531a185g80d988ih4d000044e7ah94cfgg0fe7
View image: MSP29041a185f2bi3bg12d7000016i6bf8d0i1g1ebi

2. Re: 2 easy double integrals

Hint for the second integral: change the order of integration

$\begin{gathered}I= \int\limits_0^1 {\int\limits_0^3 {x\sqrt {{x^2} + {y^2}}\,dydx} } = \int\limits_0^3 {\int\limits_0^1 {x\sqrt {{x^2} + {y^2}}\,dxdy} } \hfill \\[5pt] \left[ \begin{gathered} \sqrt {{x^2} + {y^2}} = t~~ \Rightarrow~~ {x^2} + {y^2} = {t^2}~~ \Rightarrow ~~x\,dx = t\,dt \hfill \\\int {x\sqrt {{x^2} + {y^2}} dx} = \int {{t^2}dt} = \frac{{{t^3}}}{3} + C = \frac{1}{3}{({x^2} + {y^2})^{3/2}} + C \hfill \\ \end{gathered} \right] \hfill \\[5pt] I= \int\limits_0^3 {dy} \left. {\frac{1}{3}{{({x^2} + {y^2})}^{3/2}}} \right|_{x = 0}^{x = 1} = \frac{1}{3}\int\limits_0^3 {\left[ {{{(1 + {y^2})}^{3/2}} - {y^3}} \right]dy=\ldots=\frac{23}{8}\sqrt{10}-\frac{27}{4}-\frac{1}{8}\ln(\sqrt{10}-3)} \hfill \\ \end{gathered}$

3. Re: 2 easy double integrals

The first integral

\begin{aligned}\int\limits_1^3 \int\limits_0^x \frac{2\,dydx}{\sqrt{x^2+y^2}}&= 2\int\limits_1^3 dx\! \left. {\ln\Bigl(y+\sqrt{x^2+y^2}}\Bigr)} \right|_{y = 0}^{y = x} = 2\int\limits_1^3 \Bigl[\ln (x + \sqrt 2 x) - \ln x\Bigr]dx=\\[2pt] &= 2\int\limits_1^3 \Bigl[\ln x + \ln (1 + \sqrt 2 ) - \ln x\Bigr]dx= 4\ln (1 + \sqrt 2 ) \end{aligned}

4. Re: 2 easy double integrals

thank you so much :-))))) ! just solved myself the second one but really thanks for both ! didn't expect such a quick answer !
but concerning first integral my book gives us answer (pi/2)*ln3 ...

5. Re: 2 easy double integrals

but concerning first integral my book gives us answer (pi/2)*ln3 ...
I checked in Maple. Check carefully.


text:
# Integrating a linear polynomial approx

• May 23rd 2010, 08:56 AM
matlabnoob
Integrating a linear polynomial approx
hi,

thanks for reading.can anyone help me out?(Crying)
see the attached file.i am working out for many hours how to integrate P1(t). does anyone know?
how do you get -1/2hf(..... ) + 3/2hf(....) ?
thats all i need to know!
i tried integrating MANY times. it looks easy.so what am i not understanding?

thanks all
• May 23rd 2010, 09:42 AM
running-gag
Hi

The first integral is

$\left[\frac{t^2}{2}-t_it\right]_{t_i}^{t_{i+1}} = \frac{t_{i+1}^2}{2}- t_i t_{i+1}+\frac{t_i^2}{2}$

Substituting $t_{i+1} = t_i + h$

$\left[\frac{t^2}{2}-t_it\right]_{t_i}^{t_{i+1}} = \frac{t_{i}^2}{2} + h t_i + \frac{h^2}{2} - t_i^2 - h t_i + \frac{t_i^2}{2} = \frac{h^2}{2}$
• May 23rd 2010, 10:02 AM
matlabnoob
Quote:

Originally Posted by running-gag
Hi

The first integral is

$\left[\frac{t^2}{2}-t_it\right]_{t_i}^{t_{i+1}} = \frac{t_{i+1}^2}{2}- t_i t_{i+1}+\frac{t_i^2}{2}$

Substituting $t_{i+1} = t_i + h$

$\left[\frac{t^2}{2}-t_it\right]_{t_i}^{t_{i+1}} = \frac{t_{i}^2}{2} + h t_i + \frac{h^2}{2} - t_i^2 - h t_i + \frac{t_i^2}{2} = \frac{h^2}{2}$

aha...! thats where confusion happened... h can be anything =S
a difference between one point and another. thanks! =D