Easy way to prove the earth is indeed flat

Caus there are plenty pictures from the moon of a ROUND earth!!!.

Here is a simple mental exercise:

The Moon is quite bright as seen from here and is visually about the same size as the Sun.
The Moon's diameter, we are told, is 3,500km.  Earth's diameter, we are told, is 12,750km.
Based on this, Earth's apparent diameter would be about 3.64 times the Moon's diameter.
So Earth, seen from the Moon, should be 3.64 times the size of the Moon seen from Earth.

The Moon's albedo (the reflecting power of its surface), we are told, is 0.12 or 12% (regolith).
Earth's average albedo, we are told, is about 0.3 or 30% (most of that coming from the clouds).
So Earth reflects 2.5 times more sunlight than the Moon (if you believe the Moon reflects the Sun).

Based on this, a photo of Earth as seen from the Moon should have the following properties:
- Earth should be 3.64 times the apparent size of the Moon as seen from Earth
- Earth should be 3.64 times the apparent size of the Sun as seen from the Moon
- Earth should be 2.5 times more reflective, i.e. brighter, than the Moon from here
(probably more since there is no atmosphere to scatter or dim Earth's brightness)

SUN ------------------------------------------------------------------------------------------------------------------EARTH?

Quote from: Nik7ki on August 19, 2011, 02:42:51 PM

Caus there are plenty pictures from the moon of a ROUND earth!!!.

Here is a simple mental exercise:

The Moon is quite bright as seen from here and is visually about the same size as the Sun.
The Moon's diameter, we are told, is 3,500km.  Earth's diameter, we are told, is 12,750km.
Based on this, Earth's apparent diameter would be about 3.64 times the Moon's diameter.
So Earth, seen from the Moon, should be 3.64 times the size of the Moon seen from Earth.

The Moon's albedo (the reflecting power of its surface), we are told, is 0.12 or 12% (regolith).
Earth's average albedo, we are told, is about 0.3 or 30% (most of that coming from the clouds).
So Earth reflects 2.5 times more sunlight than the Moon (if you believe the Moon reflects the Sun).

Based on this, a photo of Earth as seen from the Moon should have the following properties:
- Earth should be 3.64 times the apparent size of the Moon as seen from Earth
- Earth should be 3.64 times the apparent size of the Sun as seen from the Moon
- Earth should be 2.5 times more reflective, i.e. brighter, than the Moon from here
(probably more since there is no atmosphere to scatter or dim Earth's brightness)

SUN ------------------------------------------------------------------------------------------------------------------EARTH?

Good job on not providing any evidence piper. The sun is causing bad camera flare. Also, comparing sizes of objects in two different photos without knowledge of the various zooms or scalings of the two photos is impossible. I suggest that you gather better data.

Quote from: Nik7ki on August 19, 2011, 02:42:51 PM

Caus there are plenty pictures from the moon of a ROUND earth!!!.

Here is a simple mental exercise:

The Moon is quite bright as seen from here and is visually about the same size as the Sun.
The Moon's diameter, we are told, is 3,500km.  Earth's diameter, we are told, is 12,750km.
Based on this, Earth's apparent diameter would be about 3.64 times the Moon's diameter.
So Earth, seen from the Moon, should be 3.64 times the size of the Moon seen from Earth.

The Moon's albedo (the reflecting power of its surface), we are told, is 0.12 or 12% (regolith).
Earth's average albedo, we are told, is about 0.3 or 30% (most of that coming from the clouds).
So Earth reflects 2.5 times more sunlight than the Moon (if you believe the Moon reflects the Sun).

Based on this, a photo of Earth as seen from the Moon should have the following properties:
- Earth should be 3.64 times the apparent size of the Moon as seen from Earth
- Earth should be 3.64 times the apparent size of the Sun as seen from the Moon
- Earth should be 2.5 times more reflective, i.e. brighter, than the Moon from here
(probably more since there is no atmosphere to scatter or dim Earth's brightness)

SUN ------------------------------------------------------------------------------------------------------------------EARTH?

This is a half baked plan to make a half baked argument based on photos with absolutely no indication of their origin.

The photo on the right is believable, but you gave no information about its source. I have no reason to sit down and see if it passes a close inspection or to see if it came from a NASA archive. But assuming it is not faked, what lens was used to take it? A telephoto? a wide angle? What exposure settings were used? Where is your calculation of the expected size of the Earth for a given focal length? Where is your calculation of the expected brightness of the Earth? Why do you even claim that it is not about the same brightness of the astronaut's suit?

The photo on the left is quite unbelievable. If I see it in the NASA archives I will consider seriously your claims. But as a serious amateur photographer I can tell you there are lots of things in that photo that just don't add up. This is a montage made for the purpose of having people doubt the real photos taken on the Moon.

If you want someone to even look closely at these photos the least you have to do is show that NASA took them (or manufactured them). Show me the link to NASA's site or another credible evidence of their source.

"Cause there are plenty pictures from the moon of a ROUND earth!!!"

yes, we can photograph the moon from the Earth and determine that it is spherical. 

please excuse the FAQQERS on this site, it is their  bible and is silly as it is, to sway from it would be blasphemous.

Quote from: Piper on August 19, 2011, 07:18:31 PM

SUN ------------------------------------------------------------------------------------------------------------------EARTH?

This is a half baked plan to make a half baked argument based on photos with absolutely no indication of their origin.

The photo on the right is believable, but you gave no information about its source. I have no reason to sit down and see if it passes a close inspection or to see if it came from a NASA archive. But assuming it is not faked, what lens was used to take it? A telephoto? a wide angle? What exposure settings were used? Where is your calculation of the expected size of the Earth for a given focal length? Where is your calculation of the expected brightness of the Earth? Why do you even claim that it is not about the same brightness of the astronaut's suit?

The photo on the left is quite unbelievable. If I see it in the NASA archives I will consider seriously your claims. But as a serious amateur photographer I can tell you there are lots of things in that photo that just don't add up. This is a montage made for the purpose of having people doubt the real photos taken on the Moon.

If you want someone to even look closely at these photos the least you have to do is show that NASA took them (or manufactured them). Show me the link to NASA's site or another credible evidence of their source.

My apologies, I thought most people here would be familiar with NASA's catalog of alleged moon shots.

The well-known photo on the right is AS17-134-20384 from Apollo 17, which shows a tiny and not very bright Earth above the U.S flag (how the other astronaut took a photo at that angle with a chest-mounted camera would be interesting to figure out).

The photo on the left, from Apollo 12, is officially designated 69-HC-1341.  It appeared in the December 12, 1969 issue of Life magazine (with a strangely glowing astronaut on the cover).  Another good shot showing the size of the sun from the moon (and not "camera flare") taken from Apollo 12 is the image displayed in the first post of this thread started by Tom Bishop.

Obviously there was no change in the distance to Earth or the Sun between Apollo 12 and Apollo 17.  Both photos (and most "lunar surface" photos taken by astronauts) were taken with a Hasselblad 70mm EDC camera.  The camera "is equipped with a specially designed Biogon lens with a focal length of 60 mm, with a polarization filter mounted on the lens" (source).

Thanks for the camera data, piper. Now, these cameras are equipped with reference crosshairs over the film. With the 60mm lens, the angular separation of these cross hairs is 10.3 degrees.

The earth is about .205 the size of the separation of the cross hairs, so the angular size of the earth is about 2.09 degrees.

Divide by 3.64 to get the theoretical angular size of the moon as seen from earth.

Multiply by 60 to get the results in arcminutes to get 34-35 arcminutes for the size of the moon as seen from earth

Given the tolerances of my measurements, this is very close to the actual angular diameter of the moon as seen from earth. (29.3-34.1 arcminutes)

Therefore, your argument from angular size of the earth fails.

As for the sun, it is to bright to find the actual diameter in the photo. And yes, there is lens flare. This is easily repeated.


Sun arguments fails.

As for the brightness argument, I see no quantitative evidence for the earth being dim or bright.

Brightness argument fails.

All your arguments are invalid. Please quantitate any further arguments to avoid the mistakes you made this time.

Thanks for the camera data, piper. Now, these cameras are equipped with reference crosshairs over the film. With the 60mm lens, the angular separation of these cross hairs is 10.3 degrees.

The earth is about .205 the size of the separation of the cross hairs, so the angular size of the earth is about 2.09 degrees.

Divide by 3.64 to get the theoretical angular size of the moon as seen from earth.

Multiply by 60 to get the results in arcminutes to get 34-35 arcminutes for the size of the moon as seen from earth

Given the tolerances of my measurements, this is very close to the actual angular diameter of the moon as seen from earth. (29.3-34.1 arcminutes)

Therefore, your argument from angular size of the earth fails.

Thanks for the info on the fiducials.  It seems your numbers show the Earth's image is actually too large in this photo, not too small, once again showing that it cannot possibly be a real photo of Earth taken from the Moon's surface.  Using your own number of 10.3 degrees of angular separation between cross hairs ('fiducials') I obtain an angular size of 2.14 degrees for Earth, when measured across:

120 pixels for Earth / 575 pixels between fiducials = 0.208 x 10.3 degrees = Earth's angular size is 2.14 degrees
(Note that your own calculation for the angular size of Earth is incorrect and should be 2.11 and not 2.09 degrees)

The numbers I have for the Moon's angular diameter seen from Earth give it a range between 29.43 arcminutes to 33.50 arcminutes.  Divide 2.14 degrees by 3.64 and multiply by 60 and the result is 35.27 arcminutes, almost 2 arcminutes more than its maximum possible visual appearance.

From the Moon, Earth should have an average angular size of 1.90 degrees - much lower than my calculated 2.14 degrees and even your own 2.09 (or 2.11) degrees.  At perigee, or closest approach, it would be 2.01 degrees at most.  The image was allegedly taken on December 13, 1973, so the Moon was still six days away from perigee (according to this calculator).  Even allowing for slight distortion/focus issues, that is quite a discrepancy.

NASA fail

As for the sun, it is to bright to find the actual diameter in the photo. And yes, there is lens flare. This is easily repeated.
http://upload.wikimedia.org/wikipedia/commons/thumb/3/3d/Lens_Flare.JPG/800px-Lens_Flare.JPG

Sun arguments fails.

As for the brightness argument, I see no quantitative evidence for the earth being dim or bright.

Brightness argument fails.

All your arguments are invalid. Please quantitate any further arguments to avoid the mistakes you made this time.

Actually I did make an error in my initial calculations.  Since luminosity ratio is in fact proportional to the square of radius ratio, the increased size of Earth in comparison should equate to much more than a 2.5 times increase in brightness as compared to the Moon from here, most likely washing out any blue coloring that can be seen in the image provided - as well as in all other images purporting to show Earth from the Moon.

Ah, good to finally have a FE'er who understands geometry.

Minor note on discrepancies, we are the same to 1 part in 100, pretty good. But I didn't count pixels, so I'll go with yours.

Ok, so the perigee of the moon closest to the photo time was 358041 km.

The shoot was six days away from the perigee. I will find the approx. distance using the eccentricity of the moon (.0549) and the polar form of an ellipse.

Distance = 358041*(1.0549)/(1+cos(6/27.3*2pi))=37805 km
subtract radius of the moon to get 372068 km

angle is about 2*6378/372068 radians, which is 1.96 degrees.

Now, translate that to pixels:

1.96/10.3*575=109 to 110 pixels.
This would be true for a sharp focused picture of the earth.

However, a 9% increase in apparent size, which 5 pixels on each side of earth, is not significant evidence for any thing but having the camera out of focus with the earth, especially considering that the black of space is around the earth.

The photo on the left, from Apollo 12, is officially designated 69-HC-1341.  It appeared in the December 12, 1969 issue of Life magazine (with a strangely glowing astronaut on the cover).  Another good shot showing the size of the sun from the moon (and not "camera flare") taken from Apollo 12 is the image displayed in the first post of this thread started by Tom Bishop.

NASA's photo archive is in www.nasa.gov, not on www.apollomissionphotos.com

Also, I could not find any information on that photo, I could, however, find a more credible version of the photo:



Here you can see that details are not perfect, as you would expect from the real photo. The crosshairs are sometimes visible, sometimes partially visible, sometimes totally missing. The sun's flares look different when in front of a light background compared with a dark background. And the sun is not even inside the frame, which would have caused so much glare that the photo would have been ruined.

Get a copy of the photo in a NASA source, not in a "Moongate" site.

And no, a huge ice-wall around earth's perimeter prevents people from 'falling off'.

So if global warming is real, we will all wash off the side of the Earth?

Well, I will look forward to this massive water slide in the near coming future.

Are you, people, really trying to prove that NASA is true with NASA pictures?

That's insane.

Are you, people, really trying to prove that NASA is true with NASA pictures?

That's insane.

Are you, people, really trying to prove that NASA is part of a conspiracy by discounting all their evidence and using your own illogical arguments?

That's insane.

hello im new here so ill introduce myself ,
im marty, im a 34 years old adventurous, and intellectually vivacious individual, im here to see and try to understand why there is a debate on that subject matter ,
and quite frankly im facinated by the fact that people go to so much extend, some really even quoting einstein, and maths argument to proove the earth is flat,
First, my motivation for a spehrical earth come from the fact that in 2004 i went around it completly making a perfect circle, starting in montreal, in 3 months iv crossed canada, i use a gps to track my whole travel coordonate, and draw a neet little map for my presentation afterward :-) then i went to japan, crossing europe, and finally coming back in boat to montreal, of course on this matter you can argue that at some point the boat just made a round circle, and that all the gps data were just wrong part of the conspiracy well, i dont know how you explain this i would be curious to see anyway,
in fact all of this doesnt matter much
i guess my question to you all, is this one, you go to amazingly complex argument to explain why the earth is flat,
id like to know, where is the end of it, have you even bother to draw a map of this thing? and then if you did, why dont we go to the end of this thing and trave at the end of the earth all member of this society, with tv reporter your pick them or you record this,
and we can settle this thing once and for all,
cause there is one point we all agree is that there can be only one truth?? do we, and one of us is wrong, so intellectual masturbation isnt good for anyone,
sincerely
marty

in 2004 i went around it completly making a perfect circle, starting in montreal, in 3 months iv crossed canada, i use a gps to track my whole travel coordonate, and draw a neet little map for my presentation afterward

i wanna see the neet map

Ok, Im gonna go back to the argument stating that the world speed record for going around the circumference of Antarctica was done at 22.6 knots. Lets assume here that the earth is flat and the circumference is 125,896 km.

1 knot = 1.852 km/h
There are 24 hours in a day.
He made the trip in 102 days.

So, 1.852 * 22.6 = 41.8552 km/h
41.8552 * 24 = 1004.5248 km/d (Assuming he was moving constantly 24/7)
1004.5248 * 102 = 102461.5296 km

102,461 =/= 125,896

And this is with IDEAL CONDITIONS, and he had to be moving 24/7!
Either this record is false, or Flat Earth Theory is false.

But again from ENaG

and the circumference, 52,800 statute miles

The FAQ is wrong. There are a bunch of threads about us all asking for it to be changed. But Rowbotham calculated the earth to be 52,800 miles in circumference (85,000km). You can read how in the second chapter of Earth Not A Globe.

So ...

1 knot = 1.852 km/h
There are 24 hours in a day.
He made the trip in 102 days.

So, 1.852 * 22.6 = 41.8552 km/h
41.8552 * 24 = 1004.5248 km/d (Assuming he was moving constantly 24/7)
1004.5248 * 102 = 102461.5296 km

102,461 > 85,000

So the guy is sailing for about 19.5 hours a day. Is that so unbelievable? Why would you doubt this man's integrity?

But round Earthers say the circumference of Antartica is approx 14600km.

That means you think he only bothered sailing for 3.5 hours a day. On a world record attempt? Why would be put in so little effort?

In fact whilst 4.5 hours rest a day seems little with the FE calc, it does fit nicely with what sailors do.
Of Dame Ellen MacArthur's world record sailing ...

Her time of 71 days, 14 hours, 18 minutes 33 seconds beat Joyon's then world record time by 1 day, 8 hours, 35 minutes and 49 seconds. Ellen had no more than 20 minutes' sleep at a time, having to be on constant lookout day and night

And

"I think the boat needs a good rest - she's done a few hard miles and I've had twenty minutes sleep in the last twenty-four hours so yeah that's been a bit tough.

Allowing for icebergs and other obstacles the FE numbers seem very, very reasonable.

If as you say the total distance he travels is 102461km, what the hell was he doing on a 14,600 mile trip? On a 85,000km trip it seems very plausible with 1/5^th of the total distance added on for diversions. World record sailors do not stop for 20+ hours a day to look at the penguins.

In fact your example has only highlighted how absurd the RE numbers are. Good job. 

Where is this figure of 22.6 knots coming from? As far as I can see, the figure was a guess at the speed needed to circumnavigate the flat Earth using the circumference of the flat earth and the race time. Now it seems it's being used with the race time to justify the circumference of the flat Earth!

I had a look through the log poster earlier in the thread, and the only mention of the boat speed I can find in it is here:

Day 25: Position - 52° 43’S 160° 28’E
“Massive waves still threaten, but I have increased sail area to keep in rhythm with the waves. These are not normal waves, but small hills with white foamy crests. Each wave is 100m from the next – perfect for surfing on this 24m surf board. If I catch the waves right, we are running down them at 15-16 knots.”

There we have a speed well shy of the 22.6 knots suggested elsewhere in this thread, even when riding down the side of a wave.

Where is this figure of 22.6 knots coming from? As far as I can see, the figure was a guess at the speed needed to circumnavigate the flat Earth using the circumference of the flat earth and the race time. Now it seems it's being used with the race time to justify the circumference of the flat Earth!

I had a look through the log poster earlier in the thread, and the only mention of the boat speed I can find in it is here:

Quote

Day 25: Position - 52° 43’S 160° 28’E
“Massive waves still threaten, but I have increased sail area to keep in rhythm with the waves. These are not normal waves, but small hills with white foamy crests. Each wave is 100m from the next – perfect for surfing on this 24m surf board. If I catch the waves right, we are running down them at 15-16 knots.”

There we have a speed well shy of the 22.6 knots suggested elsewhere in this thread, even when riding down the side of a wave.

Something is very wrong with all the figures in this thread. James Clarke Ross did a journey around 5/6 of Antarctica around 1840 and his estimation was that, including the trip from his port of origin, the trip to his port of origin,  a lot of back-and-forth trips and trips to New Zealand and South America for provisions, he took about 60,000 miles.

So, if they needed to travel more than some 40,000 kilometers they did not make a simple circumnavegation of Antarctica, they did a lot more than that.

Yes, the Ross case again matches Rowbotham's numbers for a Flat Earth circumference, and makes the RE numbers seem very very wrong. It was the Ross case that helped Voliva conclude the earth's circumference could be no more than 60,000 miles.

What measurements made during Ross's 1839-43 expedition make the RE model seem wrong? How long should the journey have been instead under a RE model?

It should have been 14600 miles on a round earth. 60,000 miles matches Rowbowtham and Voliva's estimates instead.

It should have been 14600 miles on a round earth. 60,000 miles matches Rowbowtham and Voliva's estimates instead.

I still don't fully understand. Why should his journey only be 14,600 miles on a round Earth? Looking at the maps of his voyage, Ross easily covered more that that just going from Australia to South America when plotted on a round Earth. I still see no reason why the whole journey should be significantly less than 60,000 miles.

Quote from: Thork on September 09, 2011, 11:03:53 AM

It should have been 14600 miles on a round earth. 60,000 miles matches Rowbowtham and Voliva's estimates instead.

I still don't fully understand. Why should his journey only be 14,600 miles on a round Earth? Looking at the maps of his voyage, Ross easily covered more that that just going from Australia to South America when plotted on a round Earth. I still see no reason why the whole journey should be significantly less than 60,000 miles.

The 60000 mile estimate is for a whole lot more than circumnavigating Antarctica by the shortest possible route. He moved back and forth exploring the coast of Antarctica, so he even made some of the circumnavigation twice, for example while he researched the magnetic South Pole and its surroundings.

Most importantly, he made day by day records of initial and final latitude and longitude, speed and distance traveled, so you can see that his journey matched, day by day, the predictions of real science. If he had suddenly found errors in every day's record of a factor of 3, seeing that what should have taken one day took 3, he would have come back telling us all that navigation as we know it is useless or that the Earth is flat. But his journal is totally consistent with a spherical Earth. Go figure...

Quote from: NTheGreat on September 09, 2011, 02:10:32 PM

Quote from: Thork on September 09, 2011, 11:03:53 AM

It should have been 14600 miles on a round earth. 60,000 miles matches Rowbowtham and Voliva's estimates instead.

I still don't fully understand. Why should his journey only be 14,600 miles on a round Earth? Looking at the maps of his voyage, Ross easily covered more that that just going from Australia to South America when plotted on a round Earth. I still see no reason why the whole journey should be significantly less than 60,000 miles.

The 60000 mile estimate is for a whole lot more than circumnavigating Antarctica by the shortest possible route. He moved back and forth exploring the coast of Antarctica, so he even made some of the circumnavigation twice, for example while he researched the magnetic South Pole and its surroundings.

Most importantly, he made day by day records of initial and final latitude and longitude, speed and distance traveled, so you can see that his journey matched, day by day, the predictions of real science. If he had suddenly found errors in every day's record of a factor of 3, seeing that what should have taken one day took 3, he would have come back telling us all that navigation as we know it is useless or that the Earth is flat. But his journal is totally consistent with a spherical Earth. Go figure...

Please stop making up your own versions of events. He did voice his opinions. The Powers that be dismissed them and claimed he was lost. They even sent out search parties when he didn't arrive back within RE deadlines.

A nice passage about the ice-wall

On reaching nearly to Mount Erebus Ross encountered a feature he called the "Great Ice Barrier" at 78°4'S, a wall of ice rising 160 feet out of the sea and "...extended as far to the east and west as the eye could discern" that prevented any further sailing south. He called the sea at the point where he met the barrier "McMurdo Sound" after the first lieutenant on the Terror.

The ice barrier could not be circumvented nor seen over, landing was impossible as the barrier's ice wall was vertical. What was not known, but has been discovered since is that the barrier represents  the front edge of a vast ice shelf flowing from the Antarctic interior 1,000 feet deep and extending for about 1,000 miles.

Please stop making up your own versions of events. He did voice his opinions. The Powers that be dismissed them and claimed he was lost. They even sent out search parties when he didn't arrive back within RE deadlines.

What? I can't see anything about search parties being sent out to search for him.

A nice passage about the ice-wall

I throught the FE ice wall was a monstrous thing rising tens of kilometers into the sky, not something a couple of hundred feet high. I suspect he's talking about the Ross ice shelf, which doesn't really tell us much about the shape of the planet either way.

The fact the ice-wall cannot be circumvented tells you a great deal about earth's shape and nobody said the ice-shelf was thousands of feet high. I often think most people have no intention of listening to any of the evidence put forward, no matter how compelling. No wonder you all think the earth is round.

The fact the ice-wall cannot be circumvented tells you a great deal about earth's shape and nobody said the ice-shelf was thousands of feet high. I often think most people have no intention of listening to any of the evidence put forward, no matter how compelling. No wonder you all think the earth is round.

I'm not sure how you derive the shape of the Earth from about 500 miles of ice. The ice shelf is not thousands of feet high, but most flat Earth models have the ice wall as some huge structure holding back the atmosphere. That's why I think the quote refers to an ice shelf, not the FE ice wall.

The reason nobody is listening to your evidence is because apparently it's not very reliable. Using made up figures such as the speed 22.6 knots, or claiming search parties were sent out to find Ross and that is opinions were dismissed while presenting nothing suggesting these things are true doesn't do much good to your credibility.

Quote from: Thork on September 09, 2011, 04:11:52 PM

The fact the ice-wall cannot be circumvented tells you a great deal about earth's shape and nobody said the ice-shelf was thousands of feet high. I often think most people have no intention of listening to any of the evidence put forward, no matter how compelling. No wonder you all think the earth is round.

I'm not sure how you derive the shape of the Earth from about 500 miles of ice. The ice shelf is not thousands of feet high, but most flat Earth models have the ice wall as some huge structure holding back the atmosphere. That's why I think the quote refers to an ice shelf, not the FE ice wall.

The reason nobody is listening to your evidence is because apparently it's not very reliable. Using made up figures such as the speed 22.6 knots, or claiming search parties were sent out to find Ross and that is opinions were dismissed while presenting nothing suggesting these things are true doesn't do much good to your credibility.

And claiming I made up things that are right there, in James Clarke Ross' journals is death to Thork's reliability. Anyone can read the journals and see how the day by day records of his progress are right there. And anyone can see with his own eyes how Ross himself inserted drawings where there is no ice wall at all. He might have been a bit pedantic when he called the ice walls unsurmountable (or something like that) but he was exact and truthful in his journal and showed how the ice wall only spreads over a portion of the coast of Antarctica.

At least for some time the first of the journals was available free through a link that Tom Bishop provided. I don't know if it is still available, but the one thing that seems unavoidable is that Thork is inventing whatever he wants instead of reading the books.

Why do you think it took humans so long to navigate the Northwest Passage?

Why do you think it took humans so long to navigate the Northwest Passage?

Could it be because of the ice and the cold?

I have no idea what the Northwest Passage has to do with circumnavigating Antarctica or with the OP.