1821

Flat Earth Investigations / Re: Incredible CGI / Parabolic Flight / Space Travel

« on: August 08, 2018, 02:54:09 PM »

I challenge you at this very time to show me even five minutes of uninterrupted footage of anyone sailing around in a weightless state aboard a space craft.

Try this one.



or this





Honestly, there's shedloads of them on YouTube. You could look for yourself if you REALLY WANT TO KNOW. Do you?

1822

Flat Earth Investigations / Re: Incredible CGI / Parabolic Flight / Space Travel

« on: August 08, 2018, 02:48:56 PM »

No, it would not necessarily be an orbit as in traveling in a larger circle around a sphere.

... but all the available data and observations confirm that it is


It could simply be traveling in a large circle over the heads of people occupying land on the surface below.

So how would you go about proving that it does this?


Planes do this all the time.

It don't behave like no plane.

You write as if there are cameras on 24/7 depicting the type of scenes in question on a constant basis.

Maybe not, but anytime anyone photographs it, watches it with the naked eye, sets up a radio receiver and transmitter to communicate with the astronauts .... it's right there. Where it's expected to be. Predictable. Dependable. Never late. Never early.

Planes don't do that.  Not 24/7/365, they don't.

1823

Flat Earth Investigations / Re: Incredible CGI / Parabolic Flight / Space Travel

« on: August 08, 2018, 02:43:26 PM »

Everyone on your side of the aisle clearly states the object in question is in a constant state of such free fall. Any humans occupying such a craft would be in the same constant exhibition.

Exactly. They call it weightlessness.

That's why there is hour, upon hour, upon hour, upon hour of footage of astronauts sailing around spacecraft cabins, unfettered by gravity, as well as hour upon hour upon hour of them directing various objects and substances around the cabin, all of which behave in ways that would not occur at ground level.

1824

Flat Earth Investigations / Re: Incredible CGI / Parabolic Flight / Space Travel

« on: August 08, 2018, 12:48:10 PM »

Ever occupy a car that speeds over a cresting hill and then suddenly drops? Your hair displays levitation and you feel your gut rise up.

Yes, but how long can you sustain that for?

1825

Flat Earth Investigations / Re: Incredible CGI / Parabolic Flight / Space Travel

« on: August 08, 2018, 07:35:35 AM »

For some reason NASA is an expert consultant at Hollywood special effects for space movies:

So what?

If Hollywood is making a movie about (say) a concert pianist, they get a pianist to coach the actor(s). If any movie has a specialist subject, they get specialists in from that field to advise.

What's the big deal here? They asked NASA WHAT THEY DID so they could get the film closer to it.

1826

Flat Earth Investigations / Re: Incredible CGI / Parabolic Flight / Space Travel

« on: August 08, 2018, 07:32:30 AM »

NASA only really needs a small zero-g room or container and they can just superimpose the water effects, floating pen effects, or human effects onto footage of the bigger sets.

Considering the many videos pointing out the harnesses, green screen effects, AR, and bubbles-in-space, they are likely using multiple methods.

Supposition. "Could do this", "Likely using that" ....

There may well be many videos, but look closely at the comments on such videos, and you'll find a horde of people who don't agree that the "pointing out" is proof of anything....

Small zero-g rooms don't produce an object that can be seen from Earth, in isolation, or in transit over Moon and Sun. Small zero-g rooms don't produce radio signals which can be received and responded to by radio amateurs on Earth. etc. etc. So many proofs that are not negated by speculation and supposition.

1827

Flat Earth Theory / Re: The Crescent Moon

« on: August 08, 2018, 07:03:53 AM »

I'm Australian and haven't fallen off yet, so what do you say to that

I say you should post a picture of the Moon taken from Australia. Let's compare it with one taken from the Northern Hemisphere.

The thread topic IS the Moon, after all ...

1828

Flat Earth Theory / Re: The Crescent Moon

« on: August 08, 2018, 06:29:33 AM »

Model it. Use the Round Earth Distances to explain it.

What is there to model?  If you tilt your head/camera 45 degrees to the right while looking at something (insert distance of your choosing here) away, and then tilt your head/camera 90 degrees to the left while looking at the same object, it will appear to tilt, or rotate if you prefer, in your field of view.

I started down the path of setting this out, step by step, for Tom, but we faltered at the first or second hurdle;

If you were to the rear of all the observers, looking past the Earth toward the Moon, their views would vary, because of the difference in their LATITUDE. The observer at the equator will have the crescent toward the bottom of the Moon, and the observers at 45N or 45S will have it inclined to this aspect by 45 degrees each.

I don't see how that makes much sense. That doesn't use Round Earth Geometry. In fact, it seems to be assuming that the moon is very close to the earth.

.. despite further explanations, Tom seems set to abdicate from this thread.

1829

Flat Earth Theory / Re: The rotation of the sun

« on: August 08, 2018, 06:24:26 AM »

The speed of a needle on a record on a record player also changes when it is closer or further from the center. Would you say that the record needs to "change speeds" to achieve that?

Needles died out with acoustic gramophones.

The stylus on a modern record player does experience a speed change in terms of linear velocity as it moves across the record.

The rotational speed is constant, at 33.33 RPM, but the distance travelled in each revolution varies.

So, in the diagram referenced, what do you think is going on? It shows the Sun in two circles, one smaller than the other. Do you think the rotational speed is constant? Once per year? If so, the linear speed must change.

Or do you think the linear speed stays the same? In which case, the rotational speed must alter.

Or do you think the Earth is rotating under the sun, as others have asked? If so, at a constant linear speed, or rotational?

1830

Flat Earth Theory / Re: Why would the government lie about the earths shape

« on: August 08, 2018, 06:17:24 AM »

I more or less agree, with the exception of the guy with the gap in his tooth.

But that's a new topic. The topic here is the shape of the Earth, and the truth or not of whether or not Challenger was staged has absolutely no bearing on this.

The earth is portrayed as round in NASA media because NASA thinks it's round.

It's portrayed as round in Russian, Japanese, European, Chinese and Saudi-Arabian media too. As I said elsewhere, simplistic claims about "NASA lies", "NASA fakery" etc. don't hold water when only a small portion of the media and data is actually generated by NASA. Loose claims about NASA having multiple contractors in order to expand the 'conspiracy' to others don't hold water either.

It was portrayed as round in media hundreds of years before NASA was even thought of, too ....

1831

Flat Earth Theory / Re: The Crescent Moon

« on: August 07, 2018, 08:29:52 PM »

I didn't change the time, I just changed the location. If Mooncalc changed the time, then I wouldn't be surprised. It is an incredibly buggy web application.

By all means, list all the other bugs that you have found, IF you have found any.

It expects you to enter the LOCAL time for the place you're interested in. Why would anyone want to look at the Moon in Perth, Australia, for a time in London?


I've shown that it cannot be the moon that is rotating.

Did anyone claim that it was?

The only out is that it is the earth that is shifting.

Or it's the orientation of the observers ....

It doesn't seem that the Mooncalc can reproduce the Moon Tilt Illusion.

Did anyone claim that it could?

This thread is about the crescent moon. Moon Tilt Illusion is in another thread....

1832

Flat Earth Theory / Re: vsauce on flat earth

« on: August 07, 2018, 06:36:20 PM »

Yes, but what ABOUT 3.20 ... ?

1833

Flat Earth Theory / Re: The Crescent Moon

« on: August 07, 2018, 03:28:15 PM »

Do you know what you have not taken into account in your cartoon?

The Sun.

Create a proper scale-model figure which includes the thing that is illuminating the Moon in the first place (that would be the Sun, Thomas), and re-submit.

In addition, whilst he may have indicated the POSITION of the observers, he hasn't made any allowance for their orientation ...

1834

Flat Earth Community / Re: Global Positioning System

« on: August 07, 2018, 01:48:23 PM »

These chaps verify the height of various orbital satellites by bouncing lasers off them. Do you have any reason to disbelieve them?

http://sgf.rgo.ac.uk/

They might not be doing this for the GPS satellites, possibly, but they've accumulated a shedload of data on others.



"with all you round earthers on this forum, you would think a few of you would team up for a fun exercise to do this...but nothing."

And do ... what?

1835

Flat Earth Community / Re: They plan to remove windows from passenger planes

« on: August 07, 2018, 12:25:39 PM »

Go ahead, watch it again...

Done that. I uploaded a YT video which compares what was seen from the Tesla with the output from the Japanese Himawari-8 weather satellite, and - guess what? - it matched exactly.

I've also seen the output from more than one astrophotographer who captured it, exactly where expected, once it was out beyond the Moon.

Apart from scepticism and disbelief, what else do you have?

1836

Flat Earth Community / Re: They plan to remove windows from passenger planes

« on: August 07, 2018, 11:57:13 AM »

You are now claiming a sphere of supposedly 25000 miles in circumference can be depicted from a camera positioned anywhere from 190km to 7000 km?

You don't see a sphere, you see a Spherical Cap of varying size, depending on your distance from it.

At best, far from the sphere, you see close enough the hemisphere nearest to you. You might see 179, 178 degrees of the 180 in a full hemisphere.

Get closer, and what you can see reduces.

https://en.wikipedia.org/wiki/Spherical_cap



As a for instance, if we assume the craft to be on a continuation of line h, then the limit of visibility will be the rightmost end of line r, and the matching point on the opposite side of the plane which forms the base of the cap.

You can see how this will vary with the length of continuation of line h, can't you?


Also;

https://www.metabunk.org/debunked-blue-marble-photos-show-a-changing-earth.t6616/

See the illustration immediately after "You can verify this yourself with a household globe and your eyes (or a camera)" ...

1837

Flat Earth Community / Re: They plan to remove windows from passenger planes

« on: August 07, 2018, 10:32:43 AM »

... Because if plastic Tesla car suffered no damage at all in the deep space

You didn't see the Tesla in "deep space". The cameras were on whilst it was in Earth orbit, minimum 190km, max 7000km approx. It was in Earth orbit for barely a couple of hours, and spent one-third to one-half of that time in the dark, on the night side of Earth.

I'm not surprised there was no damage evident in the footage. It hadn't had enough time to get damaged.

LOL!!!  You are stating the Tesla was an altitude less than that of the ISS yet provided us the footage presented?

LOL Yes. Why not?

minimum 190km (over USA and thereabouts), max 7000km (over Australia and New Guinea). Way higher than the ISS when over Australia.

What do you see as the issue with the footage?

1838

Flat Earth Theory / Re: The Crescent Moon

« on: August 07, 2018, 07:49:03 AM »

While researching this, I found several other inexplicable Round Earth items.

- The Moon is above the horizon for observers on the Prime Meridian and the Antimeridian simultaneously, when it should not be

- The Moon is pointing in the wrong direction on the horizons of the PM and AM. While the Moon will never "tilt" or "rotate" in the Round Earth model, it is possible to look at it upside-down when standing on opposite sides of the earth.

- The Moon is above the Western Horizon for observers on the Prime Meridian and the Antimeridian, when it should be on the Western Horizon for the Prime Meridian and the Eastern Horizon for the Antimeridian

Here is what the Mooncalc shows for the Prime Meridian and the Antimeridian:

IMGs

You've shown 17.58 for both, but the Antemeridian is 12 hours ahead or behind this. This is local time.

If it's 17.58 at Greenwich, and the antemeridian point is 12 hours ahead, then it's not 17.58 there. It's 05:58 on the next day.

You've shown how the Moon looks for Greenwich, and how it looked for Australia 12 hours earlier.

Put in the correct times, and the arrows you've put on the Moon points will point from opposite directions.

You need to account for the time differences, as I did in Reply #22

1839

Flat Earth Theory / Re: The Crescent Moon

« on: August 07, 2018, 07:24:09 AM »

If we could look at these lines of sight from a position out in space behind the observers, not in side, plan or underside views, they will look like this




We're roughly aligned with the central observer (who we've taken as being on the equator), and the lines of sight from 45N
and 45S would start out 2799 miles away from this observer (at each end of the chord which has just been calculated), and
converge on the Moon, 240k miles away.

The Moon is decidedly not close to the baseball.

Tom, do you agree?  Y/N

 

1840

Flat Earth Theory / Re: The Crescent Moon

« on: August 07, 2018, 07:05:11 AM »

This is the first step of modelling this with RE distances.



The three observers are at the points indicated by the green squares. Their lines of sight to the Moon (orange) are not to scale, the Moon would be far over to the left.

If the diameter of Earth is taken as 7926 miles, then the radius is (7926/2 =) 3959, and the length of chord C (the orange vertical) is 5598 miles, using the formula on this page ("So the length of the chord is: ")

https://www.ck12.org/trigonometry/length-of-a-chord/lesson/Length-of-a-Chord-TRIG/



If the length of this chord is 5598, and we take the distance from the 45N and 45S observers to be exactly 240k miles to the Moon, that gives us an isosceles triangle, or two right-angle triangles back-to-back, one with hypotenuse to 45N, one to 45S.
Let's solve for the angle at the Moon for the right-angles, and double it.

Right-triangle calculator;

http://www.cleavebooks.co.uk/scol/calrtri.htm

Side a = half the length of the chord, so 5598/2 = 2799, and side c = 240,000. Angle A = 0.668 degrees, so the angle at the Moon made by taking the two lines of sight from 45N and 45S is 2 * 0.668 = 1.336 degrees.

This gives us the angle at the Moon made by taking the difference between lines of sight from 45N and 45S (indicated by two orange squares).

The chord length will not vary, as long as the observers are at 45N and 45S, they are on the side of the Earth facing the Moon, and we take the distance to the Moon as a fixed 240k miles (I'm excluding any allowance for Earth's axial tilt at present, and discounting any change in the 240k as they move nearer or further to the Moon along the line of latitude)



Tom, look at the above and tell me/us - do you agree?  Y/N

Do you also agree that this is another way to calculate the angle that you were trying to calculate with your huge circle around the Moon, isn't it?