How far away is the moon?

'bout two city blocks?

Nikon p900. I want one.

No!!

Are you being deliberately stupid? In the spherical earth, spherical sun model, the sun is 152 million Km away. It's only in the flat earth, small sun version that it's about 4828 Km away.


I have no idea what that equation is supposed to be, or where you got it from. It makes no sense at all.

What Hydra was meaning was a right angle triangle with the sun at 4828 Km height, and a distance fromt the sub-solar point of 5934 Km, you get an angle of 39 degrees.

Go to this page - [link to www.cleavebooks.co.uk] - and try plugging the numbers in...


It's rounding error. The actual angle is 39.132369 degrees (to 6 dp) and if you plug that in instead of 39.0, you'll get your original 4828 Km.


Whut?!?! Again, I can't even begin to understand your approach to the calculations here. (And why use 5734? Is that a typo? Using 5934 gives 4272 as the result, but I still don't know what it's supposed to mean.)

The point is, if the sun and moon are flat discs, as you claim, then someone viewing from 5934 Km away should see the disc as an ellipse. You haven't addressed that point at all...

0) I well know the ~150 million kilometers tale, I read about it in that lovely red encyclopedia that I received as a present on my birthday, I was about 7 years old. No, what I said, but you didn't consider it carefully, is, whether Hydra meant that the distance from the Sun's center to the infinite plane passing through the Earth's equator was 5934km. We are talking about a point's (perpendicular) distance from a plane.

1) According to you, not Hydra, a right triangle with height 4828km and with hypotenuse 5934km (your distance to sub-solar point?) should have an angle of 39 degrees?
This is not the case:

asin(4828/5934) / PI * 180 ~ 54.4deg. Hence the other non-right angle would be 90 - 54.4 ~ 35.6deg. None being equal to 39 degrees.

A further disproof is that this would imply an observer at distance cos(54.4 / 180 * PI) * 5934 = 3454km from the equator, this again is not what Hydra said:

"My home town is 5934 km from the equator."

3) Yes it was a typo. But what does it change? 4272 is about as significant as 42 here.

I maintain that all the luminaries are flatissimes disks less than a few hundred kilometers above our heads, our difficulties in providing a consistent model notwithstanding.

OK - that pretty much confirms you're just trolling.

You're deliberately using the wrong equation, and trying to confuse the issue. You should be using arctan, not arcsine, and I'm sure you know that all too well!

And you still haven't answered the problem that for the angles and distances involved, your hypothetical disc-sun or disc-moon should appear distinctly elliptical when viewed from distant points.

Anyway, if we're on the subject of mathematics, and you mentioned what you thought were inconsistencies in Hydra's calculations, maybe you can explain a fundamental problem with the calculation of the height of the sun.

One of the basic "commandments" of the flat earth belief is that the calculation of the earth's circumference by Eratosthenes, back around 240 BC can also be interpreted as a measurement of the sun's height, assuming the earth is flat, rather than spherical. The claimed height varies according to who's providing it, and seems to be between 2,000 and 3,000 miles. Your figure of 4828 Km corresponds to the higher end 3,000 miles.

However, there's a very significant problem with this claim, in that if you repeat Eratosthenes' experiment from different places, you get different results for the distance to the sun.

For example, on the equinox, we know the sun is exactly over the equator. If I go to Greenwich ("Home of the Prime Meridian") on that date, and measure the angle to the sun at local noon, I get about 38.5 degrees. Greenwich is also measured to be 5,720 Km from the equator Plugging these into Eratosthenes' calculations (simple geometry) it tells me the sun is 4,550 Km above the equator - a bit different from your 4828 Km.

Let's say there's someone in Spain doing exactly the same observation, and positioned on the prime meridian. He's at 40 degrees north, or about 4,445 Km from the equator. He sees the sun at an angle of 50 degrees, which gives a height of 5,297 Km!

There's another observer, this time on the prime meridian in Algeria, at 30 degrees north - 3,334 Km from the equator. His angle is 60 degrees, and that gives a sun height of 5,774 Km.

How can the sun be at different heights for different observers at the same time?

I can repeat the calculation on the summer and winter solstices, when we know the sun is exactly over the tropics of cancer and Capricorn. From Greenwich, on the summer solstice, the calculated height of the sun is 5,332 Km, and on the winter solstice 1,232 Km.

How can the sun be at different heights on different dates, when observed from the same location?

In essence, one of the most basic principles of the flat earth belief is complete and utter nonsense, and easily proven wrong. The fact that nobody who claims to believe in it is willing to try these kinds of simple observations proves to me they're either trolls, or following a religious belief (so unlikely to be persuaded).

Simple science can provide an answer, if you're willing to try it and learn...

Just to summarise, in case it's a bit tl;dr...

Measuring the sun's height on the equinox :-

- from Greenwich = 4,550 Km
- from Spain = 5,297 Km
- from Algeria = 5,774 Km

Measuring the sun's height from Greenwich on different dates :-

- Summer solstice = 5,332 Km
- Equinox = 4,550 Km
- Winter solstice = 1,232 Km

And these all use the same principle as Eratosthenes' original experiment, which the flat earth believers claim has been misiniterpreted, and which actually gives the height of the sun.

Any of you want to try to explain these inconsistencies?

Are you jesting or something?

Atan? You just said that the 5000+km are the hypotenuse! The other case was dealt with previously!
S
/ |
/ |
/ |
/ |
/ |
H------E

H is Hydra, S is Sun and E is equator.
As per you, SH = 5000+km, SE = 4000+km and angle SHE = 39+deg.
This cannot be!

sin(SHErad) = SE / SH => SHErad = asin(SE / SH)
cos(SHErad) = HE / SH => HE = cos(SHErad) * SH
ESHdeg = 90deg - SHEdeg

Oh no, can't do ASCII art here...

Nope!!

I think if you read back through the posts, you're the one who said it was the hypotenuse distance.

I've been following Hydra's model that the given distance was from his location to the sub-solar point, i.e. not the hypotenuse...

For honesty's sake, one of my first answers considered HE = 5000+km and SE = 4000+km and you said that it wasn't so (your sub-sun thing).

To untangle the knot:

The hypotenuse is of no interest in this case.

:fesundist:

S..A = the sun's height above the equator = 4828 km
A..B = distance from the equator to my home town = 5934 km.
This results in a viewing angle of 39.13 degrees and I would see a disc shaped sun as an ellipse.


It gets worse when the sun is above the Tropic of Capricorn:
S..A = the sun's height above the Tropic of Capricorn = 4828 km
A..C = distance from the Tropic of Capricorn to my home town = 8530 km.
Viewing angle = 29.51 degrees, resulting in an even more elliptical sun.


But how about sun rise and sun set?
When the sun is above the equator at noon at the equinoxes in the middle of Africa, it rises in Sumatra and it sets in Colombia.
The distance from Africa to Sumatra and Colombia is 14,142 km each on a flat earth (line A..D in the drawing).
Thus at sun rise or sun set I would look at the sun at an angle of 18.85 degrees - that's not an ellipse, that's almost a dash.

Fun fact:
If I look at the sun at an angle of 18.85 degrees when it rises or it sets, the sun is 18.85 degrees above the horizon.
Do we observe that the sun is 18.85 degrees above the horizon when it rises or sets?

Ooops:

:slsunset:

.

I dont know but whats wierd to me is that the moon is the exact same size as the sun from our perspective from Earth

Luck?

Pure Asimovian cohencidence!

Please, let us reason.
Suppose Earth is a globe.

"For example, on the equinox, we know the sun is exactly over the equator."

Thus the center of the Sun is on the plane passing through Earth's equator.

"If I go to Greenwich ("Home of the Prime Meridian") on that date, and measure the angle to the sun at local noon, I get about 38.5 degrees. Greenwich is also measured to be 5,720 Km"

51.4826deg / 360 * 12.742km * PI ~ 5724.6km Ok.

"from the equator Plugging these into Eratosthenes' calculations (simple geometry) it tells me the sun is 4,550 Km above the equator - a bit different from your 4828 Km."

That you see the sun at 38.5deg from the floor at noon means that the sun is at 38.5deg from the tangent plane to the Earth at Greenwich and in the plane passing through both Greenwich's longitude and the center of the Sun. (That's when the angle reaches its maximum)

But this results in, according to the prevalent system, a ridiculously small distance of the Earth's center from the Sun's center:

12.742 * .5 * cos(51.4826deg / 180 * PI) + 12.742 * .5 * sin(51.4826deg / 180 * PI) * tan((51.4826deg + 38.5deg) / 180 * PI) ~ 16418km.

So..much..coincedence....

Yep.

Some million years ago it appeared much bigger in the sky and some million years from now we won't have any more total solar eclipses because the sun drifts away from earth and thus appears smaller.

.

Another thang is how the human eye works.

You couldn't even see something as small as the Moon at a supposed 30-Earth-diameter distance.

Apologies - I seem to have gotten you mixed up with the Tunisian poster in my replies.

No, the distance SE (sun to earth) is 4,828 Km, and the distance from Hydra to the sub-solar point, HE, is 5,934 Km.

Solving for this triangle gives an angle of just over 39 degrees (and a hypotenuse of just under 7,650 Km).

At this angle any supposed disc of a sun or moon would be distinctly elliptical. This is the point the Tunisian A/C is refusing to answer or understand.

My supplementary posts addressed the problem that calculating the sun or moon's height gives different results depending on positions and times of observations. This points to a major flaw in the flat earth theory.

One hundred millions of years.

Judging the sun's location is virtually impossible, considering refraction in earth's air moisture, esp. at low angles at sunset and sunrise.

Crepuscular rays, seen from a side perspective, are more reliable when finding a right triangle. Since ACs can't embed images, it's necessary for us to paste videos...

Whut?!?!

That makes no kind of sense whatsoever!!

It's like saying you couldn't see your car from half a mile away (approximately the same subtended angle, give or take, depending on what kind of car you have).

For honesty's sake, these are computations, not sense data.
We are told that d(E, S) ~ 150*10^6km, not ~ 16*10^3km this is a 4 orders of magnitude difference.

384,400 potatometres

Look at the Sun's size, another miraculous deed of the all saint atmosphere?

Sorry I made an errot: it's not 12.742 but 12742 (Earth's diameter in meters)
It should read: 12742km * .5 * cos(51.4826deg / 180 * PI) + 12742km * .5 * sin(51.4826deg / 180 * PI) * tan((51.4826deg + 38.5deg) / 180 * PI) ~ 16 418 193.4km

Still not the 150 000 000km. Hope I didn't mess up something again.

Please, notice that the distances from the equator that you apply to the flat earth are bogus in that they are derived from the official radius of the non-existent globe + the mythical latitude. Did you actually measure said distances? Where did you get them from?

Kilometers argh! (A typo, the formula and result are correct, hopefully).
Ok, I'm outta here.

Stay away from Room 237.

parallax all objects from different places are visible @ different angle. atmosphere creates roundness of light, because in some places light reflects from the sky.