Above us: The firmament. It is there, we see it every day and every night.

I think I remember when I first saw the Milky Way in her full glory. It was in Switzerland, as I cannot see it as clear in the place where I live. due to much light pollution.Nowadays I cannot see it at all.

And it was amazing and mesmerising.

My father had been a sailor, and loved explaining the sky to me. I was told the names of Deneb, Altair and Vega that were used to navigate at night, way before there was gps. And about the different constellations to get your direction. And I got fascinated and always have been. One of the first things I did when I was in Kenya, just a few years ago, was look up. Because the sky was different. I could see stars I had not seen before; The southern cross for instance. At that time I was lucky to have an app on my phone with the night sky, which made things much easier.

And there off course are the planets and moons. Some you can see just looking up, most are indistinguishable from the stars with the naked eye.

The moon is easiest to find.. easy and fascinating in itself, as you only can see one side of the moon. How much easier it would have been if we could see the moon spinning.

And finally: The Sun, or very own star.

But what are they and what can we tell about them. This is where flat earth and global earth are different in the view of things. Because if one believes the earth is flat, stationary and the centre of the universe, then what do we see when we look up.

And looking up, how can we begin to explain what we experience: Day and Night, Polar night and days, Seasons, Eclipses. The movement of the planets, spinning themselves. And why can we only see one side of the moon from earth? And finally: Why, if we are moving at giant speeds across the galaxy, can't we see that movement. Constellations have not changed in centuries, or so it seems.

And this is the fascination every astronomer has.

So.. two models totally different.

The global model

The flat earth model

An animation of a flatearth model. This model of flatearth seems to be the most accepted version. It is based on the map created by gleason. (See map section on this site). | |

Another animation of a flatearth model, with paths of other planets explained. Note that this is not an accepted model, just an impression |

Not going into the specifics of the different earth model (this will be done in the map section), there is still the issue of what we can see in the sky. And how it fits into our view of the world.

So, on flat earth the earth is in the centre of the universe, and everything in the sky is not millions of kilometres from earth, but much closer. For instance in most models the sun is about 5000 km from earth surface (quoting the wiki "Later researchers have estimated the sun to be at about 3000 miles above the surface of the earth, with the stars at about 100 miles above that"

So there are a number of things in the sky we can observe. And how are these phenomena explained in a flat earth model.

- Sun rise and set.
- North pole star and inclination when moving south.
- South Pole area (as there is no star), but lets take the southern cross as a reference.
- Seasons
- Path of the planets in the sky.

## Sunrise and sunset

On a flat earth, the sun cannot go below the horizon. Instead, due to the effect of perspective, the sun will appear so small and low to the horizon, that light from the sun will fail to reach your point on the surface.

## North pole star and inclination when moving south.

## South Pole area (as there is no star), but lets take the southern cross as a reference.

## Seasons

## Path of the planets in the sky.

How would they work on a flat earth?

A number of ideas are based on the size of the sun, and the distance between the sun and the earth.

The distance to the sun is (according to the globe model) 149.597.870 km and the diameter of the sun itself 1.392.000 km. These are numbers that are actually mind blowing.

How do and did they calculate these vast distances, as you cannot take a simple ruler to measure it?

Distances can be calculated by triangulation. For instance

When the height of the tower is 70m, and using a theodolite (an instrument to measure angles) the angle is measured as 18^{o} than with tan(α)=a/d it follows that d=70 / tan(18) = 215,4m.

You can obviously see the difficulties. The distance to the base of the tower is actually based on the centre of the tower, below the peak.

When moving to larger objects (like mountains) it gets more difficult to measure the exact numbers, however not impossible.

So can this simple method be used to calculate the distance to the sun?

Well not immediately. First of all, we cannot measure the size of the sun. Secondly the angle you would get would be so small (the sun has a 0,5^{o }angle) the margin of error would be huge. In the picture above, using 18.5 degrees would calculate to 209 m for d.

So how to do it then? Well, you could pick two places on earth, with a known distance, and calculate the angle to the sun at those places, at the same time.. But again, the angles would be minute.. actually the distance between the points would be less than the diameter of the sun. And off course there is the curve of the earth. For every 111 km the angle would change 1^{o} , but the curve of the earth cannot be taken into account, at the moment.

We need to make the distances bigger, way bigger, bigger than the diameter of the earth, if possible.

The greeks were the first to attempt this. Anaxagoras (500 b.c.), Aristarchus (300 b.c.) used the angle to the moon to calculate relative distances and sizes. And found a relative distance of the sun being 20 further away than the moon (See the ancient experiment explained)

Along came Johannes Kepler some 2000 (around 1600) years later. He established his three laws of planetary motion (https://www.britannica.com/science/Keplers-laws-of-planetary-motion)

- All planets move about the Sun in elliptical orbits, having the Sun as one of the foci.
- A radius vector joining any planet to the Sun sweeps out equal areas in equal lengths of time.
- The squares of the sidereal periods (of revolution) of the planets are directly proportional to the cubes of their mean distances from the Sun.

These were based by observing the planets and their circumnavigation of the sun.

But even though Kepler at first could only make the relative assumptions. He knew the time it took for mars to circumvent the sun, and as this is 687 days he calculated that mars would be 1,5 times closer to the sun than the earth. The same goes for Venus, which is 0.72 times closer than earth (off course on average).

Based on those observations and the work of Kepler Cassini (1672) used parallax to measure the distance to mars. Is is good to be noted (on this site) that Cassini believed the earth was the center of the solar system!.

First let me explain parallax.

As the earth moves around it's orbit around the sun, the distant stars will appear to be stationary. At least in respect to each others position. Some stars actually change position, like Alpha Centauri, which shifts 0,75" (which is 0.000208^{o}) over half a year.

But if you watch an object (like a planet) which is much closer, you will see it moves, relative to the background. Based on the angles, one could calculate the distances, using basic trigonometry. Off course both observer and observed are moving and oh, but wait.. you still would need one distance, either the diameter of the earth's path around the sun, or the distance to the object.. so it is not as simple as it sounds, but still, one step closer...

In 1771 a French astronomer called Jérôme Lalande thought of using the sun and venus as the two objects to view, like this:

Venus sometimes (They occur in a pattern that generally repeats every 243 years, with pairs of transits eight years apart separated by long gaps of 121.5 years and 105.5 years [1])) traverses the sun, as seen from the earth.

Now the angle that has been traversed by venus across the sun is like

$$ \alpha = \sqrt{ (x)^2 - \frac { (t) * (v) } {2}} - \sqrt{ (x)^2 - \frac { (tB) * (v) } {2}} $$

References:

[1]: https://en.wikipedia.org/wiki/Transit_of_Venus

These videos I made with the program solar system scope and show from several viewpoints how Venus passes in front of the sun from june 2^{nd} 2012 until June 6^{th} 2012, during the solar / venus eclipse

These videos shows venus passing infront of the sun, from a position on earth (actually the Netherlands) keeping Venus in the focus of the view. The righthand video is without the earth blocking the view | |

This video is like the first one, only without the earth blocking the view | |

This video has the viewpoint just outside the solar system, but with the | |

Same as the above, only with real size sun and planets and distances | |

Here the view is slightly less angled. This show how the orbit of Venus is different from that of Earth, explaining why a solar eclipse is so very rare |

The greeks were the first to attempt this. Anaxagoras (500 b.c.), Aristarchus (300 b.c.) used the angle to the moon to calculate relative distances and sizes. Thinking because the moon is lit by the sun they would be able to measure the angles and therewith calculate the relative distances between earth-moon and earth-sun.

What they did was the following:

Imagine watching the moon at first quarter. The angle to the sun (to light only half of the moon) would have to be at a 90^{o} angle. The smaller the angle (a) the moon was viewed at, the closer (d) the sun would have to be. Of course the lines in this picture originate from each centres of the sun, moon and earth.

Aristarchus measured an angle of 87^{o} between the moon and the sun which results into the sun having to be 20 further away than the moon. Of course this was done without a telescope, and it was both hard to find the exact moment of first quarter, and actually measure the angles themselves. But the effort is to be applauded. Especially since trigonometry did not exist (sine/cosine or π had not been invented yet).

Using the same methods we would now be able to do the same experiment and see that the angle is in fact 89^{o}50', which means the sun is actually 400 times further away than the moon.

Another experiment Aristarchus did, was using the lunar eclipse. (edit...)

Now would this work on a flat earth. Yes, because nowhere in his experiment he was using the shape of the earth into his measurements or calculations.

But there are several things to keep in mind.

- The angular size of the moon is (about) the same as the angular size of the sun.This has to mean that the distance to each object is exactly relative to each comparative size.

If the sun if bigger than the moon, because the have the same angular size, the sun must be further away. - We need the angle of the sun at 90 degrees, but also see exactly half of the moon lit.

In the above picture, the sun has the same angular size. The angle of the moon vs the sun is 90^{o}, but the angle of the shadow line on the moon is not (b>0)

So this picture would be better:

But, lets put in the real angle for a (between the red lines), of 87^{o }and see how that works out.. Only that would never fit in a readable picture: We know the distance to the sun is 400x the distance to the moon. In this picture with the moon at 35mm that would place the sun at 140cm, not sure how that would fit on your screen...

So lets do some math to figure out this 400x multiplier

In the above picture there are 3 important (red) lines em (earth-moon), es (earth-sun) and ms (moon-sun)

We know the angles of a (87^{o}) that was measured by Aristarchus, b (90^{o}) which is the angle between earth, moon and sun, and is 90^{o} because we have a half moon.

Now In this picture the sun is set too close, therefore the dashed lines, to indicate a larger distance, the two red-dashed lines will cross at the sun's position. Also the whole system is aligned with the surface of the earth, as that will make the calculations easier to follow.

In general a line is defined by ## y = \phi * x + b ##, however em is not defined as such, we must take an arbitrary value. This does make sense, as we do not try to calculate the distance to the moon, only the relative distance between the moon and the sun.

So for es ## y_{es} = \phi_{es} * x_{es} ##, now where ##\phi## is the corresponding slope with the angle α. We now know ##\phi = \alpha * \pi / 180 = 3 * 3,14 / 180 = 0,052##, but trigonometry had not been invented yet. We will continue this line of thought for now.

For ms it will be a horizontal line (the angle must be 90^{o}), going through both the sun, and the moon, ## y_{ms} = \alpha_{ms} * x_{ms} + b##, and because ##\alpha_{ms}=0, y_{ms}=b ##, where b is not zero, but we can substitute it with the value of ##y_{em}##.

This will lead to the following

## y_{em} = y_{es} = y_{ms}##

## y_{es} = 0,052 * x_{es} ## or answering the question on what value of x, will the lines cross ## x_{es} = y_{es} * 0,052 = y_{es} * 19,1##. This implies that for every distance to the moon (em) the distance to the sun is 19 times that.

The method used however was not based on trigonometry, but using tables en euclid calculations. And that resulted not in a specific value, but an approximation of

$$ 18 < \frac {es}{em} < 20 $$