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Flat Earth Theory / Re: Wiki - Tom Bishop Experiment
« on: May 26, 2019, 03:40:34 PM »
I need to admit, there are several similarities on both texts, the use of same words and figurative expressions.   I also question about being flat on the ground on stomach, with a good telescope of 500x magnification? Telescope body was touching the ground?  I would not dare for the world to approximate my "good" optics from beach salty sand spray, even over a towel.  People don't do that even with photo cameras.  Tiny salty sand particles in suspension will stick to the lens, it needs special washing solution afterwards to remove it without scratching the lens coating, irreparable damage.  Of course, an astonishing finding like that is like front facing a hovering alien spaceship, when you knew it was there, took a telescope to see the aliens better, but simply forgot photo camera to take some evidencial pictures.   If I would experience such features, I would become millionaire for taking only 10 good focused sharp pictures, each with a different eyepiece to prove the sequential objective lens optical resolution, camera date/time on pictures and a GPS compass to prove my position.  Anyone would admit, a picture from domestic telescope showing kids playing with frisbee across a 48km patch of water would worth a fantastic reputation and at least good money.  Not even talking about the water spray, waves, evaporation, moisture in the air.  Of course, a picture is better than a million words. 

I recalculate here, to get into the optical resolution and visibly discriminate a very tiny person from a light pole 48km away, you need to "bring" that person close to 200m at naked eye, this means a minimum magnification of 240x, using a smaller possible eyepiece of 9mm it will require an objective with focal distance of 2160mm, being refractor telescope, it will have a body length of more than 90 inches (2.3m), with a minimum aperture of 150mm (6"), 1 arcsecond resolution, focal ratio f/14, that would be a really heavy and bulky tube.  The best powerful refractor Celestron produced recently was the Advanced VX6", just the tube is 8.3kg (19 lbs), computerized, $1500+, even so the objective focal distance (1200mm) is too short for this endeavor, it will need a 5mm eyepiece, trust me, you didn't have that.  I know dozens of astronomers with green expensive optical equipment, only few have an eyepiece like that, last weekend my neighbor Barney bought a 4.7mm TeleVue Ethos SX, cost more than $600.  This eyepiece with low focal point are only used in deep space observation, something that I guess Tom is not really interested.  The most common eyepieces delivered along with regular telescopes are 24, 25 or 40mm, even with the VX6 it means an image magnification of 50, 48 and 30x.  The 24mm would bring the kids playing frisbee to 960m (2880ft or 6/10 of a mile) at naked eye, think about it, can you discriminate a frisbee at 6/10 of a mile away? that is around 9 city blocks away.  Really, a picture taken directly at the ocular would be fantastic.

Flat Earth Theory / Re: Wiki - Tom Bishop Experiment
« on: May 25, 2019, 07:59:47 PM »
>>Thermal distribution north and south of Equator?
Common misunderstanding. It's not equal. The SH is hotter.

How come SH can be hotter in FE, if the land area the Sun must cover during its same 24h is much wider than the NH, where the radiation concentration  per km² would be higher?

Total FE area = r²xPI = 20000²xPI = 1256636000 km²
NH area = r²xPI = 10000²xPI = 314159000 km²
SH area = TotalArea - NHArea = 1256636000 - 314159000 = 942477000 km²

Total FE Area = 1.256 E+9
NH Area  = 3.141 E+8
SH Area  = 9.424 E+9

FE SH area is in fact 30 times larger than NH
How come SH can become hotter than NH, or even the same?

For FE SH temperature to be the same as FE NH, the FE Sun would need to be 30 times hotter in January.   If you are referring to perihelion, when Earth is closer to the Sun in January, it is not very significative, as a matter of fact, the atmosphere temperature is opposed, it even helps FE with 4°C...  ;)  when global temperature is even lower in January.  So, the FE solar temperature is not logical, not true to the real thing.

For the ones that didn't get it yet, if you pass your hand very close over a candle flame it may burn, or not, it only depends on how fast your hand moves.  In the FE January, the Sun needs to cover 30 times more km² per second than it covers the Northern Hemisphere in July, so it moves faster, radiating less energy per km² to the land on January than on July when it moves slower.  Even needing to cover 30 times more land, the land temperature is almost the same on both hemispheres.

Lets calculate:
FE say on July the Sun is circling Earth every (lets round to) 24 hours, right over the Tropic of Cancer, at 23°26' North.
FE say on January the Sun is circling Earth during the same period of time, over the Tropic of Capricorn, at 23°26' South.

Considering FE disc to have 180° from North Pole to Ice Wall:
Tropic of Cancer is at 23°.26' (23.4333°) North from Equator, means 66.566° from North Pole.
Tropic of Capricorn is at 23.4333° South of Equator, means 113.433° from North Pole.

The circumference the Sun must travel when over the Tropic of Cancer on July will be the radius x 2 x PI.
The Tropic of Cancer radius is FE radius x 66.566/180, 20000 km x 66.566 / 180 = 7396km
Tropic of Cancer circumference = 7396 x 2 x 3.14159 = 46471 km.

The circumference the Sun must travel when over the Tropic of Capricorn on January will be the radius x 2 x PI.
The Tropic of Cancer radius is FE radius x 113.433 /180, 20000 km x 113.433 / 180 = 12603km
Tropic of Capricorn circumference = 12603 x 2 x 3.14159 = 79191 km

the speed of FE Sun rotating over FE Tropic of Cancer is 46471 km / 24h = 1936.3 km/h
the speed of FE Sun rotating over FE Tropic of Capricorn is 79191 km / 24h = 3299.6 km/h

The speed difference is 3299.6 / 1936.3 = 1.7x
(I will not even question what makes the Sun accelerate or break speed and change circling diameter)

Means, the FE Sun runs 1.7x faster over the Tropic of Capricorn than over the Tropic of Cancer.
It simply means that the Northern Hemisphere solar speed would spread 70% more radiation per second than to Southern Hemisphere.
This is not true in the real world, the NH is NOT 70% hotter than SH.
Also, the above calculations should take in consideration just radiation per second, not radiation per squared area of land.

On the top of this post, I considered squared area land, and came to 30x less radiation per km² in the South than in the North.
What it is again, not true in the real world. 

The actually measured solar radiation energy in average, between tropics to be 1kW/m². Based on the 70% difference of solar radiation per km² on FE, if measured 1kW/m² in Rio de Janeiro (what is real), then a person living on Central Florida would receive 1.7kW/m², that is not true, it will be cooking everything on land.

Other important thing, consider the sun speeding 1.7 faster on the Tropic of Capricorn, it means people on the Southern Hemisphere would notice the Sun moving 70% faster on January sky, shadows on ground moving 70% faster, etc.  This is also not true.

Based on FE perspective, vanish point and "can not see far due atmosphere not being transparent", the Sun would disappear from southern sky on January 70% faster than on July on North.   If on the North we can see the Sun during 12 hours on July, then on the South we will see the Sun only for 3.6 hours on January, what is also not true.

There is a lot of "Not True" on the text above.
Thanks for bringing this subject to my attention.
This is one more item FE must address and explain on Wiki, will be very difficult if not impossible.

Below a comprehensive FE map about tropics circumferences and Sun's speed over them

Below a FE horizontal view in two situations, Figure1 and Figure 2 are proportional to FE diameter, Sun's altitude, tropics diameter.

Figure 1 is the Sun circling over Tropic of Cancer (July), smaller diameter. 
Figure2 is the Sun circling Tropic of Capricorn (January), larger diameter.   
Both observers, living under such tropic can see the Sun raise from the horizon and set at the horizon.

Based on FE mechanics, the Sun at 4800km of altitude never really gets to touch the horizon, but FE explains the observer actually see the Sun setting and disappearing below the horizon due perspective, vanishing point and atmosphere not being transparent, and also possible use of atmospheric refraction, not allowing to see far. 

Based on this FE statements the Figure1 allows to understand the observer can only see the Sun on sky within an angle of only 113°, below that it will be under the horizon. But Figure2 with the Sun at the same altitude, the angle is much larger, 139° for the same effect to take place.  The only possible FE explanation is that on the southern hemisphere you can see farther, perspective and vanishing point works differently, and the atmosphere is more transparent, with less refraction.

The other complication is about the visible time of the Sun during the day.  Both observers can see the sun more than, but lets assume only 12 hours on both tropics, Cancer on July, Capricorn on January.  Now lets divide the angular view of the sun in the sky during 12 hours.   For the observer of Tropic of Cancer, the Sun travels 113/12 = 9.41° per hour in average, but we know that is not true, so it must be 9.41°/h from 10am to 2pm  and fantastically speed up at raise and set to reach the apparent position close to horizon.   For the observer on the Tropic of Capricorn, will be 139/12 = 11.6°/h, with also non linear angular speed during the day.

Interesting fact that on RE we can actually measure a very linear and steady solar angular speed of 15°/h all the way from raise to set, and the solar visibility on both tropics are the same, not different visibility angles (both are 180°), not different atmospheric visibility.

I wonder if FE could produce and post explanations on Wiki, using at least images better than mines, with scientific facts, numbers, angles.

Just to clarify, the observers on Figure1 and 2 are not on the North Pole, they are somewhere over its tropic circle.   The three suns over the observer represent the sun at 6am, noon and 6pm.

Also, we can actually predict on RE, exactly when the Sun will raise and set, within seconds, due its extreme easy mechanics of a round Earth and heliocentric system. 

The FE explanations about the sunrise and sunset rely on atmospheric conditions of visibility and refraction.  All of this can change by temperature, moisture, cold/warm wind and climatic conditions that can change at any time, would dramatically change the position of the FE sun on the horizon, not being possible to predict with precision when it would happens.

I also would like to hear from FE how those sunrise and sunset can be precisely predicted under so much optical challenging situations. 
This whole post is completely relative to Tom's statement that SH is hotter, I proved mathematically above FE SH is 30x cooler per km², not hotter.

I sincerely expect someday for some FEr to answer those questions, since I feel I am writing to a brick wall.   


Flat Earth Theory / Re: FE Wiki - Optics
« on: May 25, 2019, 04:59:49 PM »
Tom, when I said thick plate glass, it might confused you.
You are using the example (as below) of the bear behind a tick plate glass filled with water, and the bear is inside this medium.
This is a perfect example of "shape", like a lens.
Any angular difference from 90° from the lens of the camera, the plate glass and the position of the bear inside the pool (image incident rays of light), will change the flat shape of the object.
That is how a glass prism works, it needs the rays of light to be out of 90° angle in order for the high frequency wavelengths to bend sharper than the long wavelengths.  If all the photons hit the difference density medium without any angle, no bending will happens.

On the image below, see how the violet light (higher frequency wavelength) bends more pronounced.  Higher density promotes higher refractive index.

Lens magnification works the same, different angle of incidence in a different medium density, cause rays of light to bend to different directions, it can produce magnification, reduction, color separation, etc.  Again, it is not density that change apparent size of an object, it is the angle.

A flashlight face on (90°) to the window glass will not have its light rays bent in anyway, it will enter and exit the glass (denser medium) without any refraction, the only thing that will happen is a little delay (in time) for the photons to get out of the glass.

If you bend the flashlight in certain angle to go across the window, then a certain light rays bending will happen, since the rays of light are hitting the different density medium in an angle.

Let me explain why this happen:

When face on (90°) all the wave oscillations hit the glass at the same time, even with different phases of the different wavelength.  All the waves have a propagation time delayed, light speed still the same, but it has more difficult to propagate in a straight line, so photons start to colide inside the denser medium and change path, like zig-zagging in wavelength distances.  The photons momentum and the alignment of the waves when hitting the medium does not allow them to change direction in a large scale, so they bounce and take a longer path to do the same straight path they had outside the glass before entering it.  This longer path, same speed, means longer time to exit the medium.  But as they entered the glass perpendicular, they will exit perpendicular in the same angle, no refraction happens.

When entering the glass in an angle, lets imagine 45° for easy understanding, the shorter path of the light wave will hit the glass first. That side of the photons will break travel speed by starting to bounce inside the denser medium before the other side of the wave that still outside the glass by the angle of penetration.   The pronunciation of this "speed brake of photons travel time" depends on two variables, a) the frequency (and energy) of the penetrating photons, and b) the density difference (delta) between both mediums the photons is crossing from and to.   If going from air to glass in an angle, all the photons will bend direction towards the side that hit the glass first. If going from glass to air, the bending is exactly the opposite, photons speed up travel time at the side of the wave that hits air first.  High frequency waves, meaning green to blue and violet will bend more pronounced than orange to red.

This is exactly what happens with any glass lens, the form and shape of how the light enter and exit can produce all kinds of refraction. Knowing this, we produce lenses for our specific needs.

On the below pictures, if the observer and his camera moves right straight to 90° angle from the plate glass AND the bear, no light bending would happen, the bear will be visible as natural, no changes in image, size or position.

See, when I say "shape" it means mostly angles.  Density is just the raw material, the instrument is the angle in how it is used.

On the picture below, by the position (refraction) of the bear image, I can say the observer and his camera was to the right of the 90° alignment to the glass, as if its right eye was closer to the glass.  The photons of the bear image that hit the camera, came through water and glass, when they change medium, glass --> air, the right side of the waves (in the picture) exit first, they accelerate the photons travel speed and bend dramatically to their left (right of the picture), hitting the camera. As the image is composed by the width of the bear, the left side photons from the bear (in the picture) would change medium with a less pronounced angle, they are around 80cm or more to the left in the glass, so the angle they move into the air is different from the ones at the right.  The angle of acceleration will be depending on such angle of medium crossing, it will be less pronounced.  It works exactly as a magnifying glass, the bear may appears out of position and magnified.   Here, the shape of the angles between observer, glass and bear, produces this effect.

Saying that a denser medium will magnify an object is wrong, if depends basically on the shape and angles.

By last... if you are inside that water you see no changes in the image at all, since your eyes (or camera) will make part of that medium.  We, inside the bottom part of the Earth's atmosphere are deeply immersed into the denser air medium.  The only effect we can see from the Sun when closer to horizon, is the reducing of the colors of faster wavelength, blues and violets (are bent first and disappear from us), this is why we see it more redish.  But no change in size at all.   The same "color refracting effect" can be observed on the bear below, note the more refracted image is more blueish,

Observe the second image, the left side of the bear (tail) produces more refraction on the glass-->air than its head, this is based on the pronounced angle from the camera to the glass, than head.  Note that there is a refraction magnification that increases linearly to his tail, what proves my text above.  It is not only the density, it is the angle of incidence.  The bear's head, glass and camera have a better alignment to 90° than its tail.

On this image below, because the half bottom part of the glass is not round, but faceted acting like a convex lens, you can even see the straw thinner, all due angles of crossing different mediums.

Optical refraction is a vast and very interesting area of study, since you can in fact test, exercise and see results on a simple kitchen table or better prepared laboratory.  It is a rewarding study, the beauty of light rays and its physical properties fascinates all students.  Low cost tools can teach you wonders.  In all the school laboratory exercises, optics is the one that more attract students. In fact, it is a great area of work in the industry, great compensation for good professionals.

A low cost portable instrument known as "refractometer" measures the concentration (%) of solids in liquids, just using the ambient light refracted in the crystals, for example used to measure the percentage of sugar in soft drinks and other lab analysis.  Just put few drops of the liquid under the plastic lid and see two different colors bands on the viewer, showing the percentage of solids.

Flat Earth Theory / Re: FE Wiki - Optics
« on: May 24, 2019, 09:15:13 PM »
I didn't write the passage. But:

In the straw refraction pictures the straw magnifies. The rays start from a high density environment (water) and move into a low density environment (air). Presumably if the opposite occurred, light moving into higher densities, the straw would shrink.

Sorry Tom, the object appears deformed by the shape of the denser medium, not only water, but the glass. 
In optics study, it is the shape of the different medium density that refracts the photons, not only the density itself.
You can have a very tick glass at your window, if it is a flat plate, no deformation would be noticed on the external image for an internal observer.
You are using the shape of the glass to induce the thought of size change just based on density, not true.
There are plenty of scientific optical explanation about this on the internet, that you can replicate at home, if you want or need, I can point some to you.

Flat Earth Theory / FE Wiki - Optics
« on: May 24, 2019, 08:28:23 PM »
"Horizon Limits with Refraction and Opacity
Horizon limits are easily explained by the fact that air is not transparent and refraction. As light travels through a denser medium, the object will appear to be smaller because light is refracted towards the normal. Furthermore, air is not transparent so it is not possible to see past a certain distance."

I am particular interested in someone that created such text above, to explain to me the science behind such (underlined) statement.
I can push an image photons through a denser medium, for instance a glass lens, and make the object appear bigger. Your wiki statement is incorrect.
Also, what is the meaning of "because the light is refracted towards the normal"? what is "normal"?
To finish, what is the "certain distance" in kilometers that it is impossible to see through because air is not transparent?
Where all this information came from? it is the wild guessing of somebody or it is multiple times science lab tested, duplicated, recorded and published?
Please provide source evidence.

Flat Earth Theory / FE Wiki - Travelling East
« on: May 24, 2019, 08:10:02 PM »
On FE Wiki:
"As it happens, on a Round Earth you do not travel perfectly straight when traveling East or West either. Consider this thought experiment:
You are on a Round Earth standing 10 feet away from the North Pole. You are then directed to travel East and are instructed to continue to do so. What happens to your path? You end up traveling in a circle, and not in a straight line that you previously thought you would.
The exact same thing happens regardless of where are you on the Flat Earth. Your path will not be straight without you having to constantly change the direction you are traveling in reference to a compass

Except of course, if you are over the equatorial line. 
On RE you don't make any turn, you just walk straight to East. 
On FE you NEED to continuously make a left turn of 0.0057°/km (0°00'20"63/km) walking East.
That is a big difference.  So, the wiki is not correct and it is a misleading.

Also, even if you walk in a very straight line on RE, and even starting form the North Pole to ANY DIRECTION, you will end up in the same place you started after make a round trip around the globe. On FE you WILL end up hitting the ice wall, no matter where you start and the direction you take.

Reer, that is the thing.
"Cows are made of milk", now lets try to justify this statement as much as we can, and ignore the unanswerable.

Flat Earth Theory / Re: Wiki - Tom Bishop Experiment
« on: May 24, 2019, 05:32:33 PM »
Based on lots of controversy, discussion and unreliability caused by light bending over different media density (over patches of water), ships disappearing over the horizon and such, what creates more questions than answers, I created a much more reliable experiment what I would recommend to append to Tom Bishop Experiment.  It is much more clear and concise, much less variables and doubts.

This suggestion is based on what much strongly changes from the RE to FE, the equatorial line circumference.  On RE the circumference plane is perpendicular to North Pole, it literally divides the North to South hemisphere, hypothetically you could walk straight over this RE equatorial line without making any turn left or right and end up in the same location after a very long walk.  On FE this equatorial line circumference plane is a horizontal circle on the ground, it is a flat surface, to walk over this line you need to keep turning left in a very long circle if walking eastward.  It is a completely different shape and that is what basically defines RE or FE.

It is very easy to check and verify if it is one or another, without any confusion or mishap, it shows clearly the results with a simple pole shadow line angle in reference to North Pole, as proposed on the exercise I posted on , specifically asking Tom Bishop to answer, he just sent me to read Equinox on Wiki, as if would answer my questions, obviously it does not. 

Again, I kindly ask Tom Bishop to answer, as a FE representative with significance in this forum, with a straight and direct answer about the angles from the exercise. It is indeed a simple geometry calculation, and will finish this for once and for all.  Also, if he wants, he could append this exercise to his own experience on wiki, or, create a Spherical Exercise on FE wiki, I would appreciate very much.

Flat Earth Theory / Re: Wiki - Tom Bishop Experiment
« on: May 24, 2019, 02:48:35 PM »
Aristotle (not Aristotile in English) and so many others used the tools they have available at the time.
Whenever technology advances we have better tools to help us to understand and live over this planet.
We first navigate by the stars, then we found the use for magnetic compass, now computerized maps and GPS.
We don't fight technology advances, we never more need to seek wood into the backyard to make a coffee, we just press a button.
We don't need to set the saddle over a horse in order to be able to deliver a letter, we press "SEND" on the browser.

Our actual tools to make the same Aristotle experiment are a little bit more advanced, like satellites and very precise atomic clocks.
For refusing such tools, one should also refuse all other new tools, like Internet, cell phone, computers, a/c, cars, fridge, freezer, coffee maker, electricity, medicine.
You can always choose to go back to the cave world.

Ok, I'm getting it too what Tom doesn't get. If we imagine to watch Earth from the North in Space, and imagine it as an immovable round planet, the Sun would circle around it in 24 hours, but the Moon would be slower, doing it in 23 hours and 10 minutes (they'd be both travelling incredily fast in Space). During an Eclipse, the shadow on Earth would be caused by a faster Sun, making its direction going from East to West.

And of course, this is not what happens (23:10h is faster than 24h, not slower).
Even in FE, the Moon does not make a full turn in 23 hours and 10 minutes, it will be slower (as you said), will take 24 hours and 51 minutes.
But this has nothing to do with the movement of the eclipse shadow time and direction over Earth.

Like I post before, on RE the Earth could be stopped, the eclipse shadow would move eastward at 1002 m/s, moon orbital speed.
Even in FE, the Sun and Moon moves westward (east->west), as the sun moves faster, the moon shadow moves west->east.
It is really difficult to understand why this topic took 8 pages already.

Some people are confusing what they can understand, with simple orbital mechanical optical facts.

I use to say to my students, by simple proven fact water is a liquid, you can lucubrate, discuss, theorize, make papers, try to disprove, water calmly will continue to be a liquid.

Flat Earth Theory / Re: Wiki - Tom Bishop Experiment
« on: May 23, 2019, 09:38:24 PM »
You are talking about something that occurred 12 years ago. I no longer live in that area, nor do I have the telescope. It was a refracting Celestron that was advertising itself as 500x equivalent. The experiment was conducted from several different spots in that area.

Tom, I have several telescopes, Celestron CPC800 and 1100, Meade LightSwitch and LX200.  To achieve 500x magnification with humanly visible optical resolution, you need a huge aperture, never produced on a refracting unit for popular use.  It would need to have a more than 10 inches of refracting glass as objective, you will not be able to carry it handy.    Also, the smaller eyepiece you can see anything from 12 years ago technology is no less than 6mm.  To have a 500x magnification such refractive telescope must have an objetive with a focal length of 3000mm (3 meters long), the whole telescope with the focuser would be longer than 3200mm. Celestron never produced such best. 

You may had a 50~70mm objective refracting unit, handly transportable, 800 to 1000m long, but it needs a nice mount (tripod), you can not use it on the floor. Somebody may stick a label with "500x" on it, but no cigar for that. That unit can give you a maximum of 200x magnification in the limit of optical resolution.  Optical resolution formula is d/1.22Lm, where d is aperture and Lm is the green light wavelength, often used on astronomy calculations. Regular 15mm~24mm objectives (popular for that telescope) can give you a somehow visible image with 40x~60x magnification.  With that magnification, a 2m tall image at 38km distance can be seen with an aparent size of around 0.02°, what is 1/20 of the size of the Moon.  It will be like watching the details of lunar crater Langrenus by naked eye, or a person as seen by naked eye 633 meters away (38000/60).  I need to admit, based on naked eye observation I can not even tell if there is a person 6 city blocks away, not even a car. 

A 22cm ball (or freesbe) has an apparent size of 1 arcsecond at 46.5km away, that is 1/1800 the size of the Moon.  My CPC1100's aperture is able to discriminate 0.5 arcsecond, so it in fact start to lose optical discrimination of a freesbe at 90km away, image fuses with surrounding photons. It means you can not recognize it as a freesbe, the wavelength of the image is higher than the size of the object, you just can actually see a different brighness fuzzy thing, nothing else.  Considering a person has similar size head, you can not even say if that fuzzy dot on top of a very tinny little stick is a person's body or a lamp pole, and that with my CPC1100  (11" mirror, schmidt cassegrain, 2800mm focal length, 85lb of weight, resolution 0.5 arcseconds, $3k) without any extra features.  Over many kilometers of water, moisture a lot, waves spraying it becomes really difficult.

Next time take pictures, from the telescope, from the scene, from the scene through the telescope with different eyepieces so you can have a progressive image magnifications.

There is an easy experiments, not even need image or photos, it uses three lasers, two powerful (minimum 1W units) red and one green lasers.  I wonder why nobody made it before.  Mount them side by side over a wooden base, green in the middle, with screw for vertical alignment.  Shot them at night over dry land against a building wall few miles away, as far as possible.  A person close to the buildings cell phone the one with the lasers, and tell him to adjust the screws for the three lasers to be aligned, no matter if they are angled or horizontal, just aligned, green as better is possible in the middle of reds.  Then go to the beach and shoot them over the 48km patch of water against a big building on the other side.  Using a cell phone tell the person with the lasers to very slowly tilt down the front of the lasers base until the spots disappears down at the receiving side, then slowly tilt it up until they appear, so you can measure the minimum altitude the lasers hit the building wall. Then, measure or estimate, how big is the difference of alignment between the green and red spots.   There will be a difference of alignment, the green laser will be lower than the reds. Red light refracts different from green on a moisture oceanic air.  By measuring the misalignment of the beams, meaning diference of refraction from red to green wavelength, we can calculate how much refraction it is actually happening in general, air density, etc.  So we can insert this variable on the minimum height of the receiving beams on the building side and calculate the correct numbers.  Another blue laser could be used together, since blue bends even more. We can not just assume the light travels straight over a patch of moisture air, it will bend down as if going through a very low density glass. This is specially pronounced over lakes, ocean, etc.  It also happens over land with less effect, the thermal difference from the ground and the air creates this cushion of moisture and warm turbulent air, it creates havoc for visible sight.  Sharpshooters know that and compensate for plain dirt terrain, moist, water, jungle, dry, rocky, sometimes even the color of the land changes everything, dark color retain more warmth and create uplift air flow.  It is very difficult to hit a 20cm target with a bullet at 1500m away, even with a supersonic projectile, they need to know and compensate for everything. That is not only compensation for the bullet travelling, it is also visual compensation on the scope, light refracts easily.

What I was trying to get at is that the moon's orbit around the earth is completely independent of the earth's rotation.  That is, if we were to magically stop the rotation of the earth, the sun would appear to stand still in the sky, but the moon would still cross the sky at a rate of about .5 degree per hour.

Exactly, what I post previously.  Earth's rotation has nothing to do with the Moon's conical shadow projection path and eastward direction over Earth during an eclipse.  Earth's rotation only changes where (observers) physically the umbra hits and exposure time.

The further the straight lines originate from you, the more parallel they should appear to each other.

Sorry, that is not a true statement.

Photons are projected from the Sun in all directions, each form a straight line. The ones projected at 30° from each other will not appear parallel, even being the Sun very far away.  I am stating here the "observer view aperture", not distance.

I can have a small 0.2" LED projecting photons at 2cm of distance, entering a micron slit between two separated sets of two blades and obtaining extremely parallel photons. This is a very common optical lab setup for filtering narrow parallel frequency bands, in some chromatography experiments.
The "parallel rays" appearance depends greatly on the observer view aperture, distance from the source is a secondary variable and can even be discarded.

Someone please check my math, but here's the way I see it.  Because of the earth's rotation, the sun appears to move across the sky from east to west at a rate of 15 degrees per hour.  The moon moves in its orbit around the earth from east to west at a rate of just over .5 degrees per hour (13 degrees per day).  This means that if you add the speed of the earth's rotation to the moon's orbital speed, then that means that the moon should appear to cross the sky from east to west at a rate of about 15.5 degrees per hour.  Does that sound about right?

Yes, and that is why everyday the Moon rises around 52 minutes and 44 seconds later on East.   If today you have a full moon exactly on top at midnight, tomorrow it will be at the same spot almost at 00:53am, next day at 1:45am, and so on.  The Earth's rotation must be subtracted from any calculation, this is why the best way is to consider the observer sit over the Sun and seeing it as the heliocentric system as it is.   When looking from the Sun, doesn't matter if the Earth is rotating or not, the Moon will be orbiting Earth once per 27.3 days Earth's eastward direction, it WILL produce a project a conical shadow with apex at 380,000km, if the shadow hits Earth we call it solar eclipse, it will move eastward, no matter if the Earth is rotating or not. 

By the unquestionable fact Earth rotates 15°/h, and the Moon orbital speed of 27.3 days is 1022m/s, its projected shadow speed is always faster (eastward) than Earth's surface speed (463m/s on Equator line), depending on where the umbra hits Earth, a fixed observer on Earth could be under umbra for no more than 7 minutes and 32 seconds.

FE rely on a fixed huge Earth and small Sun/Moon rotating above. 

The smaller Umbra reported on RE was 120km in diameter, on FE it will create a problem as I reported in another post including formulas.  As the FE Sun and FE Moon has only 48km in diameter and the Sun is 4800km of altitude, the Moon creating the total eclipse must be significative lower than the Sun to projects a 120km umbra total shadow, also the Moon must be larger than 48km in diameter.  The final formula I calculated was "D*0.015 + 48km", where "D"  is the Moon's distance down below from the Sun in kilometers.  The Moon must have 48km plus 15 meters per kilometer far from the Sun, and it can not be very close to the Sun since its shadow (umbra) would cover the entire FE below, not only a spot. 

The formula is linear if a little away from the Sun. If the Moon is 4800km down from the Sun, it will on the Earth's ground, so, it needs to be 120km in diameter to cast a 120km diameter shadow, 4800*0.015+48 = 120. 

Worse, FE would have a solar eclipse every 27.3 days, since there is no way for the Moon closer and lower than the Sun to project its shadow out of FE, it will hit Earth down below - remember, FE is much larger than Sun/Moon. FE is 40 thousand kilometers in diameter, Sun/Moon is only 48 kilometers in diameter and only 8.33% of FE diameter in altitude.  I will produce a model with a graph showing how far horizontally and vertically the Moon from the Sun can be in order to still projecting its shadow over FE, of course it would happen every 27.3 days with low nodal lines, but in real world it is not like that. I would love to read FE optical math details about that. 

Flat Earth Theory / Wiki - Tom Bishop Experiment
« on: May 22, 2019, 10:02:30 PM »
On Tom Bishop Experiment at Wiki, he reports to have seem people playing at Lighthouse State Beach at Santa Cruz in Monterey Bay CA, from Lovers Point, 23 miles away, using a "good"  telescope, on a cold clear day.   Below a picture of the place he was, pointing directly to Lighthouse State Beach (Santa Cruz).  The picture is from Google Street View, with the maximum magnification it allows, perhaps 3 or 4 times.  Note the map at left and the compass at right, what help me to try to point to the Lighthouse State Beach.  I may be wrong with the exact location, tried my best.  Of course that camera is pretty bad, can't see anything on the other side of the bay, barely the mountains.

Can I ask you what brand and model of the telescope you used?
what aperture? eyepiece?
You said about chest on the ground, your telescope was not on tripod?
Do you have any pictures from the beach through the telescope?
It would be nice to have the pictures at the Wiki, don't you think?

The second picture is the opposite, from the Lighthouse State Beach directly to Lovers Point, at maximum magnification (3 or 4x).

This became an endless discussion about A video. Any discussion based on an observer on Earth is prone to a lot of confusion, since Earth is rotating, Moon is orbiting.  The correct way is to sit the observer on the Sun and think about the real think, heliocentric model.

I think the topic is related to the OP stating the solar eclipse shadow path moving in wrong direction, and that was already clarified, indeed it moves eastward on Earth if a little bit far from the poles.  Is there any doubt about it?

The further the straight lines originate from you, the more parallel they should appear to each other.

I wonder the middle and high school grades made by this man.

I can only answer with:
"The taller the tree that falls on a forest on the other side of the world, louder is the silence you hear here".

Flat Earth Theory / Re: Total Eclipse July 02 2019
« on: May 22, 2019, 05:15:07 PM »
If you guys want to know how prediction of the eclipses work, open any astronomy textbook or consult any astronomy source on the topic. That's how it is predicted. It's all there. If you are curious about the details, look into it. It has nothing to do with the Wiki.

I got really confused now.

1.) There is a real world where we live and observe things.
2.) There is a controversy about this world to be an oblate spheroid and related conspiracies, so Flat Earth Theory was created.
3.) Based on that (2), Flath Earth Society Forum was created.
4.) On this Forum (3) a Wiki was created to explain how this our real world (1) works based on 2.

So, according to Tom Bishop, when participants try to understand something from 1 on 2, they should NOT rely on 3 and 4 ??
If you really want to make some points on 2 upon educated and not ignorant people, it is better to be prepared to answer all kinds of questions on 4.
If when things got ugly, you say 4 is not the place for answers, what the purpose of 4 (and 3) anyway?

Tom is just playing with words and distorting reality.
He knows very well how this works, if he does not I would be very surprised.
If I am a car manufacturer, I need to understand very well the guts of my competitor's cars, right?

Sun and Moon setting in the West is a play of words caused by Earth's rotation.
When Tom says the Moon sets on the West, he is making people think the Moon is actually moving to the West, it is a trick of words, it does not.
It is the same as a passenger in a running car saying the trees are moving backwards.

1.) Earth spins towards East, CCW as seen by the North Pole.

2.) Moon orbits Earth also towards East, but it takes 27.3 days for a full orbit, so its orbital movement is 27.3 times slower than Earth's rotation.  From an observer on Earth it seems the Moon moves to the West, it is not true for an observer on Polaris.

3.) Taking 27.3 days for a full orbit, based on its orbital radius its orbital speed is on average 1022 m/s.

4.) An observer at the Sun watching Earth with North Pole up, could see the Moon crossing in front, above or below the Earth's face (diameter distance) to the right with certain angle, in 12481 seconds, 3.46h.  It would take another 27.3 (actually synodic 29.5) days for that to repeat.

5.) Due the angles alignment of Sun, Moon and Earth, the Moon shadow could be cast over Earth or projected lost to space.

6.) On average, the shadow cone has the apex at 380,000km from the Moon.  As the Moon is distancing from Earth year after year, at some point in the future the distance to Earth will be longer than the apex and there will be no more total solar eclipses, no more Umbra.

Unquestionably, solar eclipses Umbra and Penumbra are projected Eastward on Earth.
Tom knows all of this in details.
It is known that in war (common to see here), if your enemy is stronger, just confuse them.


Flat Earth Projects / Re: Wiki - Moon
« on: May 22, 2019, 01:35:19 PM »

The only thing I can think of which shows the moon to be spherical is lunar liberation.

Tom, I really don't like your idea of lunar LIBERATION, it might end up flat-face crashing somewhere.
MOON IN CHAINS!  a good song's name. 
Not a typo when used twice in the same post.

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