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Topics - Zack Bimmel

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Looking at the moon from earth three different phenomena can be observed :

1) The actual path of light from the sun to the moon is of course a straight line but visual observation of the orientation of the phases of the moon seem to contradict this when one draws a straight line between sun and moon when both happen to be visible at the same time.

2) At exactly the same time all observers of the moon see the same side - the near side - of the moon irrespective of their location on earth at least by observing with the naked eye.

3) The apparent rotation of the moon's "face" and the day-night terminator during the course of a night.

All three phenomena are distinct from each other in the sense that the underlying reasons are different. Let me give the explanation for point 1) here by bringing it down to earth - literally.

Let me start off with a statement : If we were able to see light travel from the sun to the moon the light would follow a curved path as seen from an observer on earth. This phenomena is not restricted to the sun-moon problem but very general in nature. Why is that ?

The underlying reason is that we don't have any perception of depth of an object in the sky but describe its location only in terms of two angles : the angle of elevation ( 0 when the object is at the horizon, 90 degrees when straight overhead) and azimuth angle ( 10 deg east of south or something like that ). So let's bring it down to earth.

Imagine you are standing on a big level field - it is pitch dark and all you can see is five very small lights which are NOT moving. Because of the total darkness you don't have any depth perception. You also have no knowledge about how the actual brightness of each light; one might be actually very bright but far away, the next one very dim but close by. So, you do what astronomers do when measuring the positions of objects in space, you measure azimuth and elevation angle. Here are the numbers you measured, assigning 0 deg azimuth to the left most post and 90 deg azimuth for the right most post (you will be able to verify these numbers yourself shortly) :

Post  azimuth [deg]  elevation [deg]
 1       0.00              15.00
 2      18.43             18.72
 3      45.00             20.75
 4      71.57             18.72
 5      90.00             15.00

Now, before you leave the field you mark the spot you were standing to come back to it during day light.

You go home, take a piece of paper to create a nice x-y drawing, azimuth angle along your x-axis and elevation angle in the y-direction. Connect the five points by a nice arc.

Next morning you come back to the field to the spot you marked the night before and you make the following discoveries : from where you were standing you had looked at five poles with little tiny lights on top. The center post (# 3) being exactly 100 meters away. The posts are located along a line perpendicular to the line from you to the center post. The posts are equally spaced, 25 m apart. They also turned out to be exactly the same height as verified with a laser beam going from the right most to the left most post just grazing all five light bulbs. Of course, if you could have seen the laser light at night it would have followed the nice arc you drew.

With this new information, your calculator and knowledge of trigonometry you can now not only calculate the height of the posts (quite high I might add) and verify the numbers in the above table.

Finally, some food for thought : I presented all the distance in terms of meters. Would you have to make a new azimuth-elevation drawing if they ALL were in feet or all in miles ?

Flat Earth Investigations / Equinox - daylight/nighttime across earth
« on: March 21, 2020, 11:45:22 PM »
Hi everybody,
for you enjoyment I bring you a picture which shows the daylight/nighttime distribution across earth's surface. The distribution shown occurs when it is exactly noon in Greenwich, UK (0 deg longitude) on the day of the spring equinox. I could have picked any other time on the day of the equinox - my choice was noon at Greenwich. My plotting software was not quite good enough with labeling the latitudes. In my picture 0 deg is at the North Pole, 90 deg at the equator and 180 deg in the Antarctic. Similar with the longitude, instead of east and west of Greenwich mine goes from 0 deg to 360 deg when circling the North Pole going east.

Of course, you can check this all out. Look up the times of sunrise and sunset for any city you desire which gives you approximate 12 hours of daylight. Also check out that for any two (or more) cities located at the same longitude sunrise occurs at the same time and so do their sunsets. Best is if you get all your times in terms of GMT to avoid trouble with the time zones. Also, most places on the internet and printed media define daylight as beginning and ending when the upper rim of the sun appears to coincides with the horizon adding a couple of minutes of daylight to what you get, as I did, when the center of the sun hits the horizon.

Two question arise.

1. Where is the sun located when it is noon on the day of equinox in Greenwich when everybody on the red and green line views the sun being exactly straight to the right in the picture ? Shouldn't it be exactly above the equator and south of Greenwich as I indicated ? But nobody, except those living at 0 deg longitude, looks that direction at sunrise/sunset.

2. What is the mechanism by which the sun divides the flat earth exactly in half along a perfectly straight line (Which we won't see again until the next equinox). Take into account the fact that the sun is approximately spherical radiating light in all directions like a light bulb ?

Flat Earth Investigations / Revisiting Bedford Level Experiment
« on: December 20, 2019, 09:19:29 PM »
Hello everybody,
thought you might be interested in some science concerning measurements of distances and heights in the lower atmosphere which has been applied in the real world for decades now. References given at the end of this post span the time from 1979 to 2016. Hyperlinks to the original articles are provided.

Bedford Level Experiment

It is well-known that the measurements in surveying over distances of some miles can be affected by atmospheric refraction due to the change in index of refraction of air with height. Such changes are due to change in temperature and relative humidity with height above ground. I do not conѕider the latter in this post.

References [1], [2] and [3] (at the bottom of this post) present this phenomena in terms of a coefficient of refraction, k, defined as :

k = R / r

where R is the radius of earth and r the radius of the circle describing the path of the refracted light beam. Hence k=0 represents the case when the light is not refracted and follows a straight line from the target to be investigated to the observer. For k=1 the light would follow exactly the curvature of a round earth of radius R = 6370 km. Negative values for k correspond to a case with the light path curving upwards.

In the above mentioned references a mathematical expression for the value of k is presented :

k = 503*p*(0.0343 + dT/dh)/(T*T)

p = atmospheric pressure in mbar (1015 mb = 14.7 psi)
T = absolute temperature in degree Kelvin (288 K = 59 Fahrenheit)
dT/dh = change of temperature with height in Kelvin/meter.

The values for p and T are mine representing an average atmospheric condition.
If the temperature does not change with height then dT/dh = 0 and the value of k becomes k=0.21 with the above numbers for pressure and temperature.

Close the ground dT/dh can be expected to be positive if the ground is cooler than the air above. Higher up we observe in general that the temperature decreases with height and therefore a negative dT/dh is common in that region.

In order to get a feeling for what numbers we are talking about, let's assume the air temperature changes by +0.13 deg kelvin ( = 0.234 Fahrenheit) per vertical meter ( about a yard ). We get k=1.01. In that case, on a round earth, any object a few miles away would still appear to sit on the horizon in full view regardless of distance. An observer who is not familiar with refraction will therefore conclude that the earth is flat.

Reference [2] cites a variety of typical ranges for k depending on at which height above ground geodesic measurements are taken.
In the region of 100 m (330 ft) and above the temperature gradient is fairly constant at dT/dh = −0.006 K/m resulting in values of k around 0.17 .
For heights between 20-30 m up to a 100m dT/dh = −0.01 K/m resulting in k=0.15.  For these values of k the bending of light due to refraction would be still small in comparison to the curvature of an earth with a radius of 6370 km and line-of-sight measurements would give pretty conclusive evidence about the flatness of earth's surface. But, both, target and observer and anything in between has to be more than 20m ( 65 ft ) above ground. Even if that is the case, a really good experiment will be accompanied by precise measurements of the temperature gradient at that height.

Below 20-30m the temperature gradient in the air is subject to the thermal properties of the surface underneath. [2] cites several experimental studies with values for the coefficient of refraction, k, ranging from -14 to +18 as distance to the ground decreases to below 10m (33 ft). Based on that and my calculations with dT/dh=0.13 K/m giving already a value of k=1.01 it is clear that line-of-sight experiments conducted close to the ground must be accompanied by very precise measurements of the temperature gradient along the path of light.

In summary : the accuracy of the measurement of the temperature gradient must better than, let's say, a few hundreds of a degree per vertical meter in order for an experiment to prove or disprove conclusively the flatness of earth's surface.

[1] D. Gaifillia, D
"Empirical Modelling of Refraction Error in Trigonometric Heighting Using Meteorological Parameters"
Journal of Geosciences and Geomatics, Vol. 4, No. 1, 2016, pp 8-14

[2] Hirt, Christian
"Monitoring of the refraction coefficient in the lower atmosphere using a controlled setup of simultaneous reciprocal vertical angle measurements"
Journal of Geophysical Research: Atmosphere, Volume 115, Issue D21; Nov 2010

[3] Fraser, C.S
"Atmospheric Refraction Compensation in terrestrial Photogrammetry"
Photogrammatic Engineering and Remote Sensing, Vol 45, No.9, September 1979, pp.1281-1288

Flat Earth Theory / Distance between two cities
« on: March 18, 2019, 02:13:28 PM »
Australia is a pretty big country south of the equator. Two cities, Perth and Brisbane are located on opposite sides, Perth on the west coast and Brisbane on the east coast. According to (not easy to deal with if you are not living in Australia) gives a driving distance of 2740 miles and a travel time of some 44 hr 23 min for an average speed of 61 mph (rather optimistic if you ask me). Well, the route between these two cities is not particularly straight so the true distance, as the crow flies, is probably a bit shorter. I assume though that mapquest is fairly accurate with its mileage otherwise we would have heard a lot of serious complains by motorists.

Perth is located at 32 degrees South and 116 degrees East, for Brisbane the numbers are 27 degrees South and 153 degrees East. So, in first approximation they at about 30 degrees South and have a difference in longitude of 153-116=47 degrees.

So, what do the two competing theorїes, FET and RET, have to say about the distance.

RET : 30 degrees South makes these two cities lie on a circle with a radius of 3963*cos(30) with a circumference 2*Pi as large. Hence the distance between them is 3963*cos(30)*2*Pi*47/360 = 2216 miles. That is about 80% of the driving distance. Does not sound unreasonable to me.

FET : 30 degrees South of the equator is 120 degrees south of the North Pole, the radius of the circle these cities lie on is now about 14666 miles ( assuming 22000 mile from the North pole to the Antarctic ). Hence an approximate distance between the two cities would 14666*2*Pi*47/360 = 9471 miles.

Such a difference, 9471 vs 2740 miles, needs to be explained by the FET experts, don't you think ?

Flat Earth Theory / Celebrating Equinox
« on: March 18, 2019, 02:04:19 AM »
Well, just a few day until we celebrate the spring equinox as we do every year.

I like to start with three important observations for this day (everything I say holds approximately true for days around it). Firstly, everybody on earth experiences 12 hour day-light time, please check it out wherever you live. Secondly, for everybody the sun rises exactly in the east and settles exactly in the west whether you live in northern Canada or southern Australia. Thirdly, people who are living anywhere on the equator, i.e on a circle centered on the north pole with a radius of about 11 thousand miles, the sun rises in the morning vertically upwards in the east, crosses the sky in a straight line being exactly over head at high noon and going straight back down in the west. So much for what is completely verifiable and has been observed for thousands of human generations.

A first conclusion we can draw, adhering to FE thinking, goes along the following lines :  For every minute of the day there is somebody on the equator for whom it is high-noon with the sun overhead. Therefore the sun must be traveling at this time of the year exactly above the equator, i.e. on a circle centered above the north pole with a radius again of 11 thousand miles.

But now some problems arise and they are begging for explanations for those of curious mind.

Imagine that you are one of those people living on the equator and it is high-noon with the sun over head. You observed the sun rose some 6 hours earlier directly in the east. But where was the sun 6 hour earlier according to FET ? Well, in the previous 6 hours the sun had traveled a quarter along its circle and was above a point which from your point of view is exactly in the north-east meaning 45 degree off from your own observation.

Not only that, the point above which the sun was 6 hours earlier is some 15,500 miles away from you ( just some trigonometry needed to get that number) and the sun - according to FET - some 3000 miles above the surface of the earth which makes the angle between the horizontal and a line to the sun some 11.3 degrees at time of sun rise. 

Finally, at sun rise the distance between you and the sun was 15,500 miles horizontally and 3000 miles upwards for a total distance 15800 miles at distance which shrinks to 3000 miles 6 hours later at high-noon. Shouldn't the sun's diameter appear to be over 5 times smaller at sun rise than at noon ?

Lastly, looking now at the situation at sun-set. Again, for the elevation above the horizontal and size of sun observation and FET show the same discrepancy. Same for the 45 degree mismatch in direction with one difference. When you look east in the morning at sun rise FET says the sun is actually 45 degrees off to your left, while when you look at the sun setting in the evening FET puts it 45 degrees to your right ( North-West ).

Before closing a short comment on refraction : of course, it does exist and luckily for those of us who wear glasses or contact lenses that part of science is very well established (Snell's law). Fraction for light coming from the sun through vacuum (index of refraction 1.0000) into the atmosphere ( 1.0003 ) and falling into your eyes make the sun appear a tiny bit higher than it actually is. The effect though is totally negligible even when it is at its largest, namely at sun rise and sun set, not even coming close to 11.3 degrees. Furthermore, because the atmospheric layer is as flat as the surface of the earth underneath, there is no - meaning none - side wise refraction. If you see the sun exactly in the east it is exactly in the east. I bet my glasses, the camera of my cell phone, the rear-facing camera of my car, and my microscope on it.   

Flat Earth Theory / FE sun model cannot be correct
« on: June 27, 2018, 01:45:10 AM »
According to" : The sun is a rotating sphere. It has a diameter of 32 miles and is located approximately 3000 miles above the surface of the earth."

But we know more about the sun namely that its surface temperature is about ten thousand degree Fahrenheit (yeah, no typing error, 10,000 is the number).

Of course, nobody went up there with a thermometer – it would melt and then evaporate like anything else at such high temperatures. So, how do we know ?

Well, it goes back to the 1800 when physicists realized that everything, including you and I, emits electromagnetic energy the amount of which (per unit time and surface area) depends on the material of its surface and its temperature. Many people make use of this using all kinds of night-vision goggles and cameras. Just google for “night vision” and “thermal imaging” or infrared radiation. Our eyes cannot see this electromagnetic radiation but snakes, for example, have sensors to pick up on that radiation to hunt down mammals which have a slightly higher temperature than the surrounding.
At higher temperatures materials can be seen to assume a dark-red color like the coils of an electric range, your toaster, or electric bathroom heater.  Those of us who have worked – or still work – in an steel foundry know that the color of the molten iron is a unique indication of its temperature, the whiter  the higher the temperature. No need to dip a thermometer into the liquid iron. Engineers use the equations governing electromagnetic radiation when they design and build the boilers for power plants, jet engines and even car engines. So, we know a lot about elector-magnetic radiation and apply to everyday problems rather successfully.

Hence, looking at the light coming from our sun and analyzing its content ( ultra-violet , rain-bow colors , infra-red ) we can deduce the temperature of the sun – some 10000 deg Fahrenheit, give or take.

Well, so why is that important ?

Like liquids, gases have the property of filling the available space. In this case, surrounded by empty space, a 32 miles diameter gas ball would just vanish in front of our eyes. Poof … gone. Come to think of it, it wouldn’t even have formed in the first place.

Flat Earth Theory / Southern Sky. Need explanations
« on: June 02, 2018, 03:09:31 AM »
One of the pleasures of vacationing down under, for those of us living in the North-East of the USA, is viewing the night sky with the constellation called Southern Cross as a dominant feature. But there is more to learn about the night sky down under. Below I visit two cities of the South, Buenos Aires at 35 deg South and 58 degrees West and Cape Town at 34 deg South and 18 deg East. My choices are not quite arbitrary. Firstly, although these two cities are 5 time zones apart there are a few hours during which it is night in both simultaneously at this time of the year (June). Secondly, they are both large cities so there is a good chance that we can find an FE enthusiast in either one who can confirm or refute my observations.

So, what do we observe in Buenos Aires, Argentina ? Watching the movements of the stars for some few hours you find out that they move on concentric, circular arcs in exactly the same way as stars move around the North Star in the Northern Hemisphere. The center point for this circular motion is located exactly due south and 35 degrees above the horizontal. There is actually a star close to this center point; it is called Sigma Octantis. Unfortunately, it is barely visible to the naked eye even on a clear night. Anyway, all stars move around this "South Star" in clock-wise direction when looking south. If we were to go further south Sigma Octanis appears higher and higher in the sky and once we reach 90 deg south latitude it would be directly above us.

Now, for people in Cape Town the story is exactly the same. They find, if the night is  clear enough, Sigma Octantis exactly to their south and 34 deg above the horizon. All other stars again move around it on concentric, circular arcs in clock-wise direction when looking south and at the same distances as those observed by the Buenos Aires people.

From the point of view of FE theory we have now three problems :

1. Why are there stars which can only be seen by an observer located south of the equator, and other stars can be seen only by an observer north of the equator ?

2. People living south of the equator observe that stars move in clock-wise direction when looking at the star about which all other stars rotate. Why is there such a star in the first place ? Furthermore, I thought we all live on a flat plane and all stars are moving in a counter-clockwise direction about an axis pointing vertically upwards and being located at the north pole. And now there is a second such star ? And things move around in an opposite direction ?

3. A line drawn from Buenos Aires due south and a line drawn from Cape Town due their south are not parallel but form an angle of about 58 + 18 = 76 degrees (remember the 5 time zone in between those two cities ?). So, how can both lines point to the same star Sigma Octantis simultaneously ? Or in other words : let's install a video camera in each city and aim each one directly south and upwards by 35 degrees and start recording at the same time. Both would show Sigma Octanis smack in the center of their respective pictures and other stars are moving around on circular arcs. And we all could watch that via the internet in real time.

Any FE folks want to give it a stab ? I know for RE folks this is all not the slightest problem hence no need to talk about their ideas.

Flat Earth Theory / Surface Area of Earth, a comparison
« on: May 30, 2018, 07:32:38 PM »
Hello everybody,
just got curious about how big earth is in terms of square miles. FE have a flat disk with a radius of 12250 miles  as announced on this web site. Well, pi*radius*radius results in 471435247 square miles.

Contrasting that we the RE folks : the mean diameter of the globe is 3960 miles. The surface area of a sphere is calculated by 4*pi*radius*radius =   197969797 square miles.

Ratio : 2.4 approximately.

Curiously enough, the distance from North to South pole for the round earth is pi*radius =  12440 miles matching the radius of the FE actually, there i a good reason why the two are almost equal, if you know some math. But it is comforting to know the FE and RE folks agree to distances on earth as long as we are measuring in the North-South direction.
Back to the surface area. The size of the property I am living on is known rather accurately and so is that of my neighbors. Adding them all up gives the size of my county, the sum of counties gives the size of my state and the sum of the states gives us the square miles of the good old USA : 3797000 square miles. You can do a rough check on that by travelling east-west and north to south using mapquest and multiplying : as I said a rough estimate.
In the same fashion we can verify the surface area of all the other continents of course with the exception of Antarctica. There all our methods of detecting the shape and measuring the land areas fail.

With the area covered by the oceans the story is not quite as easy. Nevertheless, they have been navigated upon and measured for a few hundred years by now over and over again and many ships are on the high seas as you read this. Before the age of GPS we have been relying on maps of the oceans which showed with sufficient accuracy not only the directions but also distances. And were the maps proved to be inaccurate and led to a disaster they were quickly improved upon - for a few hundred years by now.

Given all of that I just cannot see that the size of the flat earth, radius of 12250 miles, can be correct, not even approximately.  What do you think ?

Flat Earth Theory / Erathostenes and the North Star
« on: May 28, 2018, 01:11:34 AM »
Most of you are familiar with Erathostenes' shadow experiments in which he thought that he would measure the diameter of Earth. Alternatively, one can derive from his measurements (7.12 deg shadow with 500 miles distance between measurement points) the diameter of the flat earth to be 25000 miles ( ) and also the height of the sun above earth ( 500 miles * tan(90-7.12)) to be some 4000 miles. So who is right ?  Well, I am not going to travel to repeat and expand on his experiment, somebody else might do that.

Here is an alternative : let's measure the height of Polaris, the North Star, above earth. We all can do that without having to travel. Don't believe me ? So, here it goes.

If you were at the North pole the North Star would be vertically above you and the angle between the horizontal and the line from you to the North Star would be 90 degrees. Of course you couldn't see the North Star right now - it's daylight there right now, 24 hours a day.
Anyway, lets say you live at 70 deg northern latitude - the angle between the horizontal and the line to the North Star would be 70 deg. And similar for other latitudes. Nice thing is, everybody can check that out if you live not too far North. If you do, you will have yo wait for the winter to do the checking.
Now, we can do the calculations using the radius of earth, the number of degree latitude from north to south and trigonometry :

(12250miles*((90-70)/180)*tan(70) = 3740 miles above flat earth.

And now we can go beyond Erathostenes !!!   I am calling up my friend who lives a little bit north of San Antonia, Texas, at 30 deg latitude.

(12250miles*((90-30)/180)*tan(30) = 2357 miles above flat earth.

So, who is right ?  My challenge to everybody : do the calculations for the latitude where you live.

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