1

Flat Earth Theory / Re: Latitude and longitude - please enlighten me

« on: June 30, 2020, 01:08:52 AM »

Doesn't the altitude of the sun at it highest point during a day depend on the time of the year ? Maybe at one of the equinoxes you would be a good choice ?

Would love to hear from the FE folks too.

2

Flat Earth Theory / Re: Latitude and longitude - please enlighten me

« on: June 24, 2020, 01:41:43 AM »

Somerled,
sorry to come in to this discussion rather late. I just like to make sure that I understand your position in response the OP's original question :
a) you support the idea that a latitude/longitude coordinate system allows us to uniquely identify each and every point on earth. And we could use the elevation angle of polaris to clearly identify the latitude and the time difference w.r.t Greenwich (position of sun in the sky) as a measure of longitude.

b) if we look up on the internet latitude and longitude for a particular place on earth people living there would confirm these numbers by measuring the elevation angle of polaris and time difference to Greenwich. Well, at least you and I could do that as well as others on this thread making small allowances for our measurement errors.

Thank you for your response

3

Flat Earth Investigations / Re: RE Lunar Phases With Extreme Perspective Changes

« on: June 07, 2020, 01:01:38 AM »

Please let me try to clarify, from the point of view of a RE person, a few questions raised in this thread. And sorry for the length of my post. Please enjoy anyway.

1. It is claimed that observers at different locations on earth should see different sides of the spherical moon depending on their respective location. That statement is correct. But if any imaginable distance on earth (like its diameter) is small in comparison to the distance moon-earth the effect is hardly noticeable to the naked eye. Diameter of earth 12,740 km, distance earth-moon 384,400 km (average) , ratio of 30. You can verify this effect by looking at a building head-on but from far away. Moving side-wise by a few feet will not reveal in any noticeable way one of the sides of the building particularly when the building is round.

2. We always see only one side of the moon, called the near-side. That is correct in very good approximation. The reason is that the moon is tidally locked which is a short way of saying that the moon rotates around its own axis exactly once during a complete orbit around earth. Pluto and its moon Charon are a more advanced example thereof. They both show each other only their respective near side.

3. During the course of the night the image of the moon including moon phase seems to rotate. "seems" is the keyword. It is actually you whose horizontal plane is changing its orientation with respect to the moon. Reason : earth is spherical in shape and rotates around an axis. You could clarify that to yourself in the following way :
On a piece of paper draw a nice circle and then add a vertical line through the center of the circle. This is a crude presentation of the spherical earth with the straight line representing earth's axis. Mark the north pole at the top. Now add the equator - a horizontal line again through the circle's center. Lastly mark three points at identical distance from the equator, one on the axis and the other two on the circle on opposite sides. One could say that these three points are at the same latitude. In this scenario your face is now going to represent the moon and to define its angular orientation we consider the line connecting your eyes. In the following I will call your, personal, left eye the moon's left eye. Now imagine an observer on your sketch of earth looking at the moon, the observer being situated at the point on the circle to the left of the axis. With the known direction of earth's rotation an observer at this point would see the moon rising with the "moon's" left eye well above the horizon in comparison to its right eye. Some 6 hours later this observer would be located at your mark on earth's axis. The "moon's" eyes appear now at equal height. The moon seems to have rotated !!! I leave it up to you to explore what our fictitious observer will see additional 6 hours later when the moon is setting.     

4. Moon-tilt Illusion. How I hate the word illusion in this discussion. If you happen to look at the moon other than at full or new moon you see the day/night terminator on the moon's surface giving rise to what we call the moon's phases. There is no illusion nor distortion nor rotation involved in the image you are seeing. None whatsoever. You also see correctly that the line connecting the two "horns" where the terminator meets moon's perimeter is inclined most of the time with respect to your vertical. Actually, if you know enough math you can derive an equation for the tilt of this line given the position of moon and sun in the sky in terms of azimuth and elevation angle.
Trouble arises when you want to draw the line the light follows when it travels from the sun to the moon. Except for a very special circumstance drawing a straight line means you commit a serious error !!! Let me continue 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.

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 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 far left light and 90 deg azimuth for the far right light (you will be able to verify these numbers yourself shortly) :

Light  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

According to these numbers, if you let your outstretched arm go smoothly from the far left to the far right light you would trace out a nice arc in the night sky reaching the highest point at the center light. 


The big surprise comes the next morning when you revisit the field to complement what you have seen the night before. You discover the following : from where you were standing you had looked at five poles with little tiny lights on top. The center post ( light # 3) is exactly 100 meters away from you. 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 and exactly of 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 your outstretched arm followed.

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) but also verify the numbers in the above table.

Bottom line : The path of light traveling in a straight line from the far right to the far left post (the above mentioned laser) should be drawn as an arc in the night sky. Drawing now an analogy to the sun-light-moon problem : If you climb up on one of the intermediate posts you are putting yourself in the same position of earth being located between sun and moon simulating a lunar eclipse. If you move a little bit to the side or up/down so that the light can pass you by we have the situation of a full moon. Yikes, where is the sun when you look at a full moon high in the sky  ?

4

Flat Earth Investigations / Re: Short note on perspective : where is the sun at full moon at midnight ?

« on: May 17, 2020, 01:24:49 AM »

BRrollin,
you are good. Congrats to the hat trick. And yes, the height of the posts really does not matter except one has to check whether they are all of the same height as I claimed. If you can do it, do it. Never trust anybody's math or statements just because they seem to know what they are doing. Doesn't hurt either to brush up on your own math skills or even learn new stuff - including calculus. That's by the way where the fun really starts; before that its sometimes a real drudgery.

Anyway, concerning your comment about light from the moon being generated by something/somebody on the moon ? Haven't seen that yet. Do you know how that should work ?

5

Flat Earth Investigations / Re: A flaw with the Flat Earth model?

« on: May 14, 2020, 01:26:19 AM »

The moon must be much, much further away from earth than any distance between two points on earth. There are at least two reasons why we actually can see a little bit of the far-side of the moon. 1) The moon's orbit is an ellipse (albeit very close to a circle) and 2) the moon is not infinitely far away.
By the way, the effect you mentioned (that at the same time we all see the same face of an object in space irrespective of location on earth) applies also to the sun and planets when we can identify some distinct features. That would be sun-spots on the sun, great red spot on Jupiter, moon around planets  and phases of the planets all look the same when viewed at the exact same time from different positions on earth.

6

Flat Earth Investigations / Re: Short note on perspective : moon-sun

« on: May 14, 2020, 01:03:03 AM »

Hi BRrollin,
Well, you are right again. The mathematical proof that the perspective view of the light from the right post (the curved line) and the line connecting the two "horns" on the ball are exactly perpendicular to each other is actually quite tedious;  algebra, trigonometry and a little bit of calculus are needed.
Anyhow, an important question is now concerning the position of the observer when he/she wants to see the  ball on the left-most post in the "full moon" mode. The answer tells you where the sun is located when you watch the full moon at mid-night high up in the sky. That's really bad for those folks who think the sun is circulating above the flat earth.
By the way did you check out the height of the posts ?

7

Flat Earth Investigations / Re: Short note on perspective : moon-sun

« on: May 11, 2020, 01:15:46 AM »

Yes you are.
Here is another question. Let's say you put a basket ball on the left-most post and a spot light on the right-hand post illuminating the basket ball at night. What would the lit/unlit pattern on the basketball look like and which way would it be oriented from the point of view of the person standing at the original observation point. Two options : a) w.r.t. a straight line from one post to the other ? b) w.r.t. the curved path ?

8

Flat Earth Investigations / Short note on perspective : where is the sun at full moon at midnight ?

« on: May 10, 2020, 02:23:06 AM »

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 ?

9

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 ?

10

Flat Earth Investigations / Re: Revisiting Bedford Level Experiment

« on: January 20, 2020, 03:11:18 AM »

But WGS-84 is a reference co-ordinate system system , not a globe models. Clearly explained here.
https://gisgeography.com/wgs84-world-geodetic-system/

OP's original post shows how surveyors who believe the earth is a sphere ( he kindly quotes this in his 2nd post) use a set of equations including a constant of refraction k calculated with a value of R = 6370km .  This is valid is on a perfect sphere only . We know this is not true whichever model we choose to believe .

WGS-84 is an ellipsoid reference co-ordinate system applied to one of many geoids which in the words of that site leads to Geodesists to " believe the error is less than 2cm which is better than NAD83."

The use of that word " believe " leads me to conclude that we are encouraged to have faith in whatever they tell us. Science now has the appearance of religion in which no one is allowed to question the high priests and the faithful have no need for truth .

Dear somerled,

I find it amazing that you still think about R=6370km used by surveyors when displaying and communicating with each about refraction of light in the presence of thermal and humidity gradients. For a last time : they could have used as well the distance between Glasgow and London as a reference value or any other distance. The choice of a reference variable does NOT ever influence the calculations of the amount of refraction when expressed in terms of lengths and angles .

I realize that it is hard for you to accept that the line-of-sight experiments conducted in the past without taking thermal and humidity ' gradients into account are inconclusive. Therefore, true to the method Zeteticism, we have to strive for more up-to-date experiments, like a laser in a vaccum tube.

And for the record : I did not state that surveyors believe that the earth is a sphere. I used the term "round" which can be a sphere or an ellipsoid or a pear-shaped entity. But, as far as earth is concerned the naked eye is not good enough to discern the difference. The model ellipsoid of GPS is a pretty good approximation to earth with a semi-major and semi-minor axis of 6378137 meter and 6356752 meter, respectively. Choosing 6370 km as a reference length does seem to me a reasonable choice.

An ellipsoid, like the sphere, has no vertices. But you are correct in that the ellipsoid we are talking about is centered on the earth's center of gravity. You might want to study this web site https://www.state.nj.us/transportation/eng/documents/survey/Chapter4.shtm . I also encourage you to have somebody who has a GPS receiver with the required accuracy for vertical distance, verify the Bedford Level Experiment. Wouldn't it be worthwhile to find the truth ?

11

Flat Earth Investigations / Re: Revisiting Bedford Level Experiment

« on: January 16, 2020, 04:42:50 AM »

Comment 1 :
The pipes of the LIGO experiments are not only supported at either end, some 4 km apart, but at many intermediary support piers. Compensating for earth's curvature means that the height of theses piers would have to change quadratically with distance if measured with respect to the water level inside a 4 km long water trough running parallel to LIGO pipes. FE would result in a linear dependency.

Comment 2 :
Some of the more sophisticated GPS system measure to within a few centimeters the altitude with respect to an ellipsoid model of the earth. The value for the altitude can be positive or negative depending on whether the elevation you wish to measure is above or below the ellipsoid. Again, the elevation of various points along the length of the pipe should change quadratically. (P.S. Of course one could base a GPS system by approximating earth by a sphere. The principle is the same)

Comment 3 :
All GPS measurements ultimately lead back to having (preferably) four satellites above the horizon at the moment of measurement. All GPS related satellites orbit the round earth at an altitude of about 20,000 km (12,427 miles) and complete two full orbits every day.

Comment 4 :
In addition to millions of earthlings, NASA, NOAA and the entire military complex of NATO uses GPS rather successfully. Russia has its own GPS version (Glonass) and so do China (Beidou ?) and Europe (Galileo). All these systems are satellite based.
 

12

Flat Earth Investigations / Re: Revisiting Bedford Level Experiment

« on: January 11, 2020, 01:37:37 AM »

AllAroundTheWorld (also mentioning RonJ's post)
thanks for posting that video. But, it really doesn't help the discussion about flatness or not because the experiment again does not address the impact of refraction in a satisfactory manner. One of the FE persons tries to talk about it ( naming it mirage effect or something like that) and the "science" guy just smiles condescendingly. At the end, everybody, on both side of the debate, felt confirmed in their opinion. Come to think of it, that was utterly foreseeable and one has to blame the "science" guy.

As I said before, line-of-sight experiments like the Bedford level and Lady Bount experiments are only really useful when the atmospheric conditions including temperature and humidity gradients have been measured accurately or when the atmospheric conditions do not come into play in the first place like with LIGO with its vacuum tubes. Also, the post by RonJ about establishing microwave links must be taken into account. We have those links all over the world and they are working and they take earth's curvature into account.

Does anybody know of more things being built which have to take earth's curvature into account ?

Interestingly enough, comments from the FE side have not been definite in this thread although they should be shaking to their core. Guess I am rattling the cage here hoping that something constructive falls out.

13

Flat Earth Investigations / Re: Revisiting Bedford Level Experiment

« on: January 08, 2020, 02:40:17 AM »

Hi,
Don’t mean to hijack this conversation, but I do have two comments.
I agree with somerleds position that if you have a flat earth model, refraction doesn’t make sense as the line of sight is always perpendicular with the thermal gradient, so refraction would always be zero.

But you do have two problems otherwise.
1.   In the Wiki, it is stated that one of the reasons that ships vanish below the horizon, which is basically the same thing as this Bedford experiment, is refraction.  You can have it both ways, either refraction interferes with the path of light in the flat earth model or it doesn’t.

2.   Whenever I have seen these “Rowbotham effecst” demonstrated, it is pretty clear that the experimenter always puts the observation point and the target very close to the ground. (including Rowbotham).  The reason seem pretty obvious, when you are close to the ground, the thermal gradient is the highest and therefore the refraction is the highest and you get the illusion that the earth is curved more than it actually is.  If you see others perform the experiment, they always make sure to do it well above the surface which minimizes refraction and shows that the earth is curved.

DaveP : sorry to contradict you a little bit. Refraction of light takes place of cause when light enters a medium with a n index of refraction different from that medium the light beam comes from. In addition to that, light is also refracted when light beam and temperature gradient are perpendicular to each other. Or for that matter by gradients of absolute humidity.

But, you are correct with you last paragraph, you can use refraction to make points in favor or against the flat earth model. Unless, of course, you measure the temperature gradient exactly along the line of sight during your experiment and then correct accordingly. I am not aware of anybody in the FE community or otherwise has done that nor published about his/her/their experimental evidence.

There is at least one experiment (conducted for totally unrelated purposes) called LIGO of which we have two examples in the US and one in Italy. It is based on a laser beam traveling down a long straight tube ( 4km) and being reflected back to where it came from. No air or other gases on the inside, just a plain, high-quality vacuum. They had to take earth's curvature into account increasing the cost of a rather expensive experiment (hundreds of millions of dollars). LIGO is doing well continuously listening to the "sounds" of the universe.

It seems to me there might be other such instance where creating large structures might have to consider earth's curvature. Maybe underground subway system where two tubes cross each other; one on top of the other with little clearance in between. I am thinking of those instances where these tunnels were "dug" by those huge machine which grind their way for miles on end never to see the light of day why doing so.

14

Flat Earth Investigations / Re: Revisiting Bedford Level Experiment

« on: January 08, 2020, 02:00:55 AM »

Your experiment has already been done: check out the 'LIGO Experiment'.
It's a laser in a vacuum.  The arms are long enough so the concrete supports for the
pipes had to be adjusted to compensate for the earth's curvature. 

Maybe they should have consulted the wiki on here and saved themselves a bunch of money!

Congratulations, but you really took away my punch line.

15

Flat Earth Investigations / Re: Revisiting Bedford Level Experiment

« on: January 08, 2020, 01:58:54 AM »

Hi someled

Here is my last attempt to convey to you that indeed, in the references I provided the occurrence of the radius of the earth, R, does not indicate that anybody suggests that the mechanism of the refraction of light is influenced by R. It also should become clear to you why it pops up in the equations in the references. If that seems to you a contradiction in terms than please study the following with care to see the contradiction vanish in thin air (pun intended).


For the following see my attached image, I have it available only on my home computer and didn't know how to imbed into this post.
This image is a my counterpart to Fig. 1 in Ref.[3]. As in that figure we have the cord-length S stretching from point 1 to 2 with point 3 in the middle. The circular arc going from point 1 to 2 represents a beam of light under influence of refraction with point 2 being the target and point 1 the observer. Line 1-4 is tangential to the circular arc at point 1 and points in the direction the observer at point 1 perceives the target to be. Again, Delta_beta is the angle of refraction. Perpendicular to this tagent is the radius line 1-0. By way of similarily of right triangles you can prove that the angle Delta-beta occurs again between the lines 0-1 an 0-3.

Some trigonometry :Or with extremely small errors (less than 1%) according to the info in my previous post :

Delta_beta = S/(2*r)

Now use this relationship to eliminate Delta_beta from Eq.(7) in Ref.[3]. The resulting equation is :

S/(2*r) = -S/2 * (dN/dh) * 1.e-6

I assumed that the cord-length, line 1-3, is running horizontally. Hence cos(beta) = 1

The factor S/2 appears on both sides of the equation and therefore cancels out.

1/r = - (dN/dh) * 1.e-6

This equation still does not contain R=6370km which is what you rightfully demand to be. So, how does R=6370km come into being in eq.(10) in Ref.[3] you might ask.

Well, here is what scientist often do. In order to converse easily among each other it is better to present r with respect to some reference length, let's call that L. One can argue endlessly as to how big the value of L might be. If all the involved scientists were to live in the UK one (usually it is the one who publishes first on the subject at hand) might suggest the distance between London and Glasgow. If subsequent publishers in the field like that choice because they think that distance is relevant to the refraction of light in the context of surveying the landscape or the building of large structures the original suggestion for L will pervail, otherwise somebody else comes up with a different idea.

I suggest that L = distance between London and Glasgow is really not a good idea but maybe using the radius of the earth might not be a bad choice for two reasons. a) Everybody knows at least a good approximate value of that ( 6370km or the equivalent in miles or whatever units one prefers). b) r in above equation often comes out to be in the neighborhood of thousands of kilometers under common atmospheric conditions.

Last step to get to Eq.(10) in Ref.[3].

Multiply my last equation by R on both sides :

R/r = - R*(dN/dh) * 1.e-6

and abbreviate : k = R/r to obtain eq.(10) in Ref.[3]. Bingo.

And this is my last post trying to explain to you that in all the references I provided the radius of earth is simply used as a convenient way to display and communicate the effect of refraction of light due to temperature gradient in the atmosphere. Thank you for your time. Zack

sin(Delta_beta) = S/(2*r)

16

Flat Earth Investigations / Re: Revisiting Bedford Level Experiment

« on: January 05, 2020, 02:15:06 AM »

somerled
You are mistaken when you assume I am distancing myself from k, the opposite is the case. I will lead you to it and will also show you how R=6370km appears.

You have to be a little bit patient with me though. Math is sometimes not easy and sometimes multiple steps have to be executed flawlessly. Just by way of preliminaries : of course I know that most people measure angle in degrees ( 360 for the full circle ). I think I learned that in grade school. But then later on when we got to trigonometry radians popped up as an alternative. The relationship between the two is easy (PI=3.1415...) :

radians = PI/180 * degree

Hence, 360 deg = 2*PI , 180 deg = PI , 90 deg = PI/2 etc.

The reason why I bring this up is that later on I will make use of the fact that :

sin(Delta_beta) is in very good approximation equal to Delta_beta (in radians) if Delta_beta is small. Example :

Delta_beta = 10 degrees = 0.1745 rad ; sin(10 degree) = 0.1736

Luckely, as far refraction of light in air is concerned, Delta_beta is much smaller than 10 degrees and the error between it and its sin() becomes even smaller than the 1% error in my example.

Let me know whether what I brought so far is OK with you. Feel free to ask any question you might have and with a little bit of patience we will get to eq.(10) of Ref.[3] and see how k and R arrive on the scene. May I recommend that in your response you just comment on the math I brought in this post. Talking about the silly pear shaped orange model is as the saying goes, balking up the wrong tree.

17

Flat Earth Investigations / Re: Revisiting Bedford Level Experiment

« on: January 04, 2020, 12:13:20 AM »

somerled

I certainly admire and appreciate your tenacity in our discussion. Let me try to guide you a little bit through the math involved.

First, let's see that we find a common starting point and then go from there. Allow me to suggest eq.(7) you mentioned as a starting point. It does not contain R=6370km in any way shape or form as you so rightfully demanded.

Look at that equation in connection with Fig.1 and the subsequent list of variables to see that S stands for chord length and Delta_beta is the refraction angle. If you are not quite sure what I mean by Delta_beta look up the capital version of the Greek letter delta and the lower case version of the Greek letter beta and write them next to each other.

For the next step we need to know that in mathematics angles are measured in radians (some times abbreviated as rad). Look that up as well if you are not familiar with that. But first let's agree on the starting point and address any questions you might have about it before we take the next step.

18

Flat Earth Investigations / Re: Revisiting Bedford Level Experiment

« on: January 02, 2020, 01:09:15 AM »

Dear somerled,
thanks for response. Unfortunately a few improvements to it are in order. First of all, nobody has claimed nor is deducing the amount of refraction from the curvature of earth. To assert publicly that anybody ever would do that is akin to spreading false information. Please clarify that to yourself by reading the equations in ref.[3] I had mentioned in a previous post. If you can't do that by yourself ask somebody for help but do not classify Algebra 101 matters as "mathematical trickery". That makes FET look really bad.
Secondly, you start off correctly by stating that "refraction takes place at distinct boundaries dependent of angle of incidence and differing density of medium involved." The latter part is not exactly correct. You can have two media with equal density but different refractive index and vice versa. More importantly, your statement should be appended to include the fact that refraction ALSO takes place inside media where the index of refraction varies continuously. In connection with our discussion, feel free to google for "refraction thermal gradient" and "refraction humidity gradient". You will find ample links to these topics. Some of the linked-to articles derive the equations of how light (or in general electromagnetic waves) refract in the presence of temperature and humidity gradients in some detail. For those articles understanding calculus is often prerequisite.
Happy New Year

19

Flat Earth Investigations / Re: Revisiting Bedford Level Experiment

« on: December 24, 2019, 08:49:27 PM »

someled
Sorry that I was not careful enough in explaining the appearance of R in the equation in my submission from Dec 20. You were absolutely correct when you stated that what people perceive as the radius of a round earth should in no way affect the calculation of the amount of refraction of light in the atmosphere.

So let me try again :

As stated :

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

divide that equation 6370,000 then by definition of k=R/r you get :

1/r = 79*10^{-6}*p*(0.0343 + dT/dh)/(T*T)

with r now in meters. The value of R does not appear anymore. You can verify the validity of this division by inspecting the equations ( through (10) in reference [3]. Feel free to calculate values for r for various values of dT/dh, the vertical temperature gradient, assuming standard atmospheric conditions.

Now, this equation and similar ones incoporating humidity, wave length of the light (think of lasers) and other effects have been used for decades. It seems to me somebody would have caught on to significant errors on the part of surveyors.

All this quarrel about refraction can of course be avoided by conducting a laser-based experiment inside a vacuum-filled, long tube. Maybe, if we were to look around, somebody has built already such a tube ? Maybe they did it for another purpose, but with the condition that the laser beam coincides with the tube's central axis all the way from one end to the other ? Did they consider earth's supposed curvature ?

Happy Holidays to you and everybody else ... Zack

20

Flat Earth Investigations / Re: Revisiting Bedford Level Experiment

« on: December 22, 2019, 02:15:50 AM »

somerled,
sorry to say, you are wrong. R=6370km is merely used as a convenient reference for those surveyors who believe in a round earth. R has no influence on the amount by which a light beam gets refracted in terms of feet (or whatever units you prefer) for given atmospheric conditions and distance between target and observer. Personally, I would have chosen to not have R appear because it leads to confusion as it happened to you. It was more important to me to represent the equations in a form given by the authors.
The main point remains, experiments like the Bedford level experiment need to determine the temperature gradient with a very high accuracy and document carefully their measurements. Do you know of any such experiment ?