Show Posts

This section allows you to view all posts made by this member. Note that you can only see posts made in areas you currently have access to.


Messages - SteelyBob

Pages: [1] 2 3 ... 22  Next >
1

No, I prefer a bi-polar model and don't think the sunlight necessarily takes those shapes. That type of behavior is the general argument for the Monopole model though.

Tom - I’d be really grateful if you explain how the sun illuminates the various parts of your preferred ‘map’, and how that works with your previous explanation of EA

2
You are assuming that it's always circular for all times of the year. The wider Flat Earth community generally holds that light curves and behaves as if it were coming through a magnifying dome in the Monopole model, and that the daylight area changes shape upon the earth as the sun proceeds southwards:



This is also the reason given for the southern celestial rotation as seen by the observer:



Again, you're fixing one problem, or trying to, and creating another. If that is the case, then the same must also hold true for the stars, which you've previously stated to be at the same altitude as the sun. But the model you've described for a viewer of the stars is completely different to that which you are now relying on for an observer of the sun. You can't have both. Again, pick a horse.

No, I prefer a bi-polar model and don't think the sunlight necessarily takes those shapes. That type of behavior is the general argument for the Monopole model though.

Well if you prefer a bipolar model, why are you showing us a monopole map with the curved longitude lines on it?

Again...another horse. Show us precisely what you think the map looks like, and how you think the sun illuminates it.

3

Further, you need to think about your arguments more.The direction of North to the observer wouldn't change if the longitude lines were curved.

Indeed it would not. You can draw the lines of longitude as wiggly as you like - it's just a convention.

It's not really clear from your map what modifications you have made, exactly, or why. It sort of looks like you've curved everything, thereby keeping places on the same line of longitude, albeit curved. On closer inspection though, that isn't the case, as places that are generally considered to be on the same line of longitude, like say London, the east coast of Spain, and the previously mentioned 0N 0W point, no longer are - the curved lines they are on are different. But neither do the east coast of Spain and London line up with the North Pole, so a traveller flying towards Polaris would be flying over different places. Maybe you could clarify what, precisely, the intent of the map is - are places supposed to retain their current lat/long position, or are you proposing that our current set of coordinates is wrong? And should places that currently align with Polaris still stay aligned?

But to be honest, there's far, far bigger issues with that map. If it's real world observations you wanted, here's one. According to timeanddate (and https://rl.se/sub-solar-point, the sun is currently directly overhead a position in the pacific around 21S 135W, somewhere near French Polynesia. Hard to pinpoint that exactly on your map, but here's a 90 degree circle centred on that rough location:



Now you might take issue with my placement of the circle - it's slightly south of what I presume to be the Tropic, which clearly can't be right, but I tried to line it up on the right part of the west coast of South America...it should also be roughly half way to Australia...there's lots of challenges.

In any case, the more important point to note is that, according to your description of EA, it should be dark everywhere outside that circle. But it's currently daytime in Australia, which is miles away on your map. Miles away - it's not even close to the circle. And look, here's a webcam from Sydney:



...taken from a still from this site - https://webcamsydney.com

How can it be daylight in Sydney if your model is correct?

And before you say...how do you know that sun placement is correct...aside from the fact that it's a well-proven model, backed up from numerous websites and daily observations by millions of people...here's a webcam from the Falklands at the same time. It cannot be daytime simultaneously in both Sydney and the Falklands if your model and map are correct:



That webcam is from: http://www.webcams.horizon.co.fk

How would you explain that?






4
Actually the Wiki suggests that this is a better map for the Monopole model in regards to equinox and longitude discussions:



There's so much wrong with that it's hard to know where to start. I'll try...

Firstly, bending the lines of longitude like that hasn't fixed the problems I alluded to. The 45 degree circle centred on 0N 0W, for example, now lands on the eastern tip of Somalia. At 1200 UTC on next September's equinox in that place the sun will be around 35 degrees elevation, not 45.

The 90 degree circle still doesn't cover all of South America, so your model is saying it will be dark in places when it fact it will be very much daylight.

Next, in the process of trying to correct for some problems, you've introduced more problems. According to your EA model, you are suggesting that Polaris is 6000 miles overhead the North Pole. That means that everybody on a line of longitude running south from the North Pole should view Polaris directly north, at progressively lower elevations until the equator is reached. But now you've curved the lines, so people on, for example the 0 degree meridian won't see Polaris on the same heading. That is completely at odds with what we observe, what conventional science predicts, and what your initial explanation of EA would expect.

You need to back a horse here - you seem to be flitting from map to map as it suits. You can't have one map to explain one phenomenon and another for a different one - they all have to work the same way.

5
I completely agree on the need to get this back on track. I thought we'd got to an interesting place, particularly with WTF's point on the sun - https://forum.tfes.org/index.php?topic=18734.msg251176#msg251176.

I'd be interested to hear Tom or Pete's views on this.

For example, if you consider 12 noon UTC on an equinox (I used next September and timeanddate.com), then we would expect the sun to be directly overhead the point shown by the red cross on this monopole FE map:



The circle is 45 degrees latitude radius - so from the discussion and wiki regarding EA, we would expect all points on the circumference of that circle to observe the sun at 45 degrees elevation at 1200 UTC, and in the direction of the centre of the circle. This is true for the point, for example, at 45N 0W - the sun is indeed around 45 degrees elevation at that time, due south. But it falls apart if you stray from that point. If you go to the point in Africa marked by the green arrow, for example, the sun is at an elevation of around 60 degrees at 1200 UTC, not 45. So the sun isn't where it should be according to EA.

If you extend the circle to 90 degrees radius it gets even worse, because one would expect from EA that nighttime would be the area outside of the circle. But it isn't like that at all - all of South America, for example, should be in daylight, and yet the 90 degree circle excludes half of it.

Thoughts? Tom? Pete?

6
It all stems, I suppose, from the fact that the monopole FE map holds north-south distances equal to those on the globe, but allows east-west distances to differ, with the difference becoming progressively more marked as you journey south. Hence the locus of points observing the sun in your example at 45 degrees elevation would not be a circle, despite another aspect of FET requiring it to be just that.

The bi-polar model is a little more difficult to try to examine but still has the same problem.

The bi-polar model is so ridiculous I think it can almost be dismissed ‘by inspection’. There’s hundreds of problems, my personal favourite being the challenge of re-telling the pacific campaign of WW2 using that map…it makes no sense at all.

7
You don't need multiple stars to evidence the flaw in FE theory.  You only need one, the sun, and the easiest day to demonstrate the problem, the equinox.

FE theory dictates that everyone at an equal distance from the sun would see the sun appear at the same altitude.  This is true whether you consider bendy light or not.  This means that if you locate the sun above the plane and then draw a circle with the sun as center, everyone at the edge of the circle will view the sun at the same altitude.

Now let's look at the equinox as it is the simplest say to compare what is observed, that zetetic thing, with what should be observed.  If we place the sun at 0 lat, 0 lon. 4 observers placed at 0 lat.-45E, 0 lat.-45W, 45N-0 lon., 45S-0 lon. will each observe the sun at an altitude of 45 degrees.  However, if you place the sun above the plane at 0,0 and draw a circle that touches 45N and 45S you will notice that the circle does not reach 45E and 45W.  On FE, the observers at 45E and 45W should view the sun at an altitude of less than 45 deg.

That’s a really elegant, simple problem - much neater than my idea, so thank you. It all stems, I suppose, from the fact that the monopole FE map holds north-south distances equal to those on the globe, but allows east-west distances to differ, with the difference becoming progressively more marked as you journey south. Hence the locus of points observing the sun in your example at 45 degrees elevation would not be a circle, despite another aspect of FET requiring it to be just that.

Very interested in a FE response to this - Tom, thoughts? I’m guessing the response will be that east-west distances are somehow unknown, which seems to be the get-out-of-jail option for problems of this nature. Amazing how those poor Aussies don’t know the size of their own country…

8
They would surely also change in luminosity and angular size with the varying distance.

Yep.

9
Incorrect. If you have two golf balls, two inches apart from each other side by side, located at a distance of one foot away from your face horizontally and then increase their vertical height by two feet they are no longer one foot away from your face and the two golf balls will not maintain the same angular distance from each other.

If that's the principle you're relying on here, it's working in the wrong sense. If you consider somebody viewing a pair of stars over the North Pole from, say, 60 degrees north, and consider what happens as they change their position to, say, 30 degrees north. We both agree, albeit for different reasons, that the elevation angle will reduce from 60 to 30 degrees. Given that the stars are now significantly more distant, according to your model, one would expect them to appear closer together - a reduced angular separation. Can you explain how exactly the compensation would work? By your golf ball analogy, objects moving lower would appear to get closer together, not further apart, which is the correction your model needs to work. 




10

When coming in at a lower angle the apparent angles are different. Elevated angles look differently from the observer's vantage point and are height dependent. You assume that it would always equal, but this is not so.

This doesn't make any sense. We know what angle to the horizontal the light rays are arriving at the observer's eyes, because we know their latitude, so we have a good idea, in your model, of what the end result of the curvature is, even if you can't say for sure how far away the stars are. If they are at 30 degrees North, then light from a star over the North Pole will arrive at an angle of 30 degrees to the horizontal. It has to. But that arrival angle is all we need to know - it doesn't matter whether the light rays are straight, bendy, wiggly or whatever. We know they aren't curving laterally - viewed from above they trace a straight line. If you have two light rays arriving at your eyes from two equidistant objects at the same height, then they can move up and down all day long, but as long as they move together, the angle you see between them won't change.

The only way to get anything looking like a circle facing the observer in the video on page one is if there was compensation inherent in the geometry of this.

The earth could be globe-shaped, and the stars could be a long, long way from us. That would work, would it not?

11
Thanks for the reply Tom.


The problem is that you want to use some parts of EA as if things operate in straight line geometry and think that you have identified a view and situation where it must apply. But this is incorrect. There is also distortion from that when viewing stars on a lateral view as well.

From this top down view an observer is observing two stars:



From a "3D" view of this below we can see that the closer star would create one angle, but if the curve of one star is dropping down to a lower elevation laterally then the rays of the second star would dip to a lower elevation and the angle the observer sees between the stars would not match the prediction of straight line geometry.



The points and curves I made here are somewhat arbitrary to show a point, but we can here that one star would create a greater curve than the other. The angle between the curved lines wouldn't make the same angle in space as the straight lines and angular separation as you envision it to be.


But in the example I gave, the angle between the stars was fairly small, meaning the distance between them was small compared to the observer's displacement. This means the hypotenuse of the triangle is essentially the same length as the adjacent side, which would mean, going from your EA diagram, that the curvature of the two paths would be very similar.

And if that difference is still too big for you, consider a situation where the observed stars are either side of the pole, and our observer retreats away down a line of longitude splitting them in two - the two light rays would be identical on either side, so the effect you're relying on there wouldn't happen. This would be as if the 'Ob' in your first diagram moved to the right, to a point equidistant from the two stars. How in that case would you explain the lack of changing angular separation as the observer moves closer and further from the mid-point between the stars? That's essentially very similar to your '+' example - if you're equidistant from the stars either side, why would one star dip more than another? Surely they would just dip or rise together as you changed latitude?

12
I am also looking at it from a practical point of view. If the stars were so near then they would also have to be very small. I can aim my 20" telescope at any star and even through 300x or more magnification I cannot see any star as anything more than a point of light. Perfectly explainable if the stars are at effective infinity in terms of distance. Their angular diameter on the sky is less than the resolving limit of my telescope. Therefore I only see an Airy disk formed optically by the telescope. Yet I can see fine detail on the Moon through the same eyepiece. So if the stars are so near and so small then what has been their source of power all this time?  Why do they vary in colour and brightness?  FE promote 'questioning' things so I am questioning my observations against their theories and claims.

The Sun they claim is only 32 miles across. So likewise, what is the source of power? That has been shining for as long as the Earth has existed and we know from geological studies that is millions of years. Or are they going to insist that geologists are lying to us as well. I can see the Sun as a disk in the sky half a degree in diameter but if that is also only 3000 miles away (as FE Wiki lays claim to) then why can I not see any physical disk for any other star?

I am not going to limit myself to purely what I can see with my naked eye when I have access to my own telescopes and cameras etc. If I have equipment available to me then I will use it. As have millions of other astronomers over the world.

Bottom line is that what the FE Wiki claims simply doesn't add up to what we see in the real world. No matter how you dress it up.

All very good points - I entirely agree. And that's before we even touch on the obvious issues with the Southern Hemisphere...

13
It seems to be the case that those with a FE mindset have a problem with any kind of non-contact force. But then that kind of includes all the fundamental forces such as gravity, magnetism and even electrical force.

It seems that having a belief that the Earth is flat means you effectively have to throw out the whole of physics.  Which seems a bit ridiculous to me.  But then so too does believing the Earth is flat. It's OK though to propose some non-existent property of light which magically bends light in just the right way under just the right conditions to produce the impression of sunrise, sunset, the phases of the Moon or indeed the observed motion and positions of the stars.  That is some property!

If it were true and real you'd think there would be a mention about that somewhere other than just the FE Wiki wouldn't you?!?

I guess when you join a forum which is part of a FE website then you are naturally going to get some people who have 'unconventional' ideas about stuff.

Whilst I completely agree, I think I come at it from a completely different angle. This is a debating forum, and whilst you're absolutely right in that if something was true then it probably wouldn't only be found on the FE wiki, that is a form of an appeal to authority - it's a fallacious argument. Likewise, it is highly unlikely that light does bend so conveniently, but that's not why the concept is wrong. I'm trying to show that the idea can be shown to be wrong by simple observation - the moon wouldn't wouldn't stay the same size, nor would the stars retain the same angular separation as they rotated, or as the viewer moved north and south, if the earth was flat. Even the bendy light, convenient though it may be for explaining some things, cannot explain that observable fact.


14
There is nothing to tell me what that 'mechanism' is, how it does what it does or how it can be controlled to provide the effects that we observe.
I don't believe there is any known mechanism behind UA or EA or the magnifying affect which keeps the sun and moon a consistent angular size despite their vastly varying distances.

This is another issue I have with Tom. I've seen him attack gravity because there is no mechanism for why mass attracts mass (although I believe Relativity does actually explain this). But he readily accepts these other mechanisms which have no known mechanism or explanation behind them.

Indeed. The sun and moon apparent size issue is very similar to the star separation point I made previously - they wouldn't hold constant if they were moving the way they are supposed to according the model presented in the wiki.

15
Azimuth angles are not maintained. The azimuthal angle between two bodies will not be the same for two observers who view the celestial bodies from any position. Take an Azimuthal Grid Chart and take two points overhead which gives an azimuth angle of about 180 degrees separation and then try to put the two points at lower elevation angles on the chart.



Apologies - you're absolutely right, I've muddled the conversation by using the word 'azimuth' which is of course refers to a horizon measurement, which does indeed change with position.  I was trying to refer the to lateral angular separation between the stars - the angle an observer would see if they measured it, or held up a suitable sight marked with angular graduations etc. That absolutely does stay the same regardless of position, which is how and why star almanacs can refer to star's positions by means of RA and dec number pairs - the numbers don't change with location, just their relative position in the sky. The time and date site has quite a good night sky simulator where you can change viewing location and see this effect yourself - I played with varying latitude in the northern hemisphere and it shows the principle very nicely - Polaris just moves higher up in the sky and all the other stars around it retain their relative positions - https://www.timeanddate.com/astronomy/night/@80,-0.

If you work out the angular separation between stars, they remain the same. That's not what your model would predict, as per my diagram - you would expect the angular separation to reduce as you got further away. It doesn't, though, does it?


16

You basically appear to be asking why the stars always maintain the same angular distance away from each other

Yes, that's exactly it. As Trillion rightly points out, in conventional science we explain that by pointing out that the stars are so far away that our movements around the earth are trivial in comparison. They are effectively at an infinite distance.

like in my previous EA diagram that I provided and want to "see workings". You have conceded that the diagram I had provided would also work in three dimensions,

No, not really - I agree that it would explain the vertical movement. The problem is that you have to explain the lateral aspect of the problem as well.


 if it were split into a three dimensional symmetrical cross section, looking like a + from above, with four stars instead of two. So lets take one arm of a 2D scene and look at it:



It is easy to see graphically why the angular displacement would be the same for any two stars for the observers. As seen in the diagram, below a star the star is 90 degrees overhead to the observer. At the extremes of either arm of a graph the star is 0 degrees at the horizon of the observer. The stars and celestial bodies set relatively, but not exactly, consistently, meaning that the the angle of descent is spaced relatively consistently along the length of the diagram.

Take one arm (one half of a EA diagram) and overlay it with another arm. The angle you see between one star and another star will compensate to see the same angular displacement between each star, in a static scene, wherever the observer is within the light sources.



Moving Star 1 away a little bit:



This is why, in a graphical manner, this occurs.

That's just a rehash of our earlier discussion - I'm absolutely happy with that (although I clearly disagree with the model being proposed). You need to address the azimuth angle issue. Your explanation of EA says that light bends upwards. If we accept that premise, and then model what that would mean for an observer looking at two stars some way in the distance, we would see something like this:




If that observer moves closer or further away from the stars, then it is obvious that the azimuth angle he perceives, indicated by the green dashed arc, would change. Do you disagree with my diagram?

17
EA in the sense that FE keep on harking on about it doesn't exist. It doesn't need to exist because the conventional spinning globe of Earth can account for everything in the sky that we see without having to bend light in the fanciful ways that they say it does. 

EA does exist but it's meaning in conventional physics is not even slightly related to the way FE mention it. They have just borrowed the term and re-defined it to their own suiting. FE are quite good at mis-representing things to make them appear to suit what they believe. Toms recent 'confusion' between tracking and guiding in relation to equatorial mounts was a classic example.

It is now relatively easy to measure the distances of the nearest stars. Even with amateur equipment. So for Tom to even mention the stars are only 6000 miles away is nonsense. If they were that near they would also be very small. Very small indeed. No matter what telescopes we aim at stars, they never appear as anything other than a point of light. So what does he think has been making such tiny stars shine for as long as they have been?  What would be the energy source for such tiny stars?

There’s hundreds of things wrong with it. I’m just picking one thing, in this case perspective, and running with it until we reach some kind of conclusion. All your other points are of course entirely true, although I generally prefer to stick to those things that people can observe for themselves, as that seems to be the mantra of the FE community - much like the OP’s original request for simple experiment ideas.

18
I can't determine exactly what you are describing due to the usage of azimuth angle which appears unrelated to common definitions,

I am using azimuth in the conventional sense of the term - https://en.wikipedia.org/wiki/Azimuth .

In this case, we are measuring the angle, at the eyes of an observer, between two stars. It's pretty straightforward.

but appears you are trying to use a straight line geometry analysis on this.

Yes, I am, because that's a completely reasonably way to analyse this problem. Unless you are now claiming that light curves laterally as well as vertically?

You say "I've kept the example deliberately small so we don't stray into needing to make big corrections for EA" like you assume that the angles aren't constantly slightly changing if you even move your eye an inch.


That just doesn't make any sense I'm afraid. The EA reference was regarding the 300nm estimate for the distance between the Polaris and a star 5 degrees away. My point was that if I had chosen a larger number, like 30 degrees, then your alleged curvature of the light rays would prevent a simple conversion of degrees to distance in nm by multiplying by 60. I wasn't talking about moving the observers eyes.

If you still don't understand where I'm coming from, then put it to bed by doing your own calculations for us. We have two stars, one overhead the North Pole at your suggested altitude of 6000nm, and one at the same altitude 300nm miles away, above say 85N 00W. The question then is what angle would an observer who is at 30N 90W see between the two stars? Show your working as they say.

It's really a simple question of perspective - as you get further away from things, the apparent angle between them reduces. Your system upends that without explanation.

19
Two points above an observer do not maintain their apparent angular displacement no matter how far they are above the observer.

If we imagine that the image you provided was a three dimensional scene starting with those stars at the same altitude as the observer, if we then increase the altitude of the stars over the observer nearest the North Pole the angle the observer sees between the two top and bottom stars would become less and less as the stars get further and further from the observer.

If, instead, we imagine that the previous image I provided is three dimensionally sliced through the center with a copy of itself on other axis like a + when viewed from overhead, creating a symmetrical three dimensional scene with four stars instead of two, we can see that the consistent angles would also work on the other axis.

You're absolutely right in the sense that the angle we're interested in is the apparent azimuth as far as the viewer is concerned - the azimuth angle is tilted up at the elevation angle, as if the observer was making an azimuth measurement. So yes, my diagram is something of a simplification, in that the two pairs of red lines aren't precisely comparable, but the point I'm making is that the difference is enormous, and correcting for elevation doesn't fix the problem.

Let's take an example, with one viewer at the North Pole observing a star that is 5 degrees away from the pole, so roughly 300nm laterally, and one observer at 30 degrees north, so 3600nm away. The calculation for our polar observer is simple - it's 5 degrees, whichever way he looks, and wherever the star is on its circular track. I've kept the example deliberately small so we don't stray into needing to make big corrections for EA...difficult to do as you haven't actually got a formula for it! I hope you agree that the difference from a straight line is pretty negligible at 5 degrees.

Our more southern observer requires some maths. The position is 3600nm horizontally, and displaced 5400nm vertically according to your updated number. Pythagoras gives us a direct viewing distance for Polaris of around 6500nm. The other star is roughly 300nm displaced from the pole at the same altitude (we're assuming it's at its maximum azimuthal displacement, 3 or 9 o'clock around Polaris with respect to the ground/observer). So now our 6500nm viewing line becomes the adjacent side of a new right angled triangle, the 300nm becomes the opposite. Trigonometry gives an azimuth angle of around 2.6 degrees for our distant observer (tan 2.6 = 300/6500), so roughly half the azimuth angle for our polar observer.

That's a massive difference - that means the distant observer would see elliptical movement, with the ellipse roughly twice as high as its width. That is not what we observe, is it? Do check my maths, of course...it's late.

20
What you pointed out with that particular diagram is part ongoing debate about EA. The original diagram was made assuming a distance to the celestial bodies of about 3000 miles, the current listed distance to the celestial bodies for FET from the pre-internet FES.

Well, ok, but there doesn't seem to be much debate going on about it! Maybe you should change the wiki?

As I have mentioned a number of times in the past, I have dispute with that distance for use with EA.

25. How big is the sun and how far away from the flat earth disc is it?

If EA is considered, the celestial bodies are possibly about 6000 miles in altitude. Assuming that the distance from the NP to the Equator is correct, it takes about 6000 miles for the North Star to set when traveling from the NP where it is overhead to the Equator where it is on the horizon. Therefore, if EA causes bodies to descend consistently, the North Star would be an equal distance above the Earth.

Some of the information is from the pre-internet Flat Earth Movement done under straight line trigonometry and is held as an informal number. For a background see: https://wiki.tfes.org/Distance_to_the_Sun



Work needs to be done by the modern society before it can be updated. EAT is becoming a popular astronomical model. Building off of Voliva's 3000 miles distance for the sun at 45 degrees in the sky via straight line trigonometry, and assuming that the sun moves in the sky at around a constant pace (although I've heard that it moves slightly quicker at zenith), my own initial estimation would be that under the schema of EAT at 90 degrees zenith the Sun would be somewhere about 6000 miles in height; although this is assuming that the area of illumination is linearly related to its distance in the sky. EAT proponents can give their own ideas.

I also found some correlation with distances and moon positions here on the other forum, using an altitude of 6100 miles.

If the distance to the celestial bodies is about 6000 miles and we stretch the diagram vertically to make it about twice as high then the angles will be about the same at the end cross sections.



Well, that's great. You could actually improve that diagram by putting a graduated scale on it - we know that the North Star (or Sig Oct, in the Southern Hemisphere) appears at almost precisely the same elevation angle as the observer's latitude, so the angle of those rays meeting the observers' eyes, that we have both crudely drawn, is actually a precisely known quantity - 90 degrees at the pole, 80 degrees at 80 degrees north, 70 at 70 north etc.

The problem you have though, is that whilst that magics away the elevation angle problem, it doesn't address the azimuth issue - I'm curious to know how you would explain that. Back to crudely drawn stick men, this time in plan form:



We have one observer close to the pole, observing two stars moving in circles around the pole star. That observer perceives a particular angle between those stars, as shown by the red lines. Our other observer is much further away, at say 30 degrees north. That observer sees the same stars, but your model would have the observed azimuth angle, as shown by the red lines, very much less than that for the closer observer. That doesn't match observations - the angle between stars doesn't change with latitude. The generally accepted, conventional RE model would explain this by pointing out that this distance between the observers and the stars is massive compared to the size of the earth, hence no observable angle difference. Thoughts?

Pages: [1] 2 3 ... 22  Next >