1

Flat Earth Theory / Re: Looking for curvature is a fool's errand.

« on: May 09, 2023, 09:59:33 AM »

When I travel to the other end of Lake Ontario and view my home city of Toronto from the city of St. Catharines, why does the CN Tower appear 2/3 as tall as it should?

If the Earth was flat the CN Tower should appear as-is. In the image below, you can see a Mirage Effect (eg. where the white roofed Rogers Center is smeared) but due to its gigantic height you can see how low the CN Tower dips below the horizon.

Your superimposed image shows buildings that are in your original image therefore they have not been 'hidden' by any such curve. And as you point out the white building has taken on a mirage effect. Any reason the white building features twice (above and/or below itself) in the superimposed pic?

2

Flat Earth Theory / Re: Appearance of the sun

« on: April 20, 2023, 08:53:51 AM »

Delegating perspective shrinking to an illusion is even worse for you, as you are now stamping your feet about one optical illusion over another optical illusion.

Not at all, no stomping required. Just seemed that you were singling out that illusion is somehow "special" and unique to astronomy when it's no more special than terrestrial perspective illusion.

Although the photo of you posted was a result of 'trick' photography (using a tilt/shift lens) and not a natural optical illusion.

3

Flat Earth Theory / Re: Curvature of the Horizon

« on: April 02, 2023, 09:03:03 AM »

.... This therefore suggest that for every 1.57 metres I travelled down the global building that I dropped 1 metre in height (and is obviously a constant for a circle/globe). ....

No, it is not a constant for circles/globes.  It would be a constant for a straight slope of constant angle.  This is what you do not or perhaps refuse to understand or acknowledge.  Do you really think that if standing next to someone on the top of this building and that they take only 2 steps away from you that they will then be a full meter lower than you are?

And a circle is constant. Every section no matter how large or small has the same curve as any other section of that circle. Thats constant.

Yes but you are not talking about a drop relative to each section but to the starting point (at N) and that is NOT constant.   Someone moving 2 steps away from a person at N (on your dome) would NOT be 1.57m lower.  As others have pointed a constant rate of your "drop" results in a straight slope not a curve.

There are no straight slopes on a globe. They are all curved. If i walked 1.57 miles due south from someone at the north pole i would be 1 mile lower than them on the globe. But because that doesnt appear to occur then the earth can not be global.

4

Flat Earth Theory / Re: Curvature of the Horizon

« on: April 02, 2023, 09:00:39 AM »

You are always 'dropping' at the same rate.

My favourite question applies here - "relative to what?" If you choose a fixed point on RE, then the "drop" is not, in fact, occurring "at the same rate". If you choose the traveler's own frame of reference and measure it at some consistent interval, then it is. However, you fail to consistently apply one frame of reference to your logic, which introduces contradictions.

Your argument relies on a misrepresentation of RET. Please don't do that.

For explanatory purposes I start at one point on a fixed global earth. The north pole. I could start anywhere but its better to imagine it from the north pole. And for the example am assuming north is always at the top and south is always at the bottom of the globe.
Am not sure what am misrepresenting in relation to RET but I am demonstrating why the earth can not be a globe due to the rate of drop - folk have been relying on the 8" per Mile squared. This is a pure constant it never changes and works.

5

Flat Earth Theory / Re: Curvature of the Horizon

« on: April 02, 2023, 08:57:26 AM »

I have revised the image to hopefully better explain this.

Instead of walking from N to E1 imagine walking from N to X. This is half the distance to the equator and represents one eighth (1/8) of the earths circumference ie 3,113 miles.  Can we agree on this?
If so the drop/fall/decrease in height in relation to the north pole (call it whatever) will be equal to 1,982 miles ie one half (1/2) the radius of the earth. Can we agree on this?
If either of the above figures are incorrect please tell me how?

Accepting the above if we divide 3,113 miles by 1,982 miles we get a drop/fall/decrease in height in relation to the north pole of 1 mile per 1.57 miles travelled.

Like it or not and forget what I have called these dimensions does anyone disagree with these maths?

Hopefully not. And regardless of what others have said every single infinite point on a circle is at the 'top of the curve'. Above that point the circle curves away as does it below that point wherever that point is on the circle. And as a circle is one continuous curve there are no parts of the curve that are any different to other part. Take any two segments of the curve and they will be identical no matter where on the circle they came from.

Now instead of me walking 3,113 miles I am going to divide the circle into 360 (purely for conventional purposes - I could have chosen any figure to divide it by; 100, 125, 299 - it wouldn't make any difference). The circumference of the earth divided by 360 = 69 miles. I am now going to walk that 69 miles from the north pole. And when I have finished I will be at a point on the circle some 43 miles below the north pole. Forget linear dimensions they don't matter. The fact is I will have dropped by roughly 43 miles. Or to make it simpler 1 mile for every 1.57 miles travelled around the circumference. And if someone stood at the north pole and watched me walk 1.57 miles away from them I should be at a point 1 mile below them. These figures are irrefutable. Its down to the wording. If anyone disagrees can you please do so in layman's terms? Many thanks

Well, yes we do disagree with your maths and find the figures entirely refutable. In layman's terms, I'll try drawing out what you are actually describing. Starting at the north pole, you travel 1.57 miles and find yourself 1 mile lower than the pole:–



Another 1.57 miles and you're another 1 mile lower than the pole:–



On and on, for each 1.57 miles you travel, you're another mile lower than the pole:–



Does the path travelled bear any resemblance to the curve of a globe? No, it doesn't, it's a straight line: you are travelling down a constant slope.

If you disagree, explain in layman's terms.

What you say is almost true and in the spirit of what I am saying. But there are no straight lines on a globe. They are all curves. So i walk 1.57 miles from the north pole heading due south to the north pole. The line I have walked is a curve although it will look straight if looking from above and it will feel straight to me. It is a curve. And when i have travelled that 1.57 miles curve my rate of drop in height will be 1 mile (for each 1.57 miles).

6

Flat Earth Theory / Re: Curvature of the Horizon

« on: April 02, 2023, 08:54:22 AM »

You dont read. Am not going 'around' the building I am going vertically down it.

You've said you're going along the circumference cleaning windows - that's what I mean by 'around'.

Okay - hopefully this should explain what am saying although I would appreciate some assistance from any mathematicians out there.

This final diagram is of a quarter circle to make things simple.

Duncan - please look at the image. Its the same dimensions as my global building - it has a 250 metres diameter but I am showing it as a quarter circle therefore half the diameter is 125 metres (aka the radius). Hopefully you will see that if I travel from point A along the circumference in the direction of the arrows to point B that I will have covered 197 metres. Now if you look to the left of point B to point C you will notice that I have 'dropped' in height by 125 metres. Please tell me what it is you don't understand about that as you have already accepted that this is correct.

By dividing 197 metres by 125 metres you get 1.57. Therefore for every 1.57 metres travelled down the curve ie from A to B you will descend in height by 1 metre.

Come on guys give me a break? Its simple maths.

Well, we're trying to help. So now redo your diagram, but only travel 1.57 metres around the circumference - less than one degree of the circle. Even if you don't trust my maths, you should be able to see graphically that you won't 'drop' anything close to one metre.

What I said was I freeline halfway down the building. Not around it. If I freelined 1.57 metres down the building my rate of drop in height would be 1 metre (per 1.57 metres).

7

Flat Earth Theory / Re: Curvature of the Horizon

« on: April 01, 2023, 08:54:30 PM »

Okay - hopefully this should explain what am
By dividing 197 metres by 125 metres you get 1.57. Therefore for every 1.57 metres travelled down the curve ie from A to B you will descend in height by 1 metre.

I guess on average that’s true, but you can see that the first few meters there’s hardly any drop and the last few meters you’re dropping pretty much the whole meter (with respect to the top of the curve).
Because it’s a curve.

No way. Never mind on average its a fact all the way round; the curve is the same all the way round - it doesn't drop away any more or less than anywhere else on the circle. Choose any 2 arcs on a circle it doesn't even matter if one is longer than the other and overlay one against the other - the curve is identical. You are always 'dropping' at the same rate. I could divide a circle into any number of degrees/sections you like - 1000 for example and the ratio of circumference to 'drop' will be 1.57 every time. The reason people are finding it so hard to comprehend is because that's not what it looks like on a globe earth. Now why might that be I wonder?

8

Flat Earth Theory / Re: Curvature of the Horizon

« on: April 01, 2023, 08:30:06 PM »

Okay - hopefully this should explain what am saying although I would appreciate some assistance from any mathematicians out there.

This final diagram is of a quarter circle to make things simple.

Duncan - please look at the image. Its the same dimensions as my global building - it has a 250 metres diameter but I am showing it as a quarter circle therefore half the diameter is 125 metres (aka the radius). Hopefully you will see that if I travel from point A along the circumference in the direction of the arrows to point B that I will have covered 197 metres. Now if you look to the left of point B to point C you will notice that I have 'dropped' in height by 125 metres. Please tell me what it is you don't understand about that as you have already accepted that this is correct.

By dividing 197 metres by 125 metres you get 1.57. Therefore for every 1.57 metres travelled down the curve ie from A to B you will descend in height by 1 metre.

Come on guys give me a break? Its simple maths.

9

Flat Earth Theory / Re: Curvature of the Horizon

« on: April 01, 2023, 08:03:33 PM »

.... This therefore suggest that for every 1.57 metres I travelled down the global building that I dropped 1 metre in height (and is obviously a constant for a circle/globe). ....

No, it is not a constant for circles/globes.  It would be a constant for a straight slope of constant angle.  This is what you do not or perhaps refuse to understand or acknowledge.  Do you really think that if standing next to someone on the top of this building and that they take only 2 steps away from you that they will then be a full meter lower than you are?

And a circle is constant. Every section no matter how large or small has the same curve as any other section of that circle. Thats constant.

10

Flat Earth Theory / Re: Curvature of the Horizon

« on: April 01, 2023, 08:02:31 PM »

.... This therefore suggest that for every 1.57 metres I travelled down the global building that I dropped 1 metre in height (and is obviously a constant for a circle/globe). ....

No, it is not a constant for circles/globes.  It would be a constant for a straight slope of constant angle.  This is what you do not or perhaps refuse to understand or acknowledge.  Do you really think that if standing next to someone on the top of this building and that they take only 2 steps away from you that they will then be a full meter lower than you are?

You arent reading this correctly. Its nothing to do with being 2 steps away from someone. Its to do with being 2 steps away (your example) from someone on a perfect curve. Stand on the top of St Pauls Cathedral. Ask someone to stand 1.57 metres away from you. They will be in the region of (because its a dome roof not a circle) 1 metre lower than you.

11

Flat Earth Theory / Re: Curvature of the Horizon

« on: April 01, 2023, 08:00:17 PM »

To make it a little easier to explain. Imagine the big round circle in my diagram is actually a big round global building - a perfect sphere. And S is the base of it and N is the top of it. As I stated previously I am assuming a global stationery earth so this global building is no different. And for this example lets assume the building has a 250 metre diameter with a 786 metres circumference (all future numbers will be rounded up).
I have the window cleaning contract for the building and my anchor point is at the top (point N). I anchor myself to the top of the building and freeline halfway down the building. In doing so I have travelled one quarter (1/4) of the circumference of the building ie 197 metres. I have also dropped in 'height' 125 metres.

This is correct , so far.

Divide 197 by 125 and we come up with that magical figure of 1.57 metres. That same figure I arrived at earlier using the earth as the model. This therefore suggest that for every 1.57 metres I travelled down the global building that I dropped 1 metre in height (and is obviously a constant for a circle/globe). And I can't understand why no one agrees with this.

...but this is wrong, I'm afraid. As the others have said, it's not linear. If you go 1 metre around the circumference of your building, you'll have gone 1/768th of the way around, which is 0.458 degrees. The 'drop', as you describe it, for that first metre, is 125 metres (the radius) - (cos 0.458 x 125), which comes to 0.004 metres. Close to the equator, the relationship goes the other way - almost all of the travel is 'drop', referenced to the 'top' of the building. But of course for us on earth, wherever we are feels like the top - indeed the choice of the North Pole as the top is entirely arbitrary. So every mile we travel the 'drop' remains very small indeed.

You dont read. Am not going 'around' the building I am going vertically down it. And for every 197 metres down the curve I have dropped 125 metres in height. Wheres the confusion? You have already accepted that.

12

Flat Earth Theory / Re: Curvature of the Horizon

« on: April 01, 2023, 07:58:48 PM »

To make it a little easier to explain. Imagine the big round circle in my diagram is actually a big round global building - a perfect sphere. And S is the base of it and N is the top of it. As I stated previously I am assuming a global stationery earth so this global building is no different. And for this example lets assume the building has a 250 metre diameter with a 786 metres circumference (all future numbers will be rounded up).
I have the window cleaning contract for the building and my anchor point is at the top (point N). I anchor myself to the top of the building and freeline halfway down the building. In doing so I have travelled one quarter (1/4) of the circumference of the building ie 197 metres. I have also dropped in 'height' 125 metres.

This is correct , so far.

Divide 197 by 125 and we come up with that magical figure of 1.57 metres. That same figure I arrived at earlier using the earth as the model. This therefore suggest that for every 1.57 metres I travelled down the global building that I dropped 1 metre in height (and is obviously a constant for a circle/globe). And I can't understand why no one agrees with this.

...but this is wrong, I'm afraid. As the others have said, it's not linear. If you go 1 metre around the circumference of your building, you'll have gone 1/768th of the way around, which is 0.458 degrees. The 'drop', as you describe it, for that first metre, is 125 metres (the radius) - (cos 0.458 x 125), which comes to 0.004 metres. Close to the equator, the relationship goes the other way - almost all of the travel is 'drop', referenced to the 'top' of the building. But of course for us on earth, wherever we are feels like the top - indeed the choice of the North Pole as the top is entirely arbitrary. So every mile we travel the 'drop' remains very small indeed.

You agree am right - theres nothing wrong with the second half of my submission. Please forget your formulas - just do the maths its easier. Use my figures and see what happens. You admit that by me travelling 197 metres down the circumference i am going to drop 125 metres in height. That should be the end of it. Nothing further needs to be said. Cant you see?

13

Flat Earth Theory / Re: Curvature of the Horizon

« on: April 01, 2023, 07:56:32 PM »

You're just not getting this are you?

If you come halfway down the circumference of a dome, you are not vertically-halfway to the ground.  The relationship between circumference and "drop" is not linear. 

If you want to "drop" halfway to the ground, you have to travel 2/3 of the circumference between apex and ground

Duncan - you actually do concur with me. You just dont realise it. By travelling 1.57 units of measurement you will drop 1 unit of measurement in height. You just prefer to deal in imperial instead of metric. Try it.

14

Flat Earth Theory / Re: Curvature of the Horizon

« on: April 01, 2023, 07:53:20 PM »

You're just not getting this are you?

If you come halfway down the circumference of a dome, you are not vertically-halfway to the ground.  The relationship between circumference and "drop" is not linear. 

If you want to "drop" halfway to the ground, you have to travel 2/3 of the circumference between apex and ground

Its you who isn't getting this. I have never said am halfway down the circumference AND halfway down the dome. What am saying is for every 1.57 'units of measurement' I travel down the circumference I am going to drop (in height) by 1 unit of measurement. Try it yourself it works every time. And the relationship between the curve of a perfect circle and the drop IS relative - it has to be. Its 1.57 (or thereabouts depending on how many places you put after the point for Pi).

15

Flat Earth Theory / Re: Curvature of the Horizon

« on: April 01, 2023, 10:31:39 AM »

To make it a little easier to explain. Imagine the big round circle in my diagram is actually a big round global building - a perfect sphere. And S is the base of it and N is the top of it. As I stated previously I am assuming a global stationery earth so this global building is no different. And for this example lets assume the building has a 250 metre diameter with a 786 metres circumference (all future numbers will be rounded up).
I have the window cleaning contract for the building and my anchor point is at the top (point N). I anchor myself to the top of the building and freeline halfway down the building. In doing so I have travelled one quarter (1/4) of the circumference of the building ie 197 metres. I have also dropped in 'height' 125 metres. Divide 197 by 125 and we come up with that magical figure of 1.57 metres. That same figure I arrived at earlier using the earth as the model. This therefore suggest that for every 1.57 metres I travelled down the global building that I dropped 1 metre in height (and is obviously a constant for a circle/globe). And I can't understand why no one agrees with this. Forget the earth and its gravity and its unevenness and the fact it rotates and doesn't have an up or down per se and just think of it as one continuous curve. Now look out to see from the shore 1.57 miles. There is no evidence of the 1 mile fall away (I won't use the word height as its confusing) of the curve. Yet it happens on the building why shouldn't it happen on a similar size and shaped object.
Can we use this as something to build on as I am certain there is more to this than meets the eye? And I really would like to know what is causing such an impasse. And for those who say it isn't true simply because it doesn't manifest on earth this way then think again...is that because you were wearing your global hat at the time?
Looking forward to hearing further comments.

16

Flat Earth Theory / Re: Curvature of the Horizon

« on: March 31, 2023, 08:14:18 PM »

I am not a scientist so please bear with me. My definition is the amount by which the curve 'drops' in 'height' on an assumed non-rotating global earth from any single point on that globe. And for illustrative purposes my example would be a person standing at the north pole on that globe (the north pole being at the 'uppermost' part of that globe) would see the curve fall in height by 1 mile for every 1.57 miles of circumference.

Curvature is measured as the angular turn per unit distance. Your definition seems to be based on straight line measurements.

Ok lets forget about the actual curve itself for a moment. What I am calculating is how much does the curve drop in height (assuming a non-spinning globe earth that has a top and bottom). So if I walked a quarter of the earth's circumference from the north pole to the equator in a straight (obviously curved) line I would cover 6,225 miles (or so).  And in doing so I would have dropped in height by 3,963 miles (the radius of the earth). Therefore for every 1.57 miles I walked there is a drop in height of 1 mile. Using the ocean as an example; and again assuming a globe earth, if rowed out to sea a distance of 1.57 miles there should have been a drop in height of 1 mile. Now a physical drop of 1 mile in height is something we just do not see (in fact we see no such thing and to us it looks quite level) but we should see it if we were on a globe earth.
Looking at it another way. If I walked across the salt flats for 1.57 miles I should be 1 mile lower than when I started. And am sure we all know that this is not the case.
I am not sure if I am explaining this as I intended or indeed correctly but would welcome some genuine advice/debate/discussion on this particular matter as something just doesn't seem right and am sure I haven't miscalculated the actual maths.

The fundamental mistake you are making is an assumption that your "Rate of Drop" is linear; it isn't.  The "rate of Drop" as you call it increases as you travel south. 

Consider standing at the North Pole in your model and travel 1 mile.  Your actual drop is negligible, and you can probably still see the Pole.  The Rate of Drop is zero. 

Now stand 1 mile north of the equator and then walk to it.  Your Rate of Drop is now 1 mile per mile. 

Your formula only works if the drop is linear, as if the Earth was a cone.

I have revised the image to hopefully better explain this.

Instead of walking from N to E1 imagine walking from N to X. This is half the distance to the equator and represents one eighth (1/8) of the earths circumference ie 3,113 miles.  Can we agree on this?
If so the drop/fall/decrease in height in relation to the north pole (call it whatever) will be equal to 1,982 miles ie one half (1/2) the radius of the earth. Can we agree on this?
If either of the above figures are incorrect please tell me how?

Accepting the above if we divide 3,113 miles by 1,982 miles we get a drop/fall/decrease in height in relation to the north pole of 1 mile per 1.57 miles travelled.

Like it or not and forget what I have called these dimensions does anyone disagree with these maths?

Hopefully not. And regardless of what others have said every single infinite point on a circle is at the 'top of the curve'. Above that point the circle curves away as does it below that point wherever that point is on the circle. And as a circle is one continuous curve there are no parts of the curve that are any different to other part. Take any two segments of the curve and they will be identical no matter where on the circle they came from.

Now instead of me walking 3,113 miles I am going to divide the circle into 360 (purely for conventional purposes - I could have chosen any figure to divide it by; 100, 125, 299 - it wouldn't make any difference). The circumference of the earth divided by 360 = 69 miles. I am now going to walk that 69 miles from the north pole. And when I have finished I will be at a point on the circle some 43 miles below the north pole. Forget linear dimensions they don't matter. The fact is I will have dropped by roughly 43 miles. Or to make it simpler 1 mile for every 1.57 miles travelled around the circumference. And if someone stood at the north pole and watched me walk 1.57 miles away from them I should be at a point 1 mile below them. These figures are irrefutable. Its down to the wording. If anyone disagrees can you please do so in layman's terms? Many thanks

17

Flat Earth Theory / Re: Curvature of the Horizon

« on: March 30, 2023, 10:12:30 AM »

I am not a scientist so please bear with me. My definition is the amount by which the curve 'drops' in 'height' on an assumed non-rotating global earth from any single point on that globe. And for illustrative purposes my example would be a person standing at the north pole on that globe (the north pole being at the 'uppermost' part of that globe) would see the curve fall in height by 1 mile for every 1.57 miles of circumference.

Curvature is measured as the angular turn per unit distance. Your definition seems to be based on straight line measurements.

Ok lets forget about the actual curve itself for a moment. What I am calculating is how much does the curve drop in height (assuming a non-spinning globe earth that has a top and bottom). So if I walked a quarter of the earth's circumference from the north pole to the equator in a straight (obviously curved) line I would cover 6,225 miles (or so).  And in doing so I would have dropped in height by 3,963 miles (the radius of the earth). Therefore for every 1.57 miles I walked there is a drop in height of 1 mile. Using the ocean as an example; and again assuming a globe earth, if rowed out to sea a distance of 1.57 miles there should have been a drop in height of 1 mile. Now a physical drop of 1 mile in height is something we just do not see (in fact we see no such thing and to us it looks quite level) but we should see it if we were on a globe earth.
Looking at it another way. If I walked across the salt flats for 1.57 miles I should be 1 mile lower than when I started. And am sure we all know that this is not the case.
I am not sure if I am explaining this as I intended or indeed correctly but would welcome some genuine advice/debate/discussion on this particular matter as something just doesn't seem right and am sure I haven't miscalculated the actual maths.

18

Flat Earth Theory / Re: Curvature of the Horizon

« on: March 29, 2023, 07:14:06 AM »

Let's give you one shot at this. Define "fall of the curve".

I am not a scientist so please bear with me. My definition is the amount by which the curve 'drops' in 'height' on an assumed non-rotating global earth from any single point on that globe. And for illustrative purposes my example would be a person standing at the north pole on that globe (the north pole being at the 'uppermost' part of that globe) would see the curve fall in height by 1 mile for every 1.57 miles of circumference.

19

Flat Earth Theory / Re: Curvature of the Horizon

« on: March 28, 2023, 09:06:58 PM »

Okay I agree that curvature of the horizon from left to right is not visible from the surface of the earth.
What I am wondering is what sort of curvature would you expect to see... would it be in a north south direction? An east west direction?

If you expect to see curvature what happens when you are in the middle of the ocean (or somewhere else where you could see the horizon in all directions) and turn around 360 degrees? Would you expect to see the horizon at a lower level when you have turned 180 degrees and then rise up again as you complete your 360 degree rotation?

Just wondering what the flat earth believers expect to see when they look at the horizon and declare "It's flat, no curvature there". But especially what would you expect to see if you could turn around 360 degrees and see the horizon in all directions. Isn't a flat horizon as you rotate around 360 degrees what you would expect to see if the earth is a sphere?

Because the flat horizon is the major point which seems to persuade people that the earth is flat. But it seems illogical to me that people would expect to see a curve down to either side when eg viewing a picture of the horizon.
Yet in reality there is curvature, but just not side to side as we look toward the horizon, instead the earth curves away from you - in every direction - as you look toward the horizon and rotate 360 degrees. And the fact that you could climb the crows nest of a ship and see further is irrefutable - after all isn't that why they had crows nests in the first place? "Land Ahoy!" So that they could see further over the horizon to see other ships coming or land in the distance. And also the curvature over the horizon is the reason lighthouses are built very tall?

In the diagram I have attached below the following apply (assuming a round earth and approx. dimensions).
N = North Pole
S = South Pole
C = Centre
E1 – E2 = Equator
Circumference (N-E1-S-E2-N) = 24,901 miles
N – E1 = 6,225 miles
E1 – S = 6,225 miles
S – E2 = 6,225 miles
E2 – N = 6,225 miles
Diameter (E1 – E2) = 7,926 miles
Diameter (N – S) = 7,926 miles
Radius (E1 – C) = 3,963 miles
Radius (C – S) = 3,963 miles
Radius (N – C) = 3,963 miles
Radius (E2 – C) = 3,963 miles

Lets say I walk from the north pole (N) to the equator (E1) a distance of 6,225 miles. And when I get to the equator the curvature of the earth has fallen away by 3,963 miles (the radius of the earth).

6,225 divided by 3,963 = 1.57. Therefore for every 1.57 miles I walked the curve fell away by 1 mile. That’s an awful lot.

If I carried on to the south pole (S) I would have walked 7,926 miles (the diameter of earth) and the curve would have fallen away by 12,450 miles; which (divided by 7,926) is 1.57 0r rather every 1.57 miles the curve falls away by 1 mile.

These figures are consistent. A circle is a continuous curve. And If I divided the circumference (24,901 miles) by 360 degrees each degree would be 69 miles in length. And the fall of the curve over each 69 miles would be 44 miles; a ratio of 1 mile fall for every 1.57 miles travelled.

I would be interested to know if anyone disagrees with these figures (errors and omissions excepted) and if so for what reason?

20

Flat Earth Theory / Re: Curvature of the Horizon

« on: March 20, 2023, 12:37:29 AM »

There are multiple accounts of most RE adherents that curvature can be detected even at ground level

Do you have any relevant quotes from people to back this up?

An oldie but a goodie claim by AATW that no RE-er has ever claimed that curvature can be detected at ground level by the human eye...

yet...

Unsurprisingly, here he is (along with his choir) in this very thread, doing just that.

It seems that you are taking this out of context. You folks were talking about the curvature observed of the horizon behind the Concorde at a cruising altitude some 70k' feet above the earth.



AATW's comment was in regard to whether one can see the horizon line curve at ground level as you claimed REr's say they can see curvature at ground level. Even the OP opened up this thread with, "Okay I agree that curvature of the horizon from left to right is not visible from the surface of the earth."

Curvature is curvature.

Just stop with the equivocation.

There is no curvature.

Compelling argument. I guess if you simply say so, then it must be true. I can't think of anyone who would know better considering the level of thought and intellect you've poured into the discussion. Clearly the curve shown in the Concorde image is not a curve as you have just commanded that it isn't. My fault for not running the image by you first so that you could determine what is seen by the rest of us and what isn't. Thanks for applying your acute observation skills to an otherwise indeterminate and murky situation.

Even the fusilage of Concorde has a slight curve to it in this pic.