[Action80]

And on a clear day, with little swell or chop, lo and behold it’s still a clear line.

I really don’t understand how people can say it’s not.

You can keep making this false statement until the end of time (if you choose), but I have already pointed out why it is false.

Can you explain what you mean? Because I don’t think you have.

On a clear day with good visibility, the delineation between sea and sky is very easy to discern.

Have you lived on the coast? How often do you look out to sea on the average day?

Although I don't currently live on a shoreline of a major body of water, I have spent ample time there.

Fact of the matter is this: the traits of both mediums, such as color and reflectivity, are such that no one person can claim with certainty what it is they are looking at from such a distant point away.

Fact of the matter is?

Sorry, no, that’s your opinion.

No, it is fact.

Every seafarer and navigator would disagree.

Every seafarer and navigator know the traits of both mediums are identical in most instances when it comes to coloration.

Yes, at time, in poor visibility, one cannot distinguish the horizon. But on many other occasions it is very clear.

Are you saying that even when it is clear, you believe that the water continues on, effectively appearing above the horizon, but that it looks to us exactly the same as the sky?

I am saying no one knows what it is they are looking at from that distance.

[AATW]

if there is something you don't understand, you can try the radical approach of asking questions.

Well alright then. Why would RE make a sharp horizon impossible? I mean, it wouldn't be perfectly sharp because of the atmosphere and it wouldn't be completely straight because of waves and that, but the line between sea and sky is, on a clear day, very distinct. At a viewer height of 10 feet the horizon is just under 4 miles away. I'd suggest that's high enough to be looking over waves on a calm day, why can't you see any more sea after that?

Now, I reckon you forgot what we're talking about by now, so let me offer a quick reminder. It was your position that a sharp edge would be proof of RE, and that a gradient horizon would be indicative of FE. My position is that this is not the case - you should be expecting a gradient in both models, and therefore your argument is a waste of time. We are now stuck on you simultaneously rejecting that there is a gradient to the horizon, while repeatedly stating "well, okay, it's not mathematically perfect, but duuuuuh". In reality, there are no ifs or buts about it. Take your favourite photo of the horizon (n.b., not a wave that's less than a mile away from the photographer) and inspect the colours. They will gradually fade away, as is expected of RE and FE alike.

I think the word "gradually" is where we are stuck. The foggy day picture is a gradual fade. The other horizon pictures are not.
Now, having thought about this some more I'm not sure there would be as much difference between a FE horizon and a RE one. My gut feeling is a FE horizon would be less distinct than the RE one, but I must concede it wouldn't be like the foggy day image on a FE.

Uuuuuuuuuuuuuurgh. Are you sure you took a course on image processing? I'm getting suspcious here.

Well alright, I'll admit it was ages ago as I'm very old. I remember we did stuff around JPG compression - there was a collective gasp when the professor put the equation for a Fourier Transform on the OHP (yes, I am that old), before he said we didn't need to know that, he was just showing us. And we did code a rudimentary edge detection algorithm. But from your explanation I'll consider myself schooled. I did say I wasn't an expert.

Tbh, I don't think we are a million miles away from agreeing now. I think there are other horizon based discriminators between FE and RE, I dipped my toes in those waters and you didn't want to talk about those.

One comment on the horizon dip thing - when I first came here the claim that the horizon always rises to eye level was vigorously defended by TB and on the Wiki. When Bobby Shafto spent an inordinate amount of time doing experiments to demonstrate that where is a dip to the horizon which increases with altitude Tom bent over backwards calling black is white to deny his results. Now it seems that Wiki page has been quietly deprecated. Which is good, I guess. It does show some progress, a lack of which I have criticised you guys for. But I didn't know that had happened, so I can understand other people not realising it either.

[Pete Svarrior]

Why would RE make a sharp horizon impossible? I mean, it wouldn't be perfectly sharp because of the atmosphere

I mean, you answered your own question. The horizon in RE curves downard with distance from the observer. This, combined with refraction, will cause it to gradually blur away.

the line between sea and sky is, on a clear day, very distinct

This continues to be false, and you've provided ample evidence for that. I really wish you could stop just restating it without making a further argument.

I think the word "gradually" is where we are stuck. The foggy day picture is a gradual fade. The other horizon pictures are not.

That certainly sounds like the disagreement. I explained what I mean by the term (and I'm going with a pretty straight-forward dictionary definition of "gradual" and "gradient"), and I showed you how you can verify the presence of a gradient in an image. My reading of what you're saying is "nuh uh, it's obviously not gradual", repeated ad nauseam.

Now, having thought about this some more I'm not sure there would be as much difference between a FE horizon and a RE one.

Yup. They'd be more-or-less identical, with too many different factors to account for to make a reasonable distinction between the two in real life.

One comment on the horizon dip thing - when I first came here the claim that the horizon always rises to eye level was vigorously defended by TB and on the Wiki.

Was it just Tom, perchance? I know he used to subscribe to the "no EA, just perspective" view, and, as far as I know, he was in the minority here.

Now it seems that Wiki page has been quietly deprecated.

"Quietly"? It sounds like you're trying to put a negative spin on this. We don't usually announce changes to the Wiki, but if you really wanted to track them, the changelogs are public.

It's how things usually work here - if a good debate takes place, we* try to update the Wiki to reflect the outcomes. This is evident in the Water Level Devices page (the closest I could find to an active defence of the horizon "always being at eye level") - it explicitly refers back to the thread, and includes copies of Bobby's pictures that tinypic has since deleted. This isn't some sneaky secret effort - it's preservation of information.

* - mostly Tom these days, though I used to have a good streak a few years ago, and I hope to get back into it.

[Tom Bishop]

Suggested changes to 'Horizon is always at eye level' in the Wiki

It looks like it was originally here: https://wiki.tfes.org/Horizon_always_at_Eye_Level

If you go to that redirect and read the content it says that the horizon is not always at eye level. So why are you claiming that the wiki says that the horizon is always at eye level or that it is the official stance?

[stack]

Suggested changes to 'Horizon is always at eye level' in the Wiki

It looks like it was originally here: https://wiki.tfes.org/Horizon_always_at_Eye_Level

If you go to that redirect and read the content it says that the horizon is not always at eye level. So why are you claiming that the wiki says that the horizon is always at eye level or that it is the official stance?

Because, quite simply, I was mistaken. Let me explain.

I was reading through ENAG to see what good old Sam had to say about the horizon, atmosphere, eye level, etc. And I came across this from Chapter 'PERSPECTIVE ON THE SEA’ :
“…it is shown that the surface of the sea appears to rise up to the level or altitude of the eye…”

He refers back to FIG. 44 under EXPERIMENT 15:


And I swore that the wiki aligned with that belief being ENAG, Rowbotham and all. But I couldn’t find anything in the wiki specifically about it. So I searched the forum and found a suggestion thread from 2018 requesting that the following statement in the wiki under the page https://wiki.tfes.org/Horizon_always_at_Eye_Level be altered:

"A fact of basic perspective is that the line of the horizon is always at eye level with the observer. This will help us understand how viewing distance works, in addition to the sinking ship effect.

Have you ever noticed that as you climb a mountain the line of the horizon seems to rise with you? This is because the vanishing point is always at eye level with the observer. This is a very basic property of perspective. From a plane or a mountain, however high you ascend - the horizon will rise to your eye level. The next time you climb in altitude study the horizon closely and observe as it rises with your eye level. The horizon will continue to rise with altitude, at eye level with the observer, until there is no more land to see.”

That particular suggestion thread somewhat died seemingly without a resolution.

However, apparently that statement was deleted in 2019 and the page redirected to https://wiki.tfes.org/High_Altitude_Horizon_Dip.

It now appears that TFES no longer holds the Rowbotham position that the horizon always rises to eye level as it did back in 2018. At least FE and GE are now in agreement on one thing.

[SimonC]

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?

[Pete Svarrior]

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

[SimonC]

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.

[Mack]

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.

 

[SimonC]

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.

[DuncanDoenitz]

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. 

[ichoosereality]

....
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. 
....

Exactly.  Perhaps playing a bit with a chord calculator like this one https://planetcalc.com/1421/ would be instructive.

Go down near the bottom to the "Circular segment - complete solution".  The height here is the distance from center of the arc to the center of the chord so to have that stand in for the north pole you will need to double your arc lengths (since its arc length includes the arc on both sides of the height line). Put in radius=3963, arc length=2, increase the digits after the decimal point to 5 and see that the height is .00013 . Since that is in miles that's about 8 inches drop for being 1 mile from the start.  For the last mile set the arc length=12448  and compare that height with the full arc length=12450 and you will see the height difference is 1 mile (actually very slightly less than 1 mile if you were to do the calculation with sufficient precision).

Isn't that non-linearity pretty obvious just thinking about a circle or ball?

[AATW]

If you go to that redirect and read the content it says that the horizon is not always at eye level. So why are you claiming that the wiki says that the horizon is always at eye level or that it is the official stance?

The Wiki page is titled "High Altitude Horizon Dip"
And the text says that "according to a planar prediction the horizon should be at 0 degrees eye level"

In this thread:
https://forum.tfes.org/index.php?topic=9492.0
Which is a response to the original Wiki page. You spend 20 pages finding fault with every method which demonstrates horizon dip.
Finally on page 21 you say:

I agree that the horizon isn't always at eye level, and drops as elevation increases. I have actually been planning to update the Wiki with some of Bobby's content. I have been thinking of making a page dedicated to the water level experiment as well.

...but then later on the same page you say:

I believe that was the point of all of this. The horizon isn't "always at eye level" as asserted in the Wiki, and a change is needed. The Wiki forgot about the concept of fog and atmosphere.

So...it sounds like you do believe that the horizon is always at eye level, but atmospheric conditions can affect the observation of it.

Then in this thread: https://forum.tfes.org/index.php?topic=11721.0
You reiterate that:

In FET the true horizon is always at eye level, and moves with you. Whether you can see that is another story. The error we made was in the wiki, by leaving out that you may not be able to see it.

And

Every time I've looked at the horizon I was seeing half land and half sky, with an apparent line cutting through my vision. Take a mirror and turn around, with the horizon facing your back and, when studying the mirror, the horizon follows the level of your eye

The default is, therefore that the horizon is at eye level. If someone has a crazy theory about the earth being a ball and the horizon being imperceptibly below the eye level, in contradiction to observation, it seems that the onus is on that person to demonstrate their claim.

But you then say

None of this is to say that the horizon is always at eye level at every altitude and atmospheric condition, or that one could expect to see the same from a plane where the horizon is very foggy, just that it has been tested at sea level to be so, just as Rowbotham tested it from the third story building

I understand that the common response to this is "that's too low," or whatever, but yet, it remains a test of the horizon. If one is to argue something about the imperceptible drop, that is a claim against reality, and thus one should be burdened to show it rather than argue that it is the burden of others to prove that there is no imperceptible drop

Although as I said for 20 pages in the first thread I references you bent over backwards with multiple methods which showed the horizon drop.

So...in brief, your beliefs about this are confusing. You seem to concede that horizon drop happens but spent pages and pages refusing to acknowledge the evidence that demonstrated it. Then when you did concede the point you did so only by claiming that in cases where there is horizon dip it's because you're not seeing the true horizon.
But even in the amended Wiki page it still says that:

"according to a planar prediction the horizon should be at 0 degrees eye level"

You see how this is all confusing?

[AATW]

I think the word "gradually" is where we are stuck. The foggy day picture is a gradual fade. The other horizon pictures are not.

That certainly sounds like the disagreement. I explained what I mean by the term (and I'm going with a pretty straight-forward dictionary definition of "gradual" and "gradient"), and I showed you how you can verify the presence of a gradient in an image. My reading of what you're saying is "nuh uh, it's obviously not gradual", repeated ad nauseam.

I think the issue is by your definition the edge of every object, certainly distant objects, will be gradual. There will always be some atmospheric effect which causes a gradient. That doesn't mean that objects don't have edges, and it doesn't mean that for all practical purposes those edges aren't clear enough to determine where they are.
In RET the horizon is effectively the edge of the earth. You objected to that word some pages back. I'd agree it's not an ideal word, but I don't know what word to use. It's a circle on the earth's surface.
If you're standing on a perfect sphere then the limits of how much of the sphere you can observe are because of the ground sloping away from you. You're looking down at a spherical cap and the "horizon" is the circle formed by the slice of the sphere at the base of the cap. The radius of the circle is determined by viewer height.
That all makes sense of the observations of the increasing horizon distance with altitude and less of distant objects being hidden by the horizon (see Turning Torso video).
Obviously we don't live on a perfect sphere and the sea isn't perfectly flat. So sure, that adds some gradient. But with decent visibility the point where the sky ends and the sea starts always seems clear to me

Yup. They'd be more-or-less identical, with too many different factors to account for to make a reasonable distinction between the two in real life.

OK. I have come to think that this is more or less true. But there are other differentiators which I have hinted at above which we can move on to if you want.

Was it just Tom, perchance? I know he used to subscribe to the "no EA, just perspective" view, and, as far as I know, he was in the minority here.

I think it pretty much was just Tom. But he was always the loudest FE voice on here a few years back so his (ridiculous, IMO) arguments stick most in my mind.

And I'm not saying you're trying to actively hide changes to the Wiki, but given how vocal Tom was in his defence of "the horizon always rises to eye level", despite the wealth of evidence to the contrary, he wasn't that clear about a change of view. He does mention updating the Wiki in some of the threads I've mentioned in my reply to him above, to be fair, but even then the new Wiki page still says that "According to a planar prediction the horizon should be at 0 degrees eye level", so there doesn't seem to be that much shift in his stance. It always seemed like a silly argument because the horizon wouldn't be at 0 degrees on a FE anyway.

[Mack]

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).

There’s already a formula for computing the amount of drop along a curve. It’s  h = r * (1 - cos a). h is the amount of drop, r is the radius of the sphere, a is the angle created by the start and end points measured from the center of the sphere.

The number of degrees in the angle is the drop. The more you move along the curve, the number of degrees increases.

The radius of the earth is about 3959 miles, so in one mile the curve drops .0000137 degrees = 0.67 feet=8 in.  This is just approximate since it assumes a perfect sphere though.

From point a to point b, it drops 8 inches and from point b to point c, it drops another 8 inches. But you can’t add the 8 inches from a to b and from b to c like you were moving is a straight line. For every unit down, you also move sideways, but it isn’t a 1:1 ratio because the direction keeps changing.

[SimonC]

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

[ichoosereality]

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.

As several folks including myself have pointed out, its incorrect as the drop/fall IS NOT LINEAR.  So computing one value of drop/fall for a particular distance traveled, dividing by the distance to get an average drop per unit distance for that particular distance, and then multiplying it by another distance traveled  is mathematically wrong.  Circles and spheres are not linear.

You can see this for yourself using the chord calculator I posted.   Use that to get the drop for 1 mile walked south from the North Pole (or 1 mile walked away from any other point)  If you multiply that times 1/4 of the circumference of the earth does it come anywhere close the the actual drop that we know must be equal to the radius of the earth?  Of course not.   Its just as wrong doing it with the bigger values going to the smaller distances as you are doing.

[Mack]

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.

I’m going to start with this comment because I think it is the source of your confusion.  The “height” of a  position is measured against sea level and depends on the terrain. If the earth was a perfectly smooth sphere everything would be at sea level.  “Height” doesn’t have anything to do with the curvature of the earth.

Curvature is the rate a line is changing direction at a given point. The “drop” in the curvature of the earth is a change in direction of an arbitrary line on its surface, not a change in height.
 

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.

You have walked 3,113 miles south, not “down”.  Eventually you would be going north.

You have two errors.  The first is you try to measure curvature using straight line distances.  That doesn’t work. Second you are adding the curvature linearly.



Using the equation I gave before: h = r * (1 - cos a). One mile across the surface is .0145 ° change in direction from straight.
The cosine of a .0145 ° angle is .99999997
h=3959*(1-.99999997)
h=3959*.000000032
h=.000127 miles
h=8.04672 in

So in one mile, there is a “drop”, or change in direction of the surface of 8 in. from A to B.

By your logic, then there should be 16 in of drop between A and C, but:

Two miles across the surface is .0288 ° . The cosine of a .00288 ° angle is 0.99999987
h=3959*(1-.99999987)
h=3959*.00000013
h=.000515 miles
h=32.6304 in

To sum up, a sphere or circle isn’t a straight line so the ratio of distance to curvature isn’t 1:1. It’s close to quadratic…the curvature is proportional to the square of the distance.

[SimonC]

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.

[DuncanDoenitz]

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