Curvature (and the lack of it)

Please see attached image below. I came across an idea recently and would appreciate anyone's comments. The more positive the better.
Imagine a global non-rotating stationary earth with the north pole fixed at the top and the south pole fixed at the bottom. The 'C' on the image refers to the Centre of the diagram.

If I started at the north pole walked directly towards the south pole by the time I reached the equator I would have travelled 6,225 miles. I would also have dropped in 'height' from the north pole by 3,963 miles. Which is equivalent to the radius of the earth (ie half its diameter). If I divide 6,226 by 3,963 I get a figure of 1.57. And this figure is the relationship to miles travelled against height of descent. And it means that for every 1.57 miles travelled there will be a descent of 1 mile.
This is a constant figure for a perfect circle. So if I travelled 1,570 miles from the north pole heading south I would have descended by 1,000 miles. Or if I travelled 15.7 miles I would descend by 10 miles. It works for any unit of measurement; metres, inches, feet, kilometres.
Just to expand a little if I walked from the north pole directly to the south pole I would have travelled 12,450 miles (half the earth's circumference) in the process I would have descended by the diameter of the earth namely 7,926 miles. 12,450 divided by 7,926 = 1.57.
I appreciate that I might not have explained this too well but the issue is a simple one; if there was two people at the north pole and one set off walking 1.57 miles. By the time they stopped they would be 1 mile 'lower' than their companion who would obviously have no chance of seeing them. And we know that this could not be true on the earth that we live because we can see for over 20 miles on a clear day and even further with a telescope. This therefore must prove that we cannot be on a global earth.
I am aware there are a few methods that pertain to determine the rate of curvature including the 8" per mile squared (which we are often told is for a parabola and not a circle).
I therefore look forward to any comments and will do my best to answer any queries on this topic. Please keep them genuine though.



https://imgur.com/9UIjfO6

Your attached image doesn't work

You need to put in the embed code. It should have a .png (or the like) at the end.


But as for your argument, it doesn't work like that.
You would need Earth's surface to be flat, but sloped downwards to have that relationship.

The relationship which does work like that is for the angle.
If you want to use degrees it is 360 degrees/2*pi*r = 0.00899 degrees per km.
That means after each km, you will have curved down by 0.00899 degrees.
Quite difficult to measure.
And using radians will make the math easier as it is simply 1/r.

This also explains why the drop is not linear.
At the start, the slope is 0, so there is no drop.
But as you go further along, the angle of the ground changes meaning the rate of descent increases.
At the 1/4 way point, the slope is now 90 degrees straight down, meaning it is a 1:1 relationship.

The other way to easily see why it can't be linear is to keep on going.
Why stop at the south pole? Complete a full 360 degree trip.
If your statement was true, you have travelled the entire circumference, a distance of ~40 000 km, and so you should finish off 25484 km below where you started (4 Earth radii), yet clearly you are back where you started so it needs to be 0.

There are a few different ways to calculate it depending on exactly how you have defined it.
For the way you appear to have defined it, with the drop being the vertical distance measured from the starting point to a line parallel to a line passing level through you which passes through the final point, and the distance measured along Earth.
i.e. if we use this image here showing the different options:

You are using d1 and h3.

The angle at the centre relates to the distance by (using radians for simpler math):
d = r*a
a = d/r
The height relates to the angle by:
cos(a) = (r-h)/r
h=r*(1-cos(a))

Combining them you get h=r*(1-cos(d/r)).

This can be approximated as h=d^2/(2*r)
This comes from the small x approximation for cos, which in turn comes from the Taylor series expansion (starting from n=0):
cos(x)=((-1)^(n)x^(2n)/(2n)!) = 1-x^2/2+x^4/4!+...
The first 2 terms are the small x approximation, which gives:
cos(x)~=(1-x^2/2)
Putting in x=d/r we get:
cos(d/r)=1-d^2/(2*r^2)
And sticking that in the above we get:
h=r*(1-(1-d^2/(2*r^2)))
=r*(1-1+d^2/(2*r^2))
=r*d^2/(2*r^2)
=d^2/(2*r).

This is also where the often quoted 8 inches per mile squared comes from.

That means over a distance of 1.6 km (approximately 1 mile), where the 2 formulae cannot be distinguished by any sane means as the first error term is of the order 10^-16 km; we end up with a drop of 0.0002 km = 20cm =~8 inches.

As for how far you see, that depends a lot on altitude and how tall/high the object you are looking at is.
And it is important to distinguish between the drop due to curvature and how much should be hidden:

Consider this simple example of a person (shown in purple) looking towards the orange object in the distance.
The drop is more than the height of the object. So if they had their eyes level with Earth, their line of sight would be limited by the light grey line and they couldn't see the orange object.
But if they are standing above, they can have a line of sight like the red line and still see the object, or at least the top of the object.

Also note that the amount of the orange object hidden plus the height of the observer is not equal to the drop, the drop is much larger.

This is easiest to understand by considering the horizon as the 0 point. You have a drop to the left of the horizon towards the viewer, and you have a drop to the right of the horizon towards the object.


It is also complicated by refraction which results in light curving downwards allowing us to see more.

If you are standing on a beach, close to sea level, the horizon is typically around 5 km away, and has a drop of 2 m. Again, this is best done the other way, from the horizon there is a drop of 2 m which brings your eyes level to the horizon.
Then from the horizon, the curvature blocks the view, with the amount hidden being roughly d^2/2r, where that d is the distance from the horizon to the object.

As you gain altitude, you can see much further, as the horizon is much further away, as the curvature needs to drop more from the horizon back to you to obstruct your view.

And this matches what is observed quite well.
We see standing on the beach looking out that the horizon is relatively close, but we can see tall objects beyond it with the bottom missing.
We can see that as we gain altitude the horizon gets further away.
We also note that a telescope doesn't allow us to see beyond the horizon to see objects obstructed by the curve.
Instead, it just allows us to resolve objects that are otherwise too small to resolve, or see objects otherwise too faint to see.
Even looking at an object beyond the horizon will not bring the bottom back into view. Instead, we see the top, including portions much smaller than the part of the bottom that is hidden from view, clearly resolved, with the bottom still missing.

So what is observed matches curvature, not a flat Earth.

Please see attached image below. I came across an idea recently and would appreciate anyone's comments. The more positive the better.
Imagine a global non-rotating stationary earth with the north pole fixed at the top and the south pole fixed at the bottom. The 'C' on the image refers to the Centre of the diagram.

Simon, everyone has already run you through this on the other site for pages. You are incorrect in how you are applying whatever it is you are applying to the Globe Earth model. Not sure why you want to waste more people's time over here...

Drop: is the amount the surface at the target has dropped from the tangent plane at the surface of the observer. This amount depends on the surface distance between observer and target. This distance is dependent on the Target Distance and the Side Pos of the target via Pythagoras.

The Drop after walking 1.57 miles on the globe is 1.644'.


http://walter.bislins.ch/bloge/index.asp?page=Advanced+Earth+Curvature+Calculator

This is the Globe Earth model. Whatever you are doing is not.

Simon, everyone has already run you through this on the other site for pages.

Why do FE’s post like it’s the first time every time?  Ignoring the long history and baggage of being soundly and repeatedly refuted. Even debunked? 

Quote from: Simon2k on April 02, 2023, 01:43:04 PM

Please see attached image below. I came across an idea recently and would appreciate anyone's comments. The more positive the better.
Imagine a global non-rotating stationary earth with the north pole fixed at the top and the south pole fixed at the bottom. The 'C' on the image refers to the Centre of the diagram.

Simon, everyone has already run you through this on the other site for pages. You are incorrect in how you are applying whatever it is you are applying to the Globe Earth model. Not sure why you want to waste more people's time over here...

Drop: is the amount the surface at the target has dropped from the tangent plane at the surface of the observer. This amount depends on the surface distance between observer and target. This distance is dependent on the Target Distance and the Side Pos of the target via Pythagoras.

The Drop after walking 1.57 miles on the globe is 1.644'.


http://walter.bislins.ch/bloge/index.asp?page=Advanced+Earth+Curvature+Calculator

This is the Globe Earth model. Whatever you are doing is not.

Hi  -which other site? Do you mean TFES site? Is that related to this one or totally seperate?

Stash,

you can't assume everyone knows what you already knows.

Your attached image doesn't work

You need to put in the embed code. It should have a .png (or the like) at the end.


But as for your argument, it doesn't work like that.
You would need Earth's surface to be flat, but sloped downwards to have that relationship.

The relationship which does work like that is for the angle.
If you want to use degrees it is 360 degrees/2*pi*r = 0.00899 degrees per km.
That means after each km, you will have curved down by 0.00899 degrees.
Quite difficult to measure.
And using radians will make the math easier as it is simply 1/r.

This also explains why the drop is not linear.
At the start, the slope is 0, so there is no drop.
But as you go further along, the angle of the ground changes meaning the rate of descent increases.
At the 1/4 way point, the slope is now 90 degrees straight down, meaning it is a 1:1 relationship.

The other way to easily see why it can't be linear is to keep on going.
Why stop at the south pole? Complete a full 360 degree trip.
If your statement was true, you have travelled the entire circumference, a distance of ~40 000 km, and so you should finish off 25484 km below where you started (4 Earth radii), yet clearly you are back where you started so it needs to be 0.

There are a few different ways to calculate it depending on exactly how you have defined it.
For the way you appear to have defined it, with the drop being the vertical distance measured from the starting point to a line parallel to a line passing level through you which passes through the final point, and the distance measured along Earth.
i.e. if we use this image here showing the different options:

You are using d1 and h3.

The angle at the centre relates to the distance by (using radians for simpler math):
d = r*a
a = d/r
The height relates to the angle by:
cos(a) = (r-h)/r
h=r*(1-cos(a))

Combining them you get h=r*(1-cos(d/r)).

This can be approximated as h=d^2/(2*r)
This comes from the small x approximation for cos, which in turn comes from the Taylor series expansion (starting from n=0):
cos(x)=((-1)^(n)x^(2n)/(2n)!) = 1-x^2/2+x^4/4!+...
The first 2 terms are the small x approximation, which gives:
cos(x)~=(1-x^2/2)
Putting in x=d/r we get:
cos(d/r)=1-d^2/(2*r^2)
And sticking that in the above we get:
h=r*(1-(1-d^2/(2*r^2)))
=r*(1-1+d^2/(2*r^2))
=r*d^2/(2*r^2)
=d^2/(2*r).

This is also where the often quoted 8 inches per mile squared comes from.

That means over a distance of 1.6 km (approximately 1 mile), where the 2 formulae cannot be distinguished by any sane means as the first error term is of the order 10^-16 km; we end up with a drop of 0.0002 km = 20cm =~8 inches.

As for how far you see, that depends a lot on altitude and how tall/high the object you are looking at is.
And it is important to distinguish between the drop due to curvature and how much should be hidden:

Consider this simple example of a person (shown in purple) looking towards the orange object in the distance.
The drop is more than the height of the object. So if they had their eyes level with Earth, their line of sight would be limited by the light grey line and they couldn't see the orange object.
But if they are standing above, they can have a line of sight like the red line and still see the object, or at least the top of the object.

Also note that the amount of the orange object hidden plus the height of the observer is not equal to the drop, the drop is much larger.

This is easiest to understand by considering the horizon as the 0 point. You have a drop to the left of the horizon towards the viewer, and you have a drop to the right of the horizon towards the object.


It is also complicated by refraction which results in light curving downwards allowing us to see more.

If you are standing on a beach, close to sea level, the horizon is typically around 5 km away, and has a drop of 2 m. Again, this is best done the other way, from the horizon there is a drop of 2 m which brings your eyes level to the horizon.
Then from the horizon, the curvature blocks the view, with the amount hidden being roughly d^2/2r, where that d is the distance from the horizon to the object.

As you gain altitude, you can see much further, as the horizon is much further away, as the curvature needs to drop more from the horizon back to you to obstruct your view.

And this matches what is observed quite well.
We see standing on the beach looking out that the horizon is relatively close, but we can see tall objects beyond it with the bottom missing.
We can see that as we gain altitude the horizon gets further away.
We also note that a telescope doesn't allow us to see beyond the horizon to see objects obstructed by the curve.
Instead, it just allows us to resolve objects that are otherwise too small to resolve, or see objects otherwise too faint to see.
Even looking at an object beyond the horizon will not bring the bottom back into view. Instead, we see the top, including portions much smaller than the part of the bottom that is hidden from view, clearly resolved, with the bottom still missing.

So what is observed matches curvature, not a flat Earth.

 Your D1 and H3 are the measurements I am looking at.
To explain this a little further - imagine the 'ball' in my diagram is a commercial building. North is the top of the building and South is the bottom. I have a window cleaning contract and I anchor myself to 'N'. And lower myself 'downwards' across the curve towards S for 6,225 miles. Before I commenced this I was 7,926 miles above ground level. when I stopped my descent I had dropped in height by 3,963 miles. 6,225 divided by 3,963 miles = 1.57. So for every 1.57 miles I travelled across the circumference of the global building I dropped in height by 1 mile.
Without resorting to formulae - please tell me why these figures or which of these figures is incorrect.

Stash,

you can't assume everyone knows what you already knows.

Nor can he assume that those on another site know more than those on this site.

Quote from: Simon2k on April 02, 2023, 01:43:04 PM

Please see attached image below. I came across an idea recently and would appreciate anyone's comments. The more positive the better.
Imagine a global non-rotating stationary earth with the north pole fixed at the top and the south pole fixed at the bottom. The 'C' on the image refers to the Centre of the diagram.

Simon, everyone has already run you through this on the other site for pages. You are incorrect in how you are applying whatever it is you are applying to the Globe Earth model. Not sure why you want to waste more people's time over here...

Drop: is the amount the surface at the target has dropped from the tangent plane at the surface of the observer. This amount depends on the surface distance between observer and target. This distance is dependent on the Target Distance and the Side Pos of the target via Pythagoras.

The Drop after walking 1.57 miles on the globe is 1.644'.


http://walter.bislins.ch/bloge/index.asp?page=Advanced+Earth+Curvature+Calculator

This is the Globe Earth model. Whatever you are doing is not.

Thats your opinion. I am posting here in the hope I can get someone who can take an independent view of my findings. The figures I provide are irrefutable - just because you dont agree doesnt make them wrong. If you believe they are wrong tell me which individual calculation is wrong and why. I would appreciate that - as long as its in laymans terms.

Thats your opinion. I am posting here in the hope I can get someone who can take an independent view of my findings.

It’s been proven curvature exits.


I came across this video.  I think it is compelling and reasonable proof showing no doubt the earth is curved.

Turning Torso (190m tall) - seen from 25km - 50km

The rate the building is blocked by the horizon is reasonable proof of earth’s curvature.

Part four, the classic.  Ships disappearing bottom up.

During the video of “Turning Torso (190m tall) - seen from 25km - 50km”, the individual pans the camera across a near ship.



Then a ship farther away.




If that isn’t conclusive concerning the ship over the horizon.  There is always my go to ship video.

Proofs the earth is curved.    The video “Turning Torso (190m tall) - seen from 25km - 50km” published on YouTube.  And the classic boats disappearing bottom up as a bonus to the video “Turning Torso (190m tall) - seen from 25km - 50km”.  And in the video “Huge container ship eclipsing the horizon. Nikon Coolpix P900.”


Used from this thread.
Four cases together show beyond a reasonable doubt the earth is curved

https://www.theflatearthsociety.org/forum/index.php?topic=91626.0

A thread posted before this one you evidently can’t create a decent counter argument.

I'm neutral to both sides, honestly.
From what I heard, it's caused by light bending so the ship disappears.

I'm neutral to both sides, honestly.
From what I heard, it's caused by light bending so the ship disappears.

Ok. Bending by what?  And why isn’t it taken in account by surveying, laser range finders, and radar other than well understood and documented atmospheric refraction? 

I'm neutral to both sides, honestly.
From what I heard, it's caused by light bending so the ship disappears.

I have looked for this bending.  I have yet to find it other than atmospheric refraction.


Sorry to use this again….

Not from the way mist lies.



To..

How are these two different sets of towers parallel on the horizon?





With no sign /indication of these distortions:


https://www.vision-doctor.com/en/optical-errors/distortion.html

Where the “lensing” effect has to be powerful enough to hide the sun at sunset on the flat earth fantasy?

Or the bottom of a boat? 

Quote from: Themightykabool on April 03, 2023, 10:56:21 AM

Stash,

you can't assume everyone knows what you already knows.

Nor can he assume that those on another site know more than those on this site.

that said

can we assume you know what a circle and a triangle are?
the basics of down on a piece of paper vs towards-the-center?


(i'm being a little rude...given i know what stash knows)

I said I'm neutral to both sides meaning both theorys could be wrong or correct

Quote from: Stash on April 02, 2023, 03:20:50 PM

Quote from: Simon2k on April 02, 2023, 01:43:04 PM

Please see attached image below. I came across an idea recently and would appreciate anyone's comments. The more positive the better.
Imagine a global non-rotating stationary earth with the north pole fixed at the top and the south pole fixed at the bottom. The 'C' on the image refers to the Centre of the diagram.

Simon, everyone has already run you through this on the other site for pages. You are incorrect in how you are applying whatever it is you are applying to the Globe Earth model. Not sure why you want to waste more people's time over here...

Drop: is the amount the surface at the target has dropped from the tangent plane at the surface of the observer. This amount depends on the surface distance between observer and target. This distance is dependent on the Target Distance and the Side Pos of the target via Pythagoras.

The Drop after walking 1.57 miles on the globe is 1.644'.


http://walter.bislins.ch/bloge/index.asp?page=Advanced+Earth+Curvature+Calculator

This is the Globe Earth model. Whatever you are doing is not.

Thats your opinion. I am posting here in the hope I can get someone who can take an independent view of my findings. The figures I provide are irrefutable - just because you dont agree doesnt make them wrong. If you believe they are wrong tell me which individual calculation is wrong and why. I would appreciate that - as long as its in laymans terms.

It's not my opinion. And I'm not even arguing that the earth is a globe. I'm just saying that this is the globe model if the earth was a globe. You're fabricating a new version of the globe model.

You've already been shown on the other site by multiple people exactly how your calculations are wrong and why, even in layman's terms. Even the Mod there said you are mis-representing the globe model and politely asked not to do that:

Quote from: SimonC on April 01, 2023, 01:54:30 PM

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.

I said I'm neutral to both sides meaning both theorys could be wrong or correct

And yet spherical earth accurately predicts items important to me.  Knowing when comets are close enough to the earth to see, low tide to play in tidal pools, when the next solar eclipse will be.  Why the equatorial mount on my telescope makes amateur astronomy easier.  As others have pointed out.  How gravity and accelerometers makes useful measurements and systems.

How can FE make accurate predictions and more useful items? 

How is FE more useful to me?

How does spherical earth produce such accuracy? 

I personally believe people can believe in whatever they want even if I don't agree.
If you say the earth is round, go for it. I don't mind.
If you say it's flat, it's also fine.

I personally believe people can believe in whatever they want even if I don't agree.
If you say the earth is round, go for it. I don't mind.
If you say it's flat, it's also fine.

Yes.  But it’s another thing when one wants to make assertions in debate.

If one wants to believe in FE, knock yourself out.

But when statements come up concerning “wrong” or “right”, that’s not really the issue.  The real issue is having a model that is practical and makes accurate predictions.

Making a deck in the backyard, I guess either model could work.

Needing to know the difference between high tide and low tide to get a ship to sea past barrier islands, and knowing how it affects water depth and currents for ship safety.  FE is useless. 

Quote from: Simon2k on April 03, 2023, 11:08:40 AM

Thats your opinion. I am posting here in the hope I can get someone who can take an independent view of my findings.

It’s been proven curvature exits.


I came across this video.  I think it is compelling and reasonable proof showing no doubt the earth is curved.

Quote

Turning Torso (190m tall) - seen from 25km - 50km

The rate the building is blocked by the horizon is reasonable proof of earth’s curvature.

Part four, the classic.  Ships disappearing bottom up.

During the video of “Turning Torso (190m tall) - seen from 25km - 50km”, the individual pans the camera across a near ship.



Then a ship farther away.




If that isn’t conclusive concerning the ship over the horizon.  There is always my go to ship video.

Quote

Proofs the earth is curved.    The video “Turning Torso (190m tall) - seen from 25km - 50km” published on YouTube.  And the classic boats disappearing bottom up as a bonus to the video “Turning Torso (190m tall) - seen from 25km - 50km”.  And in the video “Huge container ship eclipsing the horizon. Nikon Coolpix P900.”


Used from this thread.
Four cases together show beyond a reasonable doubt the earth is curved

https://www.theflatearthsociety.org/forum/index.php?topic=91626.0

A thread posted before this one you evidently can’t create a decent counter argument.

Am afraid those videos cant be verified  -like you say you found them. They just arent scientific. They could be anywhere and anything. There are as many videos that show that curvature is not real as those which say it is. So there is no 100% proof yet either way.

Quote from: Simon2k on April 03, 2023, 11:08:40 AM

Quote from: Stash on April 02, 2023, 03:20:50 PM

Quote from: Simon2k on April 02, 2023, 01:43:04 PM

Please see attached image below. I came across an idea recently and would appreciate anyone's comments. The more positive the better.
Imagine a global non-rotating stationary earth with the north pole fixed at the top and the south pole fixed at the bottom. The 'C' on the image refers to the Centre of the diagram.

Simon, everyone has already run you through this on the other site for pages. You are incorrect in how you are applying whatever it is you are applying to the Globe Earth model. Not sure why you want to waste more people's time over here...

Drop: is the amount the surface at the target has dropped from the tangent plane at the surface of the observer. This amount depends on the surface distance between observer and target. This distance is dependent on the Target Distance and the Side Pos of the target via Pythagoras.

The Drop after walking 1.57 miles on the globe is 1.644'.


http://walter.bislins.ch/bloge/index.asp?page=Advanced+Earth+Curvature+Calculator

This is the Globe Earth model. Whatever you are doing is not.

Thats your opinion. I am posting here in the hope I can get someone who can take an independent view of my findings. The figures I provide are irrefutable - just because you dont agree doesnt make them wrong. If you believe they are wrong tell me which individual calculation is wrong and why. I would appreciate that - as long as its in laymans terms.

It's not my opinion. And I'm not even arguing that the earth is a globe. I'm just saying that this is the globe model if the earth was a globe. You're fabricating a new version of the globe model.

You've already been shown on the other site by multiple people exactly how your calculations are wrong and why, even in layman's terms. Even the Mod there said you are mis-representing the globe model and politely asked not to do that:

Quote from: Pete Svarrior on April 01, 2023, 02:38:30 PM

Quote from: SimonC on April 01, 2023, 01:54:30 PM

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.

Because other people think they are wrong and as a result you follow suit by believing them doesn't mean they are wrong. Thats not a scientific method to start with so your credibility just took a tumble. If you want to disagree then do so using science not hearsay or opinion. My figures are irrefutable in relation to the calculations i have made. If you dont agree then please say which particular calculation you do not agree with and why. But please leave formulae out of this. All I am demonstrating is basic maths. If you dont understand then please ask and i will try and explain it to you.

Quote from: JackBlack on April 02, 2023, 02:38:51 PM

Your attached image doesn't work

You need to put in the embed code. It should have a .png (or the like) at the end.


But as for your argument, it doesn't work like that.
You would need Earth's surface to be flat, but sloped downwards to have that relationship.

The relationship which does work like that is for the angle.
If you want to use degrees it is 360 degrees/2*pi*r = 0.00899 degrees per km.
That means after each km, you will have curved down by 0.00899 degrees.
Quite difficult to measure.
And using radians will make the math easier as it is simply 1/r.

This also explains why the drop is not linear.
At the start, the slope is 0, so there is no drop.
But as you go further along, the angle of the ground changes meaning the rate of descent increases.
At the 1/4 way point, the slope is now 90 degrees straight down, meaning it is a 1:1 relationship.

The other way to easily see why it can't be linear is to keep on going.
Why stop at the south pole? Complete a full 360 degree trip.
If your statement was true, you have travelled the entire circumference, a distance of ~40 000 km, and so you should finish off 25484 km below where you started (4 Earth radii), yet clearly you are back where you started so it needs to be 0.

There are a few different ways to calculate it depending on exactly how you have defined it.
For the way you appear to have defined it, with the drop being the vertical distance measured from the starting point to a line parallel to a line passing level through you which passes through the final point, and the distance measured along Earth.
i.e. if we use this image here showing the different options:

You are using d1 and h3.

The angle at the centre relates to the distance by (using radians for simpler math):
d = r*a
a = d/r
The height relates to the angle by:
cos(a) = (r-h)/r
h=r*(1-cos(a))

Combining them you get h=r*(1-cos(d/r)).

This can be approximated as h=d^2/(2*r)
This comes from the small x approximation for cos, which in turn comes from the Taylor series expansion (starting from n=0):
cos(x)=((-1)^(n)x^(2n)/(2n)!) = 1-x^2/2+x^4/4!+...
The first 2 terms are the small x approximation, which gives:
cos(x)~=(1-x^2/2)
Putting in x=d/r we get:
cos(d/r)=1-d^2/(2*r^2)
And sticking that in the above we get:
h=r*(1-(1-d^2/(2*r^2)))
=r*(1-1+d^2/(2*r^2))
=r*d^2/(2*r^2)
=d^2/(2*r).

This is also where the often quoted 8 inches per mile squared comes from.

That means over a distance of 1.6 km (approximately 1 mile), where the 2 formulae cannot be distinguished by any sane means as the first error term is of the order 10^-16 km; we end up with a drop of 0.0002 km = 20cm =~8 inches.

As for how far you see, that depends a lot on altitude and how tall/high the object you are looking at is.
And it is important to distinguish between the drop due to curvature and how much should be hidden:

Consider this simple example of a person (shown in purple) looking towards the orange object in the distance.
The drop is more than the height of the object. So if they had their eyes level with Earth, their line of sight would be limited by the light grey line and they couldn't see the orange object.
But if they are standing above, they can have a line of sight like the red line and still see the object, or at least the top of the object.

Also note that the amount of the orange object hidden plus the height of the observer is not equal to the drop, the drop is much larger.

This is easiest to understand by considering the horizon as the 0 point. You have a drop to the left of the horizon towards the viewer, and you have a drop to the right of the horizon towards the object.


It is also complicated by refraction which results in light curving downwards allowing us to see more.

If you are standing on a beach, close to sea level, the horizon is typically around 5 km away, and has a drop of 2 m. Again, this is best done the other way, from the horizon there is a drop of 2 m which brings your eyes level to the horizon.
Then from the horizon, the curvature blocks the view, with the amount hidden being roughly d^2/2r, where that d is the distance from the horizon to the object.

As you gain altitude, you can see much further, as the horizon is much further away, as the curvature needs to drop more from the horizon back to you to obstruct your view.

And this matches what is observed quite well.
We see standing on the beach looking out that the horizon is relatively close, but we can see tall objects beyond it with the bottom missing.
We can see that as we gain altitude the horizon gets further away.
We also note that a telescope doesn't allow us to see beyond the horizon to see objects obstructed by the curve.
Instead, it just allows us to resolve objects that are otherwise too small to resolve, or see objects otherwise too faint to see.
Even looking at an object beyond the horizon will not bring the bottom back into view. Instead, we see the top, including portions much smaller than the part of the bottom that is hidden from view, clearly resolved, with the bottom still missing.

So what is observed matches curvature, not a flat Earth.

 Your D1 and H3 are the measurements I am looking at.
To explain this a little further - imagine the 'ball' in my diagram is a commercial building. North is the top of the building and South is the bottom. I have a window cleaning contract and I anchor myself to 'N'. And lower myself 'downwards' across the curve towards S for 6,225 miles. Before I commenced this I was 7,926 miles above ground level. when I stopped my descent I had dropped in height by 3,963 miles. 6,225 divided by 3,963 miles = 1.57. So for every 1.57 miles I travelled across the circumference of the global building I dropped in height by 1 mile.
Without resorting to formulae - please tell me why these figures or which of these figures is incorrect.

Can you?

Quote from: Alexei on April 03, 2023, 11:59:24 AM

I said I'm neutral to both sides meaning both theorys could be wrong or correct

And yet spherical earth accurately predicts items important to me.  Knowing when comets are close enough to the earth to see, low tide to play in tidal pools, when the next solar eclipse will be.  Why the equatorial mount on my telescope makes amateur astronomy easier.  As others have pointed out.  How gravity and accelerometers makes useful measurements and systems.

How can FE make accurate predictions and more useful items? 

How is FE more useful to me?

How does spherical earth produce such accuracy?

One way FE is useful to you is if you go walking up hills, down dales up mountains and anywhere in the wild. Dont forget to take your FLAT map with you. Like many Billions of people use everyday. And when you are driving your car theres always that handy pocket atlas in your glove compartment for when you get lost. Yes its printed on flat paper too. RAF pilots probably still use printed flat maps for their sorties. They did unil not long ago nothing to suggest they still dont other than technology. If you want to measure how far you walked or ran or pushed a shopping trolley you can always measure it on a flat map. All these things you cannot do with a 3D globe. But hey life goes on, we all survive we all didnt get lost and we all got home. What more could one want?

Your attached image doesn't work

You need to put in the embed code. It should have a .png (or the like) at the end.


But as for your argument, it doesn't work like that.
You would need Earth's surface to be flat, but sloped downwards to have that relationship.

The relationship which does work like that is for the angle.
If you want to use degrees it is 360 degrees/2*pi*r = 0.00899 degrees per km.
That means after each km, you will have curved down by 0.00899 degrees.
Quite difficult to measure.
And using radians will make the math easier as it is simply 1/r.

This also explains why the drop is not linear.
At the start, the slope is 0, so there is no drop.
But as you go further along, the angle of the ground changes meaning the rate of descent increases.
At the 1/4 way point, the slope is now 90 degrees straight down, meaning it is a 1:1 relationship.

The other way to easily see why it can't be linear is to keep on going.
Why stop at the south pole? Complete a full 360 degree trip.
If your statement was true, you have travelled the entire circumference, a distance of ~40 000 km, and so you should finish off 25484 km below where you started (4 Earth radii), yet clearly you are back where you started so it needs to be 0.

There are a few different ways to calculate it depending on exactly how you have defined it.
For the way you appear to have defined it, with the drop being the vertical distance measured from the starting point to a line parallel to a line passing level through you which passes through the final point, and the distance measured along Earth.
i.e. if we use this image here showing the different options:

You are using d1 and h3.

The angle at the centre relates to the distance by (using radians for simpler math):
d = r*a
a = d/r
The height relates to the angle by:
cos(a) = (r-h)/r
h=r*(1-cos(a))

Combining them you get h=r*(1-cos(d/r)).

This can be approximated as h=d^2/(2*r)
This comes from the small x approximation for cos, which in turn comes from the Taylor series expansion (starting from n=0):
cos(x)=((-1)^(n)x^(2n)/(2n)!) = 1-x^2/2+x^4/4!+...
The first 2 terms are the small x approximation, which gives:
cos(x)~=(1-x^2/2)
Putting in x=d/r we get:
cos(d/r)=1-d^2/(2*r^2)
And sticking that in the above we get:
h=r*(1-(1-d^2/(2*r^2)))
=r*(1-1+d^2/(2*r^2))
=r*d^2/(2*r^2)
=d^2/(2*r).

This is also where the often quoted 8 inches per mile squared comes from.

That means over a distance of 1.6 km (approximately 1 mile), where the 2 formulae cannot be distinguished by any sane means as the first error term is of the order 10^-16 km; we end up with a drop of 0.0002 km = 20cm =~8 inches.

As for how far you see, that depends a lot on altitude and how tall/high the object you are looking at is.
And it is important to distinguish between the drop due to curvature and how much should be hidden:

Consider this simple example of a person (shown in purple) looking towards the orange object in the distance.
The drop is more than the height of the object. So if they had their eyes level with Earth, their line of sight would be limited by the light grey line and they couldn't see the orange object.
But if they are standing above, they can have a line of sight like the red line and still see the object, or at least the top of the object.

Also note that the amount of the orange object hidden plus the height of the observer is not equal to the drop, the drop is much larger.

This is easiest to understand by considering the horizon as the 0 point. You have a drop to the left of the horizon towards the viewer, and you have a drop to the right of the horizon towards the object.


It is also complicated by refraction which results in light curving downwards allowing us to see more.

If you are standing on a beach, close to sea level, the horizon is typically around 5 km away, and has a drop of 2 m. Again, this is best done the other way, from the horizon there is a drop of 2 m which brings your eyes level to the horizon.
Then from the horizon, the curvature blocks the view, with the amount hidden being roughly d^2/2r, where that d is the distance from the horizon to the object.

As you gain altitude, you can see much further, as the horizon is much further away, as the curvature needs to drop more from the horizon back to you to obstruct your view.

And this matches what is observed quite well.
We see standing on the beach looking out that the horizon is relatively close, but we can see tall objects beyond it with the bottom missing.
We can see that as we gain altitude the horizon gets further away.
We also note that a telescope doesn't allow us to see beyond the horizon to see objects obstructed by the curve.
Instead, it just allows us to resolve objects that are otherwise too small to resolve, or see objects otherwise too faint to see.
Even looking at an object beyond the horizon will not bring the bottom back into view. Instead, we see the top, including portions much smaller than the part of the bottom that is hidden from view, clearly resolved, with the bottom still missing.

So what is observed matches curvature, not a flat Earth.

Not sure why you have so many straight lines in your diagrams. I was assuming a stationary, non rotating fixed global earth. There are only curved lines on such a thing  -no straight lines. My only one straight line was to demonstrate the 'drop' Oh and he radius and diameter hich are the same when used from North to South.

Am afraid those videos cant be verified 

Showing something you can’t refute.  That people experience in real life.

Have you proof they are “fake” in anyway? 

How about you actually exert some effort and go out and document the real world.  Like create a theory around a local landmark.  State that theory concerning the local landmark.  How it can be tested.  Test that theory for different times of the day and weather.  Then document what you find.

You have this too

Rainy Lake Experiment: Conclusion

http://walter.bislins.ch/bloge/index.asp?page=Rainy+Lake+Experiment%3A+Conclusion

Summary
All data and observations agree with the predictions of the Globe Model, which includes Terrestrial Refraction. The predictions for the Flat Earth Model, however, contradict the observations.

The Rainy Lake Experiment shows even better than the Bedford Level Experiment that the earth is a globe, since we also have GPS measurements that are not influenced by Refraction or Perspective, but are of a pure geometric nature. GPS measurements directly provide the radius of the earth.

Only one conclusion remains:

The earth cannot be flat, but is a globe with a mean radius of 6371 km!

Why not start a go fund me page, and verify the nicely laid out experiment so it can be verified or discredited.

Funny FE’s want to “debate”. When there are so many ways curvature is proven through experience and experiments.  Experiments that flat earther’s never seem to have the time to “verify”.  Yet they will whataboutism a subject to death with language when they could be doing actual experiments, data collection, and documentation.

 And when you are driving your car theres always that handy pocket atlas in your glove compartment for when you get lost.

US road grid corrections because of the Earth’s curvature

Jason Kottke   Jan 03, 2018

https://kottke.org/18/01/us-road-grid-corrections-because-of-the-earths-curvature

There are many ways to project a map.  It’s different when you actually build on the spherical earth. 

Please see attached image below. I came across an idea recently and would appreciate anyone's comments. The more positive the better.
Imagine a global non-rotating stationary earth with the north pole fixed at the top and the south pole fixed at the bottom. The 'C' on the image refers to the Centre of the diagram.

If I started at the north pole walked directly towards the south pole by the time I reached the equator I would have travelled 6,225 miles. I would also have dropped in 'height' from the north pole by 3,963 miles. Which is equivalent to the radius of the earth (ie half its diameter). If I divide 6,226 by 3,963 I get a figure of 1.57. And this figure is the relationship to miles travelled against height of descent. And it means that for every 1.57 miles travelled there will be a descent of 1 mile.
This is a constant figure for a perfect circle. So if I travelled 1,570 miles from the north pole heading south I would have descended by 1,000 miles. Or if I travelled 15.7 miles I would descend by 10 miles. It works for any unit of measurement; metres, inches, feet, kilometres.

Your claim that this is a constant figure is incorrect. For the same surface distance travelled, the corresponding distance along the N-S axis (what you call 'descent') is smaller near the pole and greater near the equator. The arrows in your own diagram illustrate this quite well.

Quote from: DataOverFlow2022 on April 03, 2023, 12:07:20 PM

Quote from: Alexei on April 03, 2023, 11:59:24 AM

I said I'm neutral to both sides meaning both theorys could be wrong or correct

And yet spherical earth accurately predicts items important to me.  Knowing when comets are close enough to the earth to see, low tide to play in tidal pools, when the next solar eclipse will be.  Why the equatorial mount on my telescope makes amateur astronomy easier.  As others have pointed out.  How gravity and accelerometers makes useful measurements and systems.

How can FE make accurate predictions and more useful items? 

How is FE more useful to me?

How does spherical earth produce such accuracy?

One way FE is useful to you is if you go walking up hills, down dales up mountains and anywhere in the wild. Dont forget to take your FLAT map with you. Like many Billions of people use everyday. And when you are driving your car theres always that handy pocket atlas in your glove compartment for when you get lost. Yes its printed on flat paper too. RAF pilots probably still use printed flat maps for their sorties. They did unil not long ago nothing to suggest they still dont other than technology. If you want to measure how far you walked or ran or pushed a shopping trolley you can always measure it on a flat map. All these things you cannot do with a 3D globe. But hey life goes on, we all survive we all didnt get lost and we all got home. What more could one want?

Most of the flat maps you mentioned probably use the mercator projection of a globe. For instance, if you're using USGS Topo maps for hiking they are mostly Mercator projections using The World Geodetic System of 1984 (WGS84)

Mercator projection:

World Geodetic System of 1984: It consists of a reference ellipsoid, a standard coordinate system, altitude data, and a geoid. It uses the Earth’s center mass as the coordinate origin.

The RAF maps use the Ordinance Survey of Great Britain, OD for the Ordnance Survey is ODN (Ordnance Datum Newlyn). OSGB36 is what might be termed a “traditional” datum. The “36” refers to 1936 when the coordinate reference system concepts were designed and adopted. It uses an ellipsoid, known as Airy 1830, that’s fitted close to the geoid just across the area of GB.

All are based upon an ellipsoid earth:

To explain this a little further - imagine the 'ball' in my diagram is a commercial building. North is the top of the building and South is the bottom. I have a window cleaning contract and I anchor myself to 'N'. And lower myself 'downwards' across the curve towards S for 6,225 miles. Before I commenced this I was 7,926 miles above ground level. when I stopped my descent I had dropped in height by 3,963 miles. 6,225 divided by 3,963 miles = 1.57. So for every 1.57 miles I travelled across the circumference of the global building I dropped in height by 1 mile.
Without resorting to formulae - please tell me why these figures or which of these figures is incorrect.

The original figures aren't incorrect. It is your application.
On average you dropped 1 archaic unit for every 1.57 archaic units.
But that is an average.
Most of that was done at the end.
The rate that you drop is not constant.

If you would like another simple example, consider if Earth was square (technically a cube).
You start at the north pole, at the centre of one face, and then walk along the surface to the centre of an adjacent face.
You have travelled 12742 km and ended up 6371 km below where you started.
Does that mean for every 2 km you travelled that you dropped 1 km?
No. In the first 6371 km you didn't drop at all.
In the last 6371 km, you dropped at a rate of 1 km per km.

It is quite similar with your claim.
The rate you drop is not constant.

This image should make it easy to see the difference between your claim and reality:

Earth is round, like the green line.
Your claim, that Earth drops 1 km for every 1.57 km is like the red line. It has to be because you are acting like the rate is consistent. (It would actually have to be a different slope, but the issue remains)
It is trivial to see that the red line is dramatically overestimating the drop.
You cannot treat it as a uniform rate of drop.
If you do treat it as uniform, you will get the numbers wrong until you are 1/4 of the way around Earth.

Am afraid those videos cant be verified

They can, quite easily. You can go and do it yourself.

There are as many videos that show that curvature is not real as those which say it is. So there is no 100% proof yet either way.

There are many videos claiming it, but often what they actually show is a claim of missing curvature, often using incorrect numbers or misapplication of theory.
For example, with your opening post where you boldly proclaim that you shouldn't be able to see someone 1.57 miles away, because you incorrectly claim that there should be a drop of 1 mile over this distance.

But the point remains, how is the bottom of the building missing? What is obstructing the view? Do you think the building has sunk into the water?

Because other people think they are wrong and as a result you follow suit by believing them doesn't mean they are wrong. Thats not a scientific method to start with so your credibility just took a tumble. If you want to disagree then do so using science not hearsay or opinion. My figures are irrefutable in relation to the calculations i have made. If you dont agree then please say which particular calculation you do not agree with and why. But please leave formulae out of this. All I am demonstrating is basic maths. If you dont understand then please ask and i will try and explain it to you.

What you are doing is not scientific.
You make a massive jump from one simple set of numbers with no explanation of why that should be the case given the curved surface.
You have it clearly explained why you are wrong and you dismiss it without resorting to formulae.
The formulae are a clear explanation of why you are wrong. They show what the expected drop is, and more importantly demonstrate that it is not linear.
A scientific approach would be accepting that you are wrong based upon these formulae, not just reasserting position and acting like it is irrefutable.

Also, it is nice that you say they are irrefutable, as if they were, that means you fully accept Earth is round and has this radius.

Can you?

Because Earth is round, there are these things called time zones.
That means that while it is around 3 pm for you, it is only now 8 am for me.
Don't expect me to respond while I'm sleeping.

One way FE is useful to you is if you go walking up hills, down dales up mountains and anywhere in the wild. Dont forget to take your FLAT map with you. Like many Billions of people use everyday. And when you are driving your car theres always that handy pocket atlas in your glove compartment for when you get lost. Yes its printed on flat paper too. RAF pilots probably still use printed flat maps for their sorties. They did unil not long ago nothing to suggest they still dont other than technology. If you want to measure how far you walked or ran or pushed a shopping trolley you can always measure it on a flat map. All these things you cannot do with a 3D globe. But hey life goes on, we all survive we all didnt get lost and we all got home. What more could one want?

And if you travel a long enough distance, you need to account for the distortions in the map, distortions which result from projecting a round surface onto a flat surface.
You can do these things on a physical 3D globe, but for small distances it would just need to be so large it would be impractical to use.
But if you had such a hypothetical globe, you could cut out a small section (such as for your city), and flattening it would introduce minimal distortion.
For example, if you take a region that is 100 km wide, and represent this as 1 m, then the budge expected at the centre would be approximately 0.2 km in reality which would correspond to 2 mm (if I did the math correctly) on the map. That will not be noticeable at all.

Not sure why you have so many straight lines in your diagrams. I was assuming a stationary, non rotating fixed global earth. There are only curved lines on such a thing  -no straight lines. My only one straight line was to demonstrate the 'drop' Oh and he radius and diameter hich are the same when used from North to South.

The straight lines are to help with measurements.
On the first line, the Earth is the curved black line.
The straight purple and blue and red lines are to show the different ways to measure the drop and distance.

On the second image, Earth is the curved black line.
The dark grey lines are to measure the drop.
The purple line indicates a person standing on Earth with their eyes at the top of the line.
The orange line represents some distant object
The red line indicates a line of sight from the person to the horizon and then to the object.
The light grey line indicates the line of sight from the horizon if their eyes were at sea level.

The third image is the same but using a different reference.

Quote from: Simon2k on April 02, 2023, 01:43:04 PM

Please see attached image below. I came across an idea recently and would appreciate anyone's comments. The more positive the better.
Imagine a global non-rotating stationary earth with the north pole fixed at the top and the south pole fixed at the bottom. The 'C' on the image refers to the Centre of the diagram.

If I started at the north pole walked directly towards the south pole by the time I reached the equator I would have travelled 6,225 miles. I would also have dropped in 'height' from the north pole by 3,963 miles. Which is equivalent to the radius of the earth (ie half its diameter). If I divide 6,226 by 3,963 I get a figure of 1.57. And this figure is the relationship to miles travelled against height of descent. And it means that for every 1.57 miles travelled there will be a descent of 1 mile.
This is a constant figure for a perfect circle. So if I travelled 1,570 miles from the north pole heading south I would have descended by 1,000 miles. Or if I travelled 15.7 miles I would descend by 10 miles. It works for any unit of measurement; metres, inches, feet, kilometres.

Your claim that this is a constant figure is incorrect. For the same surface distance travelled, the corresponding distance along the N-S axis (what you call 'descent') is smaller near the pole and greater near the equator. The arrows in your own diagram illustrate this quite well.

It is a constant. Do the maths. Instead of travelling a quarter of the circumference travel an eighth (1/8) ie 3,113 miles. From 3,113 miles draw a line parallel to the equator line (E1 - E2) to intersect the N-C line. Then measure the distance from that intersect to N (the north pole). You will find it is 1,982 miles ie half the distance between N and C or 1 mile descent for every 1.57 miles travelled . You could choose any point between N and E1. Measure it from N then draw the parallel line to join the N - C line. The distance from the intersect to N will be 1 mile for every 1.57 miles of the arc.
Regarding my arrows not sure if thats just a cheap shot but they are not to scale am sure you must have noticed and are for illustrative purposes only - perhaps i should have mentioned that for the odd person that didnt realise that.
What you fail to accept is a circle is one continuous curve. Any of the infinite points on the circumference is at the 'top' of the curve that lies behind it and in front of it. If you take any 2 arcs from anywhere on a circle and overlay one with the other the arcs will be exactly the same  -one wont be higher or lower than the other. Instead of debunking with hearsay why not try it yourself?