#### SteelyBob

##### Re: Curvature of the Horizon
« Reply #180 on: April 01, 2023, 12:53:59 PM »
To make it a little easier to explain. Imagine the big round circle in my diagram is actually a big round global building - a perfect sphere. And S is the base of it and N is the top of it. As I stated previously I am assuming a global stationery earth so this global building is no different. And for this example lets assume the building has a 250 metre diameter with a 786 metres circumference (all future numbers will be rounded up).
I have the window cleaning contract for the building and my anchor point is at the top (point N). I anchor myself to the top of the building and freeline halfway down the building. In doing so I have travelled one quarter (1/4) of the circumference of the building ie 197 metres. I have also dropped in 'height' 125 metres.

This is correct , so far.

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

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

#### ichoosereality

• 295
##### Re: Curvature of the Horizon
« Reply #181 on: April 01, 2023, 03:18:54 PM »
.... This therefore suggest that for every 1.57 metres I travelled down the global building that I dropped 1 metre in height (and is obviously a constant for a circle/globe). ....
No, it is not a constant for circles/globes.  It would be a constant for a straight slope of constant angle.  This is what you do not or perhaps refuse to understand or acknowledge.  Do you really think that if standing next to someone on the top of this building and that they take only 2 steps away from you that they will then be a full meter lower than you are?
« Last Edit: April 01, 2023, 03:20:49 PM by ichoosereality »
The contents of the GPS NAV message is the time of transmission and the orbital location of the transmitter at that time. If the transmitters are not where they claim to be GPS would not work.  Since it does work the transmitters must in fact be in orbit, which means the earth is round.

#### SimonC

• 127
##### Re: Curvature of the Horizon
« Reply #182 on: April 01, 2023, 07:53:20 PM »
You're just not getting this are you?

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

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

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

#### SimonC

• 127
##### Re: Curvature of the Horizon
« Reply #183 on: April 01, 2023, 07:56:32 PM »
You're just not getting this are you?

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

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

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

#### SimonC

• 127
##### Re: Curvature of the Horizon
« Reply #184 on: April 01, 2023, 07:58:48 PM »
To make it a little easier to explain. Imagine the big round circle in my diagram is actually a big round global building - a perfect sphere. And S is the base of it and N is the top of it. As I stated previously I am assuming a global stationery earth so this global building is no different. And for this example lets assume the building has a 250 metre diameter with a 786 metres circumference (all future numbers will be rounded up).
I have the window cleaning contract for the building and my anchor point is at the top (point N). I anchor myself to the top of the building and freeline halfway down the building. In doing so I have travelled one quarter (1/4) of the circumference of the building ie 197 metres. I have also dropped in 'height' 125 metres.

This is correct , so far.

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

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

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

#### SimonC

• 127
##### Re: Curvature of the Horizon
« Reply #185 on: April 01, 2023, 08:00:17 PM »
To make it a little easier to explain. Imagine the big round circle in my diagram is actually a big round global building - a perfect sphere. And S is the base of it and N is the top of it. As I stated previously I am assuming a global stationery earth so this global building is no different. And for this example lets assume the building has a 250 metre diameter with a 786 metres circumference (all future numbers will be rounded up).
I have the window cleaning contract for the building and my anchor point is at the top (point N). I anchor myself to the top of the building and freeline halfway down the building. In doing so I have travelled one quarter (1/4) of the circumference of the building ie 197 metres. I have also dropped in 'height' 125 metres.

This is correct , so far.

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

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

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

#### SimonC

• 127
##### Re: Curvature of the Horizon
« Reply #186 on: April 01, 2023, 08:02:31 PM »
.... This therefore suggest that for every 1.57 metres I travelled down the global building that I dropped 1 metre in height (and is obviously a constant for a circle/globe). ....
No, it is not a constant for circles/globes.  It would be a constant for a straight slope of constant angle.  This is what you do not or perhaps refuse to understand or acknowledge.  Do you really think that if standing next to someone on the top of this building and that they take only 2 steps away from you that they will then be a full meter lower than you are?

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

#### SimonC

• 127
##### Re: Curvature of the Horizon
« Reply #187 on: April 01, 2023, 08:03:33 PM »
.... This therefore suggest that for every 1.57 metres I travelled down the global building that I dropped 1 metre in height (and is obviously a constant for a circle/globe). ....
No, it is not a constant for circles/globes.  It would be a constant for a straight slope of constant angle.  This is what you do not or perhaps refuse to understand or acknowledge.  Do you really think that if standing next to someone on the top of this building and that they take only 2 steps away from you that they will then be a full meter lower than you are?

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

#### SimonC

• 127
##### Re: Curvature of the Horizon
« Reply #188 on: April 01, 2023, 08:30:06 PM »
Okay - hopefully this should explain what am saying although I would appreciate some assistance from any mathematicians out there.

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

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

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

Come on guys give me a break? Its simple maths.
« Last Edit: April 01, 2023, 08:32:02 PM by SimonC »

#### AATW

• 6438
##### Re: Curvature of the Horizon
« Reply #189 on: April 01, 2023, 08:48:08 PM »
Okay - hopefully this should explain what am
By dividing 197 metres by 125 metres you get 1.57. Therefore for every 1.57 metres travelled down the curve ie from A to B you will descend in height by 1 metre.
I guess on average that’s true, but you can see that the first few meters there’s hardly any drop and the last few meters you’re dropping pretty much the whole meter (with respect to the top of the curve).
Because it’s a curve.
Tom: "Claiming incredulity is a pretty bad argument. Calling it "insane" or "ridiculous" is not a good argument at all."

TFES Wiki Occam's Razor page, by Tom: "What's the simplest explanation; that NASA has successfully designed and invented never before seen rocket technologies from scratch which can accelerate 100 tons of matter to an escape velocity of 7 miles per second"

#### SimonC

• 127
##### Re: Curvature of the Horizon
« Reply #190 on: April 01, 2023, 08:54:30 PM »
Okay - hopefully this should explain what am
By dividing 197 metres by 125 metres you get 1.57. Therefore for every 1.57 metres travelled down the curve ie from A to B you will descend in height by 1 metre.
I guess on average that’s true, but you can see that the first few meters there’s hardly any drop and the last few meters you’re dropping pretty much the whole meter (with respect to the top of the curve).
Because it’s a curve.

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

#### SteelyBob

##### Re: Curvature of the Horizon
« Reply #191 on: April 01, 2023, 09:22:57 PM »
You dont read. Am not going 'around' the building I am going vertically down it.

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

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

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

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

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

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

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

#### Longtitube

• 226
##### Re: Curvature of the Horizon
« Reply #192 on: April 01, 2023, 09:27:03 PM »
I have revised the image to hopefully better explain this.

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

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

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

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

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

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

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

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

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

If you disagree, explain in layman's terms.
Once again - you assume that the centre of the video is the centre of the camera's frame. We know that this isn't the case.

#### Pete Svarrior

• e
• Planar Moderator
• 15990
• (◕˽ ◕ ✿)
##### Re: Curvature of the Horizon
« Reply #193 on: April 01, 2023, 09:38:30 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.
« Last Edit: April 01, 2023, 09:42:08 PM by Pete Svarrior »

If we are not speculating then we must assume

#### ichoosereality

• 295
##### Re: Curvature of the Horizon
« Reply #194 on: April 01, 2023, 10:51:17 PM »
.... This therefore suggest that for every 1.57 metres I travelled down the global building that I dropped 1 metre in height (and is obviously a constant for a circle/globe). ....
No, it is not a constant for circles/globes.  It would be a constant for a straight slope of constant angle.  This is what you do not or perhaps refuse to understand or acknowledge.  Do you really think that if standing next to someone on the top of this building and that they take only 2 steps away from you that they will then be a full meter lower than you are?

And a circle is constant. Every section no matter how large or small has the same curve as any other section of that circle. Thats constant.
Yes but you are not talking about a drop relative to each section but to the starting point (at N) and that is NOT constant.   Someone moving 2 steps away from a person at N (on your dome) would NOT be 1.57m lower.  As others have pointed a constant rate of your "drop" results in a straight slope not a curve.
The contents of the GPS NAV message is the time of transmission and the orbital location of the transmitter at that time. If the transmitters are not where they claim to be GPS would not work.  Since it does work the transmitters must in fact be in orbit, which means the earth is round.

#### Mack

• 17
##### Re: Curvature of the Horizon
« Reply #195 on: April 02, 2023, 01:53:58 AM »
Quote
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.

No you haven’t.  The length of your line will tell you the distance you have travelled (vertical plus horizontal).  Your line would look like the one on the right.  If you straighten it out, it will be longer than the one on the left and more than ¼ of the circumference. You’ve dropped ¼ of the circumference, but you’ve travelled more distance (vertical plus horizontal)

Quote
Never mind on average its a fact all the way round; the curve is the same all the way round - it doesn't drop away any more or less than anywhere else on the circle.

Wrong. Curvature can be measured by how much it deviates from straight.  Here’s a graph of a circle.

Notice how the circumference hits the grid lines at irregular intervals.  That means it deviates from straight a different amount each time it hits the grid and the rate of drop varies.  That’s why there is 8in of drop at 1 mile, but 32 in of drop at two, like I already explained. Compare that to the graph of a straight line where each coordinate hits the graph line in equal intervals. That’s a constant rate of drop

Quote
And I can't understand why no one agrees with this.
.
Because this is pretty basic well understood geometry and math. Even Rowbotham accepted the 8in. m^2 equation, which works pretty good up to a point.  But it’s the equation for a parabola not a sphere so eventually the errors start compounding.

#### ichoosereality

• 295
##### Re: Curvature of the Horizon
« Reply #196 on: April 02, 2023, 02:55:01 AM »
The reason people are finding it so hard to comprehend is because that's not what it looks like on a globe earth. Now why might that be I wonder?
Ah now we come to it.  Forget the earth just use a circle. You are claiming (at least as I read it)  that the change in height (the y axis) of coordinates on the perimitter of this circle are constant for a constant traversal of circumference i.e. for a constant angle which that circumference subtends.  If that were true then
sin(x)-sin(x-1) as x goes from 90 to 1 (i.e. looking at the difference in the Y coordinates of each end of 1 degree arcs 0-1, 1-2, 2-3,....89-90) would be a constant which is obviously not the case.  What you claim is nonsense.

I am guessing that you understand that and are just trolling a bit to try and make a FE claim.  Math is not your friend in this cause.
« Last Edit: April 02, 2023, 05:49:46 AM by ichoosereality »
The contents of the GPS NAV message is the time of transmission and the orbital location of the transmitter at that time. If the transmitters are not where they claim to be GPS would not work.  Since it does work the transmitters must in fact be in orbit, which means the earth is round.

#### Mack

• 17
##### Re: Curvature of the Horizon
« Reply #197 on: April 02, 2023, 07:17:44 AM »
Quote
If that were true then sin(x)-sin(x-1) as x goes from 90 to 1 (i.e. looking at the difference in the Y coordinates of each end of 1 degree arcs 0-1, 1-2, 2-3,....89-90) would be a constant which is obviously not the case.  What you claim is nonsense.

If I'm translating right, I think this website makes your point.  It's pretty cool.  Move the slider in the top right corner.

https://www.geogebra.org/m/hnZMkBdc

#### DuncanDoenitz

• 394
##### Re: Curvature of the Horizon
« Reply #198 on: April 02, 2023, 07:38:17 AM »
As this is a debating forum, not a geometry tutorial, I'm out.

#### SteelyBob

##### Re: Curvature of the Horizon
« Reply #199 on: April 02, 2023, 08:17:01 AM »
Quote
If that were true then sin(x)-sin(x-1) as x goes from 90 to 1 (i.e. looking at the difference in the Y coordinates of each end of 1 degree arcs 0-1, 1-2, 2-3,....89-90) would be a constant which is obviously not the case.  What you claim is nonsense.

If I'm translating right, I think this website makes your point.  It's pretty cool.  Move the slider in the top right corner.

https://www.geogebra.org/m/hnZMkBdc

That’s rather near. Thanks