"Equator" problem

Also, the circle that is used is a side profile/slice of the sphere. Not the planar circle that you are implying.

We use the Earth's radius in the formula. A circle is a circle and has a radius. By the way, if the Earth is flat there still will be a horizon.

http://www.askamathematician.com/2012/08/q-if-earth-was-flat-would-there-be-the-horizon-if-so-what-would-it-look-like-if-the-earth-was-flat-and-had-infinite-area-would-that-change-the-answer/

What I am arguing about here is that the calculators estimate the horizon distance wrongly, and even though the discrepancy is small, it happens to be in favor of the flat Earth horizon. Why are you still arguing when you don't know basic facts?

Quote from: Saros on November 13, 2014, 02:30:49 PM

Also, you might have noticed that the formula for calculating horizon distance doesn't use a sphere, but a circle. I guess it doesn't matter that a circle is flat.

Do you really think this is an issue, or are you just generating more chaff?

Your sightline and the center of the Earth define a plane, since a line and point not on the line define a plane. The intersection of any plane with a sphere is a circle. Since the plane contains the center of the sphere in this case, it's a great circle - that is, a circle with the same radius as the sphere. That's why a circle is all you need to calculate distance to the horizon.

Don't they teach geometry in school any more? Or do they teach it but the ones who never paid attention are the ones arguing the Earth is flat?

Well, your whole horizon argument proving RE due to observed curvature falls apart when we take into account the fact that on a flat Earth there still will be a horizon pretty much in the same place. How about that?

Quote from: Alpha2Omega on November 13, 2014, 02:57:33 PM

Quote from: Saros on November 13, 2014, 02:30:49 PM

Also, you might have noticed that the formula for calculating horizon distance doesn't use a sphere, but a circle. I guess it doesn't matter that a circle is flat.

Do you really think this is an issue, or are you just generating more chaff?

Your sightline and the center of the Earth define a plane, since a line and point not on the line define a plane. The intersection of any plane with a sphere is a circle. Since the plane contains the center of the sphere in this case, it's a great circle - that is, a circle with the same radius as the sphere. That's why a circle is all you need to calculate distance to the horizon.

Don't they teach geometry in school any more? Or do they teach it but the ones who never paid attention are the ones arguing the Earth is flat?

Well, your whole horizon argument proving RE due to observed curvature falls apart when we take into account the fact that on a flat Earth there still will be a horizon pretty much in the same place. How about that?

Why would there be a horizon at all on a flat Earth? Wouldn't things just fade into the distance when looking over, say, the open ocean instead of disappearing from the bottom up?

You didn't answer the question about whether your "circle" question was real or not. Were you sleeping through geometry?

Quote from: rottingroom on November 13, 2014, 02:42:32 PM

Also, the circle that is used is a side profile/slice of the sphere. Not the planar circle that you are implying.

We use the Earth's radius in the formula. A circle is a circle and has a radius. By the way, if the Earth is flat there still will be a horizon.

http://www.askamathematician.com/2012/08/q-if-earth-was-flat-would-there-be-the-horizon-if-so-what-would-it-look-like-if-the-earth-was-flat-and-had-infinite-area-would-that-change-the-answer/

What I am arguing about here is that the calculators estimate the horizon distance wrongly, and even though the discrepancy is small, it happens to be in favor of the flat Earth horizon. Why are you still arguing when you don't know basic facts?

What you said earlier implied that you think the formulas use of pi is because of a flat earths planar circle. Nothing I said implied that I don't know the earth's radius is used. It is a side profile.

What you are arguing is hearsay. You are throwing numbers around without any source. The only numbers we know are the heights of the mountains. Saying that you know the height of some point halfway down the mountain is a lie.

Quote from: Saros on November 13, 2014, 03:13:01 PM

Quote from: Alpha2Omega on November 13, 2014, 02:57:33 PM

Quote from: Saros on November 13, 2014, 02:30:49 PM

Also, you might have noticed that the formula for calculating horizon distance doesn't use a sphere, but a circle. I guess it doesn't matter that a circle is flat.

Do you really think this is an issue, or are you just generating more chaff?

Your sightline and the center of the Earth define a plane, since a line and point not on the line define a plane. The intersection of any plane with a sphere is a circle. Since the plane contains the center of the sphere in this case, it's a great circle - that is, a circle with the same radius as the sphere. That's why a circle is all you need to calculate distance to the horizon.

Don't they teach geometry in school any more? Or do they teach it but the ones who never paid attention are the ones arguing the Earth is flat?

Well, your whole horizon argument proving RE due to observed curvature falls apart when we take into account the fact that on a flat Earth there still will be a horizon pretty much in the same place. How about that?

Why would there be a horizon at all on a flat Earth? Wouldn't things just fade into the distance when looking over, say, the open ocean instead of disappearing from the bottom up?

You didn't answer the question about whether your "circle" question was real or not. Were you sleeping through geometry?

Did you read the link? There will be a horizon even if the Earth is flat.

Quote from: Alpha2Omega on November 13, 2014, 03:19:07 PM

Quote from: Saros on November 13, 2014, 03:13:01 PM

Quote from: Alpha2Omega on November 13, 2014, 02:57:33 PM

Quote from: Saros on November 13, 2014, 02:30:49 PM

Also, you might have noticed that the formula for calculating horizon distance doesn't use a sphere, but a circle. I guess it doesn't matter that a circle is flat.

Do you really think this is an issue, or are you just generating more chaff?

Your sightline and the center of the Earth define a plane, since a line and point not on the line define a plane. The intersection of any plane with a sphere is a circle. Since the plane contains the center of the sphere in this case, it's a great circle - that is, a circle with the same radius as the sphere. That's why a circle is all you need to calculate distance to the horizon.

Don't they teach geometry in school any more? Or do they teach it but the ones who never paid attention are the ones arguing the Earth is flat?

Well, your whole horizon argument proving RE due to observed curvature falls apart when we take into account the fact that on a flat Earth there still will be a horizon pretty much in the same place. How about that?

Why would there be a horizon at all on a flat Earth? Wouldn't things just fade into the distance when looking over, say, the open ocean instead of disappearing from the bottom up?

You didn't answer the question about whether your "circle" question was real or not. Were you sleeping through geometry?

Did you read the link? There will be a horizon even if the Earth is flat.

Clearly we disagree. We can at least agree that it wouldn't be 5 km away for 2 meter height. Yet it apparently is.

Quote from: Saros on November 13, 2014, 03:10:36 PM

Quote from: rottingroom on November 13, 2014, 02:42:32 PM

Also, the circle that is used is a side profile/slice of the sphere. Not the planar circle that you are implying.

We use the Earth's radius in the formula. A circle is a circle and has a radius. By the way, if the Earth is flat there still will be a horizon.

http://www.askamathematician.com/2012/08/q-if-earth-was-flat-would-there-be-the-horizon-if-so-what-would-it-look-like-if-the-earth-was-flat-and-had-infinite-area-would-that-change-the-answer/

What I am arguing about here is that the calculators estimate the horizon distance wrongly, and even though the discrepancy is small, it happens to be in favor of the flat Earth horizon. Why are you still arguing when you don't know basic facts?

What you said earlier implied that you think the formulas use of pi is because of a flat earths planar circle. Nothing I said implied that I don't know the earth's radius is used. It is a side profile.

What you are arguing is hearsay. You are throwing numbers around without any source. The only numbers we know are the heights of the mountains. Saying that you know the height of some point halfway down the mountain is a lie.

Well, it is not a lie. You're not paying attention.
http://www.udeuschle.selfhost.pro/panoramas/panqueryfull.aspx?mode=newstandard&data=lon%3A41.64471%24%24%24lat%3A41.64906%24%24%24alt%3Aauto%24%24%24altcam%3A10%24%24%24hialt%3Afalse%24%24%24resolution%3A200%24%24%24azimut%3A360%24%24%24sweep%3A60%24%24%24leftbound%3A16.55000000000001%24%24%24rightbound%3A22.55000000000001%24%24%24split%3A60%24%24%24splitnr%3A1%24%24%24tilt%3A0.47916666666666696%24%24%24tiltsplit%3Afalse%24%24%24elexagg%3A1.2%24%24%24range%3A300%24%24%24colorcoding%3Afalse%24%24%24colorcodinglimit%3A231%24%24%24title%3AZugspitze%20%5BTelescope%5D%24%24%24description%3A%24%24%24email%3A%24%24%24language%3Aen%24%24%24screenwidth%3A1366%24%24%24screenheight%3A728

Quote from: Saros on November 13, 2014, 03:34:03 PM

Quote from: Alpha2Omega on November 13, 2014, 03:19:07 PM

Quote from: Saros on November 13, 2014, 03:13:01 PM

Quote from: Alpha2Omega on November 13, 2014, 02:57:33 PM

Quote from: Saros on November 13, 2014, 02:30:49 PM

Also, you might have noticed that the formula for calculating horizon distance doesn't use a sphere, but a circle. I guess it doesn't matter that a circle is flat.

Do you really think this is an issue, or are you just generating more chaff?

Your sightline and the center of the Earth define a plane, since a line and point not on the line define a plane. The intersection of any plane with a sphere is a circle. Since the plane contains the center of the sphere in this case, it's a great circle - that is, a circle with the same radius as the sphere. That's why a circle is all you need to calculate distance to the horizon.

Don't they teach geometry in school any more? Or do they teach it but the ones who never paid attention are the ones arguing the Earth is flat?

Well, your whole horizon argument proving RE due to observed curvature falls apart when we take into account the fact that on a flat Earth there still will be a horizon pretty much in the same place. How about that?

Why would there be a horizon at all on a flat Earth? Wouldn't things just fade into the distance when looking over, say, the open ocean instead of disappearing from the bottom up?

You didn't answer the question about whether your "circle" question was real or not. Were you sleeping through geometry?

Did you read the link? There will be a horizon even if the Earth is flat.

Clearly we disagree. We can at least agree that it wouldn't be 5 km away for 2 meter height. Yet it apparently is.

Haha, so you disagree with the link I gave you? Good job. Write him an e-mail. The guy is a mathematician.

You must be drunk.

Quote from: Alpha2Omega on November 13, 2014, 03:19:07 PM

Quote from: Saros on November 13, 2014, 03:13:01 PM

Quote from: Alpha2Omega on November 13, 2014, 02:57:33 PM

Quote from: Saros on November 13, 2014, 02:30:49 PM

Also, you might have noticed that the formula for calculating horizon distance doesn't use a sphere, but a circle. I guess it doesn't matter that a circle is flat.

Do you really think this is an issue, or are you just generating more chaff?

Your sightline and the center of the Earth define a plane, since a line and point not on the line define a plane. The intersection of any plane with a sphere is a circle. Since the plane contains the center of the sphere in this case, it's a great circle - that is, a circle with the same radius as the sphere. That's why a circle is all you need to calculate distance to the horizon.

Don't they teach geometry in school any more? Or do they teach it but the ones who never paid attention are the ones arguing the Earth is flat?

Well, your whole horizon argument proving RE due to observed curvature falls apart when we take into account the fact that on a flat Earth there still will be a horizon pretty much in the same place. How about that?

Why would there be a horizon at all on a flat Earth? Wouldn't things just fade into the distance when looking over, say, the open ocean instead of disappearing from the bottom up?

You didn't answer the question about whether your "circle" question was real or not. Were you sleeping through geometry?

Did you read the link? There will be a horizon even if the Earth is flat.

It says the horizon would be at approximately the same angle (except if it isn't, then it would appear to rise until directly overhead). What he doesn't say is anything about a sharp line about 5 km away - instead he says it would be slightly higher (by almost 0.04 degree and way off in the distance).

You must be drunk.

Did you read the link?

http://www.askamathematician.com/2012/08/q-if-earth-was-flat-would-there-be-the-horizon-if-so-what-would-it-look-like-if-the-earth-was-flat-and-had-infinite-area-would-that-change-the-answer/

This is what it says, because apparently you can't read.

"Q: If Earth was flat, would there be a horizon? If so, what would it look like? If the Earth was flat and had infinite area, would that change the answer?

Physicist: There’d definitely still be a horizon if the Earth were flat.  It would be in almost exactly the same place, and look essentially identical.  "

So, you completely ignored my other arguments that even the calculator doesn't show correct results which match reality, now you don't want to accept this information. Great.

Quote from: rottingroom on November 13, 2014, 03:47:32 PM

You must be drunk.

Did you read the link?

http://www.askamathematician.com/2012/08/q-if-earth-was-flat-would-there-be-the-horizon-if-so-what-would-it-look-like-if-the-earth-was-flat-and-had-infinite-area-would-that-change-the-answer/

This is what it says, because apparently you can't read.

"Q: If Earth was flat, would there be a horizon? If so, what would it look like? If the Earth was flat and had infinite area, would that change the answer?

Physicist: There’d definitely still be a horizon if the Earth were flat.  It would be in almost exactly the same place, and look essentially identical.  "

So, you completely ignored my other arguments that even the calculator doesn't show correct results which match reality, now you don't want to accept this information. Great.

I've never commented on your link. Get a grip.

Quote from: Saros on November 13, 2014, 03:34:03 PM

Quote from: Alpha2Omega on November 13, 2014, 03:19:07 PM

Quote from: Saros on November 13, 2014, 03:13:01 PM

Quote from: Alpha2Omega on November 13, 2014, 02:57:33 PM

Quote from: Saros on November 13, 2014, 02:30:49 PM

Also, you might have noticed that the formula for calculating horizon distance doesn't use a sphere, but a circle. I guess it doesn't matter that a circle is flat.

Do you really think this is an issue, or are you just generating more chaff?

Your sightline and the center of the Earth define a plane, since a line and point not on the line define a plane. The intersection of any plane with a sphere is a circle. Since the plane contains the center of the sphere in this case, it's a great circle - that is, a circle with the same radius as the sphere. That's why a circle is all you need to calculate distance to the horizon.

Don't they teach geometry in school any more? Or do they teach it but the ones who never paid attention are the ones arguing the Earth is flat?

Well, your whole horizon argument proving RE due to observed curvature falls apart when we take into account the fact that on a flat Earth there still will be a horizon pretty much in the same place. How about that?

Why would there be a horizon at all on a flat Earth? Wouldn't things just fade into the distance when looking over, say, the open ocean instead of disappearing from the bottom up?

You didn't answer the question about whether your "circle" question was real or not. Were you sleeping through geometry?

Did you read the link? There will be a horizon even if the Earth is flat.

It says the horizon would be at approximately the same angle (except if it isn't, then it would appear to rise until directly overhead). What he doesn't say is anything about a sharp line about 5 km away - instead he says it would be slightly higher (by almost 0.04 degree and way off in the distance).

Haha, you continue trolling. So now your problem is that there is a horizon about 5 km away(according to your math only) as if the photos here were not enough proof for the contrary? Didn't you see that the calculators are off? Of course there is an observed drop, but it has nothing to do with the Earth's curvature. And by the way, go to any lake and try to find a horizon 5 km away. Stop showing off  and go out and check for yourself. As for the guy in the link, he clearly stated that the horizon will be essentially identical in both scenarios.

I mean I agree with everything in the article. I don't see where you've drawn any of the conclusions you've made.

I mean I agree with everything in the article. I don't see where you've drawn any of the conclusions you've made.

So you agree that on a flat Earth there is a horizon too, but at the same time you keep arguing that since the ships appear to 'sink' below the horizon it proves that the Earth is round? What conclusions have I drawn? What are you talking about? I showed you that objects appear higher in real life than what is predicted by the formula. You didn't agree with the numbers. I gave you a link to check it for yourself. There you can see the altitude for any given point of any mountain not only the peaks. You disregarded that. The whole time everyone here was arguing that since there is a visible dip below the horizon this proves the Earth is round. I showed you that even with flat Earth we still have a horizon and it is pretty much the same, I also showed you that the calculations for the position of the RE horizon are wrong when checked in practice.

Quote from: rottingroom on November 13, 2014, 04:32:39 PM

I mean I agree with everything in the article. I don't see where you've drawn any of the conclusions you've made.

So you agree that on a flat Earth there is a horizon too, but at the same time you keep arguing that since the ships appear to 'sink' below the horizon it proves that the Earth is round? What conclusions have I drawn? What are you talking about? I showed you that objects appear higher in real life than what is predicted by the formula. You didn't agree with the numbers. I gave you a link to check it for yourself. There you can see the altitude for any given point of any mountain not only the peaks. You disregarded that. The whole time everyone here was arguing that since there is a visible dip below the horizon this proves the Earth is round. I showed you that even with flat Earth we still have a horizon and it is pretty much the same, I also showed you that the calculations for the position of the RE horizon are wrong when checked in practice.

This is why I think you are drunk. I see links that prove the opposite of what you are saying. WTF are you talking about?

Quote from: rottingroom on November 13, 2014, 04:32:39 PM

I mean I agree with everything in the article. I don't see where you've drawn any of the conclusions you've made.

So you agree that on a flat Earth there is a horizon too, but at the same time you keep arguing that since the ships appear to 'sink' below the horizon it proves that the Earth is round? What conclusions have I drawn? What are you talking about? I showed you that objects appear higher in real life than what is predicted by the formula. You didn't agree with the numbers. I gave you a link to check it for yourself. There you can see the altitude for any given point of any mountain not only the peaks. You disregarded that. The whole time everyone here was arguing that since there is a visible dip below the horizon this proves the Earth is round. I showed you that even with flat Earth we still have a horizon and it is pretty much the same, I also showed you that the calculations for the position of the RE horizon are wrong when checked in practice.

On a FE, an object would not disappear from the bottom up on the horizon, it would get smaller and smaller until it vanished all together.

In fact, you can still see the 900 m mark of Mt Kvira from 125 km away above the horizon, and it should be well below it when observed from 2 m height.

H1=2 m
H2=900 m

Maximum horizon distance = 112 km

Now if you still don't understand that what you see in the photo simply doesn't match the calculated horizon distance, I can't help you out ...If you could explain why there are constant discrepancies in what the formula produces and the reality, I would appreciate it.

That's easy.  Refraction.... the same reason the shoreline buildings are visible in both shots of the bridge and hillside I posted.

These are discrepancies over small distances, but they add up and over a distance of 1000 km the discrepancy may be significant. Do we not know the correct Earth's radius? Why exactly is there a discrepancy all the time?

And where do you intend to place the marks for 100-800 meters?

Horizon distance calculator used: http://members.home.nl/7seas/radcalc.htm

Are you saying the calculator that is "better" is wrong?

Also, you might have noticed that the formula for calculating horizon distance doesn't use a sphere, but a circle. I guess it doesn't matter that a circle is flat.

  Comedy gold.  This is Sceptimatic level material right here.

Physicist says:

"There’d definitely still be a horizon if the Earth were flat.  It would be in almost exactly the same place, and look essentially identical. "

I cannot open the link, but above sentence will do solely....All we need here is to use a little bit logic...

This sentence looks to me as if a perpetrator has confessed to the crime.

Why?

Because, what this scientist claims in this sentence is an obvious lie!

Discrepancies between the reality and horizon calculators proves it.

Saros says:

What I am arguing about here is that the calculators estimate the horizon distance wrongly, and even though the discrepancy is small, it happens to be in favor of the flat Earth horizon.

You can rely on these horizon calculators only in the fog, because in the fog you can't see anything anyway, but as soon as fog dissapears you can discard calculator estimations because they are simply wrong.

Now, you have to ask yourself why would someone claim such an obvious lie (as all physicists do when they have to make references to "horizon issue")?

If the Earth were round nobody would ever came up to the idea to claim such a stupidity, but all physicists do exactly that all the time when they are faced with "horizon issue"!

Try to imagine quite opposite situation:

Imagine that a mainstream science finally embraces FET and admits it's trueness!

Now imagine that someone ask question like this:

If the world was actually round, what would the horizon look like?

You know, and i know that no one would ever come up to the idea to answer in this manner:

"There’d definitely still be a horizon if the Earth were round.  It would be in almost exactly the same place, and look essentially identical."

You just have to start to think, but i know that you can't do that because you cannot afford to yourself that kind of luxury, by doing that you would allow to yourself to think openly, and then you would have to discard all your prejudices, and then you would become a flat earthers, but you would rather die than confess that the Earth is flat, or that God exists! Wouldn't you?

So you are a prisoners of your closed minds who serve a life sentence!

Another perpetrator confessing to the crime:

On Earth, the land you see will drop off at a rate of about 0.000126 miles per mile (well, for the first mile anyway). That is small enough to make the Earth appear flat by human observation, at least when you consider that mountains and valleys prevent us from seeing infinitely far. The angle of the horizon and clouds would appear remarkably similar, as in both round and flat Earth they would appear to converge toward a single line in very similar ways. On a flat Earth, you would see more clouds as they shrink toward the horizon (assuming ideal cloud conditions), but that is already the way clouds appear to behave anyway, more or less.

That small distance adds up fairly quickly if you are looking over an ocean however. A normal human at sea level can see about 3 miles or so. On a flat plane, you could see as far as the height of the waves allowed, basically. Very far, in ideal circumstances. So you can tell the Earth is round at flat zones like oceans or big lakes.

http://www.reddit.com/r/askscience/comments/kjrbt/if_the_world_was_actually_flat_what_would_the/

Gotcha!!!

Quote

Another perpetrator confessing to the crime:

On Earth, the land you see will drop off at a rate of about 0.000126 miles per mile (well, for the first mile anyway). That is small enough to make the Earth appear flat by human observation, at least when you consider that mountains and valleys prevent us from seeing infinitely far. The angle of the horizon and clouds would appear remarkably similar, as in both round and flat Earth they would appear to converge toward a single line in very similar ways. On a flat Earth, you would see more clouds as they shrink toward the horizon (assuming ideal cloud conditions), but that is already the way clouds appear to behave anyway, more or less.

That small distance adds up fairly quickly if you are looking over an ocean however. A normal human at sea level can see about 3 miles or so. On a flat plane, you could see as far as the height of the waves allowed, basically. Very far, in ideal circumstances. So you can tell the Earth is round at flat zones like oceans or big lakes.

http://www.reddit.com/r/askscience/comments/kjrbt/if_the_world_was_actually_flat_what_would_the/

Gotcha!!!

Of course clouds would grow smaller and get closer to the horizon on the FE.  There's still going to be perspective, and 'flat zones' means areas with a constant elevation, no hills, mountains, obstructions, etc.  If you want to take it literally, "flat zones" means there would be opposing 'zones' that aren't flat, therefore "curved zones", which means Earth is both flat and spherical. 

Let  us  turn  to  Fig.  14  to  illustrate  this  fact.  Let  the point  E   represent  the  position  of  the  observer  on  the  sea-level  ;  his  line  of  sight  would  be  a  tangent  to  the  sphere  at the  place  of  observation,  as  shewn  by  the  line  E  H,  and  the  dip  of  an  object  at  J  would  be  represented  by  the  line  H  J. Now  raise  the  observer  to  the  height  of  the  telescope  at  F  ; his  line  of  sight  is  still a  horizontal  line  in  the  direction  of G, and  parallel  to  E  H,  therefore  the  dip  from  G  to  J  is manifestly  greater  than  that  from  H  to  J.  And  this  is  true whether  we  reckon  the  dip  towards  the  centre  of  the  globe in  the  direction  of  G  L,  or  at  right-angles  from  the  line  of sight  G  M.

Why is it that flat earthers invariably rely on "science" texts written a century or more ago?  Why is it that they have no texts written even in the 20th century?

The above extract is from a book written by an anonymous flat-earth author called "Zetetes" who actually believed in the existence of the Christian god, and that this imaginary god created the heavens and the earth.  Obviously then Zetetes was no scientist LOL.

It's sad that people like cikljamas just can't—or won't—accept the obvious fact that scientific knowledge and awareness of our cosmos has advance logarithmically since the period of this book's writing.  Most flat earthers seem convinced that the advancement of science ground to a halt in 1899.

So... the extract above—and all the contents of the book "The Sea-Earth Globe and its Monstrous Hypothetical Motions or Modern Theoretical Astronomy" are nothing more than imaginative fiction, or an amusing, historical literary artifact of times long gone.

—Science it is not.

Now  raise  the  observer  to  the  height  of  the  telescope  at  F  ; his  line  of  sight  is  still a  horizontal  line  in  the  direction  of G, and  parallel  to  E  H,  therefore  the  dip  from  G  to  J  is manifestly  greater  than  that  from  H  to  J. 

For parallel level lines of sight from two different heights, the difference between G-J and H-J is the same as the height of the telescope regardless of distance according to the diagram.

*edit for clarification- Actually, if the difference between H and G is based of a distant point that is 90 degrees to the ground/center of the globe, then yes, it will increase slightly, but not much (or even be noticeable) for an observer on the ground.

...the height of the observer should in strictness be added to the amount of the dip

No, the dip is going to be the same.  The height of the observer allows viewing of objects between H and J and reduces the amount hidden from view below the 'dip'.

Let  us  turn  to  Fig.  14  to  illustrate  this  fact.  Let  the point  E   represent  the  position  of  the  observer  on  the  sea-level  ;  his  line  of  sight  would  be  a  tangent  to  the  sphere  at the  place  of  observation,  as  shewn  by  the  line  E  H,  and  the  dip  of  an  object  at  J  would  be  represented  by  the  line  H  J. Now  raise  the  observer  to  the  height  of  the  telescope  at  F  ; his  line  of  sight  is  still a  horizontal  line  in  the  direction  of G, and  parallel  to  E  H,  therefore  the  dip  from  G  to  J  is manifestly  greater  than  that  from  H  to  J.  And  this  is  true whether  we  reckon  the  dip  towards  the  centre  of  the  globe in  the  direction  of  G  L,  or  at  right-angles  from  the  line  of sight  G  M.

"on the sea-level"? "shewn"? How quaint. Let's get our 19th-century on here, shall we!

What are you and the original author trying to show with this? Where did this come from, anyway? When you paste text like this, or quote it, it's at least a courtesy to the original author to give attribution. It also helps your readers see any omitted context or supporting information, too. For instance, where is Fig. 11? It's referenced in the text, so you need to show it.

To the accuracy of his approximation "ignoring some small decimal points", the straight-line distance EH in Fig. 14 is the same as the length of the curved line EJ. If we accept 3 digits of precision, you can go about as far as about 625 miles (9 degrees of angle) before the error in the length of HJ starts to blow up.

The height EF for "a telescope" is enormously exaggerated here - probably 10% the radius of the circle. The height EF is going to be a trifle compared with EC (C at the center of the circle) for any real terrestrial instrument, and for HJ at significant distance.

If you want to do this for an object in space, where EF becomes a significant factor (or even a better answer if earthbound), use the correct formulas, not the approximation described. We've had pocket calculators for 40 years that can easily give much more exact answers using the correct formulas instead of the approximations that were needed as a practical matter 150 years ago.

Where did this come from, anyway? When you paste text like this, or quote it, it's at least a courtesy to the original author to give attribution. It also helps your readers see any omitted context or supporting information, too. For instance, where is Fig. 11? It's referenced in the text, so you need to show it.

Check out the whole document HERE.  It's a true comedy of errors.   

What are you and the original author trying to show with this?

1. 1660 km/h(supposed speed of Earth's rotation) : 3600 = 0,46 km/s

0,46 * 60 = 27,6 km/min .... 27,6 : 1,5 = 18,4 miles

Now, using Rowbotham's (verified) formula we calculate like this:
18,4 * 18,4 = 338,56 * 8 (inches) = 2708 * 2,5 = 6771,2 cm = 67 m roughly

So, if the Earth is round the Sun should drop off behind horizon 67m/min which is equivalent of climbing uphill 67m/min...

8880 m (Everest altitude) : 67 = 132,5 min = 2 hours, 12,5 minutes

So, if the Earth is rotund one guy standing on the top of the Everest should be able to see sunrise 2,12 hours earlier/sunset 2,12 hours longer than the other guy who stands at the bottom of the Everest...

As we all very well know it is not the case, not even closely to that...

Disprove this if you can!

2.



Accompanying video : " class="bbc_link" target="_blank" rel="noopener noreferrer">

Quote from: Alpha2Omega on November 15, 2014, 07:42:35 PM

What are you and the original author trying to show with this?

1. 1660 km/h(supposed speed of Earth's rotation) : 3600 = 0,46 km/s

0,46 * 60 = 27,6 km/min .... 27,6 : 1,5 = 18,4 miles

Now, using Rowbotham's (verified) formula we calculate like this:
18,4 * 18,4 = 338,56 * 8 (inches) = 2708 * 2,5 = 6771,2 cm = 67 m roughly

So, if the Earth is round the Sun should drop off behind horizon 67m/min which is equivalent of climbing uphill 67m/min...

8880 m (Everest altitude) : 67 = 132,5 min = 2 hours, 12,5 minutes

So, if the Earth is rotund one guy standing on the top of the Everest should be able to see sunrise 2,12 hours earlier/sunset 2,12 hours longer than the other guy who stands at the bottom of the Everest...

As we all very well know it is not the case, not even closely to that...

Disprove this if you can!

2.



Accompanying video : " class="bbc_link" target="_blank" rel="noopener noreferrer">

Can you provide a citation showing that an observer on the ground sees the sun rise and set at the same time as someone at the peak? I have never heard this before.

Can you provide a citation showing that an observer on the ground sees the sun rise and set at the same time as someone at the peak? I have never heard this before.

Who said that?

Quote from: Alpha2Omega on November 15, 2014, 07:42:35 PM

What are you and the original author trying to show with this?

1. 1660 km/h(supposed speed of Earth's rotation)

What does the speed of rotation have to do with that post I was asking about? It was arguing about "angle of dip" or something.

Anyway...

1. 1660 km/h(supposed speed of Earth's rotation) : 3600 = 0,46 km/s

0,46 * 60 = 27,6 km/min .... 27,6 : 1,5 = 18,4 miles

Why the extra step of dividing by 3600 sec/hr and then multiplying the result by 60 sec/min. Just divide 1660 km/hr by 60 min/hr and arrive directly at 27.6 km/min, like this:

1660 km/hr / (60 min/hr) = 27.6 km/min

This is the tangential velocity of earth's rotation at the equator.

Now, using Rowbotham's (verified) formula we calculate like this:
18,4 * 18,4 = 338,56 * 8 (inches) = 2708 * 2,5 = 6771,2 cm = 67 m roughly

6771.2 cm is closer to 68 m than 67 m, but whatever. We're working with approximations here anyway.

OK, using the small-angle approximation, a height of 68m above datum (typically mean sea level) will be geometrically on the horizon at the distance the earth rotates in one minute at the equator.

So, if the Earth is round the Sun should drop off behind horizon 67m/min which is equivalent of climbing uphill 67m/min...

... for the first minute.

After two minutes, using the same formula, shadow height above datum is:
(18.4 mi/min * 2 min)² * 8 in/mi² * 2.5 cm/in = 27084.8 cm = 271 m roughly.

Note that this is exactly 4 times the value after one minute. That's because this value grows by the distance squared and the distance has doubled here.

I'd like to point out that this is consistent with the grab from the video you included. "Remember the ground would drop exponentially from his perspective and not at a fixed rate like on a flat surface." except "exponentially" isn't exactly correct - it should be "circularly" - but for small angles, "as the square" works well enough.

After three minutes, shadow height above datum is (cutting directly to the chase):
67.712 m * 9 = 609 m

8880 m (Everest altitude) : 67 = 132,5 min = 2 hours, 12,5 minutes

So, if the Earth is rotund one guy standing on the top of the Everest should be able to see sunrise 2,12 hours earlier/sunset 2,12 hours longer than the other guy who stands at the bottom of the Everest...

The upshot is that you need to take the square root of the number of minutes you calculated above to arrive at a reasonable estimate of the difference in time between the horizon's shadow at datum and at altitude.

I get 11.5 minutes.

Let's see how we did.

The angle to a datum horizon from a height of 8880 meters, assuming a datum circumference of 25,000 miles (since your approximation is based on this) is:

First find the radius in km:
R = 25,000 / (2 pi) = 3978.87 mi
R = 6403.37 km (call it 6400 km) feel free to check this conversion if you like.

The angle to the horizon at datum from 8.880 km above datum is then
A = cos^-1((6400 km)/(6400 km + 8.880 km)) = cos^-1(0.9986)
A = 3.01 degrees

Now, the Earth rotates relative to the Sun once in 1440 minutes (one mean solar day), so it will take
1440 min/360 degrees * 3.01 degrees = 12.065 min for the shadow to reach the specified altitude above datum.

Fairly close, especially considering the approximations used and rounding done.

Note that Everest isn't exactly on the equator, nor does it rise 8880 m above its "base", which is well above sea level, so the actual time differences there will be different and vary with the time of year since the Sun doesn't drop straight down below the horizon like at the equator, but as an exercise this is informative anyway.

As we all very well know it [132,5 min] is not the case, not even closely to that...

Do you see why now?

Disprove this if you can!

It looks like I have. Any questions?

2.



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Thanks for the supporting evidence.

Thanks for the supporting evidence.

Thanks for the acknowledgment that the Earth is flat!