Distance To The Horizon Computations : Round Earth-vs-Flat Earth

I am back again after a period of banning for the nth time . I am told it was because of improper posting on the Q&A  section and should have been on the debate section of this forum.

I hope this is the proper section of the forum. If it isn't another banning wouldn't be too bad. That's the way this website works. You have to abide with it.

The distance to the horizon seems to be one of the main points for proof of a "Round Earth".

I would just like to post this for debate to see what answers the "Flat Earthers" are going to come up with.

The distance to the horizon  in miles is simply the calculation of 1.22 times the square root of the height in feet  of the height of the observer.  This is for a flat surface on a clear day with no effects of atmospheric conditions such as smog, smoke or fog to interfere with the view. On a clear day with a calm sea is the best place for demonstration. The higher the observer the farther is the horizon.

So why would  crow's nests be  necessary on ships if the earth was flat ? it would seem there would be no difference to the horizon from an observer on the bridge of a ship than from a person in the crow's nest if the earth was flat ?

I know in advance what the "Flat Earthers" are going to post for their definition of the distance to the horizon on a "Flat Earth" would be according to their computations. But I would like to see their answers in their own words.

In all reality, the fact that the earth is the globular shape that it is is an undeniable and true fact , but just for review I would like to see what the "Flat Earthers" have in their defense that the formula for the distance to the horizon is not true and valid.

This has been debated previously but I would like to see how this thread develops once more.

I posted this in an earlier thread,  I think it's worth repeating that flat earthers just don't do well at maths.

From the time of Rowbotham, the flat earth believers quote chapter and verse on the distances to lighthouses, claiming they are visible much further out to sea than what curvature calculations predict.
Here is a plot of lighthouse visibility versus height of the light above sea level,   taking into account standard correction for refraction and a bridge height of 30 ft above the water line.



The formula for calculating distance to the horizon, including refraction correction for standard atmosphere  is  D =  3.86 * sqrt ( h )    h = height in meters,  D = distance in km.

The Planier lighthouse  is   66 meters asl,   and is visible for 43 km,   calculated distance is 43 km,   Flat Earthers claim the calculated value is 29 km
The Jeddah lighthouse      is 113 meters asl,  and is visible for 46 km,   calculated distance is 53 km,   Flat Earthers claim the calculated value is 38 km
The Ile Vierge lighthouse  is   82.5 meters asl,   and is visible for 50 km,   calculated distance is 47 km,   Flat Earthers claim the calculated value is 33 km
The Genoa lighthouse  is 76 meters asl,  and is visible for 46 km,   calculated distance is 45 km,   Flat Earthers claim the calculated value is 31 km

The conclusion is that if you do the calculations properly,  the flat earth argument,  falls flat on its face ( once again )

For details of the formula derivation go to ...
http://web.archive.org/web/20131717132700/http://mintaka.sdsu.edu/GF/explain/atmos_refr/horizon.html
http://en.wikipedia.org/wiki/Horizon
Calculation of ducting effects.
http://web.archive.org/web/20131717125400/http://mintaka.sdsu.edu/GF/explain/atmos_refr/bending.html

I am back again after a period of banning for the nth time . I am told it was because of improper posting on the Q&A  section and should have been on the debate section of this forum.

I hope this is the proper section of the forum. If it isn't another banning wouldn't be too bad. That's the way this website works. You have to abide with it.

The distance to the horizon seems to be one of the main points for proof of a "Round Earth".

I would just like to post this for debate to see what answers the "Flat Earthers" are going to come up with.

The distance to the horizon  in miles is simply the calculation of 1.22 times the square root of the height in feet  of the height of the observer.  This is for a flat surface on a clear day with no effects of atmospheric conditions such as smog, smoke or fog to interfere with the view. On a clear day with a calm sea is the best place for demonstration. The higher the observer the farther is the horizon.

So why would  crow's nests be  necessary on ships if the earth was flat ? it would seem there would be no difference to the horizon from an observer on the bridge of a ship than from a person in the crow's nest if the earth was flat ?

I know in advance what the "Flat Earthers" are going to post for their definition of the distance to the horizon on a "Flat Earth" would be according to their computations. But I would like to see their answers in their own words.

In all reality, the fact that the earth is the globular shape that it is is an undeniable and true fact , but just for review I would like to see what the "Flat Earthers" have in their defense that the formula for the distance to the horizon is not true and valid.

This has been debated previously but I would like to see how this thread develops once more.

Your formula is indeed true and valid. Here is the proof:
http://imageshack.com/a/img909/4246/NDa8Of.jpg
http://imageshack.com/a/img633/6079/B02EkU.jpg

Show us your formula and equations, Flat Earthers.

Quote from: Googleotomy on June 10, 2015, 09:07:19 PM

I am back again after a period of banning for the nth time . I am told it was because of improper posting on the Q&A  section and should have been on the debate section of this forum.

I hope this is the proper section of the forum. If it isn't another banning wouldn't be too bad. That's the way this website works. You have to abide with it.

The distance to the horizon seems to be one of the main points for proof of a "Round Earth".

I would just like to post this for debate to see what answers the "Flat Earthers" are going to come up with.

The distance to the horizon  in miles is simply the calculation of 1.22 times the square root of the height in feet  of the height of the observer.  This is for a flat surface on a clear day with no effects of atmospheric conditions such as smog, smoke or fog to interfere with the view. On a clear day with a calm sea is the best place for demonstration. The higher the observer the farther is the horizon.

So why would  crow's nests be  necessary on ships if the earth was flat ? it would seem there would be no difference to the horizon from an observer on the bridge of a ship than from a person in the crow's nest if the earth was flat ?

I know in advance what the "Flat Earthers" are going to post for their definition of the distance to the horizon on a "Flat Earth" would be according to their computations. But I would like to see their answers in their own words.

In all reality, the fact that the earth is the globular shape that it is is an undeniable and true fact , but just for review I would like to see what the "Flat Earthers" have in their defense that the formula for the distance to the horizon is not true and valid.

This has been debated previously but I would like to see how this thread develops once more.

Your formula is indeed true and valid. Here is the proof:
http://imageshack.com/a/img909/4246/NDa8Of.jpg
http://imageshack.com/a/img633/6079/B02EkU.jpg

Show us your formula and equations, Flat Earthers.

You should include the standard correction factor for refraction,  which makes the formula  1.32  not 1.22 times square root height.   
Refer to http://web.archive.org/web/20131717132700/http://mintaka.sdsu.edu/GF/explain/atmos_refr/horizon.html

For a quick geodetic surveying 101 lesson  " class="bbc_link" target="_blank" rel="noopener noreferrer"> 

Quote from: hello_there on June 11, 2015, 05:07:44 AM

Quote from: Googleotomy on June 10, 2015, 09:07:19 PM

I am back again after a period of banning for the nth time . I am told it was because of improper posting on the Q&A  section and should have been on the debate section of this forum.

I hope this is the proper section of the forum. If it isn't another banning wouldn't be too bad. That's the way this website works. You have to abide with it.

The distance to the horizon seems to be one of the main points for proof of a "Round Earth".

I would just like to post this for debate to see what answers the "Flat Earthers" are going to come up with.

The distance to the horizon  in miles is simply the calculation of 1.22 times the square root of the height in feet  of the height of the observer.  This is for a flat surface on a clear day with no effects of atmospheric conditions such as smog, smoke or fog to interfere with the view. On a clear day with a calm sea is the best place for demonstration. The higher the observer the farther is the horizon.

So why would  crow's nests be  necessary on ships if the earth was flat ? it would seem there would be no difference to the horizon from an observer on the bridge of a ship than from a person in the crow's nest if the earth was flat ?

I know in advance what the "Flat Earthers" are going to post for their definition of the distance to the horizon on a "Flat Earth" would be according to their computations. But I would like to see their answers in their own words.

In all reality, the fact that the earth is the globular shape that it is is an undeniable and true fact , but just for review I would like to see what the "Flat Earthers" have in their defense that the formula for the distance to the horizon is not true and valid.

This has been debated previously but I would like to see how this thread develops once more.

Your formula is indeed true and valid. Here is the proof:
http://imageshack.com/a/img909/4246/NDa8Of.jpg
http://imageshack.com/a/img633/6079/B02EkU.jpg

Show us your formula and equations, Flat Earthers.

You should include the standard correction factor for refraction,  which makes the formula  1.32  not 1.22 times square root height.   
Refer to http://web.archive.org/web/20131717132700/http://mintaka.sdsu.edu/GF/explain/atmos_refr/horizon.html

For a quick geodetic surveying 101 lesson  " class="bbc_link" target="_blank" rel="noopener noreferrer">

Thanks Rayzor for the correction. I have added refraction into consideration.
http://imageshack.com/a/img909/4246/NDa8Of.jpg
http://imageshack.com/a/img910/5316/dWRUy0.jpg

Now, show us your formula and equations, Flat Earthers.

Thanks very much to all concerned , but these are all "Round Earth" answers.

I was really looking for some "Flat Earth" answers.

And I have checked this out personally. When I was in the Navy I was assigned maintenance on the surface search radar. When I observed a ship on the horizon I would check the range on the radar. It always agreed with the formula.
Actually, I didn't know the formula for estimating the distance to the horizon at that time, but the radar would have agreed with the visual sighting estimate. I was sort of doing this in reverse. Checking the distance on the radar and then doing the visual sighting.

Navy lookouts had a manual showing the distances to the horizon for various heights. I understand they are very adept at estimating distances and comparing them with the radar in their drills and actual operations.

Flat Earthers" are always asking "Have you done this yourself ?" I have.

Come on, flatties. You cannot deny this, and you can't say it's bs with no evidence, just respond.

So, it's been a while, and there hasn't been any response from the Flat Earthers. Which reminds me of this:

The higher you climb the further you can see. 

That's not possible on a flat earth.  That's it,   case closed,  now can we all go off to the pub?

No Flat Earther is showing any kind of disagreement. So... yeah. We can all go to the pub then

B-b-but...... air density!

I suppose the moderator might as well put the lock on this one.

I had more fun with my very first post about how some amateur radio operators bounced signals off the moon and  calculated the distance from the earth to the moon. It went on for 28 pages and 549 replies until jroa eventually locked it for lack of response from flat earthers.
http://www.theflatearthsociety.org/forum/index.php?topic=58190.msg1554902#msg1554902

Last one out turn off the lights.

I suppose the moderator might as well put the lock on this one.

I had more fun with my very first post about how some amateur radio operators bounced signals off the moon and  calculated the distance from the earth to the moon. But it eventually got locked, too.

Last one out turn off the lights.

you prove the earth is round case closed,go home.

Quote from: Googleotomy on June 12, 2015, 09:17:54 PM

I suppose the moderator might as well put the lock on this one.

I had more fun with my very first post about how some amateur radio operators bounced signals off the moon and  calculated the distance from the earth to the moon. But it eventually got locked, too.

Last one out turn off the lights.

you prove the earth is round case closed,go home.

Done.

Quote from: modestman on June 12, 2015, 09:20:46 PM

Quote from: Googleotomy on June 12, 2015, 09:17:54 PM

I suppose the moderator might as well put the lock on this one.

I had more fun with my very first post about how some amateur radio operators bounced signals off the moon and  calculated the distance from the earth to the moon. But it eventually got locked, too.

Last one out turn off the lights.

you prove the earth is round case closed,go home.

Done.

Good,so it's your last message on board ? in this site ?

Quote from: Googleotomy on June 12, 2015, 09:32:20 PM

Quote from: modestman on June 12, 2015, 09:20:46 PM

Quote from: Googleotomy on June 12, 2015, 09:17:54 PM

I suppose the moderator might as well put the lock on this one.

I had more fun with my very first post about how some amateur radio operators bounced signals off the moon and  calculated the distance from the earth to the moon. But it eventually got locked, too.

Last one out turn off the lights.

you prove the earth is round case closed,go home.

Done.

Good,so it's your last message on board ? in this site ?

Not by a long shot. This website is a never ending source for fun and entertainment and I am really appreciative of the efforts of the Flat Earth Society for making it so.
So I will be around for a while. I have learned a lot in the process.
Just ending this as far as this particular thread is concerned although I would have liked to have heard something from a flat earther.

Quote from: modestman on June 12, 2015, 09:36:18 PM

Quote from: Googleotomy on June 12, 2015, 09:32:20 PM

Quote from: modestman on June 12, 2015, 09:20:46 PM

Quote from: Googleotomy on June 12, 2015, 09:17:54 PM

I suppose the moderator might as well put the lock on this one.

I had more fun with my very first post about how some amateur radio operators bounced signals off the moon and  calculated the distance from the earth to the moon. But it eventually got locked, too.

Last one out turn off the lights.

you prove the earth is round case closed,go home.

Done.

Good,so it's your last message on board ? in this site ?

Not by a long shot. This website is a never ending source for fun and entertainment and I am really appreciative of the efforts of the Flat Earth Society for making it so.
So I will be around for a while. I have learned a lot in the process.
Just ending this as far as this particular thread is concerned although I would have liked to have heard something from a flat earther.

Sure.

I posted this in an earlier thread,  I think it's worth repeating that flat earthers just don't do well at maths.

From the time of Rowbotham, the flat earth believers quote chapter and verse on the distances to lighthouses, claiming they are visible much further out to sea than what curvature calculations predict.
Here is a plot of lighthouse visibility versus height of the light above sea level,   taking into account standard correction for refraction and a bridge height of 30 ft above the water line.



The formula for calculating distance to the horizon, including refraction correction for standard atmosphere  is  D =  3.86 * sqrt ( h )    h = height in meters,  D = distance in km.

The Planier lighthouse  is   66 meters asl,   and is visible for 43 km,   calculated distance is 43 km,   Flat Earthers claim the calculated value is 29 km
The Jeddah lighthouse      is 113 meters asl,  and is visible for 46 km,   calculated distance is 53 km,   Flat Earthers claim the calculated value is 38 km
The Ile Vierge lighthouse  is   82.5 meters asl,   and is visible for 50 km,   calculated distance is 47 km,   Flat Earthers claim the calculated value is 33 km
The Genoa lighthouse  is 76 meters asl,  and is visible for 46 km,   calculated distance is 45 km,   Flat Earthers claim the calculated value is 31 km

The conclusion is that if you do the calculations properly,  the flat earth argument,  falls flat on its face ( once again )

For details of the formula derivation go to ...
http://web.archive.org/web/20131717132700/http://mintaka.sdsu.edu/GF/explain/atmos_refr/horizon.html
http://en.wikipedia.org/wiki/Horizon
Calculation of ducting effects.
http://web.archive.org/web/20131717125400/http://mintaka.sdsu.edu/GF/explain/atmos_refr/bending.html

Just sit or lay down on the beach  so your eyes (line of sight) just 10 cm above sea level (equal 0.1 meters),
according to your formula, the horizon just 1.2 km away, but you will see more far than that.
or inversery, climb up to the mountain so your line of sight 100 meters above sea level, are you see the object 38.6 km away?
So that formula not fit with the reality, because the earth is flat.
In fact if we climb up to top of the mountain up to 300 meters, there is no something visible beyond the horizon.

Quote from: Rayzor on June 10, 2015, 10:24:11 PM

I posted this in an earlier thread,  I think it's worth repeating that flat earthers just don't do well at maths.

From the time of Rowbotham, the flat earth believers quote chapter and verse on the distances to lighthouses, claiming they are visible much further out to sea than what curvature calculations predict.
Here is a plot of lighthouse visibility versus height of the light above sea level,   taking into account standard correction for refraction and a bridge height of 30 ft above the water line.



The formula for calculating distance to the horizon, including refraction correction for standard atmosphere  is  D =  3.86 * sqrt ( h )    h = height in meters,  D = distance in km.

The Planier lighthouse  is   66 meters asl,   and is visible for 43 km,   calculated distance is 43 km,   Flat Earthers claim the calculated value is 29 km
The Jeddah lighthouse      is 113 meters asl,  and is visible for 46 km,   calculated distance is 53 km,   Flat Earthers claim the calculated value is 38 km
The Ile Vierge lighthouse  is   82.5 meters asl,   and is visible for 50 km,   calculated distance is 47 km,   Flat Earthers claim the calculated value is 33 km
The Genoa lighthouse  is 76 meters asl,  and is visible for 46 km,   calculated distance is 45 km,   Flat Earthers claim the calculated value is 31 km

The conclusion is that if you do the calculations properly,  the flat earth argument,  falls flat on its face ( once again )

For details of the formula derivation go to ...
http://web.archive.org/web/20131717132700/http://mintaka.sdsu.edu/GF/explain/atmos_refr/horizon.html
http://en.wikipedia.org/wiki/Horizon
Calculation of ducting effects.
http://web.archive.org/web/20131717125400/http://mintaka.sdsu.edu/GF/explain/atmos_refr/bending.html

Just sit or lay down on the beach  so your eyes (line of sight) just 10 cm above sea level (equal 0.1 meters),
according to your formula, the horizon just 1.2 km away, but you will see more far than that.
or inversery, climb up to the mountain so your line of sight 100 meters above sea level, are you see the object 38.6 km away?
So that formula not fit with the reality, because the earth is flat.
In fact if we climb up to top of the mountain up to 300 meters, there is no something visible beyond the horizon.

What is this based on? I don't think simply saying 'the horizon should be 1.2 km away but you can see much further than that' really cuts it.

Can you give an example where your eye height being 10 cm off the ground allows you to view the horizon more than 1.2 km away?

Just sit or lay down on the beach  so your eyes (line of sight) just 10 cm above sea level (equal 0.1 meters),
according to your formula, the horizon just 1.2 km away, but you will see more far than that.

That will depend if the there is extra refraction at the water's surface which often occurs.
If it doesn't, then you will just see those 1.2 km of Earth.
You can still see things that are further than that distance, which are then higher than sea level.

or inversery, climb up to the mountain so your line of sight 100 meters above sea level, are you see the object 38.6 km away?
So that formula not fit with the reality, because the earth is flat.
In fact if we climb up to top of the mountain up to 300 meters, there is no something visible beyond the horizon.

The horizon is those 38.6 km away, unless there are taller things in the way.

But both of these are just your claims with no backing at all.
I haven't measured it myself, but I know that the higher I am, the further away the horizon is, and it isn't due to limited resolving ability.

Meanwhile, on a FE, your height should have no effect on the distance to the horizon.

The crow's nest is designed to look over the bow of distant ships identifying individuals aboard as friend or foe.  The height of the nest improves the ultimate safety of the crew and vessel.

Obviously, people who sail the sea are aware the Earth is flat.  They observe and commit to the fact of a flat horizon recognizing variance in the atmoplane causing inconsistency when attempting mathematical application to stated observations.

Round Earth believers are sadly mistaken in their claims relative to the horizon and have plenty of opportunities to inquire with the sea faring community to confirm FE claims.       

The crow's nest is designed to look over the bow of distant ships identifying individuals aboard as friend or foe.  The height of the nest improves the ultimate safety of the crew and vessel.

Obviously, people who sail the sea are aware the Earth is flat.  They observe and commit to the fact of a flat horizon recognizing variance in the atmoplane causing inconsistency when attempting mathematical application to stated observations.

Round Earth believers are sadly mistaken in their claims relative to the horizon and have plenty of opportunities to inquire with the sea faring community to confirm FE claims.     

Wrong on all counts!

In particular, the fact that the earth is a globe is used to determine distance off when observing the loom of a lighthouse changing to the blink of a rising light http://www.sailtrain.co.uk/navigation/rising.htm

The crow's nest is designed to look over the bow of distant ships identifying individuals aboard as friend or foe.  The height of the nest improves the ultimate safety of the crew and vessel.

Obviously, people who sail the sea are aware the Earth is flat.  They observe and commit to the fact of a flat horizon recognizing variance in the atmoplane causing inconsistency when attempting mathematical application to stated observations.

Round Earth believers are sadly mistaken in their claims relative to the horizon and have plenty of opportunities to inquire with the sea faring community to confirm FE claims.     

Really?
And we do "inquire with the sea faring community to confirm" the claims about the Globe,
but not of the day-sailors who couldn't care less about the shape of the earth, go and ask those who sail long distances if the earth is flat.

It's so strange that you never back anything up with evidence, but look at:

How to Calculate the Distance to the Horizon

Have you ever been out on a leisurely cruise and suddenly wondered, "How far it is to the horizon?" Or maybe your destination is a port that has a lighthouse and you wonder "How far away will I be when I see the lighthouse?" (Well, you're in luck, even if you are a sick unit that thinks of these sorts of things - so are we.) We have the answer!

Of course you can find tables that do the calculation for you in numerous navigation books, almost every book that talks about passagemaking, the Coast Pilot, almanacs, etc. But what if you didn't have any of these references onboard? How could you calculate the distance to the horizon or the "distance off" if you know the height of an object?

It's simple, really. If you want to know the distance to the horizon you simply have to know your height of eye. That is the distance that your eyes are off the surface of the water.

If you're in a jon boat, that would probably be about three feet (if you are sitting like you should be in a jon boat). Of course, if you were in a jon boat you probably wouldn't care how far the horizon was.

Anyway, I digress. If you are on the tuna tower of a sport fishing boat you may be 15, 20, 25 feet above the surface of the water.

Once you know your height of eye you simply plug that into the following formula:

1.17 times the square root of your height of eye = Distance to the horizon in nautical miles

For example, if your height of eye is 9 feet above the surface of the water, the formula would be:

1.17 times the square root of 9 = Distance to the horizon in nautical miles.
1.17 × 3 = 3.51 nautical miles

If you want to calculate the distance at which an object becomes visible, you must know your height of eye and the height of the object. You then do the same calculation for your distance to the horizon and the object's distance to the horizon and add the distances together. For example:                     You have the same height of eye of 9 feet so your distance to the horizon is still 3.51 nautical miles. You're approaching a port that has a lighthouse that is shown on your chart to have a height of 81 feet. Using the same formula you would find that 1.17 times the square root of 81 or (1.17 × 9) = 10.53 nautical miles (the light house can be seen 10.53 nautical miles over the horizon) By adding the two together: 3.51 + 10.53 = 14.04 nautical miles, you should be able to see the lighthouse when you are 14.04 nautical miles away.

Yes, those people the actually "So down to the sea in ships boats" seem to think that the earth is actually a Globe.
And this post (on tfes.org) shows that the "Lookout Training Handbook NAVEDTRA 12968-D" shows that the USN trained their lookouts in the same sort if thing:
Geodetic Surveyor Straightens Out The Flat Earth Reality « Reply #45 on: March 14, 2016, 12:46:32 PM ».
Maybe the USN really thought the earth to be a Globe.

The crow's nest is designed to look over the bow of distant ships identifying individuals aboard as friend or foe.  The height of the nest improves the ultimate safety of the crew and vessel.

A fine candidate for inclusion in this thread.

The crow's nest is designed to look over the bow of distant ships identifying individuals aboard as friend or foe.  The height of the nest improves the ultimate safety of the crew and vessel.

It has the more important task of allowing them to see ships at a longer distance than is capable from the deck.

Obviously, people who sail the sea are aware the Earth is flat.  They observe and commit to the fact of a flat horizon recognizing variance in the atmoplane causing inconsistency when attempting mathematical application to stated observations.

Quite the opposite.
The observe the horizon which varies as the change height (such as going up to the crow's nest).
They observe it move away from them as they move towards it.
They observe that ignoring refraction, while they can see the horizon it is irrelevant as to what the atmosphere is.
They also observe things disappear from the bottom up as they sail away from it.
All clearly indicating a round Earth.

Yes, the horizon is flat at sea, as you would expect it to be considering it is a circle centred on a point slightly below you.

Round Earth believers are sadly mistaken in their claims relative to the horizon and have plenty of opportunities to inquire with the sea faring community to confirm FE claims.     

Yet not one of these RE claims has been shown to be wrong, nor have any of the FE claims been shown to support a FE and be true.

The crow's nest is designed to look over the bow of distant ships identifying individuals aboard as friend or foe.  The height of the nest improves the ultimate safety of the crew and vessel.

Obviously, people who sail the sea are aware the Earth is flat.  They observe and commit to the fact of a flat horizon recognizing variance in the atmoplane causing inconsistency when attempting mathematical application to stated observations.

Round Earth believers are sadly mistaken in their claims relative to the horizon and have plenty of opportunities to inquire with the sea faring community to confirm FE claims.     

This is actual nonsense. I would refute your claims, but that would require what you wrote to actually make sense...

The crow's nest is designed to look over the bow of distant ships identifying individuals aboard as friend or foe.

Ha!  Good one.

 

They observe and commit to the fact of a flat horizon recognizing variance in the atmoplane causing inconsistency when attempting mathematical application to stated observations.

This isn't even parsable English.

I posted this in an earlier thread,  I think it's worth repeating that flat earthers just don't do well at maths.

From the time of Rowbotham, the flat earth believers quote chapter and verse on the distances to lighthouses, claiming they are visible much further out to sea than what curvature calculations predict.
Here is a plot of lighthouse visibility versus height of the light above sea level,   taking into account standard correction for refraction and a bridge height of 30 ft above the water line.



The formula for calculating distance to the horizon, including refraction correction for standard atmosphere  is  D =  3.86 * sqrt ( h )    h = height in meters,  D = distance in km.

The Planier lighthouse  is   66 meters asl,   and is visible for 43 km,   calculated distance is 43 km,   Flat Earthers claim the calculated value is 29 km
The Jeddah lighthouse      is 113 meters asl,  and is visible for 46 km,   calculated distance is 53 km,   Flat Earthers claim the calculated value is 38 km
The Ile Vierge lighthouse  is   82.5 meters asl,   and is visible for 50 km,   calculated distance is 47 km,   Flat Earthers claim the calculated value is 33 km
The Genoa lighthouse  is 76 meters asl,  and is visible for 46 km,   calculated distance is 45 km,   Flat Earthers claim the calculated value is 31 km

The conclusion is that if you do the calculations properly,  the flat earth argument,  falls flat on its face ( once again )

For details of the formula derivation go to ...
http://web.archive.org/web/20131717132700/http://mintaka.sdsu.edu/GF/explain/atmos_refr/horizon.html
http://en.wikipedia.org/wiki/Horizon
Calculation of ducting effects.
http://web.archive.org/web/20131717125400/http://mintaka.sdsu.edu/GF/explain/atmos_refr/bending.html

you check this:
http://fjr66.blogspot.co.id/2017/11/distance-to-specific-height-of-object.html

Quote from: Rayzor on June 10, 2015, 10:24:11 PM

I posted this in an earlier thread,  I think it's worth repeating that flat earthers just don't do well at maths.

From the time of Rowbotham, the flat earth believers quote chapter and verse on the distances to lighthouses, claiming they are visible much further out to sea than what curvature calculations predict.
Here is a plot of lighthouse visibility versus height of the light above sea level,   taking into account standard correction for refraction and a bridge height of 30 ft above the water line.



The formula for calculating distance to the horizon, including refraction correction for standard atmosphere  is  D =  3.86 * sqrt ( h )    h = height in meters,  D = distance in km.

The Planier lighthouse  is   66 meters asl,   and is visible for 43 km,   calculated distance is 43 km,   Flat Earthers claim the calculated value is 29 km
The Jeddah lighthouse      is 113 meters asl,  and is visible for 46 km,   calculated distance is 53 km,   Flat Earthers claim the calculated value is 38 km
The Ile Vierge lighthouse  is   82.5 meters asl,   and is visible for 50 km,   calculated distance is 47 km,   Flat Earthers claim the calculated value is 33 km
The Genoa lighthouse  is 76 meters asl,  and is visible for 46 km,   calculated distance is 45 km,   Flat Earthers claim the calculated value is 31 km

The conclusion is that if you do the calculations properly,  the flat earth argument,  falls flat on its face ( once again )

For details of the formula derivation go to ...
http://web.archive.org/web/20131717132700/http://mintaka.sdsu.edu/GF/explain/atmos_refr/horizon.html
http://en.wikipedia.org/wiki/Horizon
Calculation of ducting effects.
http://web.archive.org/web/20131717125400/http://mintaka.sdsu.edu/GF/explain/atmos_refr/bending.html

you check this:
http://fjr66.blogspot.co.id/2017/11/distance-to-specific-height-of-object.html

While his numbers don't match the formula, a key part which is missing is the height of the observer. I assume he is referring to some specific claim.
Your source also ignored refraction.

Using the heights and distances he has provided as "calculated distance" you need the following heights:
9.095399286
9.61262982
9.567977433
8.644971276
Yes, some variation, but all roughly 9 km.

If you use a height of 9 km, then you get the following distances:
42.93879
52.61236
46.64019
45.23070

Or rounded to the nearest km:
43
53
47
45

Matching what he said.

So the big issue is the 3.86.

If you take the radius of Earth in Mm and the height in m, giving the distance in km, the formula is:
D=sqrt(2*R)*sqrt(h).
Putting in the common value of 6.371 Mm for R, this gives:
D=sqrt(2*6.371)*sqrt(h)
=~3.57*sqrt(h).

But that doesn't account for refraction.
The simple way to account for standard refraction is to pretend R is larger (there is math supporting it which I wont go into here).  The common value used is 7/6, that is R is replaced by 7R/6, such as that you get
D=sqrt(2*(7/6)*R*h)
which means we multiply our previous number by sqrt(7/6)=1.08012345
That gives us:
D=~3.856*sqrt(h/m) km.
Which when rounded slightly more gives the 3.86 figure he had.

Any objections?

Quote from: Rayzor on June 10, 2015, 10:24:11 PM

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Here is a plot of lighthouse visibility versus height of the light above sea level,  taking into account standard correction for refraction and a bridge height of 30 ft above the water line.
<< plot missing >>
The formula for calculating distance to the horizon, including refraction correction for standard atmosphere  is  D =  3.86 * sqrt ( h )    h = height in meters,  D = distance in km.

The Planier lighthouse  is   66 meters asl,   and is visible for 43 km,   calculated distance is 43 km,   Flat Earthers claim the calculated value is 29 km
The Jeddah lighthouse      is 113 meters asl,  and is visible for 46 km,   calculated distance is 53 km,   Flat Earthers claim the calculated value is 38 km
The Ile Vierge lighthouse  is   82.5 meters asl,   and is visible for 50 km,   calculated distance is 47 km,   Flat Earthers claim the calculated value is 33 km
The Genoa lighthouse  is 76 meters asl,  and is visible for 46 km,   calculated distance is 45 km,   Flat Earthers claim the calculated value is 31 km
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

you check this:
http://fjr66.blogspot.co.id/2017/11/distance-to-specific-height-of-object.html

Your calculations are fine, though you should note that Rayzor's post included: "taking into account standard correction for refraction".
Standard refraction is included by simply increasing the earth's radius by a factor of ⁷/₆.

When "standard refraction" is included, your calculations agree quite well.

The crow's nest is designed to look over the bow of distant ships identifying individuals aboard as friend or foe.  The height of the nest improves the ultimate safety of the crew and vessel.

Obviously, people who sail the sea are aware the Earth is flat.  They observe and commit to the fact of a flat horizon recognizing variance in the atmoplane causing inconsistency when attempting mathematical application to stated observations.

Round Earth believers are sadly mistaken in their claims relative to the horizon and have plenty of opportunities to inquire with the sea faring community to confirm FE claims.     

Gotham
This is the worst bit of nonsense I have yet witnessed.
And that is saying something.
There is not one bit of truth in the whole thing.

It makes about as much sense as your photo that you post that you claim is the ice wall.

I think you would have saved  yourself some embarassment if you had talked to someone - anyone - in the  navy - any country's navy.
They know that the earth is a globe. They know that the distance that they can see to the horizon depends on their height above the sea. The higher the crow's nest is, the farther they can see. This is why crow's nests and some types of surface search radars are located on the tallest maps on ships.

The Navy has a  "Manual For Lookouts" showing distances for various heights.
A person standing up in a lifeboat at sea level might only be able to see 2 or 3 miles to the horizon.
A person on the bridge, 8 or 9 miles.
A person  in the crow's nest, 12 or 13 miles.

I suppose gotham's opening statement could be true if you have ever watched an old movie.
It was called "Titanic" and it came out in 1997.
Two characters cast as lookouts who were named Fleet and Lee could watch two characters romancing on the deck below called Rose and Jack. So I guess the  purpose of the crow's nest was to identify individuals aboard as friend or foe.

I've yet to interview Navy personnel regarding Earth shape but have met with commercial fishermen and tall ship captains who swear by a flat Earth due to their own experiences.

That last phrase is important when considering this "Manual for Lookouts." That manual could use some updating just like we recognize our Wiki may require updating.  Our motivation is our own experiences and your manual could benefit from our experiences, as well, and go through some necessary FET/Zetetic updating.

My posted ice wall photo represents "what could be" as explanation for how oceans remain viable on the Earth.  I have seen no rational argument that explains how they could remain deep and resident on a globe shaped Earth.

I recall that scene in "Titanic" and do appreciate it's inclusion as an abstraction as great as your claim.