Explained: 443 km distance- mountains visible.

FlightMuj

Active Member
Hello all!!!
Eric Dubay posted a video: [Video removed from YouTube]
In this video Eric says that the Mountain (or any of them) should not be visible as like Pic Gaspard below.


The maximum distance achieved is 443 km, and he does his calculation; evaluates observers height but he misses the point that object's height itself is 3867 m. Nonetheless I tried figuring out myself but could not find a calculator that took into account the target (object) height. There is a link of the photographers work :https://beyondhorizons.eu/2016/08/03/pic-de-finestrelles-pic-gaspard-ecrins-443-km/
I cannot get much information after an hour or two research and this is the first time I am creating a post on a flat Earth debate I otherwise solve and prove it on my own and I am kind of embarrassed that I cannot gather information.
So, simply lack of data has motivated me to write on a forum!!!
P.S. I am a new member and have personally disproved every claim about flat Earth (not bragging; I love it!!!), but I just wanted to research it even if it is a repeated claim . I am not giving my full today so this post is not that detailed, Sorry!!!
 
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Trailblazer

Moderator
Staff member
The maximum distance achieved is 443 km, and he does his calculation; evaluates observers height but he misses the point that object's height itself is 3867 m. Nonetheless I tried figuring out myself but could not find a calculator that took into account the target (object) height.
You can use the Metabunk curve calculator: https://www.metabunk.org/curve/

To take into account the target object height, all you need to do is see whether the "hidden height" is greater than the target object height.

Plugging in the numbers from this example, with the photographer 2,820 metres above sea level:

upload_2017-8-9_13-50-59.png

You can see that with standard refraction, about 3,817 metres should be hidden at that distance.

That would leave 50 metres of the top of the target mountain visible. If the refraction was greater than standard, then more of the peak would be visible.

The photographer's own description says that was the case:


You can also do the same calculation from the other end, as visibility works both ways. If you put the viewer height as 3,867m, the height of Pic Gaspard, then you get a hidden height of 2,777m, suggesting that a little over 40 metres of the 2,820m Pic de Finestrelles should be visible.
 
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Trailblazer

Moderator
Staff member
At 3:21 he writes that from the 9,251ft peak of Pic de Finestrelles the distance to the horizon is 117 miles.

That is correct (actually, with refraction it is 127 miles).

He should also do the same distance calculation for Pic Gaspard, which is 12,687ft high.

The distance to the horizon from Pic Gaspard is 138 miles (or, with refraction, 149 miles).

If you add together the two horizon distances and the total is greater than the distance between the peaks, then they are intervisible.

149 + 127 = 276.

The actual distance is 275 miles, so these peaks are right at the limit of intervisibility, and probably require a bit of an extra refraction boost to make them clearly visible.

The fact that these two peaks appear to offer the world record for visibility, and that this has only been recorded when refraction conditions are favourable, is surely excellent evidence that the globe Earth model is accurate! If the Earth were flat, then there would be nothing to stop more distant mountains being visible, as long as there was nothing in between.

I wonder what Eric Dubay would say when asked why the world record for photographing a distant mountain just happens to be exactly on the limit of visibility predicted by the globe model?
 
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FlightMuj

Active Member
Great answer, so the "Refracted Hidden" was the thing I was looking for. Just one more thing what were you trying to say about adding horizons, I did not quite get that...

Thanks in advance!!!
 

FlightMuj

Active Member
You can see that with standard refraction, about 3,817 metres should be hidden at that distance.

That would leave 50 metres of the top of the target mountain visible. If the refraction was greater than standard, then more of the peak would be visible.
So in conclusion he misses out the mathematical fact that refraction is a thing and says that Earth is flat??? Right??? Even in the beginning of his video he completely (deliberately???) misses the height of the observer, tries to solve it and forgets (pun intended) about his another blunder...
 

Mick West

Administrator
Staff member
Still I would like to hear what was that adding horizons thing you were saying.


Here the tops of h1 and h2 are just visible (from each other). The light path between them just grazes the horizon.

S1 and S2 are the distances to the horizon. If you add them together then that gives you the maximum distance that the two points are visible from each other.

If there's refraction, then the light path is bent a bit, making these distances longer.
 

FlightMuj

Active Member
Love it!!! Thanks Mick!!! The added distance i.e. 276 miles is greater than the actual distance, 275 miles, so that means the peaks can be seen but are at the limit of being seen, but then a little more refraction (like in this special case) can boost the distance more so you see more of the peak, right???
 

Mick West

Administrator
Staff member
Love it!!! Thanks Mick!!! The added distance i.e. 276 miles is greater than the actual distance, 275 miles, so that means the peaks can be seen but are at the limit of being seen, but then a little more refraction (like in this special case) can boost the distance more so you see more of the peak, right???
Exactly, yes.
 

Rory

Senior Member.
S1 and S2 are the distances to the horizon. If you add them together then that gives you the maximum distance that the two points are visible from each other.
Brief note: S1+S2 isn't quite the distance between, in this case, the two peaks, as that's the distance at sea level, and doesn't take into account the "tilt".

Doing so here, though, only adds about another 140 metres to the distance between the two peaks, and doesn't make any real difference to the results of the calculations (other variables such as refraction will no doubt have a greater influence).
 
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FlightMuj

Active Member
Brief note: S1+S2 isn't quite the distance between, in this case, the two peaks, as that's the distance at sea level, and doesn't take into account the "tilt".

Doing so here, though, only adds about another 140 metres to the distance between the two peaks, and doesn't make any real difference to the results of the calculations (other variables such as refraction will no doubt have a greater influence).
Yeah I was thinking the same, that the actual distance will be greater as it is a bigger arc.
How do you know that it will be an increment of 140 m???
 

Rory

Senior Member.
How do you know that it will be an increment of 140 m?
I was curious what difference it would make for some calculations on another flat earth debunk I was working on, so I made a spreadsheet to work it out (simplified version attached).

What I discovered: it genuinely makes no real difference to the results, so isn't worth the extra work. We're talking inches, feet, and thousandths of a degree. Even here, over a relatively massive distance, that extra 140m results in only 4m or so increased hidden amount.
 

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Trailblazer

Moderator
Staff member
Still I would like to hear what was that adding horizons thing you were saying.
Mick has explained it pretty well but you can also think of it like this:

If the horizon from peak 1 is 127 miles away then a person standing at sea level 127 miles away can just see the top of peak 1 (because lines of sight work. It's ways).

That person is therefore 148 miles away from peak 2 (275-127=148).

And the horizon from peak 2 is 149 miles away, which means that anyone 149 miles or less from peak 2 can see it.

So the person standing at that point can just see both peak 1 and peak 2. And if you can see both peaks in opposite directions from the same place at sea level, then clearly you can also see one peak from the other.

You can imagine each peak has a circle of visibility around it whose size is dependant on the height of the peak. If those circles intersect then the peaks are visible from each other. (Barring real world issues like intervening terrain, of course.)
 

Keith B

New Member
I have composed a few images that prove curvature on the original image beyond doubt. (I even used Mick's Google Earth trick in the first of them). The first two show how you are looking "through" the earth to the sea level at the points under the distant peaks, while the third one proves that the earth calculator is correct.
Here they are:


 
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