Here's a puzzle for you...

[Manc33]

and that is why drop is the wrong measure of whether something will be tall enough to be visible or not.

That has no bearing on the light needing to bend upwards by 2,200 feet from point to point.

This guy is pretty entertaining lol...

http://vocaroo.com/i/s0JZz2HNRwdS

He is the guy with the nano laser (1000MW/532nm, class 4 device) that reaches 50 miles and he is doing a few tests so we should find out soon enough assuming he can carry out the test without anything stopping him and I ain't talking about the weather.

[beardy]

That has no bearing on the light needing to bend upwards by 2,200 feet from point to point.

Which is a digression from a digression and not towards the point but yet further from it.

I have forgotten what the original point was but if the basic geometry of working out the height a building would have to be in order to clear the horizon is so drastically flawed, it kind of discredits the whole argument which was based on such figures.

It also ignores the effect that having BOTH the object and observer elevated has on the range of visibility. As in lighthouses and ships' observation posts.

[jochta]

Drop is 2,200 feet. Who knows what the other distance is (the smaller arrow) but I don't know why it is being measured or what it establishes.

The only point of contention is the two black dots. No fiddling about with numbers and measuring other parts of it can decrease the drop.

Ah but put your two black dots on a horizontal line and see the curvature the light needs to refract around, over a hump of water about 500 feet tall. This curve is incredibly shallow as the Earth is a really big thing. Someone who can be bothered to do the maths will work out angle A for you for an object 58 miles away (it'll be very small). Note that this is assuming your eyeball is resting on the Earth's surface.

[Manc33]

You're putting the two dots at the same elevation there and they aren't - if there's a curve.

[jochta]

You're putting the two dots at the same elevation there and they aren't if there's a curve.

Eh? That doesn't even make sense. All I've done is rotate your sketch. They are at the same elevation in both sketches, at sea level.

[Manc33]

Turning it is cheating.

You have to have a horizontal line through the observer and a horizontal line through the object. The difference is the drop. Its not just whats above the dotted line that counts, it all counts. All the sea under that also counts. You can't start turning the globe to "make it work" because then you've changed everything around, the Earth itself doesn't do that.

[jochta]

Turning it is cheating.

You have to have a horizontal line through the observer and a horizontal line through the object. The difference is the drop.

I can turn it any angle I want. It doesn't make any difference to the maths. The vertical difference is still 2,200 feet for a tangential line from an observer's eyeball at sea level. The hump of water isn't 2,200 feet tall as you've said several times.

[Manc33]

I can turn it any angle I want.

Which is my point, the Earth itself can't do that. You have to keep it all straight because that's the reality.

It doesn't make any difference to the maths. The vertical difference is still 2,200 feet for a tangential line from an observer's eyeball at sea level. The hump of water isn't 2,200 feet tall as you've said several times.

Yes it is because you're not stood on the middle of it, you're stood at the extreme left (in my image above).

If the guy on the left could dive down 2,200 feet then swim across, he would then reach the object. He will have had to drop 2,200 feet to do that.

What you're forgetting is that you can just turn a piece of paper, but the Earth itself does not and cannot physically do that.

[jochta]

I can turn it any angle I want.

Which is my point, the Earth itself can't do that. You have to keep it all straight because that's the reality.

It doesn't make any difference to the maths. The vertical difference is still 2,200 feet for a tangential line from an observer's eyeball at sea level. The hump of water isn't 2,200 feet tall as you've said several times.

Yes it is because you're not stood on the middle of it, you're stood at the extreme left (in my image above).

If the guy on the left could dive down 2,200 feet then swim across, he would then reach the object. He will have had to drop 2,200 feet to do that.

What you're forgetting is that you can just turn a piece of paper, but the Earth itself does not and cannot physically do that.

I'm not turning anything, it's just a matter of viewpoint. If I'm stood on the North Pole or on the Equator the diagram is the same. It's a segment of a circle. Segments A and B are identical in my crude sketch below.

Your hypothetical swimmer could just swim out horizontally straight towards the target, at the midpoint of his swim he would be at a depth of 500 feet. The difference in distance between doing this or staying on the surface doing breaststroke is miniscule.

[aspiringcyclist]

What I think we should be measuring is, for a given angle, what height the two objects have to be so that we draw a line between them, the line is a tangent of the circle at the midpoint of the arc.

So if R is the radius and h the object height and a the angle of the arc, h should satisfy ( h + R )* cos(a/2) = R. So that h = R/cos(a/2) -R. For h = 1.7m, the angle a/2 is 1.5*10^(-3). In this case the horizon is 4.7km away, which is the value I found on the internet for an object of that height.

So for arc length of 93 km, to see the top of the the tower they would both have to be 560 ft.

[Mick F]

On a clear day from Snaefell on the Isle of Man, you can see Blackpool Tower.
60odd miles.

Blackpool Tower is only 500ft tall.

[beardy]

Snaefel on the other hand is 620 metres. So you could probably see all of the Blackpool tower.

Even without the effects of refraction.

[Mick F]

Yes, that's what I meant.
If there's a 2,200ft "drop", the tower couldn't be seen.

[jochta]

Snaefel on the other hand is 620 metres. So you could probably see all of the Blackpool tower.

Even without the effects of refraction.

http://www.ringbell.co.uk/info/hdist.htm

Horizon from Snaefell is 89km.

Horizon from the ISS at an altitude of 400km is 2300km

[aspiringcyclist]

I found another more useful formula. If h1 is your height, d distance of the straight line between the top of the observer and the object, and h2 is the height of the object, then h2 = (1/2*R)*(d - Sqrt( 2*h1*R) )^2. So if the distance is 90km and the viewer height is 620m, the minimum height of the object is 0.1m.