Here's a puzzle for you...

[TonyR]

Its not my fault I ask questions that are either really hard or nearly impossible to answer.

Actually they are really easy to answer, its just that you find the answers impossible to understand. Perhaps something to reflect on.

Or you could ask yourself "What does it matter?" Whether you think the earth is flat or round makes no difference at all in your day to day life. All it does is distract far more important matters such as working out how many angels can dance on the head of a pin.

[Mick F]

........... 58 miles away when there needs to be a curve of water that is over 2,000 feet tall ............

Where do you get that figure from?

2,000ft in 58miles?

[aspiringcyclist]

........... 58 miles away when there needs to be a curve of water that is over 2,000 feet tall ............

Where do you get that figure from?

2,000ft in 58miles?

I think it's 6370km - 6370km*cos(93/6370) = 0.68km. The latter is about 2,200 feet. 6730km is the radius of the Earth and 93/6370 is the angle in radians due to a 58 mile ( 93 kilometre ) arc on the circumference.

[jochta]

when you kept it in the one thread I did nt see the problem either - that was the power crazy mod's doing...

Again, we answered your questions on lighthouses many times. The response was basically "I'm not good enough at maths to understand it"

You don't have to be "good at maths" to understand how to work out the curve once you know how.

8 inches X the miles X the miles = the drop in inches.

There's nothing wrong with asking "How can something less than 250 feet tall be visible from 58 miles away when there needs to be a curve of water that is over 2,000 feet tall obscuring it, that isn't there?"

How can you have "answered" that, when on a curved Earth the answer doesn't work? All you gave me was an answer that physically/optically doesn't work out, light bends, but not to that extent.

People get angry when they can't answer it, or they just throw out the usual "refraction" answer as though just because refraction exists in the world, oh well then it has to account for all of it. The fact that light needs to bend to ridiculous extremes doesn't put you off. It puts me off, because it is bunk. Light refracts, but not like that.

Light refracts just like that. The distances are tiny compared with the scale of the Earth, it's a negligible refraction.

[jochta]

........... 58 miles away when there needs to be a curve of water that is over 2,000 feet tall ............

Where do you get that figure from?

2,000ft in 58miles?

It's correct. If you put your eyeball at sea level and looked out tangentially to the surface of the Earth an object 58 miles away would be around 2000 vertical feet below your eyeball...

http://www.davidsenesac.com/Information ... sight.html

http://mathcentral.uregina.ca/QQ/databa ... rley3.html

Still that's less than 1% of the horizontal distance which can be overcome by refraction given the right atmospheric circumstances. Also if your eyeball isn't resting on the Earth's surface, like on top of a cliff or in a crow's nest, the vertical height changes significantly.

[Manc33]

A good way to test it would be a ship painted in a non reflective paint, I don't mean matt black I mean like a special paint that gives off zero light. That would be a good way of doing it but, unlikely to happen. If the object cannot have light shine off it, you could rule refraction out and do the experiment properly.

Wait though, wouldn't the ship be invisible if "no" light bounced off it?

It does make me wonder what colour the paint would have to be.

What if the ship was just a silhouette of a ship? I mean you'd be then observing a black looking ship, only distinguishable by its outline, which isn't the same thing as light bouncing off to view it.

Folks honestly I still can't believe with the waves there are on the sea, it would refract over a 2,200 foot hump of water @ 58 miles. Maybe doing the test on rough seas would be better, but you can't know how rough it is all the way from you to the object.

[TonyR]

A good way to test it would be a ship painted in a non reflective paint, I don't mean matt black I mean like a special paint that gives off zero light. That would be a good way of doing it but, unlikely to happen. If the object cannot have light shine off it, you could rule refraction out and do the experiment properly.

Wait though, wouldn't the ship be invisible if "no" light bounced off it?

It does make me wonder what colour the paint would have to be.

What if the ship was just a silhouette of a ship? I mean you'd be then observing a black looking ship, only distinguishable by its outline, which isn't the same thing as light bouncing off to view it.

Folks honestly I still can't believe with the waves there are on the sea, it would refract over a 2,200 foot hump of water @ 58 miles. Maybe doing the test on rough seas would be better, but you can't know how rough it is all the way from you to the object.

I haven't had such a good laugh at a thread in a long time. Thank you.



P.S. It happens by Magic*

* "Any sufficiently advanced technology is indistinguishable from magic" Arthur C Clarke.

[sjs]

........... 58 miles away when there needs to be a curve of water that is over 2,000 feet tall ............

Where do you get that figure from?

2,000ft in 58miles?

I think it's 6370km - 6370km*cos(93/6370) = 0.68km. The latter is about 2,200 feet. 6730km is the radius of the Earth and 93/6370 is the angle in radians due to a 58 mile ( 93 kilometre ) arc on the circumference.

About 2200 feet is the vertical distance below a tangent to the earth of a point on the surface 93 km away. But, the maximum height above a straight line joining those two points (hence burrowing through the earth) of a point on the earth surface is only a quarter of that. You can see that by drawing a few triangles and circles and doing a bit of Pythagoras. So the hump is only 500 feet high.

[Mick F]

Now THAT makes sense to me.
Thanks.

[ddraver]

A good way to test it would be a ship painted in a non reflective paint, I don't mean matt black I mean like a special paint that gives off zero light. That would be a good way of doing it but, unlikely to happen. If the object cannot have light shine off it, you could rule refraction out and do the experiment properly.

Wait though, wouldn't the ship be invisible if "no" light bounced off it?

It does make me wonder what colour the paint would have to be.

What if the ship was just a silhouette of a ship? I mean you'd be then observing a black looking ship, only distinguishable by its outline, which isn't the same thing as light bouncing off to view it.

Folks honestly I still can't believe with the waves there are on the sea, it would refract over a 2,200 foot hump of water @ 58 miles. Maybe doing the test on rough seas would be better, but you can't know how rough it is all the way from you to the object.

Jochta has given you the answer multiple times now. Care to discuss what he's actually writing? At the moment you re just sticking your head in the sand. Keep doing that and no we wont take you over seriosuly...

[Manc33]

So the hump is only 500 feet high.

I would say more like 1,100 feet, but regardless of that, the drop from the observer to the object is still 2,200 feet - thus there's a total elevation of 2,200 feet to take into account.

Why would you be cutting a line through it anyway?

Isn't that irrelevant when the thing being discussed is the total difference between the observer and the object?

I haven't had such a good laugh at a thread in a long time. Thank you.

P.S. It happens by Magic*

You think waves moving around will not break light up?

Since an object like a ship is actually touching the water I would say it is to be expected.

Where does the "laugh" come in?

I wish people explained why they are laughing instead of just stating that they are. Anyone can say it.

You had a good laugh but never explained why it was funny. People seem to do this more than explain anything.

[beardy]

True, but the drop from the observer to the object is still 2,200 feet, thus there's a total elevation of 2,200 feet to take into account.

On a flat earth there would be. On a spherical earth it is less.

If I have understood his point correctly the difference is that on our spherical earth the horizon itself is also falling down into the distance and as that is what is supposed to be obscuring your view of a distant object then the drop you are talking of is substantially larger than the height needed for any point beyond the horizon to come back into our line of sight.

To illustrate imagine measuring the height needed with a ruler placed on a globe with a couple of towers.
Your ruler is ALWAYS touching the globe at the point of the observer, our ruler is touching the globe between observer and object. The effect becomes most apparent when observing between two towers of equal height and the ruler is touching at the midpoint between them.

Or more simply put, you are measuring the drop below level and we are talking of drop below the horizon which is smaller and what is actually relevant.

[jochta]

You think waves moving around will not break light up?

It's all about scale. The Earth is a really really big thing. The drop over a few miles is a really really small thing. Waves are a really really really really really really tiny thing.

Refraction doesn't give a monkey's about a couple of thousand feet in several tens of miles. Air masses which cause refraction don't give a monkey's about waves a few tens of feet high.

[Manc33]

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.

[beardy]

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