Interesting Physics

[thinkandmull]

Here is a couple of physics questions I was thinking about last night. There is a poem somewhere that says that the wind was “leaning on the grass”. Is this physically possible? Leaning too heavily would cause the grass to go down all the way, but could maybe the wind lean a little bit on it? (This somewhat sounds like the question of how many angels can fit on the point of a needle)

In discussions on whether the earth was flat or round, people use to point out that a sail is
seen first on the horizon, then the bottom of the ship. I want to know about the geometry
here. How far away is the horizon of the ocean from the viewer, and does the earth really bend enough at the distance for one to see only the top of the ship?

A friend of mine told me that we can see lights bending around the curve of the earth, but this implies that the earth bends noticeably within the range of the horizon

If the earth was flat, what would be seen first in the horizon as the ship approaches? This part of the questions comes from a thought I had while driving out of town yesterday. If someone looks out at the horizon unto a large meadow, there doesn’t seem to be the optical effect that one has when seeing the road stretch out towards a point as you drive on it. So there is something optically here I’d like to know more about.

 

[Gertabelle]

Well, I can say that there are certain places here in Denver where we have such a long view that the clouds to the east seem to be sitting on the land. That’s the curve of the earth, keeping us from seeing the land below, but allowing us to see the clouds above that far-off land.

As for your question about what part of the ship you would see first if the earth were flat – that would be the front part of the ship. You’d see the whole front of the ship coming at you on the water when it was close enough to see.

 

[seagal]

I found an interesting video on YouTube that answers your first question, how far away in the horizon (about a mile if you’re in the water, about 3 miles if you’re standing on shore). I tried to find other videos about ships sails disappearing and just found a whole bunch from some deluded people who are trying to prove the earth is flat

 

[GEddie]

In a flat world, ships would not seem to go over the horizon; they would just appear smaller and smaller as they moved farther and farther and farther away.

In the non-visual realm, radio reception would be unlimited by distance in all bands as long as power was sufficient.

ICXC NIKA.

 

[hecd2]

In discussions on whether the earth was flat or round, people use to point out that a sail is
seen first on the horizon, then the bottom of the ship. I want to know about the geometry
here. How far away is the horizon of the ocean from the viewer, and does the earth really bend enough at the distance for one to see only the top of the ship?

A friend of mine told me that we can see lights bending around the curve of the earth, but this implies that the earth bends noticeably within the range of the horizon

If the earth was flat, what would be seen first in the horizon as the ship approaches? This part of the questions comes from a thought I had while driving out of town yesterday. If someone looks out at the horizon unto a large meadow, there doesn’t seem to be the optical effect that one has when seeing the road stretch out towards a point as you drive on it. So there is something optically here I’d like to know more about.

seagal:

I found an interesting video on YouTube that answers your first question, how far away in the horizon (about a mile if you’re in the water, about 3 miles if you’re standing on shore). I tried to find other videos about ships sails disappearing and just found a whole bunch from some deluded people who are trying to prove the earth is flat

1. Why not do the geometrical calculations yourselves? They are not that difficult. You can easily google the radius of the Earth and how how tall a ship’s sail is, and assume you are 6 foot 4 inches tall as a boost to self-confidence . Then use basic geometry on a circle. If you manage it yourselves then you’ll have far more insight into the answer than just reading it. So try this first before going to 2. I’m sure tandm can do it as he has a proof that Cantor’s diagonal argument is flawed and “as elegant as a zombie”.
2. If you can’t or don’t want to figure it out yourselves from scratch, try googling “horizon distance” and see what Wikipedia has to say in the Horizon article. It’s all there.

I bet tandm will do 2. (or neither 1. nor 2.). I don’t know about seagal.

 

[thinkandmull]

1. Why not do the geometrical calculations yourselves? They are not that difficult. You can easily google the radius of the Earth and how how tall a ship’s sail is, and assume you are 6 foot 4 inches tall as a boost to self-confidence . Then use basic geometry on a circle. If you manage it yourselves then you’ll have far more insight into the answer than just reading it. So try this first before going to 2. I’m sure tandm can do it as he has a proof that Cantor’s diagonal argument is flawed and “as elegant as a zombie”.
2. If you can’t or don’t want to figure it out yourselves from scratch, try googling “horizon distance” and see what Wikipedia has to say in the Horizon article. It’s all there.

I bet tandm will do 2. (or neither 1. nor 2.). I don’t know about seagal.

The earth is 24,901 miles in circumference and the horizon is about 3.5 miles. The earth would have to bend about 25 feet in order to be able to see the sail and not the ship.

The problem for Cantor is that the diagonal proof does not use numbers as based on geometry. Why do people get so touchy about Cantor? There were mathematicians who disagreed with him during his lifetime, before he lost his mind.

 

[thinkandmull]

1. Why not do the geometrical calculations yourselves? They are not that difficult. You can easily google the radius of the Earth and how how tall a ship’s sail is, and assume you are 6 foot 4 inches tall as a boost to self-confidence . Then use basic geometry on a circle. If you manage it yourselves then you’ll have far more insight into the answer than just reading it. So try this first before going to 2. I’m sure tandm can do it as he has a proof that Cantor’s diagonal argument is flawed and “as elegant as a zombie”.
2. If you can’t or don’t want to figure it out yourselves from scratch, try googling “horizon distance” and see what Wikipedia has to say in the Horizon article. It’s all there.

I bet tandm will do 2. (or neither 1. nor 2.). I don’t know about seagal.

As you said to JuanFlorencio today, “I am not on home territory… I am not a natural mathematician”. It’s been more years than I want to name since I’ve worked on complex math, but Cantor’s arguments are not complex to me. I’ve read very much on related topics, such as Zeno’s paradoxes.

 

[hecd2]

The earth is 24,901 miles in circumference and the horizon is about 3.5 miles. The earth would have to bend about 25 feet in order to be able to see the sail and not the ship.

You’ve made yourself about 8 feet tall , but leaving that aside, why don’t you go on with the geometry and calculate how far away the ship is when its hull begins to appear above the horizon?

 

[hecd2]

As you said to JuanFlorencio today, “I am not on home territory… I am not a natural mathematician”. It’s been more years than I want to name since I’ve worked on complex math, but Cantor’s arguments are not complex to me. I’ve read very much on related topics, such as Zeno’s paradoxes.

Right.

Perhaps you shouldn’t cast pearl before swine on this backwater of a forum, but write up your proof and submit it to Annals of Mathematics or Acta Mathematica who would be delighted to have Cantor’s theorem disproven.

 

[GEddie]

You’ve made yourself about 8 feet tall , but leaving that aside, why don’t you go on with the geometry and calculate how far away the ship is when its hull begins to appear above the horizon?

Wouldn’t you need her dimensions to be able to do that?

 

[hecd2]

Wouldn’t you need her dimensions to be able to do that?

Yes. tandm mentioned the top of the superstructure is 25 feet above the waterline. Or one could derive an expression for the distance of the ship as a function of its dimensions.

 

[You]

maybe someone can see where I’m going wrong;

25 feet = height of sailboat
3.5 miles = distance top of sail is visible on horizon

quadrant of earth 6225.25 miles on surface
divided by 3.5miles
= 1778.64285
x 25 feet
= 44,466.0714 feet

or in feet

32869320 feet quadrant of earth
divided by 18480 distance to horizon
= 1778.64285
x 25 feet
= 44,466.07 feet
divided by 5280 feet in one mile
= 8.4 miles radius of earth

 

[hecd2]

maybe someone can see where I’m going wrong;

25 feet = height of sailboat
3.5 miles = distance top of sail is visible on horizon

quadrant of earth 6225.25 miles on surface
divided by 3.5miles
= 1778.64285
x 25 feet
= 44,466.0714 feet

or in feet

32869320 feet quadrant of earth
divided by 18480 distance to horizon
= 1778.64285
x 25 feet
= 44,466.07 feet
divided by 5280 feet in one mile
= 8.4 miles radius of earth

I’m not sure what you are trying to calculate? Or why you are using quadrants? Think triangles

 

[You]

…or working out the radius using 3.5 miles to the horizon as the hypotenuse gives 4402 miles as the radius.
but its supposed to be 3959 miles.

 

[hecd2]

Ok, so the problem is to find out how far away the ship is when its hull just appears over the horizon. That happens when you can draw a straight line from the top of the hull to your eye that just grazes (is tangent to) the earth. Divide the distance into two parts, one from the ship to the tangent point (the horizon) and one from the tangent point to your eye. Add those two distances together. How do you find each one? Well, think about drawing triangles with one vertex at the centre of the earth. You know its radius. If you get stuck again come back.

 

[You]

the horizon is just under 3.2 miles away… oh the joy.

 

[hecd2]

the horizon is just under 3.2 miles away… oh the joy.

So if the hull of the ship is 25feet above the water and it has just come into view above the horizon, how far away is it?

 

[ProVobis]

In the movie “Titanic’s Final Mystery” it tries to explain why the ship couldn’t be seen by the closest ship to her. Interesting theory actually, had to do with the temperature of the water.

A physicist would have to consider all possible ways light can be bent.

 

[Vico]

Ok, so the problem is to find out how far away the ship is when its hull just appears over the horizon. That happens when you can draw a straight line from the top of the hull to your eye that just grazes (is tangent to) the earth. Divide the distance into two parts, one from the ship to the tangent point (the horizon) and one from the tangent point to your eye. Add those two distances together. How do you find each one? Well, think about drawing triangles with one vertex at the centre of the earth. You know its radius. If you get stuck again come back.

Light travelling horizontally is refracted downward so the actual distance to the horizon is greater than the distance calculated with geometrical formulas, using STP, the difference is about 8%. In the formulas substitute 3.86 in the metric formulas for 3.57.

(Please Note: This uploaded content is no longer available.)

 

[hecd2]

Light travelling horizontally is refracted downward so the actual distance to the horizon is greater than the distance calculated with geometrical formulas, using STP, the difference is about 8%. In the formulas substitute 3.86 in the metric formulas for 3.57.

One thing at a time old boy. We’re looking at the geometry without refraction first, and we’re trying to understand how to do the calculations not how to apply formulae.