Interesting Physics

[hecd2]

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.

yes, in this case light is refracted by the fact that the refractive index of air changes with density, density changes with pressure, and the pressure falls with altitude. There can also be an effect due to changes in refractive caused by temperature gradients (that’s how some desert and hot road mirages are caused - specifically by total internal reflection at high/low refractive index boundary - the low refractive index caused by the heating of the air near the sand or road.). But,as I said, we are doing an exercise to derive the results in the absence of refraction.

 

[Vico]

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.

Look at the formulas in the diagram which are without any refraction (using 3.57).

hB = 7.62 m (25 feet)
hL = 1.7 m (5.6 feet)

3.57 * (sqrt(1.7) + sqrt(7.62)) is 3.57 * (1.30+ 2.76) = 3.57 * 4.96 = 14.51 km (7.8 nautical miles)

 

[hecd2]

Look at the formulas in the diagram which are without any refraction (using 3.57).

hB = 7.62 m (25 feet)
hL = 1.7 m (5.6 feet)

3.57 * (sqrt(1.7) + sqrt(7.62)) is 3.57 * (1.30+ 2.76) = 3.57 * 4.96 = 14.51 km (7.8 nautical miles)

For heaven’s sake, I am trying to help a couple of guys understand how to do the geometric calculations themselves, and to derive the formulae for themselves, not just to blindly quote formulae from Wikipedia.

 

[Vico]

For heaven’s sake, I am trying to help a couple of guys understand how to do the geometric calculations themselves, and to derive the formulae for themselves, not just to blindly quote formulae from Wikipedia.

The diagram posted goes with your comment from post #15. The approximation formulas are helpful to verify the formal geometry or trig.

 

[mikekle]

It is interesting, in the year 2015, there is STILL a ‘flat earth society’, Ive been to their website, surprising to see some of the people/ scientists they have, and some of their arguments, you should be able to find the website pretty easy on Google.

 

[Protestor]

Forgive me if this was already said

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?

I spent a few years sailing so take this for what it is worth. I think you are thinking about this wrong. The distance to the horizon is dictated by how high up on the mast you are. In other words you can see farther the higher you are so the horizon is different distances depending on where you are observing it. I think this formula is correct (1.2 times the square root of your height of eye in feet= Distance to the horizon in miles) or at least close to do the full calculation you need to take into account the atmosphere refraction which adds like 10% to the total distance for my height this equation looks something like D~=1.1(1.22*(6^.5))~=3.20miles or 2.8 nautical miles. Yes there is a visible difference between a haul up and a haul down ship and the visible difference is correlated to the curvature of the earth.

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

I don’t know what your friend means since you can quoted him without any context. Your implication is false and I am not sure if this is on purpose or not.

Eratosthenes did a fun experiment which you can replicate more or less. Also another fun thing to do is look up great circles. If the world were flat or on a one dimensional plane the shortest distance between any two points will be a straight line. While in the 2-dim plane on a sphere the shortest distance between 2 points is a curved line. Curvature is not the only factor in navigation though. There are a lot of cool things that you can learn about spherical geometry online which will help you if you actually want to learn about this stuff.

 

[You]

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?

i think i might draw a triangle with a line from the centre of the earth to the surface and another line from the centre of the earth to the top of the ship and the right angle resting on the surface of the earth. knowing the length of the hypotenuse and of the another side and the right angle gives the third side which is the distance away of the top of the ship when it is first seen on the horizon. so adding my own distance from the horizon, 3.2 miles, means the ship is 8.55 miles away from me when i first see the top of it over the horizon.

 

[ProVobis]

The diagram posted goes with your comment from post #15. The approximation formulas are helpful to verify the formal geometry or trig.

His point, though, was to DERIVE your reasoning. Anyone can memorize formulas and substitute numbers.

 

[hecd2]

i think i might draw a triangle with a line from the centre of the earth to the surface and another line from the centre of the earth to the top of the ship and the right angle resting on the surface of the earth. knowing the length of the hypotenuse and of the another side and the right angle gives the third side which is the distance away of the top of the ship when it is first seen on the horizon. so adding my own distance from the horizon, 3.2 miles, means the ship is 8.55 miles away from me when i first see the top of it over the horizon.

OK, so, you have the right method, and you are getting results of the right order, but the 3.2 miles from you to the horizon is a little high unless you are assuming that you’re very tall - what did you assume?; and your distance from the horizon to the ship (5.35 miles) is definitely a bit too small. Can you show your working?

 

[You]

OK, so, you have the right method, and you are getting results of the right order, but the 3.2 miles from you to the horizon is a little high unless you are assuming that you’re very tall - what did you assume?; and your distance from the horizon to the ship (5.35 miles) is definitely a bit too small. Can you show your working?

i drew a circle for the earth. then i drew a triangle starting from my eye to the centre of the earth and from the centre of the earth to the top of the ship and then joined the two points with another line.
i then drew a line from the centre of the earth to the horizon, where the last line touches the earth.
so i have two right angled triangles back to back.
if i am 6 foot 6" then the hypotenuse of the first triangle is 6.371 million meters plus my height which makes the distance from me to the horizon 3.12 miles.

using the square on the hypotenuse thingy minus the square of the other side then the square root of the answer gives the distance in meters.

the hypotenuse of the second triangle is 6.371 million meters plus the 25 foot hull expressed in meters.
which makes the distance from the top of the hull to the horizon 6.12 miles.

ok, so its 9.24 miles from me to the top of the ships hull on the horizon.

 

[Vico]

i drew a circle for the earth. then i drew a triangle starting from my eye to the centre of the earth and from the centre of the earth to the top of the ship and then joined the two points with another line.
i then drew a line from the centre of the earth to the horizon, where the last line touches the earth.
so i have two right angled triangles back to back.
if i am 6 foot 6" then the hypotenuse of the first triangle is 6.371 million meters plus my height which makes the distance from me to the horizon 3.12 miles.

using the square on the hypotenuse thingy minus the square of the other side then the square root of the answer gives the distance in meters.

the hypotenuse of the second triangle is 6.371 million meters plus the 25 foot hull expressed in meters.
which makes the distance from the top of the hull to the horizon 6.12 miles.

ok, so its 9.24 miles from me to the top of the ships hull on the horizon.

It is very close to using the approximate formula (without diffraction):
3.57 * (sqrt(2) + sqrt(7.62)) = 14.9 km
14.9 km = 9.3 miles or 8 nautical miles

 

[hecd2]

i drew a circle for the earth. then i drew a triangle starting from my eye to the centre of the earth and from the centre of the earth to the top of the ship and then joined the two points with another line.
i then drew a line from the centre of the earth to the horizon, where the last line touches the earth.
so i have two right angled triangles back to back.
if i am 6 foot 6" then the hypotenuse of the first triangle is 6.371 million meters plus my height which makes the distance from me to the horizon 3.12 miles.

using the square on the hypotenuse thingy minus the square of the other side then the square root of the answer gives the distance in meters.

the hypotenuse of the second triangle is 6.371 million meters plus the 25 foot hull expressed in meters.
which makes the distance from the top of the hull to the horizon 6.12 miles.

ok, so its 9.24 miles from me to the top of the ships hull on the horizon.

Good.

That’s the pure geometrical approach - in reality, as others have said, the horizon is a little further away because of the bending of light rays by the air. How much further depends on atmospheric conditions but it’s typically around 8% further.

 

[hecd2]

It is very close to using the approximate formula (without diffraction):

Not diffraction but refraction which is entirely different optical phenomenon.

 

[Vico]

Not diffraction but refraction which is entirely different optical phenomenon.

Yes, thank you, refraction.

 

[ProVobis]

yes, in this case light is refracted by the fact that the refractive index of air changes with density, density changes with pressure, and the pressure falls with altitude. There can also be an effect due to changes in refractive caused by temperature gradients (that’s how some desert and hot road mirages are caused - specifically by total internal reflection at high/low refractive index boundary - the low refractive index caused by the heating of the air near the sand or road.). But,as I said, we are doing an exercise to derive the results in the absence of refraction.

It’s probably negligible being of the short distance and modest mass of the earth itself, but probably the gravitation pull affects the light as well. This would be the same principle as that which allows viewing of the stars BEHIND the sun during an eclipse, for example. (This was pointed out in the Physics of Light series (I saw it through Netflix)). This principle would be perhaps more significant on the planet Jupiter, which of course has more mass (thus gravitational pull) than Earth.

 

[hecd2]

It’s probably negligible being of the short distance and modest mass of the earth itself, but probably the gravitation pull affects the light as well. This would be the same principle as that which allows viewing of the stars BEHIND the sun during an eclipse, for example. (This was pointed out in the Physics of Light series (I saw it through Netflix)). This principle would be perhaps more significant on the planet Jupiter, which of course has more mass (thus gravitational pull) than Earth.

Yes, over that distance on the Earth the effect is tiny and quite negligible. How tiny? Well, over 15km, which is the distance between boat and observer in the example above, the light is deflected by about 5x10^-8 metres or about 1/20th of a micron. This is equivalent to only about 0.1 wavelengths of blue-green light over 15km!

On Jupiter, the effect is larger. Over a fixed distance path, say 15km, the effect is only ~2.5x that of the Earth, but for a grazing pass, the deflection angle past Jupiter is ~28x that of the deflection angle past Earth (the grazing deflection angle past the Sun, famously predicted by Einstein and first measured by Eddington is ~3,055x times that of the Earth).

 

[ProVobis]

Yes, over that distance on the Earth the effect is tiny and quite negligible. How tiny? Well, over 15km, which is the distance between boat and observer in the example above, the light is deflected by about 5x10^-8 metres or about 1/20th of a micron. This is equivalent to only about 0.1 wavelengths of blue-green light over 15km!

On Jupiter, the effect is larger. Over a fixed distance path, say 15km, the effect is only ~2.5x that of the Earth, but for a grazing pass, the deflection angle past Jupiter is ~28x that of the deflection angle past Earth (the grazing deflection angle past the Sun, famously predicted by Einstein and first measured by Eddington is ~3,055x times that of the Earth).

Thanks for the work you put into that.

 

[inocente]

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?

When you lean against a wall the force is roughly horizontal, same as the force of wind on grass, so don’t see why not as the difference is only in the rigidity of what’s being leaned on.

*does the earth really bend enough at the distance for one to see only the top of the ship? *

Imho science is about finding out for yourself. Get thee to a port!

 

[inocente]

if i am 6 foot 6" then the hypotenuse of the first triangle is 6.371 million meters plus my height which makes the distance from me to the horizon 3.12 miles.

using the square on the hypotenuse thingy minus the square of the other side then the square root of the answer gives the distance in meters.

the hypotenuse of the second triangle is 6.371 million meters plus the 25 foot hull expressed in meters.
which makes the distance from the top of the hull to the horizon 6.12 miles.

ok, so its 9.24 miles from me to the top of the ships hull on the horizon.

Are Americans required by federal law to never make up your mind which units to use?

btw this kind of calculation should always include ± X to show the uncertainty in the measurements you used, otherwise you could end up with a spurious accuracy. For instance how much does that hypotenuse vary and what if the ship was heavily laden etc.

 

[You]

Are Americans required by federal law to never make up your mind which units to use?

btw this kind of calculation should always include ± X to show the uncertainty in the measurements you used, otherwise you could end up with a spurious accuracy. For instance how much does that hypotenuse vary and what if the ship was heavily laden etc.

i considered that and it worried me a lot that i have to use somebody else’s measurement of the radius of the earth. assuming they are wrong i can blame them. i will try to correct this uncertainty by measuring the radius of the earth myself, (in my spare time).

i agree with the units, we should all use the same units.