8" x mile²
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OP
Not my formula, it's the formula that is being taught in school trigonometry class. I'm just going by what is taught in the mainstream American educational institutions.
OP
Correct! According to science being taught in our schools, I should not be able to see the boat, yet I do and so do you.
Sapient
Ambassador of Vocal Inspiration
That formula calculates the inches of drop. So in your case over 25 miles there is 416.666 inches of land drop.
OP
Exactly! So my question is, where is the drop? Why can I still see the small boat when science says that it should have gone over the curve and be totally out of my vision?
Sapient
Ambassador of Vocal Inspiration
Lol. This is getting interesting. You calculated the drop that is accumulated over a 25 mile curve span with the formula. Those boats are far from 25 miles away, therefore "science" as your stating is not saying that you shouldn't see it. It says you should see it.
mcblink
Ambassador of Riffs & Spliffs
A very specific type of mirage, which has a specific name. Here's a blurb from a wiki page about it:
"A superior mirage can be right-side up or upside-down, depending on the distance of the true object and the temperature gradient. Often the image appears as a distorted mixture of up and down parts.
Since Earth is round, if the downward bending curvature of light rays is about the same as the curvature of Earth, light rays can travel large distances, including from beyond the horizon. This was observed and documented in 1596, when a ship in search of the Northeast passage became stuck in the ice at Novaya Zemlya, above the Arctic Circle. The Sun appeared to rise two weeks earlier than expected; the real Sun had still been below the horizon, but its light rays followed the curvature of Earth. This effect is often called a Novaya Zemlya mirage. For every 111.12 kilometres (69.05 mi) that light rays travel parallel to Earth's surface, the Sun will appear 1° higher on the horizon. The inversion layer must have just the right temperature gradient over the whole distance to make this possible.
In the same way, ships that are so far away that they should not be visible above the geometric horizon may appear on or even above the horizon as superior mirages."
You were asking why you could see stuff that you weren't supposed to be able to see, according to the formula. I'm only offering possible answers.
Really I can't say that the boat itself is a mirage, I think Sapient is right and that it's not actually that far out there, but that oil rig....the thing is an absolute behemoth if that's 25 miles away and still looks that huge..
Jethro Rocker
Ambassador of Live & Loud Action
From looking at video - boats are on horizon. Oil rig is very hazy, much further away. Rig is not sitting at sea level, how high are those things? 600 feet? So you see boats at horizon, whatever distance that is, and the top part of rig that is apparently a known 25 miles out.
Makes sense to me. Physics works so without actually measuring distance to boats, you have no idea how far away they really are.
Makes sense to me. Physics works so without actually measuring distance to boats, you have no idea how far away they really are.
I’ll further complicate the discussion. One of the early things you learn in physics when dealing with equations that have different units of measure is to determine WHAT unit of measure the answer will have. The units of measure in the OP has inches times square miles. Miles can easily be converted to inches. The leaves inches times square inches. That means the answer is in cubic inches, which is a volume measurement. Not sure HOW that can relate to elevation drop (e.g. Curvature).
I’d spend more time on this but I’m sitting on a deck in Ocean City MD, sipping Chardonnay with this view.
I’d spend more time on this but I’m sitting on a deck in Ocean City MD, sipping Chardonnay with this view.
Jethro Rocker
Ambassador of Live & Loud Action
Not inches times SQUARE miles but inches times miles SQUARED. Still a measurement in inches.
Ex
Say a formula calls for y miles squared, y being 3.
Then final result for that part is 9 miles, not 3 square miles.
Also I think height above sea level affects formula more than just subtracting that distance from final result. Haven't found that part of formula yet and I coild be wrong but I think given the angles involved, it affects end result more.
gball
Ambassador of Boogie
Well, again, I don't math...but, I have spent the last 18 years living right on the beach or very close to it and see the ocean every day, and all I know is some weird stuff goes down over the horizon. What you see yesterday you may not see today or even ever again - I used to be able to see Catalina Island from the beach at the end of the street and some days you could see Avalon and other days it looked like the island was floating over the surface of the water; some days fully-loaded container ships in the channel looked like only the superstructure and containers were above the waves and others you could see the anti-foul paint along the waterline. I've seen even weirder stuff looking at the mainland from Catalina Island. Just the way it is with the ocean.
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LiveeviL2000
Ambassador of Pentagonal Pentatonics
That sentence just terrified me.
LiveeviL2000
Ambassador of Pentagonal Pentatonics
It’s almost 7am and I woke up with a science hangover.
So I guess it’s time for a little hair of the dog.
I agree with with @gball
I’ve been at sea and there are times you could see to the worlds edge, where the sky seems like the shape of a dome as it hits the horizon and you swear you can see forever with other vessels on the horizon. Other times, the horizon seems closer and not so vast.
Much of this phenomenon has to do with rarefaction. That is, light rays actually can be bent to varying degrees in different densities. The varying densities over large bodies of water can have a rather dramatic effect.
As @gball pointed out, this can vary from day to day as atmospheric conditions change. In fact, his observation of the island appearing to float in the air is not altogether unusual. Others have noted boats that appear to be floating in the air. Nothing more than a visual illusion rooted in simple physics.
To see the curvature of the earth in real time, just observe that as objects do eventually disappear over the horizon, they disappear from the bottom up, which is exactly consistent with a round earth. Of course, this is assuming other conditions, such as haze, don't obscure the object entirely.
EDIT: I should have said "refraction"!
As @gball pointed out, this can vary from day to day as atmospheric conditions change. In fact, his observation of the island appearing to float in the air is not altogether unusual. Others have noted boats that appear to be floating in the air. Nothing more than a visual illusion rooted in simple physics.
To see the curvature of the earth in real time, just observe that as objects do eventually disappear over the horizon, they disappear from the bottom up, which is exactly consistent with a round earth. Of course, this is assuming other conditions, such as haze, don't obscure the object entirely.
EDIT: I should have said "refraction"!
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"Science" does not actually say that.
Individuals say that.
The estimate of 8" x mile[Squared] at first glance suggests what you are stating, but "science" would also account for other conditions, such as refraction.
Incidentally, the equation of 8" x mile[Squared] is an equation which estimates the curvature of a parabola, not a sphere. Even with parabolas, there are more accurate equations. The 8" x mile[Squared] equation is easier to figure than other, more complex, formulas, so it does have some pragmatic value. Now, admittedly, even though the 8" x mile[Squared] formula applies to a parabola, not a sphere, it will still be reasonably accurate for a sphere, over limited distances, even up to several miles. But, eventually, the lines diverge. The equation for a sphere is h = r – r cos(s/r).
I think you meant “refraction”…the bending of light…it’s what I do…I’m a licensed, professional, light bender. My skills pale in comparison to the Earth though. Although, I suppose, rarefaction does effect refraction when dealing with atmosphere as a refractive medium.
Good morning science!
I think you meant “refraction”…the bending of light…it’s what I do…I’m a licensed, professional, light bender. My skills pale in comparison to the Earth though. Although, I suppose, rarefaction does effect refraction when dealing with atmosphere as a refractive medium.
Good morning science!
Good point!
Thanks for the clarification.