The Curve of The Earth According To NASA's Own Dimensions of The Earth

Quote from: Marciano on March 09, 2016, 09:03:18 PM

Quote from: Googleotomy on January 30, 2016, 09:43:32 AM

The curve of the earth - or the "drop" - is not linear because the earth is a sphere- the cross section is a circle - and therefore the  "drop" is not linear.

At 1 miles, the drop is .6 foot.
At 5 miles, the drop is 15 feet.
At 10 miles, the drop is 63 feet
At 15 miles, the drop is 141 feet.
https://www.quora.com/How-many-feet-per-mile-does-the-earth-curve-down-from-where-you-stand

You can prove this with a large beach ball and a yardstick.
Lay the yardstick on top of the beach ball and measure the length to the ball for each inch on the yardstick for example. You will find the increase in the drop will correspond with the figures above in the same ratio. The drop is not linear. The drop for the first mile is very small but increases, again non-linearly, as the distance increases.

Or you can just draw a circle and use a ruler to check this out.

O.k., so I did that and it doesn't work out the way you described it.  On a straight grade I see a pattern like this:  12, 24, 36, 48, 60 units of measure, but on a circle I see a pattern like this:  2.5, 5, 10, 20, 40 units of measure. 

Granted, both are different, but they are both straightforward, unlike the formula from nasa for measuring the slope of "the curve" of the earth.   

Why blame NASA, that curvature formula was around long before NASA. And, they did NOT invent the globe!
Just remember that the globe earth was around for some 2,500 years before NASA was ever thought of.

You seem to have NASAphobia, quite possibly need some CBT!

You are the "John come lately!"
And that formula is approximately correct up to a few hundred miles.

O.k, mister angry pants.  Just because an idea is old doesn't mean it is right   

Regardless of the formula, at some point nearly four thousand miles of drop has to be accounted for, over the course of around six thousand miles.  That is a boat load of drop to account for.  I mean, that is huge!  We don't observe that, hence it seems impossible!  Anyone who's been on an ocean cruise should be able to see for miles across the water.  There is no drop! 

 

O.k., so I did that and it doesn't work out the way you described it.  On a straight grade I see a pattern like this:  12, 24, 36, 48, 60 units of measure, but on a circle I see a pattern like this:  2.5, 5, 10, 20, 40 units of measure. 

Granted, both are different, but they are both straightforward, unlike the formula from nasa for measuring the slope of "the curve" of the earth.   

Then you did something wrong. Can you provide a video, or a picture of how you did it?

Regardless of the formula, at some point nearly four thousand miles of drop has to be accounted for, over the course of around six thousand miles.  That is a boat load of drop to account for.  I mean, that is huge!  We don't observe that, hence it seems impossible!  Anyone who's been on an ocean cruise should be able to see for miles across the water.  There is no drop!

And of course you ignored my post, where I provided the derivative for earth. Just add a few distances, and check how much the earth curves away from you at different distances, it will become clearer.

Quote from: Marciano on March 09, 2016, 09:03:18 PM

O.k., so I did that and it doesn't work out the way you described it.  On a straight grade I see a pattern like this:  12, 24, 36, 48, 60 units of measure, but on a circle I see a pattern like this:  2.5, 5, 10, 20, 40 units of measure. 

Granted, both are different, but they are both straightforward, unlike the formula from nasa for measuring the slope of "the curve" of the earth.   

Then you did something wrong. Can you provide a video, or a picture of how you did it?

Quote from: Marciano on March 10, 2016, 01:13:59 PM

Regardless of the formula, at some point nearly four thousand miles of drop has to be accounted for, over the course of around six thousand miles.  That is a boat load of drop to account for.  I mean, that is huge!  We don't observe that, hence it seems impossible!  Anyone who's been on an ocean cruise should be able to see for miles across the water.  There is no drop!

And of course you ignored my post, where I provided the derivative for earth. Just add a few distances, and check how much the earth curves away from you at different distances, it will become clearer.

O.k., I read your post again and gave it some thought.  Basically, it seems to me like you are saying, that if the earth is a globe, the roundness of it would be undetectable from the ground, because of the nature of curvature on such a large surface, with people being so small in comparison.  Does that about cover it? 

O.k., I read your post again and gave it some thought.  Basically, it seems to me like you are saying, that if the earth is a globe, the roundness of it would be undetectable from the ground, because of the nature of curvature on such a large surface, with people being so small in comparison.  Does that about cover it?

Nope? How'd you get that message? Sure, it's still true for a lot of occasions, but my post didn't discuss any of that. Read it again:

The derivate for a circle with the radius of the earth (in km): -x/sqrt(40960000-x^2)
Plotted in Wolfram|Alpha: http://www.wolframalpha.com/input/?i=-x%2Fsqrt(40960000-x%5E2)&rawformassumption=

As we can see in the graph in Wolfram|Alpha, at x=0km the drop off is 0km per km. At roughly x=4800km the drop off is 1 km per kilometer. At x=6400km the drop off becomes imaginary, at an x infinitely smaller than x=6400km the drop off is infinite, this is because the earth is about to curve back inwards. So the graph shows that any circle will increase their drop-off rate the further you increase the distance. The distance is a straight line that tangents the circle at x=0, so it is not ground distance.

You think that a circle's gradient (drop off per distance) is comparable to that of a triangle. Sure, at every point of a circle the drop off is equal in relation to itself. But your query was how the gradient changed depending on the distance between an observer and a point. My post gives you the derivative of a circle with the diameter of the earth. If you insert a distance X, you'll get the gradient at that point, and as X get's bigger the gradient becomes bigger. I'm just explaining how the gradient of a circle changes.

I'll edit this. Here is a simple plot showing the rate of change of the earths slope (blue) vs the one you describe (red):

As you can see, the red line is flat. This is because the rate of change is constant, it does not change. I dont know if you understand what the derivative of a function is. The derivative describes the rate of change for a function at any given point. For a straight line, the rate of change is constant, because the line is straight. It will be just as steep at any point. A curved surface will increase its drop exponentially.

So as you can see, the earths drop is very low at first. Say you draw a straight line from the earths center to the equator. Now stand at the north pole and draw an equally long line parallel to the first one, off into space. Say you can walk on this line. You go a few miles. The earth is still very near your feet. You walk a few thousand miles, and you will start to rapidly see the earths surface going away from you. Walk the radius of the earth, and you can see the equator. How is this so hard?

Edit to correct image-- jroa

You have to account for 4000 miles of drop, over 6000 miles.   I don't see that. 

You have to account for 4000 miles of drop, over 6000 miles.   I don't see that.

No shit. You'd have to have x-ray eyes to be able to see that far through the horizon!

You have to account for 4000 miles of drop, over 6000 miles.   I don't see that.

The expressions for "curvature over a distance" or "distance to horizon from a height" are just approximations that can be use over a limited distance.
This one gives height (in feet) you need to be to see a given distance (in miles):
Then the next one gives the horizon distance (in km) for a given height (in metres):
Where h is in metres and d is in km.
At a guess, these are not far off for heights of up to 100,000 m and distances of say 1,300 km. Certainly over 200,000 m altitude the error starts to grow.

Above that you need to look at whether you mean straight line distance or distance around the curve.

PS Neither the diameter of the earth, nor these expressions have anything to do with NASA!
Back in about 1000 AD, al Burini estimated the diameter of the earth as 6,339.9 km, only 16.8 km less than the current figure. He was certainly NOT in NASA, a Jesuit (FAR, FAR from it!) or a FreeMason!

One mile is equivalent to approximately one and six tenths kilometers. 

One mile is equivalent to approximately one and six tenths kilometers.

Did you read anything of what he said?

Surely this problem could all be solved by two boats at the north pole and two boats at the south pole both traveling opposite directions around the poles then compare the amount of days taken till each boat met its partner. if the earth is flat it would take a hell of a lot longer to travel the sure of the south then the north.

Surely this problem could all be solved by two boats at the north pole and two boats at the south pole both traveling opposite directions around the poles then compare the amount of days taken till each boat met its partner. if the earth is flat it would take a hell of a lot longer to travel the sure of the south then the north.

Are you expecting these boats to slide across the ice? 

Quote from: bennyc on April 02, 2016, 11:30:45 AM

Surely this problem could all be solved by two boats at the north pole and two boats at the south pole both traveling opposite directions around the poles then compare the amount of days taken till each boat met its partner. if the earth is flat it would take a hell of a lot longer to travel the sure of the south then the north.

Are you expecting these boats to slide across the ice? 

They said 'around the poles' you know? I'd say that means going through the water.

Quote from: jroa on April 02, 2016, 11:54:36 AM

Quote from: bennyc on April 02, 2016, 11:30:45 AM

Surely this problem could all be solved by two boats at the north pole and two boats at the south pole both traveling opposite directions around the poles then compare the amount of days taken till each boat met its partner. if the earth is flat it would take a hell of a lot longer to travel the sure of the south then the north.

Are you expecting these boats to slide across the ice? 

They said 'around the poles' you know? I'd say that means going through the water.

He said the boats were at the poles,   I've not known you to have reading comprehension problems in the past. 

Quote from: Jane on April 02, 2016, 01:25:56 PM

Quote from: jroa on April 02, 2016, 11:54:36 AM

Quote from: bennyc on April 02, 2016, 11:30:45 AM

Surely this problem could all be solved by two boats at the north pole and two boats at the south pole both traveling opposite directions around the poles then compare the amount of days taken till each boat met its partner. if the earth is flat it would take a hell of a lot longer to travel the sure of the south then the north.

Are you expecting these boats to slide across the ice? 

They said 'around the poles' you know? I'd say that means going through the water.

He said the boats were at the poles,   I've not known you to have reading comprehension problems in the past.

Ah, so 'at' couldn't possibly mean 'in the vicinity of,' thanks for clarifying.
I'm just going to go take a break at home: which of course means I'm going to physically phase through the walls and become one with the bricks and mortar.

Quote from: jroa on April 02, 2016, 01:31:04 PM

Quote from: Jane on April 02, 2016, 01:25:56 PM

Quote from: jroa on April 02, 2016, 11:54:36 AM

Quote from: bennyc on April 02, 2016, 11:30:45 AM

Surely this problem could all be solved by two boats at the north pole and two boats at the south pole both traveling opposite directions around the poles then compare the amount of days taken till each boat met its partner. if the earth is flat it would take a hell of a lot longer to travel the sure of the south then the north.

Are you expecting these boats to slide across the ice? 

They said 'around the poles' you know? I'd say that means going through the water.

He said the boats were at the poles,   I've not known you to have reading comprehension problems in the past.

Ah, so 'at' couldn't possibly mean 'in the vicinity of,' thanks for clarifying.
I'm just going to go take a break at home: which of course means I'm going to physically phase through the walls and become one with the bricks and mortar.

I did not realize that when you are 3000 km away from the poles, you are "at" the poles.  Perhaps I am "at" Brasília, Brazil right now.  Or, perhaps I am "at" Nuuk, Greenland.  I suppose it does not matter, since apparently, you don't have to be somewhere in order to be "at" that place. 

Quote from: rabinoz on March 09, 2016, 11:32:33 PM

Quote from: Marciano on March 09, 2016, 09:03:18 PM

Quote from: Googleotomy on January 30, 2016, 09:43:32 AM

The curve of the earth - or the "drop" - is not linear because the earth is a sphere- the cross section is a circle - and therefore the  "drop" is not linear.

At 1 miles, the drop is .6 foot.
At 5 miles, the drop is 15 feet.
At 10 miles, the drop is 63 feet
At 15 miles, the drop is 141 feet.
https://www.quora.com/How-many-feet-per-mile-does-the-earth-curve-down-from-where-you-stand

You can prove this with a large beach ball and a yardstick.
Lay the yardstick on top of the beach ball and measure the length to the ball for each inch on the yardstick for example. You will find the increase in the drop will correspond with the figures above in the same ratio. The drop is not linear. The drop for the first mile is very small but increases, again non-linearly, as the distance increases.

Or you can just draw a circle and use a ruler to check this out.

O.k., so I did that and it doesn't work out the way you described it.  On a straight grade I see a pattern like this:  12, 24, 36, 48, 60 units of measure, but on a circle I see a pattern like this:  2.5, 5, 10, 20, 40 units of measure. 

Granted, both are different, but they are both straightforward, unlike the formula from nasa for measuring the slope of "the curve" of the earth.   

Why blame NASA, that curvature formula was around long before NASA. And, they did NOT invent the globe!
Just remember that the globe earth was around for some 2,500 years before NASA was ever thought of.

You seem to have NASAphobia, quite possibly need some CBT!

You are the "John come lately!"
And that formula is approximately correct up to a few hundred miles.

O.k, mister angry pants.  Just because an idea is old doesn't mean it is right   

Earn much from this practice of yours? Are you sure it's legal?
I'm not angry, just wondering why you keep blaming NASA for the shape of the earth! I guess you haven't started your CBT yet - might explain it.

Quote from: Marciano on March 09, 2016, 09:03:18 PM

Quote from: Googleotomy on January 30, 2016, 09:43:32 AM

The curve of the earth - or the "drop" - is not linear because the earth is a sphere- the cross section is a circle - and therefore the  "drop" is not linear.

At 1 miles, the drop is .6 foot.
At 5 miles, the drop is 15 feet.
At 10 miles, the drop is 63 feet
At 15 miles, the drop is 141 feet.
https://www.quora.com/How-many-feet-per-mile-does-the-earth-curve-down-from-where-you-stand

You can prove this with a large beach ball and a yardstick.
Lay the yardstick on top of the beach ball and measure the length to the ball for each inch on the yardstick for example. You will find the increase in the drop will correspond with the figures above in the same ratio. The drop is not linear. The drop for the first mile is very small but increases, again non-linearly, as the distance increases.

Or you can just draw a circle and use a ruler to check this out.

O.k., so I did that and it doesn't work out the way you described it.  On a straight grade I see a pattern like this:  12, 24, 36, 48, 60 units of measure, but on a circle I see a pattern like this:  2.5, 5, 10, 20, 40 units of measure. 

Granted, both are different, but they are both straightforward, unlike the formula from nasa for measuring the slope of "the curve" of the earth.   

Why blame NASA, that curvature formula was around long before NASA. And, they did NOT invent the globe!
Just remember that the globe earth was around for some 2,500 years before NASA was ever thought of.

You seem to have NASAphobia, quite possibly need some CBT!

You are the "John come lately!"
And that formula is approximately correct up to a few hundred miles.

Subjects such as the curvature of the earth and the distance to the horizon were covered many , many years before NASA and have been used in such sources  as Navy Training Manuals for lookouts on ships.

Regardless of the formula, at some point nearly four thousand miles of drop has to be accounted for, over the course of around six thousand miles.  That is a boat load of drop to account for.  I mean, that is huge!  We don't observe that, hence it seems impossible!  Anyone who's been on an ocean cruise should be able to see for miles across the water.  There is no drop! 

You can't "See for miles acros the water". You can only see so far.  And you can only see as far as the horizon. I would estimate the distance to horizon in your photo as about 5 or 6 miles. This is due to the curvature of the earth because the earth is a  globe. This isn't something new from NASA. It's something that sailors have known for centuries. Check out the formula listed on a previous post. It works.

Marciani seems to have 2 problems:
(1) He is confusing gradient and curvature of the earth.Two different subjects.
(2) He is confusing "8 inches per foot" and "8 inches per mile". Of course it's "8 inches per mile."

I cant see how this is so difficult for Marciano to understand...

From the north pole to the equator there is a 4000km drop. Simple. You dont see it because when you travel along the surface of the earth you are travelling in a curve, slowly descending 8" per mile RELATIVE TO YOUR CURRENT POISTION until you get there.

If you could fix an incredibly long plank of wood (theoretically) to the north pole, that was 4000km long and perfectly flat, then walk to the end of it, then you would see the ground drop away at an exponential rate, 8" for the first mile, 16 for the second, 32 for the third etc... When you get near to the end of the plank the ground will be dropping away from you at the rate of 1 mile drop for 8" walked.

Even you can hopefully see a 4000km drop below you if you were standing at the end of this plank....

This is why if you average out this drop you get your 0.64 slope or whatever it was. BUT, this is a circle so averaging doesnt realisticaly show what is happening. If you lived on a flat surface however....

I cant see how this is so difficult for Marciano to understand...

From the north pole to the equator there is a 4000km mile drop. Simple. You dont see it because when you travel along the surface of the earth you are travelling in a curve, slowly descending 8" per mile RELATIVE TO YOUR CURRENT POISTION until you get there.

...

For the record, I think you mean 4000 miles when you say 4000 km. The diameter of the Earth is a shade under 8000 miles so its radius is just under 4000 miles, or a bit less than 6400 km.

That sort of oops(!) apparently cost a Mars mission.

Other than that, good explanation.

I have messed around with a curvature calculator. I am wondering if anyone can answer how you can look out over 100 miles and see a city when its suppose to be over 5,000 ft below the horizon?

I have messed around with a curvature calculator. I am wondering if anyone can answer how you can look out over 100 miles and see a city when its suppose to be over 5,000 ft below the horizon?

In order to answer that, we'd kinda need to know what city and how one sees it 100 miles away (that's rather far away to see, it must be well lit or the buildings must be pretty big).

And refraction counters curvature (When it comes to seeing).

Quote from: Marciano on March 10, 2016, 01:13:59 PM

Regardless of the formula, at some point nearly four thousand miles of drop has to be accounted for, over the course of around six thousand miles.  That is a boat load of drop to account for.  I mean, that is huge!  We don't observe that, hence it seems impossible!  Anyone who's been on an ocean cruise should be able to see for miles across the water.  There is no drop! 

You can't "See for miles acros the water". You can only see so far.  And you can only see as far as the horizon. I would estimate the distance to horizon in your photo as about 5 or 6 miles. This is due to the curvature of the earth because the earth is a  globe. This isn't something new from NASA. It's something that sailors have known for centuries. Check out the formula listed on a previous post. It works.

Marciani seems to have 2 problems:
(1) He is confusing gradient and curvature of the earth.Two different subjects.
(2) He is confusing "8 inches per foot" and "8 inches per mile". Of course it's "8 inches per mile."

Thanks for your reply.  I can see the downtown of the city I live near on a clear day, no problem.  It's about 12 miles away.  There are suburbs, in my area, that are around 25 miles away from downtown and when I'm traveling through those suburbs I can often see downtown and not just the tops of the buildings, I can see most the the buildings.  I think you're limiting things.  Although, the number I see all over the internet is 3.1 miles (not 5 or 6) is as far as one can see over water, because of the curve of the earth.  However, it seems that they are incorrect. 

If you truly believe in a globe earth, then you have to account for approximately 4000 miles of drop, over approximately 6000 miles.  That equates to a drop of 8 inches per foot, not 8 inches per mile!  Unless I'm looking over the side of a cliff, I don't see that much drop and there aren't that many cliffs around, so it seems as if that drop doesn't exist.  However, if you can show me that it does, I'm all ears 

In regards to NASA's figures, NASA will tell you the earth is around 25,000 MILES in circumference and around 8,000 MILES in diameter. 

So, in that regard, I am using, "NASA'S OWN FIGURES." 

So, I suppose scientists are lying when they say light can refract the other way? 

  If you truly believe in a globe earth, then you have to account for approximately 4000 miles of drop, over approximately 6000 miles.  That equates to a drop of 8 inches per foot, not 8 inches per mile!  Unless I'm looking over the side of a cliff, I don't see that much drop and there aren't that many cliffs around, so it seems as if that drop doesn't exist.  However, if you can show me that it does, I'm all ears 

Do you understand that we're talking about a sphere, (or at least a circle if a sphere is too much to grasp)?  Do you know what a circle is?  I don't believe you do.

In regards to NASA's figures, NASA will tell you the earth is around 25,000 MILES in circumference and around 8,000 MILES in diameter. 

So, in that regard, I am using, "NASA'S OWN FIGURES."

An Islamic Astronomer/Mathematician/Geodetic Surveyor named Al-Biruni measured the circumference quite accurately way back around 1,000 AD.

So Exactly how accurate was Biruni ?
With his formula Biruni arrived at the value of the circumference of the earth within 200 miles of the actual value of 24,902 miles, that is less then 1% of error. Biruni's stated radius of 6335.725 km is also very close to the original value.

From Al-Biruni's Classic Experiment
So, who was this Al-Biruni? Look in AL-BIRUNI: A MASTER OF SCHOLARSHIP.


And I believe that the "curvature" has been explained to you numerous times!
The surface of the Globe is spherical NOT a straight line from equator to poles.

The approximation horizon drop = 8" x miles² only holds over a limited distance (though not bad for up to hundreds of miles), after that the curve starts to steepen till at the point 90° away from the start the slope is (guess what) 90°.

This Al-Biruni went a lot further:

Al-Biruni continued a remarkable career in the early 1000s, doing research into fields such as:
How the earth spins on its axis
How wells and springs transport water to the surface
Combining statics and dynamics into the study of mechanics
Recording the latitude and longitude of thousands of cities, which allowed him to determine the direction towards Makkah for each city.

Same "Lost Islamic History"  site.

So stop your stupidity of blaming NASA for the Heliocentric Global model for the earth. It was well accepted centuries before NASA and space travel came along. You only make yourself look absolutely biased and indoctrinated!