[_ Old Earth _] Where Did the Idea of a Flat Earth Originate?

[Free]

If you want me to say you are right and I am wrong, due to the accuracy and deviation of my simplistic, although applicable, equation, then OK I give. You are right, I am wrong.

I'm not trying to say who is right and who is wrong. I just find that if flat-earthers can't even get basic math right, how can they be right about much else?

I have made the grave error of agreeing with people like this:
You use the Pythagorean theorem. If you are at a point P on the earth's surface and move tangent to the surface a distance of 1 mile then you can form a right angled triangle as in the diagram. Using the theorem of Pythagoras a2 = 39632 + 12 = 15705370 and thus a = 3963.000126 miles. Thus your position is 3963.000126 - 3963 = 0.000126 miles above the surface of the earth. 0.000126 miles = 1252800.000126 = 7.98 inches. Hence the earth's surface curves at approximately 8 inches per mile.

Note that the person presenting this states "approximately" 8 inches per mile. This is the point of the error in my calculation. A sphere cannot have a curve of a certain number of units per mile that changes as the distance increases. The curve of a sphere is constant. If I were able to find the exact curve per mile, I could calculate it by squaring the distance. Not for area but for the exponential increase in the curve.

No, you couldn't. That is simply not how math works. You can't just decide that you are going to square a number "for the exponential increase in the curve" and not for area. That's just arbitrary.

Here are three methods that can be used to calculate the magic number. Check this site. They show Zetetic Astronomy, Pythagorean Theorem and Trigonometry. http://flatvsround.blogspot.ca/2015/10/how-to-calculate-earths-curvature.html

And the first two are not correct, although the second is better than the first, for the following reasons:

1) As I have been pointing out, units matter. When multiplying, dividing, adding, or subtracting numbers with units, the units also have to be taken into account to ensure that the answer has the correct units one is looking for. In the first equation, the one you are fond of, we have: miles * miles * inches / inches / feet. That means the answer to the equation has the units: miles²ft, which doesn't even make sense. Of course, one could convert the inches to miles first but then we would get: miles * miles * miles, which, as I have previously stated, gives us miles³ (volume).

When you do the second and third examples, they have: miles - miles. The answer therefore has the unit: miles, which is what we are looking for. That makes perfect sense.

2) Look closely at the difference of the distance actually being calculated between the first two equations and the third. In the first two, the distance is normal to the surface of the earth at the point, which is not what we're trying to calculate. In the third, the distance we are calculating is normal to the tangency of the viewpoint. That is what we want--the vertical drop from our view.

3) I hate to point it out again, but the first equation utterly fails at a distance equal to the radius of the earth, whereas the third does not. The second fails as well. That means the first and second equations can approximate up to a point but ultimately rely on improper math.

 

[Jacks Bratt]

I do not believe math always is exact. The Atlantic and Pacific oceans are at different elevations, but water should seek it's own level. There is a reason for this, having to do with density and specific gravity. There are many phenomenons that do not have logical answers. The bending of light can cause a lot of odd and scientific infractions of Newton's Law.

I understand, totally. However, there is the example of the salt flats. I, for one, do not believe that we have a total grasp on all the mysteries of God's universe. This is why I am investigating.

 

[Jacks Bratt]

I'm not trying to say who is right and who is wrong. I just find that if flat-earthers can't even get basic math right, how can they be right about much else?


No, you couldn't. That is simply not how math works. You can't just decide that you are going to square a number "for the exponential increase in the curve" and not for area. That's just arbitrary.


And the first two are not correct, although the second is better than the first, for the following reasons:

1) As I have been pointing out, units matter. When multiplying, dividing, adding, or subtracting numbers with units, the units also have to be taken into account to ensure that the answer has the correct units one is looking for. In the first equation, the one you are fond of, we have: miles * miles * inches / inches / feet. That means the answer to the equation has the units: miles²ft, which doesn't even make sense. Of course, one could convert the inches to miles first but then we would get: miles * miles * miles, which, as I have previously stated, gives us miles³ (volume).

When you do the second and third examples, they have: miles - miles. The answer therefore has the unit: miles, which is what we are looking for. That makes perfect sense.

2) Look closely at the difference of the distance actually being calculated between the first two equations and the third. In the first two, the distance is normal to the surface of the earth at the point, which is not what we're trying to calculate. In the third, the distance we are calculating is normal to the tangency of the viewpoint. That is what we want--the vertical drop from our view.

3) I hate to point it out again, but the first equation utterly fails at a distance equal to the radius of the earth, whereas the third does not. The second fails as well. That means the first and second equations can approximate up to a point but ultimately rely on improper math.

Every site that I go to , whether globe earth based belief or flat earth belief, globe denier or flat earth denier... they all give the same math.

It's not that "flat earther's" cannot do math. The flat earth websites, with integrity in their math, are using the calculations of the supporters of the globe model...

Here is another site: http://www.smokescreendesign.com/curvature-of-the-earth.html

I'm sorry if you refuse to accept the 8 inches per mile squared and fail to understand that you can square a distance for other purposes than area. However, it is truth.

Maybe you can show me a site where the math you are using is presented and where the math shown on every site I go to is wrong.

Like I said, globe site, flat earth site, the math is the same. Over and over and over.

 

[Douglas Summers]

I understand, totally. However, there is the example of the salt flats. I, for one, do not believe that we have a total grasp on all the mysteries of God's universe. This is why I am investigating.

I believe so too. This is what God meant when He told man to (Gen. 1:27-29) Subdue the earth. (discover the mysteries, become masters of the earth and it's environment.

 

[Free]

Every site that I go to , whether globe earth based belief or flat earth belief, globe denier or flat earth denier... they all give the same math.

It's not that "flat earther's" cannot do math. The flat earth websites, with integrity in their math, are using the calculations of the supporters of the globe model...

Here is another site: http://www.smokescreendesign.com/curvature-of-the-earth.html

I'm sorry if you refuse to accept the 8 inches per mile squared and fail to understand that you can square a distance for other purposes than area. However, it is truth.

Maybe you can show me a site where the math you are using is presented and where the math shown on every site I go to is wrong.

Like I said, globe site, flat earth site, the math is the same. Over and over and over.

Then they're all using bad math. I have never said that distance can't be squared for other purposes, just that on it's own, distance squared by itself as a formula, gives area, not a curve. As I said, simply squaring a distance not gives area, it is completely arbitrary when it comes to the curvature of a sphere/circle. The formula for a circle, plotted on a graph (to actually produce a circle), is x² + y² = r², where r is, of course, the radius.

And I still cannot believe that I have to reiterate that the error of x² * 8 / 12 gets worse the farther out one goes, ending in catastrophic failure at 3963 miles, the radius of the earth. This is because the formula simply makes no sense. You'll notice that in the formula for a circle, to solve for any of the variable, one ends up with whichever units of distance one uses. That makes sense; miles²ft doesn't make any sense at all when one is looking for miles.

As for showing you a site where the math I am using is presented, I have been very, very clear that it was first given in the picture and diagram you gave and also in the previous link you provided: http://flatvsround.blogspot.ca/2015/10/how-to-calculate-earths-curvature.html. It's like you're not even reading the material you're providing.

 

[Jacks Bratt]

Then they're all using bad math. I have never said that distance can't be squared for other purposes, just that on it's own, distance squared by itself as a formula, gives area, not a curve. As I said, simply squaring a distance not gives area, it is completely arbitrary when it comes to the curvature of a sphere/circle. The formula for a circle, plotted on a graph (to actually produce a circle), is x² + y² = r², where r is, of course, the radius.

And I still cannot believe that I have to reiterate that the error of x² * 8 / 12 gets worse the farther out one goes, ending in catastrophic failure at 3963 miles, the radius of the earth. This is because the formula simply makes no sense. You'll notice that in the formula for a circle, to solve for any of the variable, one ends up with whichever units of distance one uses. That makes sense; miles²ft doesn't make any sense at all when one is looking for miles.

As for showing you a site where the math I am using is presented, I have been very, very clear that it was first given in the picture and diagram you gave and also in the previous link you provided: http://flatvsround.blogspot.ca/2015/10/how-to-calculate-earths-curvature.html. It's like you're not even reading the material you're providing.

If I take my measuring tape and use it to build a deck, it works just fine. If I were to use it to do land survey's it would be taken beyond the limits of it's tolerance and the errors would grow as did the distances measured.

The distance, in miles, squared, in my formula stands. It is the amount of drop, in inches that is where my formula loses it's accuracy.

This number may be 7.xxxxxxxxxxxxxxxxxxxxxxxx or 8.xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx I don't know and, for all intents and purposes, I don't care. The math is fine for distances that the human eye can see, or, are view able with optical devices.

I will not be calculating distances in the range that you seem to be concerned about.

You and I are viewing this from different perspectives.

All I was saying is that the earth curves more than we, or at least, I expected. Eight inches in a mile is observable and there are testable instances where this amount of curve is contradicted by repeatable, observable and testable procedures.

I used the crude simple formula, which is perfectly applicable for distances of 100 to 150 miles, to calculate the drop and confirm that some things that are seen, in a distance, should be well beyond the curve of the earth and totally out of view.

All of this was done to show that investigation of the globe earth may cause you to arrive at conclusions that only create more questions.

 

[Free]

If I take my measuring tape and use it to build a deck, it works just fine. If I were to use it to do land survey's it would be taken beyond the limits of it's tolerance and the errors would grow as did the distances measured.

Obviously. But a tape measure is accurate for it's length to begin with.

The distance, in miles, squared, in my formula stands.

No, the distance squared does not stand. I've already pointed out why this is the case. Until I see mathematical proofs as to why it is okay to square the distance in relation to the curvature of the earth, it is just arbitrary and one may as well cube it or not do anything to it at all.

It is the amount of drop, in inches that is where my formula loses it's accuracy.

This number may be 7.xxxxxxxxxxxxxxxxxxxxxxxx or 8.xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx I don't know and, for all intents and purposes, I don't care. The math is fine for distances that the human eye can see, or, are view able with optical devices.

As I have stated a couple of times, it has absolutely nothing to do with the accuracy. The entire formula is wrong. You still won't acknowledge the fact that because the error grows significantly the further out one gets, ending with catastrophic failure at a distance equal to the earth's radius, the formula is wrong. And it has little, if anything, to do with the accuracy of the 8" approximation.

Interestingly enough, as I was writing all this out, I thought of a simpler way to come up with the vertical drop, and it verifies to the nth decimal place the more complicated math (as given by you) that I was using. I stated previously that the formula for graphing a circle is x² + y² = r², where 'x' and 'y' are at the origin (0,0). If you enter that formula into a graphing calculator (or online equation "grapher"), you will get a perfect circle with what ever radius you choose, centered on the origin (which is 0,0). Make sense?

We know then that, for our purposes, the circle represents the surface of the earth, which for every horizontal distance out, the vertical drop should be a point on that circle (that is exactly what we've been trying to calculate). So we know that the horizontal distance out is 'x' and we know that the radius is about 3959 miles. Our unknown is the vertical drop, which we will call 'd'. Since in the formula x² + y² = r² the 'y' is the vertical distance from the origin (0,0), we know from basic geometry that the drop, 'd' will be equal to the radius, r', minus our vertical distance 'y', to give us: d = r - y. To figure out what 'y' is, we simply rearrange the equation for a circle to get: y = sqr root (r² - x²).

To solve for 'd' then, we just substitute our formula for 'y' into our formula for 'd', which gives us: d = r - sqr root (r² - x²). It is very simple; much more simple than using the trig equations and gives an identical answer.

I think the explanation may make it sound more complicated than it is as it really is very basic, so just use the diagram you previously gave to get a visual, except that our 'd' is their 'x' and our 'x' is their 'L': http://christianforums.net/Fellowsh...lat-earth-originate.65518/page-3#post-1270325

So, doing it my way, plotted on a graph, will give us a circle (quarter circle for our purposes). But if we use the formula you use and plot it on a graph, we get a parabola (I literally plotted it by hand and then verified online here: http://www.quickmath.com/webMathematica3/quickmath/graphs/equations/basic.jsp). A parabola immediately shows us there is a significant problem with the math.

All I was saying is that the earth curves more than we, or at least, I expected. Eight inches in a mile is observable and there are testable instances where this amount of curve is contradicted by repeatable, observable and testable procedures.

I used the crude simple formula, which is perfectly applicable for distances of 100 to 150 miles, to calculate the drop and confirm that some things that are seen, in a distance, should be well beyond the curve of the earth and totally out of view.

And I have agreed that somehow the crude, simple formula does give an approximation. Somehow but I don't know how. My only point is that the formula is nonsense. And I have also pointed out that that formula wasn't intended to calculate a vertical drop either, as shown in the link you provided.

As for places that are flat, if the surface of the earth were smooth, the surface area would be about 197,359,487 miles². Given that magnitude of surface area (actual surface area is much more), we should not be surprised if even seemingly large areas of the earth actually are flat. I have made the point before and will make it again: there is likely much more to it than these simple formulae.

 

[Jacks Bratt]

Obviously. But a tape measure is accurate for it's length to begin with.


No, the distance squared does not stand. I've already pointed out why this is the case. Until I see mathematical proofs as to why it is okay to square the distance in relation to the curvature of the earth, it is just arbitrary and one may as well cube it or not do anything to it at all.


As I have stated a couple of times, it has absolutely nothing to do with the accuracy. The entire formula is wrong. You still won't acknowledge the fact that because the error grows significantly the further out one gets, ending with catastrophic failure at a distance equal to the earth's radius, the formula is wrong. And it has little, if anything, to do with the accuracy of the 8" approximation.

Interestingly enough, as I was writing all this out, I thought of a simpler way to come up with the vertical drop, and it verifies to the nth decimal place the more complicated math (as given by you) that I was using. I stated previously that the formula for graphing a circle is x² + y² = r², where 'x' and 'y' are at the origin (0,0). If you enter that formula into a graphing calculator (or online equation "grapher"), you will get a perfect circle with what ever radius you choose, centered on the origin (which is 0,0). Make sense?

We know then that, for our purposes, the circle represents the surface of the earth, which for every horizontal distance out, the vertical drop should be a point on that circle (that is exactly what we've been trying to calculate). So we know that the horizontal distance out is 'x' and we know that the radius is about 3959 miles. Our unknown is the vertical drop, which we will call 'd'. Since in the formula x² + y² = r² the 'y' is the vertical distance from the origin (0,0), we know from basic geometry that the drop, 'd' will be equal to the radius, r', minus our vertical distance 'y', to give us: d = r - y. To figure out what 'y' is, we simply rearrange the equation for a circle to get: y = sqr root (r² - x²).

To solve for 'd' then, we just substitute our formula for 'y' into our formula for 'd', which gives us: d = r - sqr root (r² - x²). It is very simple; much more simple than using the trig equations and gives an identical answer.

I think the explanation may make it sound more complicated than it is as it really is very basic, so just use the diagram you previously gave to get a visual, except that our 'd' is their 'x' and our 'x' is their 'L': http://christianforums.net/Fellowsh...lat-earth-originate.65518/page-3#post-1270325

So, doing it my way, plotted on a graph, will give us a circle (quarter circle for our purposes). But if we use the formula you use and plot it on a graph, we get a parabola (I literally plotted it by hand and then verified online here: http://www.quickmath.com/webMathematica3/quickmath/graphs/equations/basic.jsp). A parabola immediately shows us there is a significant problem with the math.


And I have agreed that somehow the crude, simple formula does give an approximation. Somehow but I don't know how. My only point is that the formula is nonsense. And I have also pointed out that that formula wasn't intended to calculate a vertical drop either, as shown in the link you provided.

As for places that are flat, if the surface of the earth were smooth, the surface area would be about 197,359,487 miles². Given that magnitude of surface area (actual surface area is much more), we should not be surprised if even seemingly large areas of the earth actually are flat. I have made the point before and will make it again: there is likely much more to it than these simple formulae.

God Bless and Have a nice day.

 

[Free]

 

[Jim Parker]

I'm not trying to say who is right and who is wrong. I just find that if flat-earthers can't even get basic math right, how can they be right about much else?

OK
I have been ignoring this thread because I figured it was a joke.
But, now I have to ask, are there really people who think the earth is flat?
Or is this really just a joke?

iakov

 

[Free]

OK
I have been ignoring this thread because I figured it was a joke.
But, now I have to ask, are there really people who think the earth is flat?
Or is this really just a joke?

iakov

I wish it were a joke. Unfortunately, there are people who really do believe the earth is flat. There is a conspiracy (for reasons unknown) to make us believe the earth is round, and those who are Christian (maybe they all are, I don't know) believe the Bible teaches a flat earth (apparently unaware of the use of figures of speech in the Bible).

 

[Jim Parker]

I wish it were a joke. Unfortunately, there are people who really do believe the earth is flat. There is a conspiracy (for reasons unknown) to make us believe the earth is round, and those who are Christian (maybe they all are, I don't know) believe the Bible teaches a flat earth (apparently unaware of the use of figures of speech in the Bible).

There is the story of a Madison avenue advertising executive who was supposed to have said that no advertiser ever went wrong underestimating the intelligence of the American consumer.
I don't know if the story is real but the comment is accurate.
It is a demonstration that teaching people to read and write and to do arithmetic does not mean they will no longer be stupid.

oi veh

iakov the fool

 

[Free]

There is the story of a Madison avenue advertising executive who was supposed to have said that no advertiser ever went wrong underestimating the intelligence of the American consumer.
I don't know if the story is real but the comment is accurate.
It is a demonstration that teaching people to read and write and to do arithmetic does not mean they will no longer be stupid.

oi veh

iakov the fool

Yes, quite. Intro to logic should be mandatory teaching in either high school or first year university. Teaching everyone basic reasoning won't solve everything but it would go a long way in helping avoid such silliness, IMO.

 

[Jim Parker]

The Atlantic and Pacific oceans are at different elevations, but water should seek it's own level.

False.
PURE water will seek its own level.
The oceans are not pure water.

 

[jasonc]

salt water isn't pure water. its meant to be salty by the way our Maker wants it.

 

[Jacks Bratt]

False.
PURE water will seek its own level.
The oceans are not pure water.

So, are you saying that salt water doesn't seek it's own level.... I thought that all liquids seek their own level.

 

[jasonc]

So, are you saying that salt water doesn't seek it's own level.... I thought that all liquids seek their own level.

The Pacific ocean sits higher then the Atlantic

 

[Jim Parker]

So, are you saying that salt water doesn't seek it's own level.... I thought that all liquids seek their own level.

All liquids of the same density seek their own level.
So, if I put equal volumes of mercury and water in a u-shaped tube, the top of the water would be higher than the top of the mercury.

 

[Jacks Bratt]

All liquids of the same density seek their own level.
So, if I put equal volumes of mercury and water in a u-shaped tube, the top of the water would be higher than the top of the mercury.

Oh, I see, you are talking about them seeking their own level as height differences due to pressure, not a level flat surface as I assumed.

Seeking their own level, to me, means that every given point on the surface of a liquid will be exactly the same height.

What you are talking about in the "U" shaped tube, is pressure equilibrium's. The example you quoted works that way due to the mercury putting more pressure on the liquid in the very bottom of the tube. The weight of the liquid on each side of the very middle of the tube, at the bottom, must be the same.

The surface of a lake, however, will always reach a perfect level unless acted upon by a force, like wind.

Also, right at the side of a container, liquids will form a meniscus. A concave meniscus is where the liquid curves up at the edges of the container. This is because the molecules of the liquid are more strongly attached to the container material than themselves.

If the meniscus curves down and forms a convex meniscus, it means that the liquid is more strongly attracted to itself than the material that the container is made from.

Mercury will form a convex meniscus in glass while water will form a convex meniscus.

If the container is large enough in diameter, both liquids will be perfectly level in between their menisci while being at different levels in the tube from side to side.

As for ocean sea levels... I saw these explanations....

The Pacific is less salty and so less dense and so higher and various local current issues make it locally a few centimeters higher than the Atlantic right there. Sea level is not the same around the world.
Talking about the Panama Canal....
It's not really - the difference is only a matter of centimeters and basically it's due to currents and weather conditions. There is no significant difference in sea level on each end.
Or just outright denial...
There is no difference in sea level in the Pacific or the Atlantic Oceans

 

[Jim Parker]

Oh, I see, you are talking about them seeking their own level as height differences due to pressure, not a level flat surface as I assumed.

Seeking their own level, to me, means that every given point on the surface of a liquid will be exactly the same height.

I believe that is correct.
However, the earth is not a "flat" (perfectly spherical) surface. There is at least as much topography under the oceans as on land.

And, it is very curious to me as to why the Pacific and Atlantic oceans would be at different levels.......

Ah! Here! :
https://www.quora.com/Why-is-there-...ean-and-Pacific-ocean-across-the-Panama-Canal