Conflicting rate of curvature?

Hi.  I'm a newly confirmed FEer and can't conceive of a RE by logic alone.  (William Carpenter complained of having to justify a Flat Earth when it so obviously was.)  Logical thought here... How on Earth could gravity hold water in a fixed curvature around a basketball shaped object?  It just doesn't make sense and is downright incomprehensible. 

My question though is about the "rate" of curvature on the Globe version of Earth.  I've read it is 8 inches per mile and then just read a conflicting number from Wilbur Voliva (much higher if I recall).

So please someone, provide a definitive answer, mathematically correct as to the convexity of "planet" Earth.  I desperately need this information to verbally abuse knuckleheads who buy into the insane idea that this thing we live on is shaped like a basketball and water is curved.

And Thanks!

How on Earth could gravity hold water in a fixed curvature around a basketball shaped object?

If I may answer this with a question: which way do you think gravity pulls in RE terms?

My question though is about the "rate" of curvature on the Globe version of Earth.  I've read it is 8 inches per mile and then just read a conflicting number from Wilbur Voliva (much higher if I recall).

So please someone, provide a definitive answer, mathematically correct as to the convexity of "planet" Earth.

Giving the curvature in inches per mile can be misleading. If I do the calculation over 1 mile, I get 8.6 inches (close enough to your quoted value). If, however, I do it over, say, 100 miles, I get 71.6 feet per mile. You can do these calculations yourself easily enough if you understand trigonometry.

You have to take into account that from your standing point, you are curving away from the curve your are viewing ahead, so it's double each time. So 8 inches for one mile is 4 inches drop from your standing point and 4 inches from your forward view, equals 8 inches.
Move another mile and it doubles to 16 inches, then 32 inches for 3 miles, 64 inches for 4 miles, 128 inches for 5 miles, 256 inches for 6 miles, 512 inches for 7 miles, 1,024 inches for 8 miles, 2,048 inches for 9 miles and 4,096 inches for 10 miles.
So after 10 miles, any view from your point will be 341 feet, approximately, or basically, you have dipped below the curve by 170 and a half feet and the object you look at as dipped the same, so to your view, if say, it was a building that was 342 feet tall...you would see 1 foot of it and that's it.

11 miles goes to 8,192 inches, 12 miles will be 16,384 inches, 13 miles will be 32,768 inches, 14 miles will be 65,536 inches, 15 miles will be 131,072 inches, 16 miles will be 262,144 inches, 17 miles will be 524,288 inches, 18 miles will be 1,048,576 inches, 19 miles will be 2,097,152 inches and 20 miles will be 4,194,304, which would equate to 349,525 FEET or 106,535,METRES, or 66 MILES.

So basically from your point, you will drop down your dip, 33 miles and the thing you are viewing will also drop 33 miles. Do we have any building that can be viewed from 20 miles that are just over 33 miles in height so that we can see the top?

If the earth was a ball, this is what should realistically happen.

To clarify it, let's assume that a person is on a large ball and also there is a model building on it as well. Ok, we know the ball is not the earth, but the principle is the same.
Now let's assume that this ball is 24 feet in circumference.

You have to take into account that from your standing point, you are curving away from the curve your are viewing ahead, so it's double each time. So 8 inches for one mile is 4 inches drop from your standing point and 4 inches from your forward view, equals 8 inches.
Move another mile and it doubles to 16 inches, then 32 inches for 3 miles, 64 inches for 4 miles, 128 inches for 5 miles, 256 inches for 6 miles, 512 inches for 7 miles, 1,024 inches for 8 miles, 2,048 inches for 9 miles and 4,096 inches for 10 miles.
So after 10 miles, any view from your point will be 341 feet, approximately, or basically, you have dipped below the curve by 170 and a half feet and the object you look at as dipped the same, so to your view, if say, it was a building that was 342 feet tall...you would see 1 foot of it and that's it.

11 miles goes to 8,192 inches, 12 miles will be 16,384 inches, 13 miles will be 32,768 inches, 14 miles will be 65,536 inches, 15 miles will be 131,072 inches, 16 miles will be 262,144 inches, 17 miles will be 524,288 inches, 18 miles will be 1,048,576 inches, 19 miles will be 2,097,152 inches and 20 miles will be 4,194,304, which would equate to 349,525 FEET or 106,535,METRES, or 66 MILES.

So basically from your point, you will drop down your dip, 33 miles and the thing you are viewing will also drop 33 miles. Do we have any building that can be viewed from 20 miles that are just over 33 miles in height so that we can see the top?

If the earth was a ball, this is what should realistically happen.

To clarify it, let's assume that a person is on a large ball and also there is a model building on it as well. Ok, we know the ball is not the earth, but the principle is the same.
Now let's assume that this ball is 24 feet in circumference.

It's very simple. Calculating the curvature of the earth is simply of a  parabolic nature. The distance for the first mile is about 8 inches, but increases as the distance increases. It's not linear. Once again sceptimatic seems to be ignorant of the most commonly known facts. Use a ruler and a basketball for example and you can readily see how this works.

There is no confllict if you understand the basic priniciples of math, geometry, trigonometry, physics, etc. But in the words of sceptimatic, they were made up by those mad scientists just to fit a globular earth.

IMHO Simple known facts such as these are where Flat Earth fails.

I don't think you need to go into all that stuff. It's pretty self explanatory if you do it simply.
If not, show me why I'm wrong, in the usual simple terms.

Your first question about what direction I think gravity pulls in a RET situation?  We are taught that it pulls towards the center of the "globe", or in the general direction of down.  I'm not sure why you asked that or why it's relevant to my question?

Your answer as to the "rate" or curvature of the "globe/Earth":
 
"Giving the curvature in inches per mile can be misleading. If I do the calculation over 1 mile, I get 8.6 inches (close enough to your quoted value). If, however, I do it over, say, 100 miles, I get 71.6 feet per mile. You can do these calculations yourself easily enough if you understand trigonometry."

OK... a drop of 71.6 feet over 100 miles IS 8.6 inches per mile.

I guess unless we hear from anyone else, the "rate" is 8 - 8.6 inches per mile.

With THAT said it is incontrovertible proof that there is NO curvature to the Earth based on past experiments measuring "rate of fall" (or convexity) of Earth over water.  Such experiments have been discussed countless times and stated in different ways.  NO "rate of fall" can be calculated anywhere over water. 

Conclusion:  The Earth is Flat, there can be no other explanation

You have to take into account that from your standing point, you are curving away from the curve your are viewing ahead, so it's double each time. So 8 inches for one mile is 4 inches drop from your standing point and 4 inches from your forward view, equals 8 inches.
Move another mile and it doubles to 16 inches, then 32 inches for 3 miles, 64 inches for 4 miles, 128 inches for 5 miles, 256 inches for 6 miles, 512 inches for 7 miles, 1,024 inches for 8 miles, 2,048 inches for 9 miles and 4,096 inches for 10 miles.
So after 10 miles, any view from your point will be 341 feet, approximately, or basically, you have dipped below the curve by 170 and a half feet and the object you look at as dipped the same, so to your view, if say, it was a building that was 342 feet tall...you would see 1 foot of it and that's it.

11 miles goes to 8,192 inches, 12 miles will be 16,384 inches, 13 miles will be 32,768 inches, 14 miles will be 65,536 inches, 15 miles will be 131,072 inches, 16 miles will be 262,144 inches, 17 miles will be 524,288 inches, 18 miles will be 1,048,576 inches, 19 miles will be 2,097,152 inches and 20 miles will be 4,194,304, which would equate to 349,525 FEET or 106,535,METRES, or 66 MILES.

So basically from your point, you will drop down your dip, 33 miles and the thing you are viewing will also drop 33 miles. Do we have any building that can be viewed from 20 miles that are just over 33 miles in height so that we can see the top?

If the earth was a ball, this is what should realistically happen.

To clarify it, let's assume that a person is on a large ball and also there is a model building on it as well. Ok, we know the ball is not the earth, but the principle is the same.
Now let's assume that this ball is 24 feet in circumference.

I don't understand why you need to move back at all.  Why, please?  I'm just asking a simple, straight forward question that should not involve "moving backwards" from your point of focus.

Can we keep this question simple?  How far does Earth drop or "fall" (over water... for simplicity's sake) in one mile? 

It appears that the answer is 8 - 8.6 inches/mile and no one has contested that as far as I can determine.

Assuming that 8 - 8.6inches/mile IS the correct answer.  No matter which direction over water you measure the fall will be 8 - 8.6 inches/mile. 

This obviously reasoning states that "water is not always "flat" and does not seek it's own level"... or else Earth is Flat.

I was replying to Scintific Method above.

Quote from: sceptimatic on December 06, 2013, 07:19:07 AM

You have to take into account that from your standing point, you are curving away from the curve your are viewing ahead, so it's double each time. So 8 inches for one mile is 4 inches drop from your standing point and 4 inches from your forward view, equals 8 inches.
Move another mile and it doubles to 16 inches, then 32 inches for 3 miles, 64 inches for 4 miles, 128 inches for 5 miles, 256 inches for 6 miles, 512 inches for 7 miles, 1,024 inches for 8 miles, 2,048 inches for 9 miles and 4,096 inches for 10 miles.
So after 10 miles, any view from your point will be 341 feet, approximately, or basically, you have dipped below the curve by 170 and a half feet and the object you look at as dipped the same, so to your view, if say, it was a building that was 342 feet tall...you would see 1 foot of it and that's it.

11 miles goes to 8,192 inches, 12 miles will be 16,384 inches, 13 miles will be 32,768 inches, 14 miles will be 65,536 inches, 15 miles will be 131,072 inches, 16 miles will be 262,144 inches, 17 miles will be 524,288 inches, 18 miles will be 1,048,576 inches, 19 miles will be 2,097,152 inches and 20 miles will be 4,194,304, which would equate to 349,525 FEET or 106,535,METRES, or 66 MILES.

So basically from your point, you will drop down your dip, 33 miles and the thing you are viewing will also drop 33 miles. Do we have any building that can be viewed from 20 miles that are just over 33 miles in height so that we can see the top?

If the earth was a ball, this is what should realistically happen.

To clarify it, let's assume that a person is on a large ball and also there is a model building on it as well. Ok, we know the ball is not the earth, but the principle is the same.
Now let's assume that this ball is 24 feet in circumference.

It's very simple. Calculating the curvature of the earth is simply of a  parabolic nature. The distance for the first mile is about 8 inches, but increases as the distance increases. It's not linear. Once again sceptimatic seems to be ignorant of the most commonly known facts. Use a ruler and a basketball for example and you can readily see how this works.

There is no confllict if you understand the basic priniciples of math, geometry, trigonometry, physics, etc. But in the words of sceptimatic, they were made up by those mad scientists just to fit a globular earth.

IMHO Simple known facts such as these are where Flat Earth fails.

With all due respect... "Simple known facts such as these" are the reasons why Round Earth becomes and impossibility.  (Speaking of simply calculating the rate of drop or fall on a "round" Earth which to this point in the post is agreed to be 8 - 8.6 inches per mile)

I was replying to Scintific Method above.

Good, because you had me all confused for a second.

Quote from: Googleotomy on December 06, 2013, 11:41:38 AM

Quote from: sceptimatic on December 06, 2013, 07:19:07 AM

You have to take into account that from your standing point, you are curving away from the curve your are viewing ahead, so it's double each time. So 8 inches for one mile is 4 inches drop from your standing point and 4 inches from your forward view, equals 8 inches.
Move another mile and it doubles to 16 inches, then 32 inches for 3 miles, 64 inches for 4 miles, 128 inches for 5 miles, 256 inches for 6 miles, 512 inches for 7 miles, 1,024 inches for 8 miles, 2,048 inches for 9 miles and 4,096 inches for 10 miles.
So after 10 miles, any view from your point will be 341 feet, approximately, or basically, you have dipped below the curve by 170 and a half feet and the object you look at as dipped the same, so to your view, if say, it was a building that was 342 feet tall...you would see 1 foot of it and that's it.

11 miles goes to 8,192 inches, 12 miles will be 16,384 inches, 13 miles will be 32,768 inches, 14 miles will be 65,536 inches, 15 miles will be 131,072 inches, 16 miles will be 262,144 inches, 17 miles will be 524,288 inches, 18 miles will be 1,048,576 inches, 19 miles will be 2,097,152 inches and 20 miles will be 4,194,304, which would equate to 349,525 FEET or 106,535,METRES, or 66 MILES.

So basically from your point, you will drop down your dip, 33 miles and the thing you are viewing will also drop 33 miles. Do we have any building that can be viewed from 20 miles that are just over 33 miles in height so that we can see the top?

If the earth was a ball, this is what should realistically happen.

To clarify it, let's assume that a person is on a large ball and also there is a model building on it as well. Ok, we know the ball is not the earth, but the principle is the same.
Now let's assume that this ball is 24 feet in circumference.

It's very simple. Calculating the curvature of the earth is simply of a  parabolic nature. The distance for the first mile is about 8 inches, but increases as the distance increases. It's not linear. Once again sceptimatic seems to be ignorant of the most commonly known facts. Use a ruler and a basketball for example and you can readily see how this works.

There is no confllict if you understand the basic priniciples of math, geometry, trigonometry, physics, etc. But in the words of sceptimatic, they were made up by those mad scientists just to fit a globular earth.

IMHO Simple known facts such as these are where Flat Earth fails.

With all due respect... "Simple known facts such as these" are the reasons why Round Earth becomes and impossibility.  (Speaking of simply calculating the rate of drop or fall on a "round" Earth which to this point in the post is agreed to be 8 - 8.6 inches per mile)

Yes, 8 inches per mile but not per mile per mile as in 10 miles being 80 inches.

From your standing point if the curve is 8 inches per mile, then it has to be 4 inches forward and 4 inches back.
It's supposed to be a ball, so either way it has to curve both ways for any person stood on it. I mentioned a person moving back a mile because buildings do not move, but as the person moves back, he is walking down the curve as the building is disappearing down the opposite curve.

I think we are all agreed then?

The curvature/convexity/"fall" or "drop" of the Globe/Round Earth concept is thus:

For every ONE Mile of travel in any direction (over water, for simplicity) there is a "drop" or "fall" of 8 - 8.6 Inches Per Mile of Travel.  As an example, Travel one mile and the drop/fall will be 8 - 8.6 inches.  Travel One Hundred miles and the drop/fall will be 800 - 860 Inches, or 66.6 Feet to 71.6 Feet.

Anyone else want to challenge the curvature calculations? 

My hope is that we come to a definitive, mathematically proven and undeniable TRUTH as to what the drop or fall of a Round/Globular Earth is.

I think it will then be a simple matter for everyone to accept that Earth cannot possibly be shaped like a basketball or "globe".

I think we are all agreed then?

The curvature/convexity/"fall" or "drop" of the Globe/Round Earth concept is thus:

For every ONE Mile of travel in any direction (over water, for simplicity) there is a "drop" or "fall" of 8 - 8.6 Inches Per Mile of Travel.  As an example, Travel one mile and the drop/fall will be 8 - 8.6 inches.  Travel One Hundred miles and the drop/fall will be 800 - 860 Inches, or 66.6 Feet to 71.6 Feet.

Anyone else want to challenge the curvature calculations? 

My hope is that we come to a definitive, mathematically proven and undeniable TRUTH as to what the drop or fall of a Round/Globular Earth is.

I think it will then be a simple matter for everyone to accept that Earth cannot possibly be shaped like a basketball or "globe".

I don't agree with the calculations. It fails to take into account what I put in my diagram.

I agree that the earth cannot and is not a spinning ball but the calculations would not be 8 inches per mile and 8 more inches after that for each mile from a persons view point. It has to double for each mile.

Your first question about what direction I think gravity pulls in a RET situation?  We are taught that it pulls towards the center of the "globe", or in the general direction of down.  I'm not sure why you asked that or why it's relevant to my question?

It's relevant to the question it was in reply to, as it explains how water can be held in a "fixed curvature", as you put it.

Your answer as to the "rate" or curvature of the "globe/Earth":
 
"Giving the curvature in inches per mile can be misleading. If I do the calculation over 1 mile, I get 8.6 inches (close enough to your quoted value). If, however, I do it over, say, 100 miles, I get 71.6 feet per mile. You can do these calculations yourself easily enough if you understand trigonometry."

OK... a drop of 71.6 feet over 100 miles IS 8.6 inches per mile.

I guess unless we hear from anyone else, the "rate" is 8 - 8.6 inches per mile.

With THAT said it is incontrovertible proof that there is NO curvature to the Earth based on past experiments measuring "rate of fall" (or convexity) of Earth over water.  Such experiments have been discussed countless times and stated in different ways.  NO "rate of fall" can be calculated anywhere over water. 

Conclusion:  The Earth is Flat, there can be no other explanation

Pay attention. I didn't say "71.6 over 100 miles" I said "71.6 feet per mile". Over 100 miles it would be ~7,160 feet. This can be confirmed by going for a fly (which I have done countless times in my life), and noting that, when at 7,000 feet, you can see about 100 miles. Standing on the ground, you can see for about 3 miles. From about 100 feet above the ground, you can see roughly 12 miles. I've personally verified all of this, and I consider it to be pretty sound proof of curvature. I encourage you to go and verify it for yourself as well, and/or come up with a credible alternative explanation for it.

EDIT: No, I didn't stuff up my calculation, the fact that 71.6 feet divided by 100 is about 8.6 inches is just a coincidence. It threw me off when I first did the maths, so I triple checked it, and it is most definitely right. As an added note, the drop over 6,000 miles is 3,981 miles, or about 3500 feet per mile. If you can tell me why that is, I'll refrain from calling you names.

Can we keep this question simple?  How far does Earth drop or "fall" (over water... for simplicity's sake) in one mile? 

In one mile, 8.6 inches.

It appears that the answer is 8 - 8.6 inches/mile and no one has contested that as far as I can determine.

Assuming that 8 - 8.6inches/mile IS the correct answer.  No matter which direction over water you measure the fall will be 8 - 8.6 inches/mile. 

Please read and understand my explanation in my last post.

This obviously reasoning states that "water is not always "flat" and does not seek it's own level"... or else Earth is Flat.

It is not "flat", but it does seek it's own level. To put this as clearly as I can, water will settle with it's surface being at a fairly constant radius from the centre of the earth, forming a curve whichever way you look.

Quote from: sixstringthing on December 06, 2013, 03:44:52 PM

Your first question about what direction I think gravity pulls in a RET situation?  We are taught that it pulls towards the center of the "globe", or in the general direction of down.  I'm not sure why you asked that or why it's relevant to my question?

It's relevant to the question it was in reply to, as it explains how water can be held in a "fixed curvature", as you put it.

Quote from: sixstringthing on December 06, 2013, 03:44:52 PM

Your answer as to the "rate" or curvature of the "globe/Earth":


 
"Giving the curvature in inches per mile can be misleading. If I do the calculation over 1 mile, I get 8.6 inches (close enough to your quoted value). If, however, I do it over, say, 100 miles, I get 71.6 feet per mile. You can do these calculations yourself easily enough if you understand trigonometry."

OK... a drop of 71.6 feet over 100 miles IS 8.6 inches per mile.

I guess unless we hear from anyone else, the "rate" is 8 - 8.6 inches per mile.

With THAT said it is incontrovertible proof that there is NO curvature to the Earth based on past experiments measuring "rate of fall" (or convexity) of Earth over water.  Such experiments have been discussed countless times and stated in different ways.  NO "rate of fall" can be calculated anywhere over water. 

Conclusion:  The Earth is Flat, there can be no other explanation

Pay attention. I didn't say "71.6 over 100 miles" I said "71.6 feet per mile". Over 100 miles it would be ~7,160 feet. This can be confirmed by going for a fly (which I have done countless times in my life), and noting that, when at 7,000 feet, you can see about 100 miles. Standing on the ground, you can see for about 3 miles. From about 100 feet above the ground, you can see roughly 12 miles. I've personally verified all of this, and I consider it to be pretty sound proof of curvature. I encourage you to go and verify it for yourself as well, and/or come up with a credible alternative explanation for it.

EDIT: No, I didn't stuff up my calculation, the fact that 71.6 feet divided by 100 is about 8.6 inches is just a coincidence. It threw me off when I first did the maths, so I triple checked it, and it is most definitely right. As an added note, the drop over 6,000 miles is 3,981 miles, or about 3500 feet per mile. If you can tell me why that is, I'll refrain from calling you names.

"Going for a fly" is only a subjective opinion based on visual assumptions.  There can only be ONE mathematically correct number that MUST contiguous for EACH mile of travel on a uniformly shaped object.  The original formula can NOT extrapolate out to a different ratio Per Mile on a uniformly round object, right? 

Each mile travelled will have the SAME amount of fall (or drop) on a uniformly shaped "ball" or globe. 

That amount of drop or fall per mile REMAINS 8 - 8.6 inches per mile for EVERY mile travelled on a "round" Earth, and in fact it doesn't matter which direction you travel in! 

The ratio WILL remain constant.  It's simple mathematics.

sceptimatic:

You said:  "I don't agree with the calculations. It fails to take into account what I put in my diagram.

I agree that the earth cannot and is not a spinning ball but the calculations would not be 8 inches per mile and 8 more inches after that for each mile from a persons view point. It has to double for each mile."

No, I don't think so.  How can the drop/fall possibly double with each mile travelled?  No offence, but you need to re-think that one.

Scintific Method:

You said:  "It is not "flat", but it does seek it's own level. To put this as clearly as I can, water will settle with it's surface being at a fairly constant radius from the centre of the earth, forming a curve whichever way you look."

Wow, I'd really like to see water that does not lie flat, and water  which is "forming a curve whichever way you look."

Water can not possibly be anything other than flat and seeking it's own level at ALL times.

sixstringthing, your level of ignorance is staggering, I have only ever seen a couple of other that rival it, and them only on these fora. Perhaps a nice diagram will alleviate it some. Are you at least familiar with gradients?



If this were a cross-section of the earth:
- the blue line would represent the drop-per-mile calculated over 1,500 miles (about 1,060 feet per mile);
- the green line would represent the drop-per-mile calculated over 3,000 miles (about 2,040 feet per mile);
- the red line would represent the drop-per-mile calculated over 4,500 miles (about 2,870 feet per mile);
- the yellow line would represent the drop-per-mile calculated over 6,000 miles (as mentioned, about 3,500 feet per mile).

NOW can you see how it is impossible to describe a curve with a single gradient?

Scintific Method:

You said:  "It is not "flat", but it does seek it's own level. To put this as clearly as I can, water will settle with it's surface being at a fairly constant radius from the centre of the earth, forming a curve whichever way you look."

Wow, I'd really like to see water that does not lie flat, and water  which is "forming a curve whichever way you look."

Water can not possibly be anything other than flat and seeking it's own level at ALL times.

It's not flat, it's curved, following the curve of the earth, and settling to the lowest points it can (aka the points closest to the centre of the earth, thanks to gravity); finding it's own level so that the surface of any single water mass forms a curve (albeit a very slight curve, not easily perceived with the unaided eye) with a pretty constant radius, the origin point for which is the centre of the earth.

Did you follow that? or do I need to dumb it down even more?

sceptimatic:

You said:  "I don't agree with the calculations. It fails to take into account what I put in my diagram.

I agree that the earth cannot and is not a spinning ball but the calculations would not be 8 inches per mile and 8 more inches after that for each mile from a persons view point. It has to double for each mile."

No, I don't think so.  How can the drop/fall possibly double with each mile travelled?  No offence, but you need to re-think that one.

If the earth was a globe, which it isn't and you were to drive round it, then yes, every mile you covered you would assume the 8 inch curve. I understand that.

What I'm talking about is by viewing a building and moving back a mile at a time. Each mile moved, creates more curvature BOTH ways as the globe would be uniform, or basically, to save argument of the so called bulge.

Ok, I've re-evaluated. It would only double if two objects were moving away from each other by a mile EACH, not one object being stationary.

Water can not possibly be anything other than flat and seeking it's own level at ALL times.

put a small drop of water on a flat surface.

sixstringthing, your level of ignorance is staggering, I have only ever seen a couple of other that rival it, and them only on these fora. Perhaps a nice diagram will alleviate it some. Are you at least familiar with gradients?



If this were a cross-section of the earth:
- the blue line would represent the drop-per-mile calculated over 1,500 miles (about 1,060 feet per mile);
- the green line would represent the drop-per-mile calculated over 3,000 miles (about 2,040 feet per mile);
- the red line would represent the drop-per-mile calculated over 4,500 miles (about 2,870 feet per mile);
- the yellow line would represent the drop-per-mile calculated over 6,000 miles (as mentioned, about 3,500 feet per mile).

NOW can you see how it is impossible to describe a curve with a single gradient?

Scintific Method:

You have devolved into attacking me instead of the argument.  Calling me ignorant (or any deviation from the discussion at hand) only proves you are lost and confused and can no longer provide any logical arguments.

Your sketch neither proves nor illustrates anything practical.  You are playing the "fuzzy math" game.  That is a fail.

This discussion of convexity/curvature/fall or drop on a globe is really quite simple.  Let me help you out here...

Imagine selecting a point (Point A) on a large ball (say 10 feet in diameter) and turn the ball so that point A is on the top of the ball.  Now imagine selecting another point (Point B) on that ball about 3 feet away - along the curvature of the ball, following the curvature of the ball.  Now draw a horizontal line straight out, horizontally from Point A, past Point B. 

Now simply measure from the point where your Horizontal line from Point A intersects with the Perpendicular line of Point B.

Measure the distance from the intersection of those lines (this is the Height of Point A) down to the dot of Point B.  This is the amount of fall or drop from Point A to Point B.  Hopefully you can now understand that there are NOT "multiple gradients" when calculating fall on a globe.  It's very, very, very simple.

Go get yourself a ball of some sort and two rulers and you can hopefully understand "rate of fall" on a globe from one point to the next.

Best of luck with your studies.

Quote from: sixstringthing on December 07, 2013, 01:28:28 AM

sceptimatic:

You said:  "I don't agree with the calculations. It fails to take into account what I put in my diagram.

I agree that the earth cannot and is not a spinning ball but the calculations would not be 8 inches per mile and 8 more inches after that for each mile from a persons view point. It has to double for each mile."

No, I don't think so.  How can the drop/fall possibly double with each mile travelled?  No offence, but you need to re-think that one.

If the earth was a globe, which it isn't and you were to drive round it, then yes, every mile you covered you would assume the 8 inch curve. I understand that.

What I'm talking about is by viewing a building and moving back a mile at a time. Each mile moved, creates more curvature BOTH ways as the globe would be uniform, or basically, to save argument of the so called bulge.

Ok, I've re-evaluated. It would only double if two objects were moving away from each other by a mile EACH, not one object being stationary.

sceptimatic: 

Right on man!  You solved your own math problem, and you are using critical thinking!  That is indicative of intelligence.  It is also a great indicator that you say:  "Oops, I was looking at that from a slightly different view." 

I hope you take the time to look at my response to Scintific Method.  I believe the "equation" is spelled on quite clearly there.  Again, now matter where on a ball (representing Earth) you place a dot (Point A), it is actually the "top" or highest point of the ball.  Any travel in any direction along the curvature of that ball is DOWN, and the intersection of the Horizontal "A" and the Perpendicular of "B" marks a measurable point.

I hope you can "picture" that.  I'd love to have you argue WITH me rather than against me. 

Quote from: sixstringthing on December 07, 2013, 01:32:48 AM

Water can not possibly be anything other than flat and seeking it's own level at ALL times.

put a small drop of water on a flat surface.

Spank86:

Ha!  That is an excellent argument except for one thing, we are not discussing a single drop of water.  We are discussing oceans, lakes and rivers. 

I could say in response to you:  OK, pour one cup (a very small amount obviously) of water on a ball.  Does it seek it's own level?  Yes.  Pour one cup of water on the slightest grade on earth.  RET says that "gravity" will hold that water in place and it will not run down the slight gradient to seek it's own level.

WHY doesn't "gravity" hold that one cup of water on a slight gradient when it allegedly holds oceans of water in place with it's "magical" strength of attraction? 

Quote from: sixstringthing on December 07, 2013, 01:32:48 AM

Scintific Method:

You said:  "It is not "flat", but it does seek it's own level. To put this as clearly as I can, water will settle with it's surface being at a fairly constant radius from the centre of the earth, forming a curve whichever way you look."

Wow, I'd really like to see water that does not lie flat, and water  which is "forming a curve whichever way you look."

Water can not possibly be anything other than flat and seeking it's own level at ALL times.

It's not flat, it's curved, following the curve of the earth, and settling to the lowest points it can (aka the points closest to the centre of the earth, thanks to gravity); finding it's own level so that the surface of any single water mass forms a curve (albeit a very slight curve, not easily perceived with the unaided eye) with a pretty constant radius, the origin point for which is the centre of the earth.

Did you follow that? or do I need to dumb it down even more?

Scintific Method:

Yes, please dumb it down further for me.

How about showing me a pan of your curved water, OK?  Just put it on YouTube and the whole world can watch and be awed.   

Imagine selecting a point (Point A) on a large ball (say 10 feet in diameter) and turn the ball so that point A is on the top of the ball.  Now imagine selecting another point (Point B) on that ball about 3 feet away - along the curvature of the ball, following the curvature of the ball.  Now draw a horizontal line straight out, horizontally from Point A, past Point B. 

Now simply measure from the point where your Horizontal line from Point A intersects with the Perpendicular line of Point B.

Measure the distance from the intersection of those lines (this is the Height of Point A) down to the dot of Point B.  This is the amount of fall or drop from Point A to Point B...

Now add a third point, point C, repeat the measurement from A to C, and you will get a different gradient. That's what I have been trying to explain.

Here's another diagram, following your suggestion and expanding on it:



"O" is our origin point (your "Point A"), the distance OA is 1/16 the circumference of our globe, let's call this distance "x". This lets us put the other distances in terms of x, so OB is 2x, OC is 3x, and OD is 4x. "a", "b", "c", and "d" are the respective drops from the line drawn tangent to point "O", with a = 0.194x, b = 0.746x, c = 1.572x, and d = 2.546x. With me so far? I hope so! This stuff is not particularly complicated, and can be demonstrated easily using the method you suggested.

So, to finish off:
- gradient of OA is 0.194 (0.194 / 1);
- gradient of OB is 0.373 (0.746 / 2);
- gradient of OC is 0.524 (1.572 / 3);
- and the gradient of OD is 0.637 (do I need to show it? I guess so: 2.546 / 4).

NOW do you follow what I've been saying??

PS. If you were thinking "but those lines are not like what I was saying!", think about this: if i made a, b, c, and d perpendicular to the surface of the globe, they would be even longer (d would be infinite), and the gradients even more drastically different.

---

Oh, and for the sake of tidying this up a bit:

Quote from: Scintific Method on December 07, 2013, 02:45:08 AM

Quote from: sixstringthing on December 07, 2013, 01:32:48 AM

Scintific Method:

You said:  "It is not "flat", but it does seek it's own level. To put this as clearly as I can, water will settle with it's surface being at a fairly constant radius from the centre of the earth, forming a curve whichever way you look."

Wow, I'd really like to see water that does not lie flat, and water  which is "forming a curve whichever way you look."

Water can not possibly be anything other than flat and seeking it's own level at ALL times.

It's not flat, it's curved, following the curve of the earth, and settling to the lowest points it can (aka the points closest to the centre of the earth, thanks to gravity); finding it's own level so that the surface of any single water mass forms a curve (albeit a very slight curve, not easily perceived with the unaided eye) with a pretty constant radius, the origin point for which is the centre of the earth.

Did you follow that? or do I need to dumb it down even more?

Scintific Method:

Yes, please dumb it down further for me.

How about showing me a pan of your curved water, OK?  Just put it on YouTube and the whole world can watch and be awed.   

I'll go you one better. This was on YouTube:

#" class="bbc_link" target="_blank" rel="noopener noreferrer">The earth is round DERP

Explain how that could happen on a flat earth.

Not this dodgy video again!

#ws" class="bbc_link" target="_blank" rel="noopener noreferrer">LightHouse Rosslare. Panasonic HC-V520 Zoom Test.

Explain how this could happen on a round earth.

scintific method:

Your video of the ship sailing away over the curvature of the Earth was disproved over 100 years ago.  This is an optical illusion that is EASILY disproved with use of a telescope. 

The ship the sailed "over the hill" was not falling off a slope, it was disappearing from naked eyesight and only APPEARED to sail off down-hill. 

If the ship were at it's maximum distance and had seemed to disappear down the hill, it would return immediately in FULL VIEW if you looked at it through a telescope. 

Did it pick up speed when it started going down hill?  Did the Captain need to apply the brakes?

I think now would be a good time to read some introductory facts, FAQ's and critical documents to FET.  For example if you had read "100 Proofs Earth Is Not A Globe", you would NEVER have tried to prove the Earth was a globe using a video whose premise was proven false before 1900 a.d.

tappet's video is PROOF the Earth can NOT be a globe.  According to the long distance views through those telescopic lenses, there is NO convex angles or fall to the water. 

tappet's video disproves your video.