News & Current Events
will_scholten — 2016-12-02T20:03:01-05:00 — #1
Flat Earth Support
Has anyone been researching the "Flat Earth" vs "Planet Earth" lately?
I know the USA is inot "Planet" how about other countries?
What do you think?
bill_coley — 2016-12-03T00:21:53-05:00 — #2
Some comments on your video, Will:
1) You refer to the earth's 8" drop per mile. If that's true for the entire 6,225 miles of one quadrant of the earth's circumference, you seem to argue, then the total drop would be far, far less than the nearly 4,000 miles that is the radius of the earth.
The challenge to your explanation is in your math. Remember your high school math days and the Pythagorean Theorem. When you do, you'll recall that the length of the sides of a right triangle are related to the square of the sides, not the actual value of the sides. So, the earth drops 8" in the first mile, but not in subsequent miles. Look at the bottom of the page at THIS WEB SITE where a flat earth loyalist acknowledges the mathematical problem with his 8" per mile postulate. For example, you don't have to travel nine miles to drop 72". You need only travel three miles. The effect of squared sides increases exponentially as the distance from the observer increases.
This is not a math forum, so I will cut to the core: Do the math using the Pythagorean Theorem, and the expected drop is the result.
2) I'm sure you've witnessed a lunar eclipse, when for a short period of time the earth blocks part of the light of the sun hiting the moon, producing an image of the earth's shadow on the moon's surface. What's the shape of the earth's shadow on the moon's surface? Round, not square.
3) Do a Google search on the discoveries of the ancient Greek mathematician named Eratosthenes, who figured out that the fact that shadows cast by the sun at the exact same time in two cities in Egypt were not the same length confirmed the spherical nature of the earth (were the earth flat, those shadows would have been the same length). He used the angles created by those shadows to make an amazingly accurate estimate of the earth's circumference.
4) Think about the fact that the sun's not up everywhere at once. Where I am, it's after 11PM; the sun set here six or seven hours ago. But the sun is up in India, for example.
5) And speaking of the sun, if the earth is flat, why isn't the sun seen at the same location in the sky at all locations on the earth?
6) What about the moon? We've been there and flown around it, so we know it's spherical. We also know from astronomical observations and unmanned spacecraft visits that the planets are spherical. Why only the earth flat, then?
7) Finally, consider the testimony of iconic images such as this....
will_scholten — 2016-12-03T08:06:56-05:00 — #3
The actual drop is the same, for example if you measure a mile, 8" drop, walk anlther mile measure it.....
I think this link will describe what you are talking about, how much your vision is obscured over the horizon with multiple miles. If you take a compass and draw a circle, the radius will be the same everywhere, so any actual unit measure will be the same.
As in my basket ball example, the ball is 30" circumfference, 4.77 radius, a quater of the circumference is 7.5 inches, I can measure 7.5" down anywhere on the ball and it will bring you to the core of the ball's radius 7.5" drops 4.77", if it works on a basket ball, then if I am at the north pole, drive a straight line down 6250 miles, I will be at the Equator which is 3959 miles straight down to the core (middle ) of the earth.
All said and done, do we believe man, or the Creator?
18 can you join him in spreading out the skies,
hard as a mirror of cast bronze?
The New International Version. (2011). (Job 37:18). Grand Rapids, MI: Zondervan.
Do you think man went through the vault/firmament? I don't think so!
My guess would be the Van Allen Belt is the Firmament YHWH is talking about.
bill_coley — 2016-12-03T11:15:53-05:00 — #4
Will, the link you provide makes the point I tried to make about the exponential math related to the Pythagorean Theorem involved here.
When the "target distance" from a six foot tall observer is ten miles, the target's "hidden height" - the drop - is 32.68 feet (I'm rounding the decimals of these results). When the target distance increases ten-fold, to 100 miles, the "hidden height" doesn't increase ten-fold, to 326.8 feet; it increases nearly 200-fold, to 6,273.68 feet! And when the target distance increases to 622.5 times, to 6,225 miles - the distance of a quadrant of the earth's surface - the "hidden height" increases approximately 550,000 times, to 18,035,549.58 feet, or 3,418 miles.
I don't have enough time or interest to discover the roots of the 540 mile discrepancy between the link's calculation and the radius of the earth, but clearly, the drop is much, much, much greater than eight inches per mile for each of 6,225 miles (or about three-quarter's of a mile).
The bottom line here is that the math doesn't lie.... As in your example of the basketball.
Remember from school years' math days that the radius of a circle or sphere is one-half its diameter, and its diameter is the straight-line distance between any two points on the circle or sphere when the line between them runs through the circle's or sphere's center. So by definition, the straight-line distance from surface to center of a sphere whose radius is 4.77" (the basketball) is 4.77." In your basketball example that means if you travel 7.5" in a straight line from any point on the ball toward its center, you will travel PAST the sphere's center by 2.73." Similarly, since the radius of the earth is 3,959 miles, if you drive a 6,250 mile straight line from the surface toward the center, you will go PAST the center by 2,291 miles.
Perhaps some of the confusion here is grounded in the fact radius and diameter are straight-line distances, while the circumference travels the curvature of the surface.
I believe the results of the work produced by the eyes, hands, and minds the Creator has given to humanity, work that understands circles and spheres very well, and that produced the Pythagorean Theorem to which I referred in my previous post.
I offered you several points about your video, to most of which you did not respond. To reduce the burden of my request, I'll repeat and ask for your response to only one of those points: During a lunar eclipse, the earth casts a rounded shadow on the surface of the moon, not a straight-lined one. If, as you contend, the earth has "four corners," how does it cast a rounded shadow?
wolfgang_schneider — 2016-12-03T11:45:08-05:00 — #5
Instead of "flat earth", perhaps a totally different - and certainly highly unusual universe model could be considered as a possibility
The question to answer would be: Which of the various "universe models" that have been around in various ages of man's history actually is correct? It seems that there have been revisions of various models ... some accompanied by scientists and researchers being severly persecuted by the institutional church ....
will_scholten — 2016-12-03T14:17:33-05:00 — #6
Actually, the reason I divided the circumference by 4, is to tell us the curve, on the BB, radius of 4.77" and the 7.5" circumference , the difference is produced from the arch . If you make a triangle, the distance is less than 7.5".
I didn't say I have all the answers, but I did not show that the scriptures prove the world was square, but 4 corners could mean due N,S,E,W. My intention is not to explain everything.
These could be answered by ; how large and far away is the sun from earth? A closer , smaller sun could explain a lot of that away, that would also explain how on a hot sunny day, when a cloud blows over, you instantly feel relieved, how far do they say the sun is?
Have we made it out of the "lower earth's atmostphere" I am seeing the real answer is no.
Speaking of ironic images, why is the United Nations flag a map of the flat earth?
May Phillip can answer some of your questions, he has researched it more than I have!
bill_coley — 2016-12-03T16:38:39-05:00 — #7
In all candor, Will, I don't know what your point is here. The issue isn't how far one travels across the surface of the sphere to get from its 12 o'clock position to its 3 o'clock position. The issue is the vertical drop from one position to the other. Because of that, the Pythagorean Theorem IS valid, and you CAN treat the problem as a triangle. The result of the application of that theorem to calculate that drop, as I have indicated, IS the expected result for a sphere of the earth's size.
In my view, there's nothing to explain here, Will. BY DEFINITION, a square surface cannot cast a round shadow. If the earth is square or rectangular in shape, then BY DEFINITION it cannot cast a round shadow. If the earth casts a round shadow - and it does - then BY DEFINITION it must have a round shape.
"The four corners of the earth" is a euphemism for the entire planet. It is not, to my reading of the text, a literal description of the earth's shape. Similarly, the phrase "to the ends of the earth" is usually a euphemism for one's willingness to go to great lengths in order to accomplish an objective. It basically is never a reference to the planet's geometry. (Notice my phrase "to great lengths," which in context is clearly not a reference to measurements, but rather to effort.)
Again, I don't know what your point is, Will.
We know the average distance to the sun is about 93 million miles (the earth's orbit around the sun is almost, but not exactly, circular, which produces variations in the earth-to-sun distance - therefore, it's most accurate to talk about an average distance to the sun). There is nothing in that distance which explains any of the outcomes I asked you to explain.
Of course we have. We've been to the moon multiple times. We've sent unmanned space vehicles into the solar system, with the legendary Voyager I spacecraft now, amazingly, having moved beyond that system.
As for the Van Allen Radiation Belts, NASA decided that the exit velocities of its spacecraft would limit astronauts' radiation exposure to acceptable levels. Monitoring devices worn by astronauts on the lunar missions proved NASA correct.
Because a two dimensional map, by definition, cannot accurately depict a three dimensional object. Such two dimensional projections of the earth's surface are commonplace and mean simply to give a visual representation of every part of the earth in one view. To view them in their accurate physical context would require a three dimensional globe, which is obviously not possible on a two dimensional flag.
Think about the last picture you saw of a family member. Does the fact that your loved one appeared flat on the picture mean that he or she is actually flat? No, because a two dimensional image (the picture) by definition can't fully depict a three dimensional object (your loved one). The same is true with two dimensional map projections of the earth's surface.
will_scholten — 2016-12-04T15:37:01-05:00 — #8
This is very interesting, we have President Obama and people from NASA, telling us we NEVER made it out of lower space orbit.
bill_coley — 2016-12-04T17:14:04-05:00 — #9
This video is badly, badly uninformed and/or misleading, Will.
OF COURSE we can get beyond low earth orbit (LEO) for week-long, single task missions to the moon and back. The issue with LEO is not whether we have rocket ships with enough power to get them beyond the 99-1,200 mile altitude zone called LEO; it's whether we have the capacity TO GO BEYOND THAT ZONE AND THEN STAY BEYOND IT FOR EXTENDED PERIODS OF TIME.
That's why the NASA people in the video talk about exploration of the solar system. Travel to Mars and back, for example, would not be a week-long event, such as were the lunar missions; it would be a year, and that just for travel. When the president and people at NASA talk about going beyond LEO, they're talking about space flight VERY different from that ongoing in the International Space Station, for example, which orbits in LEO. They're talking about human exploration of locations FAR, FAR, FAR more distant than the moon, missions that would require technology much advanced beyond the technology necessary for space shuttle missions, stays in the International Space Station, or even week-long trips to the moon.
As a amateur science enthusiast, I am amazed and disappointed at the amount of misinformation presented in this video. Either its creators don't know what they're talking about... at all... or they know what they're talking about, and in reality are trying to hoodwink their audience. Either way, it's not good.
will_scholten — 2016-12-04T20:44:52-05:00 — #10
My point of the 12:00 and 3:00 is very important! If you are on the N. pole and take the circumference of the earth down to the equator, you will travel 6250 miles. If you are at the N. pole and could travel insid the earth to its core, you would travel 3959 miles. Now that we know the distance both ways, take 3959 and divide it into 6250 and tell me what the curve of the earth is?
This is why I used the basket ball for the example, something I could see real easy.
bill_coley — 2016-12-04T22:31:33-05:00 — #11
Okay, let's do that math. (I hope I make this math class stuff clear! )
During a 6,250 mile journey along the surface of the earth (from 12-3 o'clock), we lose a total of 3,959 miles (the straight-down distance). In fraction form, that's...
3,959 miles dropped
6,250 traveled miles along the surface
Or an average of approximately 0.63 miles dropped per every one mile traveled along the surface.
That result raises an interesting question. In the video to which you linked in this thread's OP you said we lose 8" per mile. Then in your reply to my first response to you, you claimed the loss is 8" in the first mile, and 8" every time the traveler moves an additional mile....
If the drop is actually 8" for every mile traveled along the surface of the earth, then the average drop per mile for the entire 6,250 mile journey along the surface would have to be 8" per mile. But the result of your suggested fraction - 3,959 miles dropped in 6,250 miles along the surface - produces an average drop of 0.63 miles per mile along the surface. That's more than 40,000 inches dropped per mile, not eight!
How do you explain an average drop of 40,000+ inches per mile traveled along the surface, when your claim is that the average drop is eight inches for every mile? I propose the explanation is the Pythagorean Theorem to which I referred early in this thread....
Exponentially larger drops in all miles following the first mile, particularly over a 6,250 mile journey, DO explain an AVERAGE drop of more than 40,000 inches per mile when the drop over the first mile was just eight inches.
If you dispute my math, I hope you will show me the math by which you come to a different result using the values of 3,959 miles dropped over a 6,250 mile journey across the surface. In particular, I would much appreciate seeing how, given those two values, you calculate a drop of 8" for each mile traveled along the surface.
will_scholten — 2016-12-05T06:47:16-05:00 — #12
These numbers come right from NASA,8" per mile and the radius of 3959 miles and a 25,000 mile circumference, this is my point, they do not add up!
3959 divided into 6250=1.572327044
This tells me, for every 1.572327044 miles I cover one the circumference I will go straight down 1 mile Which is probablly the same as your .63 miles per mile) and this drop is the same for every mile!
You went one step farther than me! thank you!
So now we "know the curve" of the sphere, go to a large body of water and prove the curve to your self, it is not there.
How far do you think we would see in the horizon if the curve was 1 mile over 1,57 miles?
Why are we told both of these equations that do not add up to the same result?
Why have we been lied to?
bill_coley — 2016-12-05T12:24:01-05:00 — #13
But Will, the numbers DO add up if you apply the applicable mathematics!
We know from tests that the drop over the FIRST mile from an observer IS 8".
Because we know the radius of the earth, we know the drop over the 6,250 miles along the surface of the earth from the 12 to 3 o'clock positions IS 3,959 miles.
- The math I showed in my last post demonstrates that the average drop per mile over that 6,250 miles IS over 40,000" per mile.
Those results mean the drop over the miles AFTER the first mile MUST BE more than 8" per mile. And that's the expected result when you apply the Pythagorean Theorem! (It's NOT the expected result ONLY if you assume the earth is a square, a rectangle, or some other non-spherical shape.)
NO!! Your 1.572327044 per mile number and my 0.63 per mile number are both AVERAGES not absolute rates. (By the way, our two numbers ARE completely compatible with each other.)
If I eat one apple the first day, then two apples the second day, and finally three apples the third day, ON AVERAGE I ate two apples per day. But did I eat two apples EVERY day? NO! That's because an average doesn't tell you what I did on any particular day.
To calculate the average number of apples I ate, we divide the total number of apples I ate by the total number of days it took me to eat them. That formula makes the "average" a big picture number, descriptive of the entire three day period during which I ate apples, but NOT descriptive of any particular day that I ate apples.
Similarly, your 1.57.... number and my 0.63 number are averages, calculated by dividing one total distance by another total distance. As averages, they tell us about the big picture of the drop. But they DO NOT tell us the drop for any specific mile!!!
There is NO lie here, Will. It is totally consistent with the mathematics of the situation for the drop to be 8" for the first mile, but more than 40,000 inches per mile over 6,250 miles because that 40,000+ figure is an AVERAGE, not a rate per mile.
will_scholten — 2016-12-05T19:41:14-05:00 — #14
Now my basket ball is smooth, 2/3rds of the earth is water (my BB is also 1.577... curve) surely you have some pictures of the ocean to share, where I can see a mile drop in 1.57 miles can't you? (No tsunomi don't count) I would love to see it, I always see water run off the slightest slope. I don't know if white water rapids have that kind of a drop!
I live in Michigan, Lake Michigan borders our whole western shore (I'm guessing 300 miles or more)I don't see it anywhere.
Historic High WaterThe lake fluctuates from month to month, with the highest lake levels typically experienced in the summer. The normal high-water mark is 2.00 feet (0.61 m) above datum 577.5 ft (176.0 m). In the summer of 1986, Lakes Michigan and Huron reached their highest level during the period during which records have been kept, at 5.92 ft (1.80 m) above datum. The high water records began in February 1986 and lasted through the year, ending with January 1987. Water levels ranged from 3.67 ft (1.12 m) to 5.92 feet (1.80 m) above Chart Datum. On February 21, 1986, the waters neared the all-time maximum for the period during which records have been kept.Historic Low WaterLake levels tend to be the lowest in winter. The normal low water mark is 1.00 foot (0.30 m) below datum 577.5 ft (176.0 m). In the winter of 1964, Lakes Michigan and Huron reached their lowest level at 1.38 feet (0.42 m) below datum. As with the highwater records, monthly low water records were set each month from February 1964 through January 1965. During this twelve-month period water levels ranged from 1.38 feet (0.42 m) to 0.71 feet (0.22 m) below Chart Datum.
Lake Michigan and Huron meet at the top of Michigan's lower pennisula and Huron goes over to the east side of the state, how can they have the same elevation of water on both side of the state.
Also Lake Superior I beleive is the worlds largest lake and it has the same elevation of water , it has 1300 miles of shoreline.
So if I found 6250 people to do an experiment to line up from the N. pole to the Equator , everybody would have a "first"mile of 8",you are saying some how that will total an 0 63 mile per mile curve?
All the 8"x sq I find ,ALWAYS SAYS the first mile is 8" the rest is sq because of the obscured view on the horizon.
I don't buy that at all!!
I never see a 0.63 mile drop in a mile in a normal drop anywhere!
bill_coley — 2016-12-05T22:01:01-05:00 — #15
I took the basketball we have around our house and made some best-effort, but admittedly casual approximation, measurements. Here's what I found:
- Distance along the ball's surface from its 12 o'clock to 3 o'clock positions: 7.4"
- Radius of the ball: 5.9"
- The drop in the first inch along the ball's surface from the 12 o'clock position toward the 3 o'clock position: Immeasurably small - perhaps 1/16 or 1/8 of an inch, but NO WAY IN HEAVEN OR EARTH anywhere close to 1/7.4 of the 5.9" total drop - about 3/4" - that it would have to be were the drop consistent for every inch of the surface.
- The drop in the second inch along the ball's surface: The same as for the first inch, immeasurably small - perhaps 1/16 or 1/8 of an inch, but NO WAY IN HEAVEN OR EARTH anywhere close to 1/7.4 of the 5.9" total drop - about 3/4" - that it would have to be were the drop consistent for every inch of the surface.
- So after two of the 7.4 inches along the ball's surface from its 12 o'clock to 3 o'clock positions, the total drop as calculated from the sum of the drops of the first two inches is 1/8-1/4". PLEASE NOTICE!! We are now more than a quarter of the way to the 3 o'clock position, but seemingly have dropped no more than 1/4" out of a total drop of 5.9". That means the total drop over the last 5.4" of the ball's surface will have to be more than 5-1/2". But how can that happen when...
- The drop in the third inch along the ball's surface is ALSO immeasurably small, perhaps about 1/16" or 1/8"??
BUT!!!!!!!!!!!!!!!!!! (and this is the kicker) the actual total drop measured after three inches along the ball's surface from the 12 o'clock position toward the 3 o'clock position: 2-1/4"!!!!!! That, when the drop from each of the three individual inches was no more than 1/8".
Please try this experiment on your basketball. Your numbers won't match mine exactly, but the drops you measure after each of the first, second, and third inches along the ball's surface WILL all be about the same AND they will be MUCH, MUCH, MUCH smaller than you expect given the magnitude of the total drop at the 3 o'clock position. In addition, you will find that after the third inch, the actual total drop is MUCH larger than the sum of the three individual drops. How can that be? Remember Mr. Pythagorus and his theorem.
When you confirm my findings, remember the earth is a sphere whose distance from 12 o'clock to 3 o'clock is 6,250 miles, not 7.4". So the effect will be the same, simply greatly magnified.
gao_lu — 2016-12-05T23:50:45-05:00 — #16
Not to complicate matters, but...Just for added adventure, you should remember that the earth is not actually a sphere. The poor thing is squished by about 13 miles, fattened at the equator by centrifugal force (?), and is flexing its fat at that.
will_scholten — 2016-12-06T07:12:38-05:00 — #17
When you take a compass and make a circle, you set the radius for the entire circle, is not the radius the same around the whole circle, the drop has to be the same.
Water ALWAYS self levels it self, if you don't believe me look for your self!
I like that!
Maybe NASA has to think things through better, I don't see that in the"blue marble" do you?
The story has to be the same.
Whoops here is the one from NASA!
bill_coley — 2016-12-06T14:01:34-05:00 — #18
Will, our exchange has been respectful - which I always appreciate - but has not produced much clarity. So I have turned to the illustration software I use to create graphics for church and other purposes to create the following two diagrams, which I hope will clarify how the drop is the same from any given point on a sphere, but is NOT the same in the bigger picture. (The diagrams show distances to scale. Distances are rounded to two decimals, and, other than the diameter and radius values which are precise, are very close to, but not necessarily exactly, accurate.)
First, I created a circle (the effect is the same on a sphere)...
- Diameter = 5.0"
- Radius = 2.50"
- Circumference = 15.71"
- Distance from 12 o’clock to 3 o’clock = 3.93"
Then I drew three 1" curves along the surface of the circle, starting from the 12 o’clock position and in the direction of the 3 o’clock position. The remaining curve - from the end of the third curve to the circle’s equator - is 0.93" in length because the total distance from 12 o’clock to 3 o’clock is 3.93", not 4". I gave the first three curves distinct colors.
Then, to indicate the drop in each inch along the surface of the circle, I drew horizontal lines from the circle’s vertical center dividing line to the leading ends of the first three curves, and used the illustration software to measure the distances between those lines. Those distances, along with the distance from the third horizontal line to the circle’s equator, are included in the diagram.
Finally, I placed letters at the 12 o’clock position and at the end of each curve. Those letters are A-D.
Here is that figure....
Notice that the drop in the first inch - from position A to position B - is 0.20". The drop in the second inch - from position B to position C - is 0.56", more than double the drop in the first inch. And the drop in the third inch - from position C to position D - is 0.86", more than four times the drop in the first inch. (The drop from the fourth curve - from position D to the equator - is only slightly larger than the drop in the third inch because the fourth curve is shorter than the first three curves.)
I hope you will agree with me, Will, that the drops in the first three curves are CLEARLY NOT THE SAME. In fact, those drops INCREASE as they move away from the 12 o’clock position toward the 3 o’clock position. My drawing and its measurements might be off by a few hundredths of an inch either way, but there’s NO WAY they’re off enough to explain away the significant difference in the curves’ lengths. I contend that even without the provided measurements, the difference in drop lengths is visually obvious.
BUT WAIT!!! What about your point, that after you walk a mile (or inch, in these diagrams) and measure again, the drop is still the same, that the drop is always 8" per mile on the surface of the earth? For the answer, I needed to create a second figure.
What your point misses is the fact that when we move to position B on the circle and measure again, position B becomes the new 12 o’clock position! That is, to accurately depict the drops that arise as we move away from position B, we must rotate the circle so that position B is at the 12 o’clock position. Look what happens when we do that....
The drop in the first inch - from position B to position C - is the same 0.20" drop we found in the first inch away from position A! And succeeding drops in the inches after the first inch away from position B increase in length just as they did moving away from position A.
The secret here is that the drops we’ve been talking about throughout our exchange have been measured from the 12 o’clock position to the 3 o’clock position. But in the first diagram above, position B is NOT at the 12 o’clock position! So, the drop from position B to position C in that diagram is larger. But when we move position B to the 12 o’clock position - to give it the same perspective as measurements starting from position A in the first diagram - then the drop in the first inch (or mile, or whatever) is identical to the first drop measured from position A, and in fact is identical to the first drop measured from ANY starting point on the circle AS LONG AS WE ROTATE THE CIRCLE TO PUT THAT STARTING POINT AT THE 12 O’CLOCK POSITION.
The conclusion I draw from the diagrams presented in this post is that the first mile drop from any point on the surface of the earth will be the same - 8" according to NASA. But the drops in all subsequent miles, IF VIEWED FROM THE ORIGINAL STARTING POINT, will increase dramatically, a fact that, in my view, clearly demonstrates the truth of claims about the earth’s spherical shape.
I will welcome your observations of the two diagrams I offer here.
gao_lu — 2016-12-06T17:50:17-05:00 — #19
But Bill! I see a serious problem. From your diagram one can clearly see that from horizontal movement from say 5 o'clock to 6 o'clock relative to 4 o'clock to 5 o'clock would actually be climbing. And that is breaking all the rules to give that much credit. Actually, at exactly 3:00 PM, moving horizontally through time and descending vertically along a ladder, one would step off the earth and descend into outer darkness. You wouldn't suggest that negative time exists, would you?
Unless you want an infinity of turning your trig bit by bit on it's head, you need to convert to polar coordinates and apply a little simple parametric calculus. Otherwise you aren't taking into account the full affect of infinitesimal steps along your arc of the covenant. Can your software create those equations, please? We want pu-roof, and simple trig won''t cut it. Poor Pythagoras would be spinning in his grave over this dreadful matter. Newton? Newton! Where are you?
I suspect you are on to something. I just don't see a fat wobbly belly in the NASA picture at all. Certainly not a 13 mile wide one! In fact, I suspect in reality the earth isn't a blue marble at all!
will_scholten — 2016-12-06T18:36:25-05:00 — #20
Thanks Bill, this explains a lot. We are not comparing apples to apples. You probably saw that when I suggested using 6250 people to help me.
I am not a math major, but I believe I remember you cannot figure out circles with rectangles. And we can see it does not agree with the 1.57....
My head is telling me, that the first mile would still be wrong with the 8" equation.
I think you have to figure a circle with a pie chart.
Like I said I'm not a math major, but I have to figure things out for volumes and half circles and area.
But thank you Bill for the time you spent with making your chart.
next page →