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### Messages - Mack

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##### Flat Earth Theory / Re: Curvature of the Horizon
« on: April 02, 2023, 01:37:08 PM »
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There are no straight slopes on a globe.

With all due respect, that’s the point you don't understand.  Literally, the definition of a straight line is that it has a constant rate of change.  You are arguing that the circumference of the globe has a constant rate of change, but isn’t a straight line.

Put another way, if the circumference of the globe has a constant rate of change, then it is a straight line. Do you see the contradiction in your own argument?

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##### Flat Earth Theory / Re: Curvature of the Horizon
« on: April 02, 2023, 07:17:44 AM »
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If that were true then sin(x)-sin(x-1) as x goes from 90 to 1 (i.e. looking at the difference in the Y coordinates of each end of 1 degree arcs 0-1, 1-2, 2-3,....89-90) would be a constant which is obviously not the case.  What you claim is nonsense.

If I'm translating right, I think this website makes your point.  It's pretty cool.  Move the slider in the top right corner.

https://www.geogebra.org/m/hnZMkBdc

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##### Flat Earth Theory / Re: Curvature of the Horizon
« on: April 02, 2023, 01:53:58 AM »
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I anchor myself to the top of the building and freeline halfway down the building. In doing so I have travelled one quarter (1/4) of the circumference of the building ie 197 metres.

No you haven’t.  The length of your line will tell you the distance you have travelled (vertical plus horizontal).  Your line would look like the one on the right.  If you straighten it out, it will be longer than the one on the left and more than ¼ of the circumference. You’ve dropped ¼ of the circumference, but you’ve travelled more distance (vertical plus horizontal)

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Never mind on average its a fact all the way round; the curve is the same all the way round - it doesn't drop away any more or less than anywhere else on the circle.

Wrong. Curvature can be measured by how much it deviates from straight.  Here’s a graph of a circle.

Notice how the circumference hits the grid lines at irregular intervals.  That means it deviates from straight a different amount each time it hits the grid and the rate of drop varies.  That’s why there is 8in of drop at 1 mile, but 32 in of drop at two, like I already explained. Compare that to the graph of a straight line where each coordinate hits the graph line in equal intervals. That’s a constant rate of drop

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And I can't understand why no one agrees with this.
.
Because this is pretty basic well understood geometry and math. Even Rowbotham accepted the 8in. m^2 equation, which works pretty good up to a point.  But it’s the equation for a parabola not a sphere so eventually the errors start compounding.

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##### Flat Earth Theory / Re: Curvature of the Horizon
« on: April 01, 2023, 04:47:35 AM »
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Accepting the above if we divide 3,113 miles by 1,982 miles we get a drop/fall/decrease in height in relation to the north pole of 1 mile per 1.57 miles travelled.

I’m going to start with this comment because I think it is the source of your confusion.  The “height” of a  position is measured against sea level and depends on the terrain. If the earth was a perfectly smooth sphere everything would be at sea level.  “Height” doesn’t have anything to do with the curvature of the earth.

Curvature is the rate a line is changing direction at a given point. The “drop” in the curvature of the earth is a change in direction of an arbitrary line on its surface, not a change in height.

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Instead of walking from N to E1 imagine walking from N to X. This is half the distance to the equator and represents one eighth (1/8) of the earths circumference ie 3,113 miles.

You have walked 3,113 miles south, not “down”.  Eventually you would be going north.

You have two errors.  The first is you try to measure curvature using straight line distances.  That doesn’t work. Second you are adding the curvature linearly.

Using the equation I gave before: h = r * (1 - cos a). One mile across the surface is .0145 ° change in direction from straight.
The cosine of a .0145 ° angle is .99999997
h=3959*(1-.99999997)
h=3959*.000000032
h=.000127 miles
h=8.04672 in

So in one mile, there is a “drop”, or change in direction of the surface of 8 in. from A to B.

By your logic, then there should be 16 in of drop between A and C, but:

Two miles across the surface is .0288 ° . The cosine of a .00288 ° angle is 0.99999987
h=3959*(1-.99999987)
h=3959*.00000013
h=.000515 miles
h=32.6304 in

To sum up, a sphere or circle isn’t a straight line so the ratio of distance to curvature isn’t 1:1. It’s close to quadratic…the curvature is proportional to the square of the distance.

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##### Flat Earth Theory / Re: Curvature of the Horizon
« on: March 30, 2023, 10:11:21 PM »
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What I am calculating is how much does the curve drop in height (assuming a non-spinning globe earth that has a top and bottom). So if I walked a quarter of the earth's circumference from the north pole to the equator in a straight (obviously curved) line I would cover 6,225 miles (or so).

There’s already a formula for computing the amount of drop along a curve. It’s  h = r * (1 - cos a). h is the amount of drop, r is the radius of the sphere, a is the angle created by the start and end points measured from the center of the sphere.

The number of degrees in the angle is the drop. The more you move along the curve, the number of degrees increases.

The radius of the earth is about 3959 miles, so in one mile the curve drops .0000137 degrees = 0.67 feet=8 in.  This is just approximate since it assumes a perfect sphere though.

From point a to point b, it drops 8 inches and from point b to point c, it drops another 8 inches. But you can’t add the 8 inches from a to b and from b to c like you were moving is a straight line. For every unit down, you also move sideways, but it isn’t a 1:1 ratio because the direction keeps changing.

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##### Flat Earth Theory / Re: Curvature of the Horizon
« on: March 30, 2023, 03:04:45 AM »
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I am not a scientist so please bear with me. My definition is the amount by which the curve 'drops' in 'height' on an assumed non-rotating global earth from any single point on that globe. And for illustrative purposes my example would be a person standing at the north pole on that globe (the north pole being at the 'uppermost' part of that globe) would see the curve fall in height by 1 mile for every 1.57 miles of circumference.

Curvature is measured as the angular turn per unit distance. Your definition seems to be based on straight line measurements.

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