How many of you believe the Earth is flat? Go.......

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I still don't understand the "logic":

If the horizon dips 8" in a mile, why not 16" in 2 and so on.

To suggest there's a logarithmic increase defies my lil' ol' brain's capacity for reason. :confused:

I'd really appreciate it if someone could explain this to me in laymonkey's terms.


I made the same error as you referencing the Joe Rogan video actually, in terms of the total “drop” at 100 miles. I simplified it too much (Though I do not believe it to be 6000 whatever feet)

When someone solves for the drop after one mile, we use the radius of the earth, and one mile as the constants, and are solving for the 3rd line which is the drop, in this case the hypotenuse.

However if you change it up and use the radius and the drop as the constants, and solve for the sight distance, it shows that it isn’t exactly 8” every mile. It just happens to be 8” for one mile

Sorta make sense? Why that is, Couldn’t really tell ya man lol


All this gets more confusing because line of site depends on a lot of different environmental variables as well
 
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I still don't understand the "logic":

If the horizon dips 8" in a mile, why not 16" in 2 and so on.

To suggest there's a logarithmic increase defies my lil' ol' brain's capacity for reason. :confused:

I'd really appreciate it if someone could explain this to me in laymonkey's terms.
Stick a straight edge on a round ball and observe for yourself. The contact point of the straight edge on the ball is the perspective of the person viewing the curve, i.e where you are standing. Say you are standing at the north pole looking out over you round ball, eventually the curve eventually reaches what is a straight downward drop (from your perspective as a viewer) as it approaches the equator. Because of the circle/globe and your perspective from a stationary postion, it grows exponentially, not in a linear fashion. .Maybe that will help.
 
Stick a straight edge on a round ball and observe for yourself. The contact point of the straight edge on the ball is the perspective of the person viewing the curve, i.e where you are standing. Say you are standing at the north pole looking out over you round ball, eventually the curve eventually reaches what is a straight downward drop (from your perspective as a viewer) as it approaches the equator. Because of the circle/globe and your perspective from a stationary postion, it grows exponentially, not in a linear fashion. .Maybe that will help.

Logical but not accurate
 
A quadrant of a circle has a constant curve but an exponential function isn't constant
 
Stick a straight edge on a round ball and observe for yourself. The contact point of the straight edge on the ball is the perspective of the person viewing the curve, i.e where you are standing. Say you are standing at the north pole looking out over you round ball, eventually the curve eventually reaches what is a straight downward drop (from your perspective as a viewer) as it approaches the equator. Because of the circle/globe and your perspective from a stationary postion, it grows exponentially, not in a linear fashion. .Maybe that will help.
This explains it just-fine mate. Thank you!

Thank you to Dan for trying to explain it to me too.
 
If you were to scale your height onto a round ball so that it was proportional to the earth's diameter and your height on the earth you wouldn't perceive a curve.
 

I believe this uses a similar formula that VB posted. His link had a proper “note” on the bottom saying that it is not accurate for long distances. The reason being is that it is a formula for a parabola and not a circle. A parabola does follow a circle for a little bit at its peak, but only for that little bit. Which is why it notes that it isn’t accurate for long distances
 
Logical but not accurate
The squaring is accurate to a point but long distances it becomes wildly inaccurate and requires additional calculatons, so I have learned.

Regardless i find the whole thing to be asinine. Extreme curve is extremely ridiculous. One way or another you are still sailing downhill everywhere you take your boat.
 
The squaring is accurate to a point but long distances it becomes wildly inaccurate and requires additional calculatons, so I have learned.

Regardless i find the whole thing to be asinine. Extreme curve is extremely ridiculous. One way or another you are still sailing downhill everywhere you take your boat.
Ya had me, then you lost me
 
It's a logical conclusion when sailing around on a ball earth. Please explain how it would not be so, if you can or if you care to.
Even gravity creates an even surface. The surface is measured to the center of the earth, so all across the ocean is basically the same distance from the center of the earth, hence smoothing sailing
 
Ya had me, then you lost me
100 miles is not a “long distance” relative to the alleged circumference of the earth. And accurate is a relative term — the note says the calculation is nonetheless “practical.”
 
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