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Flat Earth Theory / Re: Curvature of the Horizon
« on: March 30, 2023, 10:42:06 AM »QuoteI 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.
Ok lets forget about the actual curve itself for a moment. 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). And in doing so I would have dropped in height by 3,963 miles (the radius of the earth). Therefore for every 1.57 miles I walked there is a drop in height of 1 mile. Using the ocean as an example; and again assuming a globe earth, if rowed out to sea a distance of 1.57 miles there should have been a drop in height of 1 mile. Now a physical drop of 1 mile in height is something we just do not see (in fact we see no such thing and to us it looks quite level) but we should see it if we were on a globe earth.
Looking at it another way. If I walked across the salt flats for 1.57 miles I should be 1 mile lower than when I started. And am sure we all know that this is not the case.
I am not sure if I am explaining this as I intended or indeed correctly but would welcome some genuine advice/debate/discussion on this particular matter as something just doesn't seem right and am sure I haven't miscalculated the actual maths.
The fundamental mistake you are making is an assumption that your "Rate of Drop" is linear; it isn't. The "rate of Drop" as you call it increases as you travel south.
Consider standing at the North Pole in your model and travel 1 mile. Your actual drop is negligible, and you can probably still see the Pole. The Rate of Drop is zero.
Now stand 1 mile north of the equator and then walk to it. Your Rate of Drop is now 1 mile per mile.
Your formula only works if the drop is linear, as if the Earth was a cone.