Dinsdale wrote:Poptard, you keep citing the view of Toronto from 30 miles away as "proof." The problem is your horribly flawed geometry.
Rather than continue to join the well-deserved pile-on (OK, I lied -- I'll certainly continue), let me help where your junior high teachers failed:
Looking across a body of water at a point 30 miles (measured over the curve of the Earth)... let's even use your (quite incorrect) formula for Earth's curvature (and we'll even fudge the numbers in your favor)...
Let's call our location across from Toronto "Point A." Toronto will be "Point C." The halfway point (on the curve) will be "Point B." Our line-of-sight is between Point A and Point C.
From Point A, go 1 mile towards Point B. The curve has gone UP 8"(according to your incorrect formula), down our line-of-sight. We're not looking for Toronto on the surface of the Moon, right... we're looking down the line-of-sight. There's a "bump" in between.
Now, let's go one more mile towards Point B. The horizon just ROSE a bit LESS than 8". We're at 15" or less of RISE.
Let's do this 15 times, until we get to Point B. Point B is the apex of the arc, and is what is blocking the view of Toronto (Point C).
And Point B is well under 10 feet above the line-of-sight. Add about 5 feet for the level of the camera, and this rise becomes 6 feet or less.
But there's a catch -- Lake Erie is a large body of water, and is thus subject to tidal influence and wind-waves. The swells at Point B further occlude the view of Toronto.
BASIC FUCKING GEOMETRY (which none of your linked-to loons seem to grasp, or are intentionally playing parlor tricks) tells us that only the lowest few feet of Toronto's skyline should be occluded by the curvature of the Earth.
Not hundreds or thousands of feet... again, BASIC FUCKING GEOMETRY. It's not that complicated for anyone who got at least a D in junior high school math.
A few feet.
So sorry you had to have your latest moonbat "theory" so clearly thrashed this way.
This is truly incredible.
Are you honestly coming in here to claim to be the smart man, telling me that over 15 miles, the earth curvature is less than 10 feet?
THINK, Duncedale, THINK!!
CN Tower, is 1,800 ft tall.
It is 250 miles (as the crow flies) from CN Tower to Pittsburgh.
By advanced Duncedale math (10 ft of curvature over 15 miles distance) only 167 ft of CN Tower would be under the horizon if we traveled away 250 miles to Pittsburgh.
In Pittsburgh, a person can see 90% of CN Tower, according to advanced Duncedale math.
rotf...
In fact, according to advanced Duncedale math, we would have to travel all the way to Venezuela, South America before the entire tower would finally be hidden from our view.
Get with the program.
You won't look like such a total fool next time.
The earth curvature rate on your ball earth model is slightly less than 8" per mile squared.
The rate of curvature on a sphere COMPOUNDS.
See the diagram and -----> THINK about a sphere!
Notice how the rate of curvature naturally increases as you move further from your target?
Notice how line 2 is longer than line 1?
Yes, says a child of 6.
lol
It is 8" for the first mile, or 8" x 1 = 8"
For the 2nd mile, it is 8" x 2 x 2 = 32"
For the 3rd mile, it is 8" x 3 x 3 = 72"
etc...
https://dizzib.github.io/earth/curve-calc/
Over 30 miles, there is 600 ft of curvature -- from an eye height of 0 ft.
If we are at 6ft eye height, there would be 486 ft of curvature.
This was all covered and known (should have been) on this board over 2 months ago.