Well, since no one has answered that question as yet, allow me to reply with a question and a comment.

First, my comment : this shot with the building-below-waterline is pretty funny because it would rather seem to confirm the earth curvature – instead of its clearly stated intent to prove a flat earth:

Secondly, my question: what exactly is this expected curvature of earth so frequently referred to (and ‘demanded’) by flat earth proponents? The thing is, 1° out of a 360° circumference of 40,075km equals about 111km (1°=111km). For instance, London and Paris (distant about 333km) would only be ‘offset’ by 3°. If, by magic powers, one could see London from Paris (and vice versa), well – people would probably think the world is – indeed – flat : septclues.com/VARIOUS%20MATERIAL/flat_earth_LONDON_PARIS_01.JPG

The link below contains a tool for calculating corrections in the earth’s curvature for measuring high points in a particular geographical area, as well as a separate tool for correcting for the height of the observer. (Unfortunately, the native unit of measurement is in feet/miles)

According to the calculator on that page 341 miles (550 km) yields a distance correction of 77,800 feet or 14.73 miles. Add in the correction for a 6 foot tall observer and subtract 3 miles for a total of 11.73 miles of curvature between New York and Toronto (topographical irregularities notwithstanding).

For 25 miles (or 22 to account for the 6 foot tall observer) the correction is 323 feet, all of which should be obscured by the horizon. If the building is indeed 10 floors, even if it were a very fancy building with incredibly generous 30 foot ceilings, the entire building should be obscured by the curve of the horizon according to conventional standards.

Ab asks:

“Where’s the curve”?Well, since no one has answered that question as yet, allow me to reply with a question and a comment.

First, my comment : this shot with the building-below-waterline is pretty funny because it would rather seem to confirm the earth curvature – instead of its clearly stated intent to prove a flat earth:

septclues.com/VARIOUS%20MATERIAL/flat_earth_building_02.JPG

Secondly, my question:

whatexactly is this expected curvature of earth so frequently referred to (and ‘demanded’) by flat earth proponents? The thing is, 1° out of a 360° circumference of 40,075km equals about 111km (1°=111km). For instance, London and Paris (distant about 333km) would only be ‘offset’ by 3°. If, by magic powers, one could see London from Paris (and vice versa), well – people would probably think the world is – indeed – flat :septclues.com/VARIOUS%20MATERIAL/flat_earth_LONDON_PARIS_01.JPG

Here’s what Toronto > New York would look like:

septclues.com/VARIOUS%20MATERIAL/flat_earth_TORONTO_NEWYORK_01.JPG

Just asking.

The link below contains a tool for calculating corrections in the earth’s curvature for measuring high points in a particular geographical area, as well as a separate tool for correcting for the height of the observer. (Unfortunately, the native unit of measurement is in feet/miles)

cohp.org/hand_levels.html

550 km X .62 + 341miles

According to the calculator on that page 341 miles (550 km) yields a distance correction of 77,800 feet or 14.73 miles. Add in the correction for a 6 foot tall observer and subtract 3 miles for a total of 11.73 miles of curvature between New York and Toronto (topographical irregularities notwithstanding).

For 25 miles (or 22 to account for the 6 foot tall observer) the correction is 323 feet, all of which should be obscured by the horizon. If the building is indeed 10 floors, even if it were a very fancy building with incredibly generous 30 foot ceilings, the entire building should be obscured by the curve of the horizon according to conventional standards.