*I do not agree with modern physics as you do *
It so happens, that I do not agree AT ALL with modern physics...

*So what keeps us grounded on Earth? How about our mass and weight.*
It cannot be mass and weight.

Here is the Allais effect, explained in detail:

http://www.davidicke.com/forum/showthread.php?t=282058
Here is the Barometer Pressure Paradox:

http://www.davidicke.com/forum/showthread.php?t=282177
Here is the Biefeld-Brown effect:

http://www.davidicke.com/forum/showthread.php?t=281821
And the Lamoreaux effect:

http://www.davidicke.com/forum/showp...21&postcount=2
Terrestrial gravity is the result of PRESSURE, exerted by the telluric currents.

*Your reasoning seems contradicting; a dreamworld subject to forces? Would not your spin theories only reinforce curvature? *
Our dreamworld is absolutely subject to forces.

The spin theories are totally incompatible with any curvature of the Earth.

Since terrestrial gravity is a force of pressure, it cannot take place on a spherical Earth, but only on a flat surface.

*Everything you have stated except for curvature is mainstream science so how is this hidden?*
No, it is not mainstream.

Please read:

http://theflatearthsociety.org/forum...521#msg1639521
*I don't see how a photograph can prove curvature based on its limited depth or a video showing cartoon images of earth. *
The photograph in my previous message shows that there is no 3.35 meter midpoint visual obstacle across the Strait of Gibraltar.

No ascending slope either.

Let us increase the distance to 55 km, and the curvature to 59 meters (Lake Toronto, Grimsby to Toronto)

http://www.flickr.com/photos/suckamc/53037827/#
Again, no curvature whatsoever across a distance of 55 km, no 59 m midpoint visual obstacle.

Photo by Ms. Kerry Ann Lecky-Hepburn, no curvature whatsoever across a distance of 55 km.

From the very same spot, Ms. Lecky-Hepburn used a reflector telescope for this zoom:

Another photograph signed Mrs. Lecky-Hepburn:

http://www.flickr.com/photos/tundrab...hy/312939439/#
No 59 meter curvature whatsoever, a perfectly flat surface of the water.

http://www.flickr.com/photos/planetrick/487755017/#
http://www.flickr.com/photos/planetr...in/photostream
No curvature whatsoever, from Hamilton to Lakeshore West Blvd: no visual obstacle, just a perfectly flat surface of the water all the way to the other shoreline.

Some formulas for you...

Curvature of a spherical Earth...

R = earth radius, 6378.164 kilometers

@ = s/R

s = arclength between the two points measured on the surface

C = curvature

Earth Curvature

C = R(1 - cos[@/2])

For example:

s = 13 km (Strait of Gibraltar), C = 3.31 meters

s = 34 km (English Channel), C = 22.4 meters

s = 53 km (Lake Ontario, Grimsby - Toronto), C = 55 meters

s = 1000 km (Irkutsk - Tungusk), C = 19.5 km

A more complex formula, taking into account the height of the photographer, and the height of the visual target:

** BD = (R + h)/[[RAD[2Rh + h^2](sin s/R)(1/R) + cos s/R]] - R **
RAD = SQUARE ROOT OF [] that is, the square root of [2Rh + h^2]

For example, s = AB = 53 km, AE = 2 meters, we get BD = 180 meters, that is, we could see nothing under the altitude of 180 meters, standing on the beach at Grimsby or St. Catharines...

AE = h = height of observer

BD = height of visual target

No curvature across lake Michigan, best proofs:

http://www.davidicke.com/forum/showp...59&postcount=2