According to the Flat Earth Society, the Bishop Experiment demonstrates that it is possible to see Lighthouse Beach from Lovers Point and vice versa on a very clear and chilly day.
IF the earth is a globe, and is 24,900 English statute miles in circumference, the surface of all standing water must have a certain degree of convexity--every part must be an arc of a circle. From the summit of any such arc there will exist a curvature or declination of 8 inches over the first statute mile. Over two miles the fall will be 32 inches; by the end of the third mile, 72 inches, or 6 feet, as shown in this chart.
Correcting for the height of the observer of about 20 inches, when looking at the opposite beach over 23 miles away there should be a bulge of water obscuring objects up to 300 feet above the far beach. There isn't. Even accounting for refraction, the amount hidden should be around 260 feet - seeing down to the shoreline should be impossible
I'm aware of Bedford Level experiment, which states:
Atmospheric refraction causing an object below the horizon to be visible.
however the above experiment already takes into the account atmospheric refraction.
There was a similar experiment done using a laser test at Lover's Point CA which claims:
The complete video as taken from Lover's Point, Pacific Grove, CA. to Sand City, CA.
Distance is 4.07 miles and should show an Earth curvature of 10 feet. With a camera at 30'' and 3 foot waves, the laser is still seen at below 6'' from 4 miles away.
Here is another laser test over 7.5-8 mile distance on a frozen lake.
And here is another mirror test, shore to shore, over a distance of 13.23 miles, where, according to the Earth's Curve calculator at metabunk.org, the destination point should not be visible, even after taking into account a standard atmospheric refraction.
Distance = 13.23 Miles (69854.4 Feet), View Height = 3 Feet (36 Inches) Radius = 3959 Miles (20903520 Feet) Results ignoring refraction Horizon = 2.12 Miles (11199.16 Feet) Bulge = 29.18 Feet (350.15 Inches) Drop = 116.72 Feet (1400.62 Inches) Hidden= 82.29 Feet (987.52 Inches) Horizon Dip = 0.031 Degrees, (0.0005 Radians) With Standard Refraction 7/6*r, radius = 4618.83 Miles (24387440 Feet) Refracted Horizon = 2.29 Miles (12096.47 Feet) Refracted Drop= 100.04 Feet (1200.53 Inches) Refracted Hidden= 68.4 Feet (820.74 Inches) Refracted Dip = 0.028 Degrees, (0.0005 Radians) Tilt Angle = 0.191 Degrees, (0.0033 Radians) Horizon Curve Fraction = 0.00007 Horizon Curve Pixels = 0.23 Horizon Curve Angle v1= 0.00475 Horizon Curve Angle v2 = 0.00475
Based on the above experiments, should the surface of all standing water have a certain degree of convexity? If so, why is the point below the horizon still visible over the distance of 13-23 miles?