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I've been thinking
WilburHooksHands
Member Posts: 6,801
in Tug Tavern
California Monterey Bay is a relatively long bay that sits next to the Pacific Ocean. The distance between the extremes of the Monterey Bay, Lovers Point in Pacific Grove and Lighthouse State Beach in Santa Cruz, is just over 23 statute miles.
Distance between the two points, courtesy of Google Earth
On a very clear and chilly day it is possible to see Lighthouse Beach from Lovers Point and vice versa. With a good telescope, laying down on the stomach at the edge of the shore on the Lovers Point beach 20 inches above the sea level it is possible to see people at the waters edge on the adjacent beach 23 miles away near the lighthouse. The entire beach is visible down to the water splashing upon the shore. Upon looking into the telescope I can see children running in and out of the water, splashing and playing. I can see people sun bathing at the shore and teenagers merrily throwing Frisbees to one another. I can see runners jogging along the water's edge with their dogs. From my vantage point the entire beach is visible.
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.
Suppose that the earth is a sphere with a radius of 3,963 miles. If you are at a point P on the earth's surface and move tangent to the surface a distance of 1 mile then you can form a right angled triangle as in the diagram.
Looking over a distance of 1 mile, we can use the theorem of Pythagoras:
a2 = 3,9632 + 12 = 15,705,370
and when we square root that figure we get a = 3,963.000126 miles
Thus your position is 3,963.000126 - 3,963 = 0.000126 miles above the surface of the earth.
0.000126 miles = 12 in * 5,280 ft * 0.000126 mi = 7.98 inches
Hence after one mile the earth drops approximately 8 inches.
"Whenever I have doubts about the shape of the earth I simply walk outside my home, down to the beach, and perform this simple test. Provided that there is no fog and the day is clear and calm, the same result comes up over and over throughout the year."
—Tom Bishop
Ergo, looking across 23 miles the Pythagorean theorem becomes:
a2 = 39632 +232 = 15,705,898
and when we square root that figure we get a = 3,963.06674 miles
Thus your position is 3,963.06674 - 3,963 = 0.06674 miles above the surface of the earth
0.06674 miles = 5,280 ft/mi * 0.6674 mi = 352.3872 feet
Hence after 23 miles the earth drops approximately 352 feet.
There are a number of different methods to calculate the drop of the Round Earth. Go ahead and look a few up to try out. You will find that the drop while looking over 23 miles is on the order of 300-400 feet.
Distance between the two points, courtesy of Google Earth
On a very clear and chilly day it is possible to see Lighthouse Beach from Lovers Point and vice versa. With a good telescope, laying down on the stomach at the edge of the shore on the Lovers Point beach 20 inches above the sea level it is possible to see people at the waters edge on the adjacent beach 23 miles away near the lighthouse. The entire beach is visible down to the water splashing upon the shore. Upon looking into the telescope I can see children running in and out of the water, splashing and playing. I can see people sun bathing at the shore and teenagers merrily throwing Frisbees to one another. I can see runners jogging along the water's edge with their dogs. From my vantage point the entire beach is visible.
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.
Suppose that the earth is a sphere with a radius of 3,963 miles. If you are at a point P on the earth's surface and move tangent to the surface a distance of 1 mile then you can form a right angled triangle as in the diagram.
Looking over a distance of 1 mile, we can use the theorem of Pythagoras:
a2 = 3,9632 + 12 = 15,705,370
and when we square root that figure we get a = 3,963.000126 miles
Thus your position is 3,963.000126 - 3,963 = 0.000126 miles above the surface of the earth.
0.000126 miles = 12 in * 5,280 ft * 0.000126 mi = 7.98 inches
Hence after one mile the earth drops approximately 8 inches.
"Whenever I have doubts about the shape of the earth I simply walk outside my home, down to the beach, and perform this simple test. Provided that there is no fog and the day is clear and calm, the same result comes up over and over throughout the year."
—Tom Bishop
Ergo, looking across 23 miles the Pythagorean theorem becomes:
a2 = 39632 +232 = 15,705,898
and when we square root that figure we get a = 3,963.06674 miles
Thus your position is 3,963.06674 - 3,963 = 0.06674 miles above the surface of the earth
0.06674 miles = 5,280 ft/mi * 0.6674 mi = 352.3872 feet
Hence after 23 miles the earth drops approximately 352 feet.
There are a number of different methods to calculate the drop of the Round Earth. Go ahead and look a few up to try out. You will find that the drop while looking over 23 miles is on the order of 300-400 feet.
Comments
me on the second paragraph