I just did this and computed a 4.6 ft drop over 60 mi. Seems pretty reasonable to me. San Jose is 50 miles from San Francisco, so the drop would be slightly less than 4.6 ft.

( My precision is probably pretty low, because I used my phone calculator without any paper, and I used the circumference of the Earth rounded to the closest mile. Rough equation: drop = R*(1-cos(2pi60/24901)), where R is the earth's radius, and 24901 is the earth's circumference in miles. 2 * pi * 60 / 24901 is the angle of a 60 mile arc (in radians); cosine of that angle gives adjacent over hypotenuse, so multiplying that by the Earth's radius gives us the distance from the center of the earth up to the point adjacent with the object 60 miles away; subtracting this number from the Earth's radius gives us the total drop.)

I just did this and computed a 4.6 ft drop over 60 mi. Seems pretty reasonable to me. San Jose is 50 miles from San Francisco, so the drop would be slightly less than 4.6 ft.

( My precision is probably pretty low, because I used my phone calculator without any paper, and I used the circumference of the Earth rounded to the closest mile. Rough equation: drop = R*(1-cos(2pi60/24901)), where R is the earth's radius, and 24901 is the earth's circumference in miles. 2 * pi * 60 / 24901 is the angle of a 60 mile arc; cosine of that angle gives adjacent over hypotenuse, so multiplying that by the Earth's radius gives us the distance from the center of the earth up to the point adjacent with the object 60 miles away; subtracting this number from the Earth's radius gives us the total drop.)

I just did this and computed a 4.6 ft drop over 60 mi. Seems pretty reasonable to me. San Jose is 50 miles from San Francisco, so the drop would be slightly less than 4.6 ft.

( My precision is probably pretty low, because I used my phone calculator without any paper, and I used the circumference of the Earth rounded to the closest mile. Rough equation: drop = R*(1-cos(2pi60/24901)), where R is the earth's radius, and 24901 is the earth's circumference in miles.)

I just did this and computed a 4.6 ft drop over 60 mi. Seems pretty reasonable to me. San Jose is 50 miles from San Francisco, so the drop would be slightly less than 4.6 ft.

( My precision is probably pretty low, because I use my phone calculator without any paper, and I used the circumference of the Earth rounded to the closest mile.)

I just did this and computed a 4.6 ft drop over 60 mi. Seems pretty reasonable to me. San Jose is 50 miles from San Francisco, so the drop would be slightly less than 4.6 ft.

I just did this and computed a 4.6 ft drop over 60 mi. Seems pretty reasonable to me.

I just did this and computed a 4.6 ft drop over 60 mi. Seems pretty reasonable to me. And in all honesty, that's too large of a number. That is the distance below you that is parallel with the point 60 miles away. The actual drop would be slightly different.