4 ms·

Conveniently done already in Louisiana, where mankind has placed a line of concrete pillars twenty three miles across a lake. :) https://en.wikipedia.org/wiki/

Conveniently done already in Louisiana, where mankind has placed a line of concrete pillars twenty three miles across a lake. :) https://en.wikipedia.org/wiki/Lake_Pontchartrain_Causeway#/media/File:Lake_Pontchartrain_Causeway_From_Three_Lakeway_Center.png https://en.wikipedia.org/wiki/Lake_Pontchartrain_Causeway#/m...

deleted 4y ago
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burnte 4y ago
I've driven on that one, and the bridges across the Chesapeake Bay, they're fantastic experiences, personally.
gowld 4y ago
I don't understand how that image (And this simulation[1]) is caused by curvature. 23 miles is ~0.001 of the Earth's circumference. How can that cause a visible difference in tower height? Even more so, the visible difference is in the last few miles, even smaller fraction. cos(0.001*2pi) = 0.999980 [1] http://walter.bislins.ch/bloge/index.asp?page=Finding+the+Curvature+of+the+Earth%3A+Stand%2DAlone+App&demo=TrnsmLine#App http://walter.bislins.ch/bloge/index.asp?page=Finding+the+Cu...
- modo_mario 4y ago
  >23 miles is ~0.001 of the Earth's circumference. How can that cause a visible difference in tower height? It seems fairly intuitive that a tiny amount of an insane amount can still be a lot no? I don't want to get too deep in the math right now since it's morning where i live nor am I good at it.... but if we'd just assume the earth was a nice sphere and a quick google was correct about a 12742 diameter then every degree down from the northpole to the equater would be on average a 70.79km difference in "height". Take a thousand of that and you're still dealing with an easily perceptible >7 meter difference. Now all of this is shoddy quick thinking since the earth is not a nice sphere, this heigh difference is better not measured perpendicular to the equator but to the ground or center of the earth, etc but you get the idea i hope.

deleted 4y ago

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burnte 4y ago

I've driven on that one, and the bridges across the Chesapeake Bay, they're fantastic experiences, personally.

gowld 4y ago

I don't understand how that image (And this simulation[1]) is caused by curvature. 23 miles is ~0.001 of the Earth's circumference. How can that cause a visible difference in tower height? Even more so, the visible difference is in the last few miles, even smaller fraction. cos(0.001*2pi) = 0.999980 [1] http://walter.bislins.ch/bloge/index.asp?page=Finding+the+Curvature+of+the+Earth%3A+Stand%2DAlone+App&demo=TrnsmLine#App http://walter.bislins.ch/bloge/index.asp?page=Finding+the+Cu...

modo_mario 4y ago
>23 miles is ~0.001 of the Earth's circumference. How can that cause a visible difference in tower height? It seems fairly intuitive that a tiny amount of an insane amount can still be a lot no? I don't want to get too deep in the math right now since it's morning where i live nor am I good at it.... but if we'd just assume the earth was a nice sphere and a quick google was correct about a 12742 diameter then every degree down from the northpole to the equater would be on average a 70.79km difference in "height". Take a thousand of that and you're still dealing with an easily perceptible >7 meter difference. Now all of this is shoddy quick thinking since the earth is not a nice sphere, this heigh difference is better not measured perpendicular to the equator but to the ground or center of the earth, etc but you get the idea i hope.