Tie a rope around the Earth's equator. Now untie it, and add one meter of rope. Tie again. How loose will it be? In other words: what will be the distance between the ground and this rope going around the Earth? Most people will either answer "close to 0" or (a few) "one meter". Do the math, and the answer may surprise you.
^mind was blown
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I think you stated the problem backwards, because the actual answer is 1/2pi meters which is pretty close to 0 compared to 1 meter. My mind was not blown.
D=C/pi
D2-D1 = C2/pi - C1/pi = (C2-C1)/pi = (C1+1+C2)/pi) = 1/pi = ~0.24
so about 12cm?
Not sure what the point of this was, unless it's your homework.
less than 1mm
It will still be on the ground because the mass of the rope falling into depressions as it goes over land will pull it taught.
Ce= re*2pi; (Ce + 1)/2pi = r1; (re*2pi + 1)/2pi = r1; re - (re*2pi + 1)/2pi = d; re - re + 1/2pi =d; d =1/(2pi) = 0.16 m
>1/pi = ~0.24
Almost 16 cm
Funny thing tho, it doesn't matter
if you do the wrap around a tennis ball
or the world, the numbers are the same:
a 15.92 cm increase of radius will result in the
rope lengthening by 1 m.
I did this in primary school when we learned bout circles for the first time.
A more interesting one I calculated was
>Tie a rope around the Earth's equator. Now untie it, and add one meter of rope. Tie again. pull the rope to a peak as in the pic, how high is the peak?
about one dick
Thats very cool
obviously
>2pi(r+dr) = 2pir + 2pidr
WOW SO AMAZING!
The farthest I can get is:
1=2r*acos(r/(r+h))+2sqrt(2r^2+rh+h^2)
How do you solve an equation like that?
>pidr
cyka blyat
Russians, Russians everywhere
Alright I actually messed up that equation a bit. There should be no r^2 term inside the radical. And I figured that the angle would be small enough to use paraxial approximation, so the acos term is also 0. I get a height of approximately 39 mm, amazing.
c=2r π
Δc=1
Δr=1/(2π)=0.1592
op pic looks like boobies tee hee
The answer is -1/12
looks like boobs
literraly this u fagot
Assuming the earth is a perfect sphere (it isn't)
the answer would be:
(22/7d + 22/7) - 22/7d
measure diameter in meters and provide answer in meters, this is the distance between the proportionately stretched out rope and the fictional perfect circumference of the earths entirely smooth equator.