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# The fastest Travel from A to B

Travelling with normal distributed velocity?

v=(1/sqrt(2pi))e^(-.5*t^2)?

easier

v(t)=vmax*e^(-t^2)

Differentiation gives the acceleration curve?

Integration of the velocity bell curve gives the total distance?

from -T/2 to T/2 integrate vmaxe^-(4t/T)^2

brings us in time T along a distance of the integral by reaching a peak velocity of v in the middle of the distance.

from -144 to 144 integrate 75*e^-(t/72)^2 dt

for example brings us in 288 seconds along a distance of the integral about 10 km by reaching a peak velocity of 75 m/s = 270 km/h in the middle of the distance.

What now must be the highest velocity vmax for travelling a given distance in a distinct time?

What is the peak acceleration for this travel and which acceleration is bearable for all humans?

In a tube with diameter 18 y we could pack 2 to 7 tubes with 6 m diameter, which can be used for the railway and or other purposes. At least the middle tube could be used as emergency access and shelter.