On the truly massive designs like the O'Neill Island 3, we're going to want an elevator for hauling both people and freight from axis to rim and back. If I understand it correctly, as you ascend, you'll feel a sideways force pushing you... spinward, I think? Because as you get nearer to the axis, your angular velocity is slowing. And if you take the elevator down, you'll feel the opposite, because your angular velocity is increasing.

You only experience the acceleration if you're moving up or down; as the elevator slows down, the sideways force will decrease, and if you stop along the way, it'll cease as well. And I think the amount of acceleration you experience is related somehow to the velocity of the elevator, right?

I'm interested in the scale of this effect: what's it actually like to go up or down in an elevator on one of these contraptions? How do I figure the acceleration?

So inside our Island 3 cylinder with a radius of 3.2km and an angular velocity of 0.52rpm, let's say we've hired the engineers from the Shanghai Tower to build ourselves an express elevator that tops out at 18 m/s (~40 mph) for a 6-7 minute trip from rim to axis or vice versa. (We'll assume good pressure control so we don't blow out anybody's eardrums on the way down.) Once the lift is up to speed (at 1.5m/s^2, 12 seconds or so), what sort of sideways acceleration will a passenger on the elevator experience, and what would that feel like?

  • $\begingroup$ There is a brief description here: en.m.wikipedia.org/wiki/Space_elevator $\endgroup$
    – user167453
    Sep 30 '17 at 16:12
  • $\begingroup$ Is there some reason you aren't simply computing the Coriolis acceleration from your givens? I mean, this is a simply plug-n-chug from here so why ask us? $\endgroup$ Sep 30 '17 at 18:56
  • $\begingroup$ @dmckee Mainly because I'm not sure exactly how. :) I believe we established last time that I would fail your mechanics class. I guess this is our mutual friend $a_{cor} = -2 \omega v$ again? I don't really remember my vector math from high school physics. What sign does 18 m/s need to be to plug in to $v$? And can 0.52 rpm plug directly into $\omega$? Surely I need to convert it. $\endgroup$
    – David
    Sep 30 '17 at 22:18
  • $\begingroup$ And I just now figured out that the equation uses a cross product, and I'm lost. I could perhaps blindly "plug-n-chug" by pulling things out of the nether, but I certainly wouldn't trust the output. Isn't this Physics Stack Exchange, where one asks questions about physics when one is out of one's depth? $\endgroup$
    – David
    Oct 3 '17 at 14:10

After getting some help on my vector math, I think I've got the answer. Please pardon the Python:

import math

def cross_product(a, b):
    c = [a[1]*b[2] - a[2]*b[1],
         a[2]*b[0] - a[0]*b[2],
         a[0]*b[1] - a[1]*b[0]]

    return c

def get_coriolis_acceleration(angular_v_vec_rad_p_s, v_vec_m_p_s):
    cross = cross_product(angular_v_vec_rad_p_s, v_vec_m_p_s)
    return [-2 * i for i in cross]

angular_velocity_rpm = 0.52 # rpm
angular_velocity_rad_p_s = ((angular_velocity_rpm * 2 * math.pi) / 60., 0., 0.)
velocity_m_p_s = (0., 18., 0.) # m/s

coriolis_acceleration_m_p_s_2 = get_coriolis_acceleration(angular_velocity_rad_p_s, velocity_m_p_s)
coriolis_acceleration_gees = [i / 9.80665 for i in coriolis_acceleration_m_p_s_2]

print(coriolis_acceleration_m_p_s_2, 'm/s^2')
print(coriolis_acceleration_gees, 'g')

And the result:

[-0.0, -0.0, -1.960353815840031] m/s^2
[-0.0, -0.0, -0.19990045691852276] g

So .2g is going to be a noticeable effect as we ascend, especially as centripetal acceleration tapers off.

If we halve the velocity to 9 m/s, the coriolis acceleration halves to 0.1g.

If we increase the angular velocity, say to the 1.9 rpm needed to provide the 1g on a Bernal Sphere, our 18 m/s elevator starts feeling like an amusement park ride, with 0.73g of coriolis acceleration.


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