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Following upon of the question Throwing a ball on a rotating space station, I'm interested in understanding how to actually calculate the result of the Coriolis effect in a particular situation. I've seen the equation defined as:

$a_{cor} = -2\omega v$

But I'm really not sure how to solve it, or what the result would mean practically.

So, when I wake up in the morning on the inside of a spinning cylinder with a radius of 3.2km and an angular velocity of 0.52rpm (to simulate ~1g of gravity), and I pour my first cup of coffee, how much is the stream deflected?

Then, when I take the space taxi to the spinning Bernal sphere next door, with a radius of 0.25km and angular velocity of 1.9rpm (to simulate ~1g of gravity in the valley), when I pour my second cup of coffee, how much is the stream deflected?

Then, when I hop onto the Discovery for my journey to Jupiter, as I pour my third cup of coffee in the spinning habitat, with a radius of 30m and an angular velocity of 5.5rpm (again, simulating ~1g of gravity), how much is the stream deflected?

We can probably simplify the coffee bit to assume a spherical 1 gram droplet dropped 20cm, for a foreshortened, but dramatic, pour. :)

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    $\begingroup$ This is a problem I give my mechanics students on a regular basis, and you've written down what you need to know to at least get a first approximation. What's stopping from simply plugging in a couple of values and computing? $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Aug 17, 2017 at 20:43
  • $\begingroup$ @dmckee High school physics was a long time ago. :) Angular velocity I have, but I'm not sure what to make of the velocity vector there. Where does that come from? And which direction is the acceleration? Antispinward, I guess. $\endgroup$
    – David
    Commented Aug 17, 2017 at 20:47
  • $\begingroup$ $\vec{v}$ is the velocity of the affected object in the rotating frame. And you get the direction from the cross-product $\vec{a}_\text{cor} = -2 \vec{\omega} \times \vec{v}$ notice that the version you wrote assumes that the angular velocity and object velocity are perpendicular—but they are in the pouring problem. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Aug 17, 2017 at 20:51
  • $\begingroup$ See, here's where I start to get lost, and start needing Physics SE. :) I've been relying on online calculators to get this far. Is that $\vec{v}$ equivalent to the tangential velocity here? I don't know what other velocity we'd be talking about, but tangential velocity isn't perpendicular to angular velocity... is it? And I'm afraid I don't know how to figure a cross product. I would fail your mechanics class quite completely, I'm afraid. $\endgroup$
    – David
    Commented Aug 17, 2017 at 21:25
  • $\begingroup$ The $\vec{v}$ there is the one that a person standing in the habitat measures. If you're talking about pouring coffee it is the velocity of the coffee according to you. So it is approximately 'down' (i.e. radially outward from your current location) and has a magnitude that starts near zero as the coffee flows over the lip of the spout and increases as it falls. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Aug 17, 2017 at 21:28

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In another forum I've been told that the angular momentum and velocity doesn't actually matter for figuring the deflection, and it actually breaks down to a trigonometry problem involving ratios of radii. In a faster or slower rotation, everything just happens faster or slower. Because I'm a software engineer and not a physicist, please pardon the Python:

In [1]: import math
   ...:
   ...: def measure_coffee_deflection(dropped_from, landed):
   ...:     ratio_x = dropped_from / landed
   ...:     print("Ratio", ratio_x)
   ...:
   ...:     trajectory = math.acos(ratio_x)
   ...:     print('Trajectory', trajectory, 'radians')
   ...:
   ...:     habitat_rotation = math.sqrt(1./(ratio_x**2) - 1.) # radians
   ...:     print('Habitat rotation', habitat_rotation, 'radians')
   ...:     effective_rotation = habitat_rotation - trajectory # radians
   ...:     print('Effective rotation', effective_rotation, 'radians')
   ...:
   ...:     return effective_rotation * landed
   ...:

In [2]: # On the O'Neill Cylinder, radius 3.2km
   ...: radius = 320000 #cm
   ...: dropped_from = radius - 10 #cm
   ...: landed = radius
   ...:
   ...: print('Deflection', measure_coffee_deflection(dropped_from, landed), 'cm')
   ...:
Ratio 0.99996875
Trajectory 0.007905714738315722 radians
Habitat rotation 0.007905879445677281 radians
Effective rotation 1.647073615586303e-07 radians
Deflection 0.0527063556987617 cm

In [3]: # On the Bernal Sphere, radius 0.25km
   ...: radius = 25000 #cm
   ...: dropped_from = radius - 10 #cm
   ...: landed = radius
   ...: print('Deflection', measure_coffee_deflection(dropped_from, landed), 'cm')
   ...:
Ratio 0.9996
Trajectory 0.028285214141364843 radians
Habitat rotation 0.028292759782811733 radians
Effective rotation 7.5456414468898225e-06 radians
Deflection 0.18864103617224556 cm

In [4]: # On the Discovery, radius 30m
   ...: radius = 3000 #cm
   ...: dropped_from = radius - 10 #cm
   ...: landed = radius
   ...:
   ...: print('Deflection', measure_coffee_deflection(dropped_from, landed), 'cm')
   ...:
Ratio 0.9966666666666667
Trajectory 0.08167235558059345 radians
Habitat rotation 0.08185443645833014 radians
Effective rotation 0.00018208087773669002 radians
Deflection 0.54624263321007 cm

So:

  • on the O'Neill cylinder, the coffee deflects about half a millimeter.

  • on the Bernal sphere, just shy of 3mm

  • on the Discovery, about half a centimeter

Looks like I can continue to pour my coffee with confidence, no matter the rotating frame of reference.

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    $\begingroup$ The claim you've been given that there is no dependence on angular velocity is only valid for certain cases. In particular, the landing location of an object dropped from rest (in the rotating frame) is independent of $\omega$ because both the downward and anti-spinward accelerations are linearly dependent on $\omega$ and there is zero initial velocity. Give the object an initial velocity (in the rotating frame), and the dependence kicks in. Furthermore, you can only make a linear approximation of the effect for small deflections (which is generally the case for pouring a drink). $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Aug 19, 2017 at 19:48
  • $\begingroup$ That makes perfect sense; adding in other vectors would certainly complicate things. Thankfully, for my purposes, this approximation is enough. $\endgroup$
    – David
    Commented Aug 20, 2017 at 19:19

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