coriolis bicycle question

For extra marks, I'll include the equation for the acceleration.

a(coriolis) = -2 w x v

where v is the velocity vector and w is the rotation vector (aligned
with the rotation axis, of length the rotation speed in radians per
second), x is the cross product.

For somebody on an x% gradient at a speed of v, going north south in a
habitat of radius r and an equatorial gravity of g

|w| = SQRT(g/r)

The vertical component of the speed is x*v

so stripping away all the vector stuff the sideways acceleration is 2
SQRT(g/r)*(x*v)

g = 10 (the engineer decided not to match the earth precisely!)
r = 1000
x = 0.333 (1 in 3)
v = 50 km/h (31 mph, 13.9 m/s... wheee!)

acoriolis = 2*SQRT(10/1000)*(0.333*13.9) = 0.92 m/s^2

Now the vertical g varies proportional to distance from the axis, but
at the bottom of the hill that's an angle of ARCTAN(0.92/10) = 5.3
degrees not as much as I expected.

I *think* I did it right. It's going to be more at the top of the hill
because the g is lower. It's also going to be more significant in
smaller habitats.

--
-Ian Woollard

We live in an imperfectly imperfect world. If we lived in a perfectly
imperfect world things would be a lot better.