As the ring turns

You might notice that I disassociated the Ring Of Iron from the rest of Paul Birch. Today I've been going over some of Birch' maths. As of now I've got I.2.1 and the second appendix so, you know, not all that much.

Birch was writing in 1982 and not only underestimated the temperature of his superconductor, but insisted on Kevlar for his rope. I hasten to butt in: he had a gut understanding of "unobtanium" insofar as it showed what advances we might get, in the next fifty years; which - in part - we did get, in five.

Here in 2021 tough men use T1100G (pdf); I'll go 1991, with Zylon. For Table 1, that's tensile (Y) 5.8 against density (ρ) 1.7, thus raising his Y/gρ by a factor of over 1.5. Given Martian g, that third column is a whopping 917 km. Fourth column, which "1/P" is the mass of rope needed per unit of freight (at 110 km), is 1.12. I like it!

I didn't understand section I.2.3 "system throughput". I got as far that on Mars, especially coming up from Mars to a (likely) Earth or Venus voyage, we don't mind having total g of 8.8 ms^-2, which is the g they got in Venus' clouds. Acceleration is then 5 ms^-2, so equation 13's time = √(2H/a) = 210 seconds from bottom to top - not counting deceleration at the top. Top to bottom takes a different amount of time because that'll be an initial freefall followed by a deceleration mostly aiming for Martian 3.7 ms^-2 toward the end. Birch didn't apply to become Mars' elevator-maintenance schlub so I shan't either.

Skipping to the second appendix, we got Coriolis. If you drop something at the top of the 110 km ladder, it falls to the east of the ladder's bottom. Ω/3 x √(2H³/g), plugging in equatorial Ω = 2π/(the Martian day in seconds). 633 meters east.