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"The answer to the problem of information overload on the Net is reputations… engineer a system called a reputation server."
- Neal Stephenson
 |
| Skew-Flip Turnover | ||
| Used halfway to the destination by torch ships to slow for one's destination. |
Heinlein is the first one to describe it as a "skew-flip", but he's not the first one to describe the maneuver.
| “But right now ! think we had better hang on. We ought to be at half way point in a few minutes — and a skew-flip is disconcerting even if you are strapped down.”
I had read about skew-flip turnovers, but only as a theoretical maneuver; I had never heard of a ship that could do one. If this was a ship. The floor felt as solid as concrete and as motionless. “I don’t see anything to hang on to.” |
| Technovelgy from Have Space Suit - Will Travel,
by Robert Heinlein. Published by Scribners in 1958 Additional resources -
|
The novel was published in pieces in F & SF; here's a bit more from the second installment:
Was Pluto in opposition? Or quadrature? Or conjunction? Nobody looks at Pluto — so why remember where it is on the ecliptic? Oh, well, the average distance was 30 A.U.s — that would give a close-enough answer.
Half that distance, in feet, is: 1/2 × 30 × 93,000,000 × 5280. Eight gravities is: 8 × 32.2 ft./sec./sec. — speed increases by 258 feet per second every second up to skew-flip and decreases just as fast thereafter.
So — 1/2 × 30 × 93,000,000 × 5280 = 1/2 × 8 × 32.2 x t2 — and you wind up with the time for half the trip, in seconds. Double that for full trip. Divide by 3600 to get hours; divide by 24 and you have days. On a slide rule such a problem takes forty seconds, most of it to get your decimal point correct. It's as easy as computing sales tax.
Compare to negative acceleration from Skylark of Space (1928) by E.E. 'Doc' Smith and turnover from Off the Beam (1944) by George O. Smith.
It's a two-piece problem; accelerate for half time (and half distance); do a skew-flip and decelerate the other half time (and distance). You can't use the whole distance in the equation, as "time" appears as a square — it's a parabolic.
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