Alternate Jump Drive: Request for Comment Jeff Zeitlin (26 Nov 2016 23:58 UTC)
Re: [TML] Alternate Jump Drive: Request for Comment Rob Davenport (27 Nov 2016 03:26 UTC)
Re: [TML] Alternate Jump Drive: Request for Comment Jeff Zeitlin (27 Nov 2016 17:53 UTC)
Re: [TML] Alternate Jump Drive: Request for Comment Greg Nokes (27 Nov 2016 22:10 UTC)
Re: [TML] Alternate Jump Drive: Request for Comment Jeff Zeitlin (01 Dec 2016 01:18 UTC)
Re: [TML] Alternate Jump Drive: Request for Comment Jeff Zeitlin (01 Dec 2016 01:14 UTC)
Re: [TML] Alternate Jump Drive: Request for Comment Rob Davenport (30 Nov 2016 00:36 UTC)
Re: [TML] Alternate Jump Drive: Request for Comment Jeff Zeitlin (01 Dec 2016 00:32 UTC)
Re: [TML] Alternate Jump Drive: Request for Comment Rob Davenport (01 Dec 2016 03:20 UTC)
Re: [TML] Alternate Jump Drive: Request for Comment shadow@xxxxxx (29 Nov 2016 18:49 UTC)
Re: [TML] Alternate Jump Drive: Request for Comment Jeff Zeitlin (01 Dec 2016 00:41 UTC)

Alternate Jump Drive: Request for Comment Jeff Zeitlin 26 Nov 2016 23:58 UTC

Depending on the reaction here, I may work this up into a full
alternate-tech article for Freelance Traveller, much like I did with
the Lyman Drive
(http://www.freelancetraveller.com/features/rules/tech/lymanjd.html,
reprinted from 2002 in the April 2013 issue).

The core of the idea: A Jump-n drive can (usually) do Jump-m (m<n) in
less time than it takes to do a Jump-n. However, it takes more fuel to
do so than it would take a Jump-m drive, and the time required to do
multiple jumps, even if there is zero delay between emergence from one
jump and entry into the next, is greater than the single jump would
be.

A first cut at numbers:

Assume that N is the rated jump capability of the drive, and that it
uses fuel according to the standard rules (e.g., for CT, 10% of hull
rate * N). Then, for a Jump M where M is strictly less than N,...

   Fuel usage = M/(N-1) times the standard fuel usage.

Time: 168 hours is annoying to work with - it just happens to match up
nicely with "a week". Let's tweak that: A basic jump isn't "a week",
though it's generally treated as such on the calendar; it's 150 hours.
Given that Jump N takes 150 hours, for a Jump M, where M is strictly
less than N,...

   Time required for jump = M/(N-1) times 200 hours.

In tabular form:

Fuel (% of hull rate)

Drive     Jump 6   Jump 5   Jump 4   Jump 3   Jump 2   Jump 1
J6 Drive  60       60       48       36       24       12
J5 Drive  --       50       50       38       25       13
J4 Drive  --       --       40       40       27       13
J3 Drive  --       --       --       30       30       15
J2 Drive  --       --       --       --       20       20
J1 Drive  --       --       --       --       --       10

Time (hours)

Drive     Jump 6   Jump 5   Jump 4   Jump 3   Jump 2   Jump 1
J6 Drive  150      200      160      120       80       40
J5 Drive  ---      150      200      150      100       50
J4 Drive  ---      ---      150      200      133       67
J3 Drive  ---      ---      ---      150      200      100
J2 Drive  ---      ---      ---      ---      150      200
J1 Drive  ---      ---      ---      ---      ---      150

Note that using a Jump N drive for Jump N-1 is really not something
you want to do unless it's unavoidable; it takes *longer* without
saving you any fuel. Also, the drive that uses the _least_ fuel for
jump N is the Jump N drive - but at the cost of being the
second-slowest.

Comments/Discussion?

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Jeff Zeitlin, Editor
Freelance Traveller
    The Electronic Fan-Supported Traveller® Resource
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