Richardatnos
25th Jan 2006, 12:54
For A320, somebody says that during climb phase when CI=0, it gets to the best rate climb, is that correct? and why?
Many thanks
enicalyth
25th Jan 2006, 13:40
In general the lower the CI the steeper the climb path, the shorter the climb distance and the nearer the ToD basically because the speed is lower.
Cost index is the ratio of time-related costs to fuel-related costs. In climb it is fuel related costs that are going to predominate and when CI is 0 you are going for minimum fuel, maximum range.
Up to you to say what you mean by best climb but there is not that much in it (joke! It is an Airbus)
(usual caveats to FL330 at 75T MTOW ISA).
CI Fuel kg Time mins Dist nm Final RoC fpm
0 1760 22 150 580
20 1840 23 160 570
40 1900 24 165 550
60 1980 25 175 500
Be kind to donkeys. Be different.
Old Smokey
26th Jan 2006, 02:55
If Boeing and Airbus philosophy are the same here (I suspect that they are) that CI = 0 = Minimum fuel for the sector, then the scheduled speed for climb will be slightly above that for the best rate, and, for practical purposes, the best rate.
Best Rate climb speed in isolation offers good fuel use per mile, but by surrendering just a little bit of Rate of climb by flying at a higher speed, more ground miles are covered for almost the same fuel burned during the climb, thus eliminating the short incremental cruise that would otherwise have been needed if climb was at exactly the best rate, thus decreasing sector fuel used. (You lose a little of one, you gain a lot of the other).
CI = 0 on descent is a different matter altogether, where the scheduled speed is quite close to Vmd, a much lower than normal descent speed. The object here is to minimise rate of descent whilst still at a 'respectable' speed, and making the most of the very low fuel flows at idle thrust.
Regards,
Old Smokey (Who now types double space because the new format eliminates the single space):rolleyes: