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Old March 28th 06, 08:05 PM posted to rec.aviation.military,rec.aviation.military.naval,sci.military.naval
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Default The F14 vs what we are doing now


"B" wrote in message ...
Kevin Brooks wrote:
"B" wrote in message ...
Fred J. McCall wrote:
"Andrew Chaplin" wrote:

:"Harry Andreas" wrote in message
...
:
: Damn right.
:
: SLAMRAAM is a box launched AIM-120.
: As such, it has less range and a smaller engagement envelope than
: the airborne version, but still a bugger to go up against.
:
:Why the loss of performance?

Because you're not starting with a 600+ knot velocity when you light
the motor like you are when you launch from an aircraft and because
what you're shooting at is 'up' from where you're shooting from.

Ye kenna violate the laws o' physics, Captain! :-)

Altitude is probably more important.


I doubt that. Altitude alone yields increased range only in a look
down/shoot down scenario (yeah, decreased drag alone due to increased
altitude will have some effect regardless of the aspect of engagement,
but it will not be significant); OTOH, adding that 600 knots to the
missile at launch is imparting a heck of a lot of energy--you know, that
whole vee-squared part of the kinetic energy equation?

Brooks



Wait until you see this (check my figures please).

AIM-120 (150kg) at 10,000m
Ug=150kg x 10km x g
=150,0000 x 10,000 x 9.8
=14.7x10^9 J

AIM-120 at 600knots (300m/s)
=150,000 x 300^2
=13x10^9 J

Much closer than I thought. Of course it's a complex problem but the raw
figures are interesting.


A bigger difference than I would have thought, but a key factor to
consider--the land based system is not intended to be going after high
flyers; that would probably remain the territory for the Patriots (which do
habitually see a number of batteries get sent forwards into the
division-level sectors, they don't all remain back at corps and theater
level). As described, this is supposed to be an anti-helo, anti-UAV, and
anti-cruise missile system, so again, the altitude issue is probably not as
great as one would think. And shouldn't that KE equation use *one-half the
mass* times velocity squared? Which would make the result half of what you
calculated?

Brooks