From: Adrian Hurt <adrian@cee.heriot-watt.ac.uk>
Subject: Re: Anti-aircraft
Organization: Dept of Computing & Electrical Engineering, Heriot-Watt 
Date: Mon, 25 Jan 1993 17:26:59 GMT


From Adrian Hurt <adrian@cee.heriot-watt.ac.uk>

In article <C143J8.2wt@law7.DaytonOH.NCR.COM> creps@silver.ucs.indiana.edu (Stephen A. Creps) writes:
>In article <C124wI.7op@law7.DaytonOH.NCR.COM> "John M. Wu" <johnwu@netcom.com> writes:
>>1)  When anti-aircraft shells go up, they have to eventually come
>>back down.  With a gun shooting 90 deg., the bullet will go up to
>>around 9000 ft. and then start coming down up to terminal
>>velocity (300 fps?).

"Coming _down up_ to ..." wonderful phrasing! :-)  (Yes, I do know the
meaning that was intended, but I had to chuckle.)

>>	Do anti-aircraft weapons like those in Iraq cause large
>>ground damage and numerous casualties due to what eventually
>>happens to fired shells (eg. the Baghdad hotel damage)?

If the shells don't have time fuses to make them explode while in the air,
probably.  If they do have such fuses, the shell splinters will still come
down on the city, but at lower terminal velocity and minus the explosive
warhead.

>>Wouldn't not firing anything cause less damage to Baghdad?

Yes, but not just for the above reason.  What I heard on TV was that the U.S.
admitted it was a cruise missile which hit the hotel, after the missile had
been hit and knocked out of control by AAA.  Not firing means that all the
missiles go where they are intended; firing AAA means that most of the missiles
go where they are intended, and a few go somewhere else.  Of course, if that
"somewhere else" happens to be a civilian building, then Iraq gets some free
anti-U.S. propaganda (well, not free, there's the cost of the shells and the
cost of the building, neither of which bother Iraq much).

>   Hmm, this is kind of an old topic here, but I might be able to add
>something that I don't remember seeing here before.  Terminal velocity
>does in fact come into play when shooting into the air, and it is in
>fact a way that much of a projectile's energy can be lost.  I would
>assume that the angle at which the most energy is lost is 90 degrees.
>
>   However, while terminal velocity affects the Y-vector (vertical) of
>an object's velocity, it should not affect its X-vector (horizontal).

Time for me to get a lesson in physics by posting what I think, and seeing
someone else correct me.  (Without flames, please. :-)

Terminal velocity is reached when the force of gravity is cancelled out by
the force of air resistance, which means that the Y-vector is not being
reduced by air resistance any more.  The X-vector doesn't have the help of 
gravity, and will continue to be reduced until it reaches zero, either by
air resistance or by collision with something harder.  It is air resistance
which takes away the projectile's kinetic energy - terminal velocity is when
the projectile stops losing kinetic energy, at least from that part of its
kinetic energy given by its vertical velocity.

>   Therefore, is it valid to assume that this kind of damage is not
>caused so much by the projectile _coming down_, but by it _continuing
>in the same horizontal direction_, where terminal velocity does not
>come into play?

Therefore, damage will be caused by whatever energy is left in the projectile
when it hits something.  That may be chemical energy, i.e. an unexploded
warhead, or it may be kinetic energy due to remaining horizontal velocity,
or it may be kinetic energy due to vertical terminal velocity.  (Those are
not exclusive "or"s - combinations are allowed.)  Given that the projectile
in this case is AAA, which was probably fired at a high angle, most of that
kinetic energy will be from the vertical velocity.  If it exploded while in
the air, the projectiles which hit the ground will be the shell's splinters,
whose high drag will result in a low vertical terminal velocity and very little
remaining horizontal velocity.

-- 
 "Keyboard?  How quaint!" - M. Scott

 Adrian Hurt			     |	JANET:  adrian@uk.ac.hw.cee
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