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From:
gherbert@gw.retro.com (George William Herbert)
Newsgroups: sci.military.moderated
Subject: Re: Advanced tank main guns
Date: 26 Nov 2001 22:12:47 -0800
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Bob Keeter wrote:
>George William Herbert wrote:
>> Bob Keeter wrote:
>>> [...]
>>> The bigger bore gives you more "push". Lets assume that around 60,000 psi
>>> is the "top end" for chamber pressure. Make that pressure applied over a
>>> greater surface area (i.e. the back end of a sabot) and there are more
>>> pounds of force exerted on a 6 inch dia sabot than on a 5 inch dia sabot.
>>> take the SAME penetrator and give it more push, and it comes out the front
>>> end faster! Take a HEAVIER penetrator and you might be getting the same
>>> speed. Either way, its more energy when that penetrator runs into
>>> something!
>>
>> That's oversimplified and incorrect.
>> Two factors are working against you.
>> One, cannons don't have an infinite supply of 60,000 PSI gas to
>> work with: they have a fixed supply (PV = nRT) produced over a
>> variable but relatively short time period. If we simplify and
>> assume all the propellant were burned immediately, the work done
>> on the projectile and sabot (PV work) will be roughly the same no
>> matter what the bore is, for the same initial propellant load.
>> You can't get 2 times the work out by doubling projectile base
>> area, because the gas expands roughly twice as fast and loses
>> energy roughly twice as fast. That can be useful to shorten
>> the barrel, though. The work done is a function of barrel volume
>> and you can adjust the parameters to adjust the weapon envelope.
>
>Well, yeah I did simplyfy a tad. . . .not EVERYONE is into interior
>ballistics, but if we must. . . . . . . .
>
>I guess that I thought that the flame in the muzzle blast indicated that
>powder was still burning as the projectile was leaving the barrel. You want
>to minimize that of course but at the same time you want to USE as much of
>the barrel as efficiently as possible. . . .
The flame is because there's hot gas left inside which expands,
and will be luminous until it cools significantly. Unfortunately,
typical propellant chemical / projectile mechanical energy transfer
is 20-25%, so you get 75-80% of the powders energy in hot gases
which are too cool to effectively propel but still pretty hot,
which means big fireball.
>With modern progressive powders, and a known projectile rate, its fairly
>easy to "tune" the powder charge to take advantage of the entire barrel
>length (for example the powder charge in a short barreled 155 howitzer would
>need to be much quicker burning than the powder charge in a 155 field gun,
>assuming the same weight and "fit" of a projectile!) Bet the charge in an
>M-1s 120mm gun is still producing really close to its peak pressure when the
>sabot leaves the muzzle (afterall , speed is king with a penetrator!)
You would lose that bet. Pressure has already dropped noticably
after a meter's projectile travel, and significantly after 2 meters.
Look at the muzzle end of a 120mm gun; it's got about oh, 20mm
thick walls? Pc is around 65,000 PSI; if the chamber pressure
were still applied down there, that would be gun tube stress
of around 195,000 PSI. That's real close to the best ultimate
tensile strength of any known steel, and they don't make tank
gun tubes out of that maraging steel. And it's not just there;
the tubes are tapered, but not much. By the time the projectile
reaches the muzzle, the pressures way, way down.
See Ogorkiewicz "Technology of Tanks", Vol 1, pp 101.
He gives a chart of breech and projectile base pressures
for a 105mm gun over a projectile travel from 0 to 4.85m.
Unfortunately calibrated in MN/m^2 (multiply by 10 to get bar).
Peak projectile base pressure is around 275 MN/m^2 at travel
around 0.6 meters. Peak breech pressure is around 415 MN/m^2
at projectile travel of around 0.75 meter. By the time the
projectile has travelled 1 meter, projectile base pressure is
down to 250 MN/m^2 and breech pressure down to 375 MN/m^2.
At 2 meters, projectile base pressure is around 165 MN/m^2
and breech pressure is around 250 MN/m^2. At 3 meters,
it's around 115 MN/m^2 and 175 MN/m^2. At 4 meters, it's
about 115 MN/m^2 and 95 MN/m^2 respectively. At the muzzle,
4.85 meters it's about 100 MN/m^2 and 85 MN/m^2 respectively.
The peak pressure of around 4150 Bar is less than the most modern
120mm guns (which are 5000 to 6000 bar) but the curves will be similar.
>> The parameter of interest is "expansion ratio", ratio of total
>> bore+effective chamber volume to effective chamber volume
>> (Vb+Vceff)/Vceff
>> Effective chamber volume is the chamber volume minus
>> the casing volume, projectile volume intruding into the
>> "chamber" area, etc.
>
>Yep. We agree!
>
>> Two, the sabot mass will scale up as D^3, and sabot mass
>> fractions are already significant parasitic penetrator KE
>> losses... As an example, let's look at what happens to a
>> sample modern western DU round as you go from 120mm to 140mm
>> bore but use the same penetrator. Let's assume the penetrator
>> weighs 4.5 kilos and a modern 120mm carbon fiber sabot weighs
>> around 1.5 kilos. Going to 140mm with the same sabot tech
>> leads to a sabot weight gain of around 59%, to about 2.4 kilos.
>> We've already gained 0.9 kilogram going from 6 to 6.9 kilos for
>> total in-bore projectile.
>>
>> That suggests we need about 15% more propellant to just keep
>> KE the same, and then additional propellant beyond that will
>> be 65% utilized in accellerating the penetrator as opposed ot
>> 75% for the 120mm version.
>
>Checking my handy dandy Sovietskii Collection catalog, I see that a
>penetrator dart for a 120mm round (dont know what model) runs about 8 lbs
>so. . .4.5 kg is in the ballpark, will trust you on the sabot.
>
>Your calculation of the increase in powder charge is again about right I
>think.
>
>Of course if we assume that the "length" of the powder charge (from the base
>of the shell to the bottom of the sabot stays the same, (tank cartridges are
>already pretty long to handle inside a turret!) we have
>
>3.1416 X (120/2)^2 X the length (for the 120mm round) and
>3.1416 X (140/2)^2 X the length (for the 140mm round)
Ahh... ok. You're assuming that tank guns have straight cartridges.
That is unfortunately untrue. Tank guns have had various forms of
bottlenecked cartridge since WW 2. The more modern ones are quite
distinctly bottlenecked; the 120mm cartridge is nearly 200 mm diameter
in the cylindrical case body section. The rounds would be about 10
feet long if you had to have them straight.
Once you start bottlenecking cartridges, the chamber volume and
bore diameter are largely decoupled, though you encounter problems
with ridiculous throat ratios if you go too far.
>If I look at this long enough I think that the ratio of the possible powder
>charges would be something like 1.36 more powder by volume in the 140mm
>shell case. Now if I diddle around with the grain size and maybe the
>composition of the powder I can work it so that there is a bit over 1/3 more
>energy available, which wwill translate into almost exactly the same speed
>for the heavier projectile or a considerable increase in speed if the actual
>projectile were kept at the same size and weight.
>
>Guess Im still confused about where I missed the ball on this one?
Bottlenecked cartridges mostly. And even that is really a misnomer,
to some degree. Chamber volume and propellant loading factor are
two key ones. Basically, though, the most key ingredient is the
unfired round weight. It has to be light enough (20 kilos, roughly)
to be quickly unracked, lined up with the breech, and rammed
by the loader.
>> To really visualize all of what's going on, you have to work from
>> both ends at the same time: the mass and energy budget, and the
>> cannon physical dimentions, converging on optimal solutions.
>> Inputs into that include the projectile density for both KE
>> and HEAP rounds, the types and characteristics of propellant
>> available, etc. If you start tweaking dimentions in one axis
>> without considering the others then you end up with inefficient solutions.
>
>As I said, diddle a bit. . . . . . .
>
>> For example, the high velocity 105mm rifled tank guns used in the west
>> (L7, M68, etc) were developed from a British 83.8mm gun, using the
>> same case shape and volume, because of propellant evolution and
>> optimal KE projectile density had changed with the advent of
>> more advanced discarding sabot rounds.
>>
>> Ogorkiewicz's "Technology of Tanks" lists the following calculation
>> for first-order tank gun round performance estimation:
>> V0 = 1500 * (Mcharge/Mprojectile)^0.45 (eqn 4.8, pp 103)
>>
>> Note that bore diameter doesn't even factor in. That assumes
>> roughly optimal chamber conditions, expansion ratio, etc.
>
>I think that is a factor in there somewhere when the weight of the
>projectilie does not go up in cubic proportion to the bore! In the case of
>sabotted rounds, the weight of the projectile is relatively independent of
>the bore, so. . . . .put 45,000 psi (or whatever pressure you care to use)
>behind a 6 inch sabot with an 8-lb dart and put the same pressure behind a 5
>inch sabot with an eight pound dart, and which one will exit the muzzle
>fastest!?!?!?!
>
>Does my madness make any more sense now?
The weight of the sabot will be exactly a factor of D^3, so the
6 inch sabot will weigh 44% more. If the 5-inch Sabot was 25%
of the total round weight, that's a 2.66 pound sabot in the
5 inch gun and a 3.83 pound sabot in the 6 inch gun.
Corresponding total projectile weights are 10.66 lbs
and 11.83 lbs, with corresponding base areas of 19.63 in^2
and 28.27 in^2 and corresponding sectional densities
of 0.543 lbs/in^2 and 0.418 lbs/in^2. So the 6 inch gun
will accellerate roughly 30% more rapidly.
However, that's not the whole story. As I mentioned,
gun tube weight is porportional to contained volume.
If we set as one ton the weight of the tube past
the chamber in the 5 inch gun, and say a 5 meter length,
then the 6 inch gun for the same gun weight will only
have a 70% as long bore, or 3.5 meters long. Strangely
enough, the volume of barrel is the same either way...
So if we assume that the work on the projectile base
is as it should be PV work, then the work should be
the same in either barrel. The higher accelleration
in the 6 inch barrel will be offset by the more rapid
gas expansion and more rapid lowering of pressure;
the 5 inch barrel will have the same amount of work
done on the projectile, with a lower average accelleration
over a longer distance.
Either way, barrel volume is roughly 3960 in^3.
If effective chamber volume is 550 in^3, then total
expansion ratio will be around 7. You can work out
the pressure curve yourself, assume the propellant
burns effectively instantaneously, work the numbers
on what the pressure will be over fractions of the
barrel length for each gun... PV work is easy to
do the math on.
Except for the sabot mass fraction. In the 5 inch gun,
it's 8/10.66 or 0.750; in the 6 inch gun, it's 8/11.83,
or 0.676 ... so we did the same amount of work, but
more of that work accellerated the sabot in the 6 inch
gun than in the 5 inch gun. Oops. 8-)
-george william herbert
gherbert@retro.com