Ballistics: my math fails

those of you who know modern field guns know this: an automatic cannon/howitzer can reload and fire quick enough so as to lob two or even three shells at a stationary target at the same time.

just my off-the-cuff guess, there are only two possible elevations (angles of fire,) one steep the other shallow, for a stationary gun to hit the same stationary target. but then a mathematician friend said it’s actually infinite. assuming this is so, care to explain?

With the same muzzle velocity, your right, there’s two trajectories. If you can vary the muzzle velocity continuously, its infinite.

so > 2 is only possible if you regulate either the shell weight or the amount of propellant charge. that’s what i suspected. thanks.

Just to make sure I’m understanding, if you vary the velocity, I assume you guarantee simultaneous arrival of the shells by varying the angle that the barrel is elevated. That must be the case, right?

never mind

If nothing else, your rate of fire will keep it much less than infinite.

Lets say the fastest ballistic path you can take to get to the target is 5 seconds. Lets say the slowest path you can take is 10 seconds. Lets say your firing rate is 1 per second. So, back of the envelope says N will be 5 give or take in that case.

Also, there is nothing special about N=2 except its more than N=1 (i think).

From the same gun? I’m not sure, but I wouldn’t think two shells landing simultaneously is possible from the same barrel.

A flat trajectory straight at the target is the fastest path and shortest distance.

If you angle up the barrel, you have a significantly longer distance, but the shell must also be in flight long enough for gravity to arrest the upward velocity and and start it accelerating down toward the target. And the more “powder” you add, the higher you have to angle the gun to prevent an overshoot.

Maybe someone else will come along but it seems to me that with a parabolic arc, the faster your initial shell velocity, the longer it will take to reach a given target at a given range.

Come to think of it, this is wrong. Theoretically, I suppose you could fire a very high angle shot, and while that shot is stll in its arc, load and fire a point blank shot that could arrive at the same time or sooner. Whether that can be done as a practical matter given modern artillery, and min and max ranges, I don’t know.

A high arc will take longer for a faster muzzle velocity (since you’re spending more time getting to the apex and coming back down), but a low arc will take a shorter time for a faster muzzle velocity (since the up-and-down isn’t as relevant).

the first problem, making all shots of different elevations hit the target, is possible with varying propellant powder. you will have several solutions but not infinite, since regulation of propellant charge is likely to be in discreet amounts, not continuous.

the second problem, making 2 or more shots from different elevations hit the same target at the same time, adds the variable of time to reload, adjust elevation, and fire. it’s actually being done with the 155mm crusader gun.

Time on Target

yes, that’s the term. i wanted to discuss it dope-wise. :slight_smile:

Fair 'nuff.

It’s actually not the term you are looking for. “Time on Target” is a technique for making multiple guns impact at the same time. Multiple-Round Simultaneous-Impact (or MSRI) is what you’re after. It’s actually the next paragraph down in chacoguy’s link.

my first time to see that site. must be awful for the crew to have casings ready with different propellant contents should the order come for MSRI. loadings would also vary depending on target range. ah, computers.

The propellant is adjustable, not fixed. They are prepped immediately before (or during) the loading procedure. So there is no need to have different rounds with varying charges laying around before a mission comes in.

I think you could use standardized propellant amounts, and just have the computer vary the time-lag between successive firings depending on the distance (or reading Bear’s post, maybe not.)