My airships in combat come upon each-other and initiate combat with their 12-foot long-tom guns. My problem is, I don't know what range they would be at when they decide to fire, because the post-apocalyptic world has a thick and toxic air. The air density is 65kg/m$^3$ (about 52x normal air density), greatly reducing effective range of projectiles. So the gun has a modification, which I hope will improve the range.

  • The cannonballs are molded with a cone-tapered nose and boat-tailed back, reducing the drag to 80% of a comparable ball shot at supersonic velocities.

  • The long tom gun has a 12-foot rifled barrel and weighs 6 tons. The 650 caliber cannonball is 36 lbs. and reaches a muzzle velocity of 560 m/s due to the evacuated bore described below.

Cannon anatomy

The gun is only slightly modified by placing a light-weight plate over the lip to seal the bore. The sealed bore allows a crank near the trunnion to pump air out of the cannon, reducing the air density to 20 kg/m$^3$

What is the longest lateral range this weapon can achieve in my world?

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – L.Dutch
    Commented Feb 25, 2022 at 3:09

2 Answers 2


So, it is quite difficult to work out what the drag coefficient of your projectile might be, let alone worry about how it changes as the velocity of your projectile changes. From wikipedia's simple coefficient of form approximation, I got a (relatively high, for a bullet) drag coefficient of 0.5, but that's not a bad number for a supersonic projectile. I found a paper looking at base-bleed artillery shells (https://iopscience.iop.org/article/10.1088/1742-6596/2054/1/012013/pdf) which measured a drag coefficient of ~.274 for base bleed off and ~.238 with it on... but that's a rather more sophisticated projectile than you're considering.

I'll go with a drag coefficient of .3, which doesn't seem that great but keeping it at both sub- and supersonic speeds is pretty good. You've quoted a 6.5" calibre, and that, unfortunately, is going to make life difficult for you. As Zeiss Ikon pointed out elsewhere, a good reason for having a bullet-shaped bullet is that you can squeeze more mass into the same cross-sectional area without unduly affecting drag or aerodynamic stability. Your projectile has the mass of a 6.5" diameter iron sphere, not a 6.5" diameter iron bullet. This results in a lower ballistic coefficient, which in turn leads to worse performance.

So, here are some trajectories with various elevation angles, showing maximum possible range. The 0-vertical-height line shows the altitude of the firing airship.

Projectile altitude relative to gun vs horizontal range for various elevations

As you can see, you can't realistically reach even as far as 300m... that's clearly a maximum range, but it isn't the same as an effective range.

As you didn't define effective range, I can't really give you a definitive answer. In the real world it depends on your ability to hit your target, and once you've hit it, actually do damage. I'm not going to tackle accuracy here, but the striking power of your projectile is going to be strongly influenced by its speed.

Here's a plot of projectile speed vs the horizontal distance travelled:

Plot of projectile speed vs horizontal distance travelled

I've generated similar charts in one of my answers to your related previous question, and as you can see it doesn't look great. It would be up to you to decide how fast a projectile has to be in order to be going fast enough. Your ~16kg round travelling at ~100m/s has ~80kJ of kinetic energy, so it will still pack quite a punch against softer targets.

If 100m/s was taken to be the standard of effectiveness, then an initial elevation of 0.5° will give you an effective range of 135m, with a projectile drop of only 15cm (so the trajectory is basically flat).

The conclusion I came to in the other question was that you shouldn't be firing balls or even bullet/shell shaped things, but much more aerodynamic things if you want long range or high impact energies. Rocketry might also be a good idea, but that is definitely a subject for a separate question.

  • $\begingroup$ “Longest lateral range” I think defines effective range as hitting a lateral target? $\endgroup$
    – Vogon Poet
    Commented Feb 23, 2022 at 21:59
  • $\begingroup$ @SurprisedSeagull the simulation simply isn't suited for water. Fluid dynamics is exceptionally complex, after all. I can trivially increase the density of the working fluid in the simulation, but this does not affect the viscosity of the fluid, and the kinematic viscosity of air and water differ by an order of magnitude. I'm not simulating drag in water here, so I'm not particularly interested in continuing this line of conversation, but you might ask a separate question about it in physics.SE if you were interested. $\endgroup$ Commented Feb 23, 2022 at 22:02
  • $\begingroup$ @VogonPoet the point at which the trajectories cross the 0-vertical-height line would show you the maximum possible range at which a target at the same altitude could be hit. Its a little under 300m, and the shots are plunging at that distance. Did you mean something else by "lateral range"? $\endgroup$ Commented Feb 23, 2022 at 22:04

Air drag is proportional to air density. Very optimistically such a cannon could fire at 2 km in our world. And only 40 m in your world.

Evacuating the barrel from 50 to 10 atm pressure will improve the nozzle speed only by a few % and doesnt worth the effort. Most of acceleration happens at pressures of 100-1000 atm.

You can extend your range by x10 if you use a rocket instead of a cannon, to avoid peak speed. Keep rocket's speed always low, some 30 m/s instead of 300 m/s or more. 400 m

You can increase your range by another x10 if you add wings to your rocket and optimize its shape to be more like a plane. Keep in mind wings are small, more like a wing in ground effect. 4 km

You can increase your range by another x10 if you use maximum altitude advantage. As rocket falls it can glide forward. If you can climb at 4 km higher than your opponent, range is 40 km.

You can increase your range by another x10 if you use an engine instead of a rocket. At this pressure internal combustion engines are extremely powerful, as you get a lot of air. So even a tiny engine will provide extreme power, be it piston or jet. Most of the benefit is for use of gasoline instead of gun powder. In particular pulse jet, that requires no moving parts, and can be done with rather low metallurgy compared to all other jet engines. 400 km.

You can increase your range by another x10 if you use a bypass for the air and use a fan or a propeller, rather than directly being pushed by hot gasses, both piston and jet engines can use this. For piston engine it is a propeller, for jet engine it is a turbofan. 4000 km.

Your air battles are more like submarine battles. With torpedoes. And helium or hydrogen filled chamber in your 'torpedo' can make its range extreme. Slower the speed - longer the range, and the floater provides all the bouyoncy needed. Also fire is way more dangerous, and hydrogen tank explosion are also way more dangerous, this way floater and warhead might be the same.

And all the anti-torpedo things, like guns, have the same problem with their range being tens of meters at the very best, so slow torpedoes cant be easily shot down.

Cannons, as you showed them, are completely useless in your world.

  • $\begingroup$ 2km on Earth equates to 277m on the OP's world. Range = Earth Range ÷ sqrt(pressure), not earth range ÷ pressure $\endgroup$
    – Monty Wild
    Commented Feb 23, 2022 at 11:31
  • $\begingroup$ @MontyWild why so? en.m.wikipedia.org/wiki/Drag_(physics) , drag force is directly proportional to density, why sqrt? Time*force is momentum, also implies linear dependency. Even en.m.wikipedia.org/wiki/Impact_depth newton appoximation gives linear relationship with density. $\endgroup$ Commented Feb 23, 2022 at 15:19
  • $\begingroup$ grc.nasa.gov/www/k-12/airplane/Images/ballflght.gif if we assume drag is like g, acceleration, and height is like distance, it is still linear, not sqrt. $\endgroup$ Commented Feb 23, 2022 at 16:08
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    $\begingroup$ @JustinThymetheSecond do you mean if the shot will hit the airship's bouyancy structure, light gas baloon? It is not pressurised, it will lose gas, but not similar to a rubber baloon. So, no, even in this case ship wont spin much before falling. If you ment something else, i dont get it. $\endgroup$ Commented Feb 23, 2022 at 16:46
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    $\begingroup$ Actually, on ships the cannons were low below deck for a reason - they fired at the plane of the center of rotation of the ship, once all forces were balanced. It was definitely a design criteria. The ship rigging had a lot to do with balancing it. They also fired cannon from both ends of the ship at the same time, to balance the rotational force. $\endgroup$ Commented Feb 25, 2022 at 17:28

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