# Airship combat in dense atmosphere - What charge for 26-lb cannons?

For an airship combat scene with cannons on a world with a dense atmosphere, what would be the powder charge for effective beam-to-beam attacks at a 100-yard range?

The projectiles are 26-lb cannon balls, The charge is black powder, the cannon weighs 200x the cannonball. Assume the cannon's max elevation is 15°

Air density for combat is a thick 65 kg/m$$^3$$

• Really? Your airships only shoot when they're almost rubbing envelopes? Feb 22 at 17:36
• I believe that was the common tactic with side guns, yes. Envelopes are smaller in this density, but gun tactics had to consider very low accuracy (aside from the one or two long toms they may be carrying) Feb 22 at 17:43
• Only takes one chainshot through the gas bag, and someone's ship is done... Feb 22 at 17:46
• Lose 10% of your lift and you're still going down, just not as fast. Feb 22 at 17:52
• " I intended to be closer to a football field, my 30-yard was a typo error." edit away, I'll adjust. Feb 22 at 17:53

Drag force $$F_D$$ increases linearly with the density $$\rho$$ of the medium producing it.

# $$F_D = \frac{1}{2}C\rho A v^2$$

With a 0th order approximation:

• your atmosphere will produce 65 times more drag than our Earth atmosphere,
• the drag will dissipate 65 times more energy
• to keep the same energy on the impact, you will need to supply that more energy at the cannon (e.g. if with normal atmosphere drag would dissipated 100 J, your atmosphere would dissipate 6500 J, therefore you would need to supply 6400 J more)
• neglecting all the secondary effects (combustion, recoil, deformation, etc.) the powder charge will need to be increased of the amount that supplies the energy lost in the increased drag
• I don't think an earth cannon can contain a sixtyfivefold increase in powder charge, even before it's ignited. Given the upper limit on the mass of the cannon, do you think that will restrict the maximum amount of powder that can be used safely? Feb 22 at 17:42
• Okay, but how much is that? How much energy does a cannonball lose in a hundred yards? Hint: not much compared to their muzzle energy. 65x not much is still not much. Feb 22 at 19:31
• This is a recipe for killing your own gunners when the cannon burst. Also, supplying ke×65 means that v will be higher and drag increases with the square of v. Feb 22 at 20:31
• This doesn't answer the question. Range is proportional to 1/sqrt(Delta Density), not 1/(Delta Density). Feb 23 at 2:14

According to this article (see table, scroll down some), the common powder charge for a 26 lb. naval gun in the 16th century was 14 lb -- or just over half the weight of the ball. Charge sizes were often determined (over multiple generations of test articles) by how big they could be without blowing up the gun, so if your guns are patterned after a historical piece, I'd recommend using the same charge.

A little later in history, a cannon (used by both sides) in the American Revolution (1770s) and again in the War of 1812 had an effective range of 1000 yards (this is at comparable elevation; higher angle would be a howitzer). With 65 times the air resistance, this range would surely be shorter, likely a bit less than half, but a couple hundred yards is easily practical. Also note that while these were land cannon, they were actually smaller than the 26 lb. units you're asking about -- and with black powder, large guns usually had longer range (pressure was limited by the propellant and barrel, but a bigger ball had more driving force from that pressure and larger guns generally had longer barrels as well).

At close ranges, it was much more common to fire "canister" -- essentially making the tube into a huge shotgun by firing lots of balls smaller than a pound. Howitzers would fire "shrapnel" which added a timed-fuse bursting charge to spray the balls after delivering them all together to the burst point.

• You wouldn't want canister in an atmosphere this dense... the falloff would be terrible. Feb 23 at 2:12
• @MontyWild Since canister is point defense (i.e. very short range) anyway, why not? You wouldn't hunt deer or elk, shooting from one ridge to another, with birdshot, but inside 30-40 yards, birdshot is just right for smaller game (especially on the wing). Horses for courses. In the American Civil War, canister was usually used at around 1/10 of the effective range of ball in the same cannon -- that is, about the same as the effective range of smoothbore muskets, or around 150-200 yards. Even cut in half for the air, that's plenty for this kind of engagement. Feb 23 at 13:05
• @MontyWild Not to mention canister does much more damage to the gas bags. Feb 23 at 13:05
• I like the part about many small projectiles. But i dont like the part about the distance being little less than half. Feb 23 at 18:16
• @SurprisedSeagull I agree with the ball or shot losing energy 65x as fast as on Earth -- but canister usually used balls ranging upward from 30 grams; those will still carry with lethal energy well past a hundred yards even in this atmosphere. The bigger the ball, the slower it loses energy, because drag goes as the area, but mass as the volume. Feb 23 at 18:25

TL;DR: you're asking the wrong question. You can't overpower the thickness of the atmosphere by using a bigger charge, you need to use much more streamlined projectiles instead.

Having spent 0 effort, I couldn't find information about a 26lb cannon, but I did find this 24lb cannon factsheet that gives a muzzle velocity of ~525m/s. I'll ignore the (probably serious) muzzle speed reduction caused by the dense, high pressure atmosphere for now, but consider that actual muzzle velocities will be substantially reduced compared to real-world historical guns. I'm also going to give your projectile a constant drag coefficient of 0.4, because fluid dynamics is hard and the one place where using a sphere makes everything worse.

I've used an elevation of 0.5°. Given the density of the atmosphere, there's little to gain by larger elevations. Here's a graph showing the projectile velocity vs horizontal distance... muzzle velocity is where distance is 0.

I've done four simple models:

1. a cannon firing a projectile in Earth's atmosphere.
2. a cannon firing the same projectile in your high-pressure atmosphere.
3. a cannon firing the same projectile in your high-pressure atmosphere, but with a significantly higher muzzle velocity (to simulate a massive gunpowder charge).
4. a cannon firing a highly streamlined (probably dart-shaped) projectile with a tenth of the drag coefficient, with the slower muzzle velocity, into your high-pressure atmosphere.

.

As you can see, the unmodified cannon simply doesn't perform at all well. Its velocity is massively lower at your desired range than all the other options, making its damage potential significantly lower.

Interestingly though, you can see that massively boosting the muzzle velocity just doesn't help very much... that $$v^2$$ term in the drag equation just bites you that much harder. If you somehow managed to make a black-powder weapon which had a muzzle velocity equivalent to a modern smoothbore APFSDS round, it still only manage about 60% of the realistic earth-based cannon's velocity at at 100m, and it is dropping fast enough that its range will still be quite limited. All that crazy engineering for little reward!

The final purple line shows you what you should be doing. Clearly the thick atmosphere is still bad news, but reducing the drag force by 90% has significantly helped the projectile's velocity at range, helping improve both damage at close range and maximum effective range.

• This is a good answer. Especially the graph looks nice. How did you make it? What was the model behind it? (Step-by-step simulation? Some fancy way to generate the whole curve at once?) Feb 23 at 18:18
• @SurprisedSeagull the code I used to generate it was written in julia, with DifferentialEquations.jl as the solver and Plots.jl for the graph. For all I know the underlying solver is powered by the screaming souls of the damned, but a simple iterative solution with a small enough timestep would work just fine, too. I keep meaning to put my code in a shareable form, but that day is not today. Feb 23 at 18:31

In order to maximise the effective range of a black powder gun under conditions of high atmospheric density, simply increasing the powder charge is not going to do anything but increasing the risk of bursting the guns, and increasing wear.

When looking up the characteristics of black powder guns on the internet, it is also easy to equate effective range with maximum range. Effective range is the range (and elevation) at which a gun is effective in combat. Maximum range is the range that the gun can project a shot on level ground.

Obviously, to increase the ranges of a gun, we need to increase the velocity. To do that, we can increase the powder charge, and/or increase the length of the gun in order to increase the use of the existing charge. The Culverin, a particularly long gun, had a point blank range (similar to effective range) of 1700 yards and a maximum range of 6666 yards. In an atmosphere 65 times as dense, we can reduce those ranges to 1/65th: 26 yards and 102 yards... can't we?

No. Drag does not affect range in this manner. Drag simply robs a projectile of kinetic energy by a process of friction. drag will be 65 times higher, but that doesn't mean that the ranges will be 65 times shorter, just that the projectile will lose kinetic energy faster... and the formula for kinetic energy is e=½mv². As kinetic energy is lost, v will be reduced, but not in a linear fashion. Additionally, while drag is proportional to atmospheric density, it is also proportional to the square of velocity. This means that velocity will be lost most rapidly as the shot exits the muzzle, but the rate of velocity loss will fall off. Also, when a projectile hits the ground (or at least returns to the altitude from which it was fired, it still carries significant energy. Obviously, if drag is higher, it will carry less, but it will still not be zero.

Empirically, using this calculator, increasing the density of the medium through which the projectile travels by 65 times reduces the range that the projectile travels to .12753 of the total distance, which is roughly 1/sqrt(delta density), not 1/(delta density).

So if nothing else changed, that would make the effective range of our culverin 217 yards and its maximum 850 yards... still within the OP's requirements.

However, we also need to consider the coefficient of drag (Cd), or how smooth the projectile is. Black powder ammunition was spherical and often rough due to the fact that it was made from chiseled stone or rusty iron from which the worst of the rust might have been chiseled. A smooth sphere is reckoned to have Cd=1, so a rough sphere would have a higher Cd. However, an elongated shape can have a lower Cd. As drag is proportional to Cd, reducing Cd would be very important.

Also, when we're talking about muzzle energy, there is still the ½m in e=½mv². By increasing mass, we can increase muzzle energy. By using more dense substances, we can increase mass without increasing drag, and therefore increase the range of the weapon.

Combining an increase in mass and a decrease in the coefficient of drag would get us to something like an ogival shell. If we're still using smoothbore black powder guns, we could put tiny stabilising fins on the back (in atmosphere this dense, you wouldn't need much fin) and shape it like a fish, relatively blunt at the nose, and tapering toward the tail. Make it from steel and fill it with lead, and polish it, and you'd reduce the Cd and increase the range a lot... and make it harder for armour to stop.

With drag effects this strong, don't tell me that people wouldn't work out more ways to overcome them than simply "more powder".

As for the amount of powder, at first, 30-caliber-length Culverins were loaded with powder equal to the weight of the shot, or if they were shorter, (calibre lengths)/30 times the weight of the shot. However, over the years as powder improved and gunners realised that having more powder gave diminishing returns, powder loads decreased from equal weight to 1/3 or even 1/4 of the weight of the shot.

So, since a long culverin with a standard load of powder would meet the OP's requirements, we could say that a 26lb 30-calibre culverin with late-era black powder would require 1/3 of the shot weight, which is 8 2/3 lb of powder.

• Now that I think about it, there is a Chinese pirate society in-world so a rocket-assist, even if rudimentary, would be logical. Great post. Feb 23 at 0:34
• @VogonPoet I think that a rocket might give up a lot of mass in exchange for a moderate reduction in velocity loss. You'd have to crunch the numbers carefully to see if it would be effective. Also, have a look at my edit. Feb 23 at 2:26
• @VogonPoet With 65x the atmosphere, will rockets even work? Doubtful... Feb 23 at 12:01
• @ZeissIkon there's a lot of interesting possibilities, though, especially considering the nature of the target. Consider a sort of rocket-propelled glider, for example. You're not constrained by gun barrel shapes, after all. Feb 23 at 12:19
• Question: wouldnt the higher air pressure also increase the maximum pressure the gun can withstand? Feb 23 at 12:36