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I don't know if there is a word for a ballistic projectile with buoyancy, but feedback from the weapons in my dense atmosphere suggests that I should consider changing the shape of my projectile, and possibly consider propellant as well, making it a torpedo.

However, I am considering instead to harden the tip, evacuate the core, and render the projectile both aerodynamic and somewhat buoyant to improve range. Torpedoes are cool, and may be used in limited capacity, but everyone needs a rapid loading low-cost way to lay cover fire at a minimum.

Buoyancy is a function of relative mass to volume compared to the suspension fluid, and this does mean reducing the mass. Given an equal firing charge, this gives it vastly greater muzzle velocity.

Q: Can a buoyant ballistic projectile deliver equivalent kinetic energy at an equal lateral range than a solid equivalent?

Same weapon bore, same boat-tailed cone-nose projectile, but the core evacuated to give less weight. The suspension fluid is the same air at a density of 65kg/m$^3$ from my alien world.

The control round would be a 36-lb iron shot from a 450 caliber 12-foot cannon, using an 18-lb black powder charge. The test round would have 70% of the weight of the control due to an evacuated core. The hypothesis here is that a similar kinetic impact can reach the same lateral range in a shorter time, which may improve accuracy to boot.

note: A 650 caliber shot would be a cannonball, so the smaller bore allows shaping.

Buoyancy numbers are:

  • Air = 65 kg/m$^3$
  • Control round = 7508 kg/m$^3$
  • Evacuated round = 5255 kg/m$^3$

Assume control surfaces for tumbling

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    $\begingroup$ Hi Vogon. Your title question and the question posed in the body of the post are vastly different. They're supposed to be (at least materially) the same. $\endgroup$
    – JBH
    Commented Mar 7, 2022 at 7:30
  • $\begingroup$ I think I am missing something, the evacuated round is lighter for the same volume, but the density still seems higher than for your Air. So it would still sink? $\endgroup$
    – UVphoton
    Commented Mar 7, 2022 at 12:10
  • $\begingroup$ @UVphoton It would not be possible to make a neutrally buoyant round because it would have such a low mass, and likely just implode. So it is only slightly buoyant, yes. $\endgroup$
    – Vogon Poet
    Commented Mar 7, 2022 at 13:34
  • $\begingroup$ @JoinJBHonCodidact How so? I want to use a buoyant projectile. I called it a "torpedo" in the title because they are buoyant fired weapons but I don't know what other word there is for a strictly ballistic buoyant weapon. I know torpedoes are technically self propelled, but also buoyant. $\endgroup$
    – Vogon Poet
    Commented Mar 7, 2022 at 13:38
  • $\begingroup$ @VogonPoet I think you mean a lighter projectile. If the volume of the two projectiles are the same the amount of displaced air and buoyant force from the surrounding air is the same. If the projectile is positively buoyant, then a net upwards force, if it weighs more than the amount of the displaced air then negatively buoyant and will sink. For the same shape and volume, the heavier projectile will have a higher terminal velocity and more kinetic energy. Equal velocities heavier projectile more K.E. If the lighter projectile is faster since KE = 1/2mv^2, then it could have more KE. $\endgroup$
    – UVphoton
    Commented Mar 7, 2022 at 16:36

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No

There are two factors working against the "evacuated" rounds:

  1. Air resistance (fluid resistance) is generally proportional to the square of the velocity. So if the evacuated rounds are fired with a muzzle velocity that is 30% greater than the control then they are experiencing 169% of the air resistance (= decelerating force) that the control round would.
  2. Force = mass x acceleration, which can be rearranged to acceleration = Force / mass. Combining this with the increase in air resistance due to the increased velocity, this means that the negative acceleration on the round at the moment it leaves the muzzle will be 1.69/0.7 = 240% (approximately) the deceleration experienced by the control round. Once it (very rapidly) is reduced to the speed of the control round, at all equivalent speeds it will experience deceleration 40% (approx) greater than the control round.

A lighter round with the same drag will have a tiny engagement window just in front of the muzzle in which it will have a higher velocity than a heavier round, but it will always have much worse performance at all longer ranges. (This principle has been used deliberately to develop some weapons systems for use by air marshals which (ideally) will still have sufficient stopping power at very short range but within a few metres will hopefully lack the energy to penetrate and depressurise an aircraft. Note the qualifiers in the last sentence - fortunately they have never been used by any air marshal against any hijacker.) Looking at Starfish Prime's graphs in the linked question, my guess would be that 30-50 metres would be the crossover point within which the evacuated cannon round might be more dangerous and after which the heavier round is far superior.

The way that modern weapons use lighter rounds to achieve high penetration at range is by using discarding sabots. The key aspect of the penetrators is that they are much smaller calibre than the weapon they are fired from. These also require some reasonably advanced understanding of aerodynamics and materials technology - first experiments were in the period between WW1 and WW2 in our world - so they may not be appropriate for the world you are developing.

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  • $\begingroup$ I did not know most of this. Necessity is the mother of invention, however. Food for thought. $\endgroup$
    – Vogon Poet
    Commented Mar 9, 2022 at 17:51

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