NOTE - To the person who edited the title. I was not asking about recoil in the sense of the gun travelling in the opposite direction to a bullet. Please read the question carefully. There is no bullet - just a fixed rod that is set in concrete. Effectively therefore the Earth is the bullet and might as well be considered to have infinite mass. As has been pointed out in the answers, the result is actually a mortar and specifically a spigot mortar. (I didn't know this before I asked).

Firing guns at friendly forces

We are besieged in the castle. Friendly forces have arrived to attack the besiegers from the rear. Unfortunately our friends are terribly short of weapons although they have plenty of ammunition but no guns. On the other hand we have plenty of guns but are rapidly running out of ammunition.

Therefore we wish to project guns over the heads of the besiegers to our friends where they can be loaded and used in battle.


The gun is loaded with a bullet-less cartridge and the gun-barrel is slid over a fixed rod that closely fits into it. The rod is angled at 45 degrees to the horizontal and set in concrete at the other end The gun is fired via a thread that pulls its hair-trigger and breaks as soon as the gun flies off towards its target.


Is shooting a gun in this way even possible or would the barrel burst? If so, how do I calculate how far the gun would travel? I am not clear how to convert the range of a bullet fired from a gun (with recoil) into the range of a gun fired from an immovable 'bullet'. Will any spin imparted to the gun owing to it not being symmetrical affect the range adversely?

All assistance gratefully received.

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    $\begingroup$ Any search engine will give you plenty of pages with the ballistic equation, if you just make the effort to search. $\endgroup$ – L.Dutch - Reinstate Monica Mar 29 '19 at 10:07
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    $\begingroup$ Rig up a catapault; tie your guns in a bundle; load and launch $\endgroup$ – nzaman Mar 29 '19 at 11:15
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    $\begingroup$ Of course there's an xkcd for this $\endgroup$ – Separatrix Mar 29 '19 at 11:29
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    $\begingroup$ You are in effect loading the gun with a projectile hundreds of time heavier than the projectile it was designed to shoot. In real practice, double-loading the gun was considered to shorten its life, triple-loading was dangerous, and quadruple-loading suicidal. (As for how to compute: $m_b v_b^2 = m_g v_g^2$, hence $v_g = v_b m_b/m_g$, where $m_b$ is mass of the ball, $m_g$ is the mass of the gun etc. The magical non-bursting gun will start with about 1% of the muzzle velocity of the ball, or about 3 m/s. It will travel one meter.) $\endgroup$ – AlexP Mar 29 '19 at 11:30
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    $\begingroup$ Even if this worked, the weapons are unlikely to be usable after they impact the ground at the other and of the ballistic arc. Broken stocks, damaged mechanics, etc. This falls into my "launching an ICBM horizontally" class of questions. Can you do it? Probably - but the bad outweighs the good. But, to answer your question, we're missing details. (a) how far the guns must travel, (b) the height of the wall, (c) the location of the wall along the horizontal path, (d) the weight of the guns, (e) the composition of the guns. $\endgroup$ – JBH Mar 29 '19 at 16:25

This would not generally be practical. The chamber is designed to handle the impulse of making the bullet start moving into the barrel. That is the inertia of the bullet and forcing the bullet into the rifling. In your concept we can assume that the "rod" would be designed to have roughly the same friction but the inertia would be that of moving the entire gun. Guns are generally significantly heavier than the bullets they are designed to launch specifically to prevent recoil from moving the gun too much.

So the gun would probably break even if you figured out how to not destroy the rifling or the barrel or how to have the gun land safely. Both of which would take time to solve you probably would not have.

Also since the guns are not designed to be launched they have inferior aerodynamics so the range would be very bad for the force. And given that the range would start as a fraction of what a bullet from the gun would have... Well, if your friends are that close just shoot the enemy in between and have them bring you ammo.

These issues are probably solvable with time. You are talking about an improvised mortar here so if you make an improvised projectile with the gun safely contained within, combine propellant from multiple rounds of ammunition and launch this from an improvised mortar, you can probably launch this over the enemy.

But you probably should not. You have more pressing needs for you time and precious ammunition. You should go for something more efficient.

First, shooting excess ammunition to the castle would be more effective. Ammunition is more compact and robust than guns are, so if you are going to shoot something that is what you should shoot. You'd still need improvised mortar and containers but, well, you have people without guns out there, right?

Second, if you want to move something without breaking it, something like a hot air balloon or even a kite is much more practical solution than shooting it and hoping it survives. Even if you do not have an acute shortage of propellant.

Third, generally shoot the enemy not your allies. Ammunition is better used against the enemy unless your soldiers are proud graduates of the imperial stormtrooper markmanship academy. Shooting the enemy is what bullets were designed for, using them for anything else will be less efficient, and will be a bad idea unless you have more ammunition than you need.

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  • $\begingroup$ The people in the castle are so short of powder that they can't afford to shoot the enemy with bullets. That's why they want to send the guns to their friends who have all the ammo but no guns. I'm not sure how you could shoot ammo to the castle without any guns and of you did they would receive spent ammo. $\endgroup$ – chasly from UK Mar 29 '19 at 18:29

The principle you describe is the spigot mortar. So it works in theory, but not -- as Ville Niemi pointed out -- in practice.

  • A 36-pounder was called that name because it fires a ball that was 36 (French) pounds. For all practical purposes, 36 pounds.
  • The barrel was roughly 200 times the weight of the ball.
  • The ball would fly about 3,700 metres.

It would be incorrect to simply multiply the range by the weight ratio, but it should give you a feeling that the barrel won't go very far on any practical powder charge.

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    $\begingroup$ It should be noted that this (the way a spigot or any mortar works) does not resemble the method described by the OP. It does answer the question title in the sense of "make the gun into a shell then put it on a mortar that would set off the primer and propellant launching the gun over the wall". Converting a gun into a bullet or shell (including fins and weight re-distributions if you wanted to aim at all) would pretty much guarantee problems trying to use them after (not to include the problem with hitting the ground at high velocity) $\endgroup$ – LinkBerest Mar 31 '19 at 0:23

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