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I don't understand much about physics and ballistics, so I hope I don't make too many incorrect assumptions here.

So, I just saw this video where a 4 mm tungsten sphere hits a 85 mm steel armor, and it hits the target with so much force it simply melts the steel. No deformation or fragmentation, just melts like butter.

So, how much gunpowder (the "nitropowder" used on guns, or cordite and so on) it would be required to accelerate this projectile (a 4 mm tungsten sphere) to 50 km/s? Or at this point, it would be "easier" to just use a small charge of C4?

Well, I would guess that this is not very logistically efficient, since you would be able to shot a lot of regular sized projectiles with the same amount of gunpowder.

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    $\begingroup$ I think you're trying to achieve (super-) hypervelocity with the wrong method. Gunpowder, or nitropowder, are chemical means. There have been hypervelocity experiments yielding similar results your video shows, achieved with only 2.5 km/s (max 8.5), actually 20 times less than your requirement. en.wikipedia.org/wiki/Hypervelocity Shooting high velocity projectiles is achieved using a Light gas gun in a laboratory, a huge apparatus, comparable to a super air gun en.wikipedia.org/wiki/Light-gas_gun $\endgroup$
    – Goodies
    Commented Dec 29, 2021 at 13:21
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    $\begingroup$ A quick note: it is considered best practice to wait between 24-48 hours before accepting an answer. $\endgroup$
    – rytan451
    Commented Dec 29, 2021 at 14:06
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    $\begingroup$ It seems likely that the video does not accurately describe what it depicts. The most powerful light gas guns fire small projectiles around 7 km/s. The US Navy's experimental railguns only achieve about 2.5 km/s, albeit with much larger projectiles. $\endgroup$
    – Tech Inquisitor
    Commented Dec 29, 2021 at 21:09
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    $\begingroup$ This is off-topic for this stack. It should be asked on physics.stackexchange.com instead. $\endgroup$
    – computercarguy
    Commented Dec 29, 2021 at 22:12
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    $\begingroup$ @computercarguy This question works just fine on WB.SE. Since the OP does not even know what propellent he should be using, this is not purely a physics question. WB.SE works a lot with answering questions about bleeding-edge, near future, and theoretical technologies; so, the familiarity of this community with what propellants should or could be used may be more helpful than one that specializes at crunching the numbers. $\endgroup$
    – Nosajimiki
    Commented Dec 29, 2021 at 23:00

5 Answers 5

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You want your bullet to go from standing still to leaving the muzzle at 50 km/s, it means that it must receive an energy of $E=1/2 m_b \cdot v^2$.

The velocity is given, the mass of a 4 mm radius tungsten ball is given by its volume, $2.68 \cdot 10^{-4}\ dm^3$, times its density, 19250 $g/dm^3$, resulting in 5.16 g. Thus we are looking at a kinetic energy of E = 6450 kJ.

According to Wolphram Alpha, that't the amount of energy released by the 1.5 kilograms of TNT, which occupies about 1 $dm^3$.

This is a first, rough approximation. Once you take into account losses due to imperfect combustion, friction in the muzzle, imperfect sealing and so on, that figure goes very quickly up to the point of becoming unrealistic.

Escape velocity figures by means of explosions can be achieved with nuclear explosions, but there you will need to deal with how to make the "bullet" survive the explosion itself, and the massive drag with the atmosphere.

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    $\begingroup$ Worse, expansion in the chamber and bore as the projectile travels introduces huge pressure losses. Google "pump gun" to see a way this has been approached, once in an attempt to shell faraway targets, and once (ostensibly) in an attempt to shoot into space. $\endgroup$
    – Zeiss Ikon
    Commented Dec 29, 2021 at 13:19
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    $\begingroup$ Blast waves move at a maximum speed much lower than 50 km/s. It should be impossible under normal circumstances to shoot a projectile any faster than about 8 km/s using a cannon-like weapon. $\endgroup$
    – rytan451
    Commented Dec 29, 2021 at 14:08
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    $\begingroup$ @rytan451 is correct. Traditional guns have an absolute max speed as determined by gas expansion speed. To shoot faster still using chemical propellants, you would need to lower the pressure in the barrel to a vacuum. $\endgroup$
    – Dragongeek
    Commented Dec 29, 2021 at 14:54
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    $\begingroup$ "You want your bullet to go from standing still to leaving the muzzle at 50 km/s, it means that it must" be propelled by a propellant in which the speed of sound is not very much slower than 50 km/sec. (Because a gas cannot expand faster than the speed of sound in the gas; even with nifty tricks you cannot get more than a few times faster the speed of sound in the gas. And the speed of sound in hydrogen is only about 1.5 km/sec.) $\endgroup$
    – AlexP
    Commented Dec 29, 2021 at 16:28
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    $\begingroup$ @JustinThymetheSecond A projectile fired through a barrel continues to accelerate as long as the pressure behind it can overcome the pressure in front of it and its friction with the barrel. Longer barrels are not intrinsically more accurate. They're only indirectly more accurate because they accelerate the projectile to a higher speed, leading to a flatter trajectory; and because of a longer sight radius, if aimed with traditional iron sights. $\endgroup$
    – Tech Inquisitor
    Commented Dec 29, 2021 at 22:27
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You may want to look at the Voitenko compressor. It accelerates a disc, not a ball, and uses high explosives, not gunpowder, but can achieve shock wave velocities of over 40 km $s^{-1}$.

Initially, hydrogen gas is confined in a hemispherical dome covered by a metal plate and with a barrel at the apex of the hemisphere. The projectile seals the barrel. A shaped charge is exploded against the metal plate. Initially, it offers some resistance, allowing pressure to build; then it fails suddenly, compressing the gas behind to approximately $10^5$ atmospheres. The gas propels the projectile along the tube.

An experiment at Ames Research Centre, using a 3cm thick glass tube as the barrel (the barrel does not survive the experiment), produced a shock wave travelling at 67 km $s^{-1}$.

https://en.wikipedia.org/wiki/Voitenko_compressor http://www.islandone.org/LEOBiblio/SPBI134.HTM

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    $\begingroup$ This is defiantly along the right line of thinking, but should be expanded on further with an actual explanation of what a Voitenko compressor is and how it works; otherwise, this is answer is dangerously close to being a "link-only" answer which is against WB.SE's polices. $\endgroup$
    – Nosajimiki
    Commented Dec 29, 2021 at 22:54
  • $\begingroup$ Note that this requires high explosives; gunpowder would not be suitable. $\endgroup$
    – TLW
    Commented Dec 30, 2021 at 1:28
  • $\begingroup$ OK, I will expand. As it wasn't strictly what OP was asking for, I didn't want to go into too much detail. $\endgroup$
    – AlDante
    Commented Dec 30, 2021 at 8:32
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The speed of sound of regular air is 343 m/s. Blast waves move between 4-8 km/s. 50km/s is quite a bit above that. Projectiles fired from cannons can move at most the speed of sound of the propellant gas. I'm not quite sure of the speed of sound of the propellant gas resulting from a C4 or cordite explosion, but I'm pretty sure it's much closer to 8km/s than 50km/s. In short, even using high explosives like C4, it's impossible to move a projectile at 50km/s using a cannon.

However, all is not lost. Using a rocket, you can move faster than that limit, since you're not relying on a blast wave to push you.

A 4mm ball of tungsten has a mass of 5.17g. Military grade solid rocket propellant (which includes RDX, the active ingredient in C4) has a specific impulse of 268 seconds in a vacuum (lower in atmosphere), per Wikipedia.

The rocket equation is: $$\Delta v = I_{sp}g_0 \ln\frac{m_0}{m_f}$$

Where $\Delta v$ is the desired change in velocity (which is 50 km/s), $I_{sp}$ is the specific impulse (268 seconds), $g_0$ is acceleration due to gravity (9.81), $m_f$ is the dry mass of the rocket (we shall assume it to be the minimum of 5.17g, the mass of the tungsten ball), and $m_0$ is the mass of the rocket including fuel. We're looking for the value of $m_0 - m_f$.

Solving for $m_f$:

$$5\times10^4ms^{-1} = 2.68\times10^2s\times9.81ms^{-2}\ln\left(\frac{m_0g}{5.17g}\right)$$ $$m_0=5.17\times\exp\left(\frac{5\times10^4}{2.63\times10^3}\right)$$

So $m_0 \approx 9.33 \times10^8$ (rounding to 3 significant figures). That means you'd take about 933 thousand kilograms of military grade rocket fuel to accelerate your 5.17 grams of tungsten. This is probably too much to be practical, and this is already assuming conditions extremely favourable to the launch (you're in a vacuum, there's no rocket hull, only the tungsten ball). In practice, you'll probably need a few more times rocket fuel, rendering this concept even less practical.

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    $\begingroup$ Sounds like you'd better consider switching to a rail gun... $\endgroup$
    – Zeiss Ikon
    Commented Dec 29, 2021 at 14:27
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    $\begingroup$ According to a research paper I found (Thermal decomposition of RDX from reactive molecular dynamics), the maximum particle velocity in a computationally simulated RDX detonation was 3 km/s. However, the detonation velocity of C4 is 8 km/s. In any case, it seems unlikely that C4 is even capable of accelerating the tungsten ball to 25 km/s, much less 50 km/s. $\endgroup$
    – rytan451
    Commented Dec 29, 2021 at 14:53
  • $\begingroup$ The projectile does not need to carry the propellant with it in a "gun"-type physics problem, so the rocket equation isn't really relevant. $\endgroup$
    – Dragongeek
    Commented Dec 29, 2021 at 14:55
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    $\begingroup$ @Dragongeek, earlier in the answer I explained why a gun could not possibly shoot a tungsten ball at that speed, which is why I'm using a rocket instead. $\endgroup$
    – rytan451
    Commented Dec 29, 2021 at 15:00
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    $\begingroup$ @rytan451 Laboratory light-gas guns have shot projectiles at >8 km/s using rather conventional propellants and two-stage solutions have shot projectiles to speeds >11 km/s. Blast waves don't move projectiles down barrels, pressure does, and I suspect that while there is a cap somewhere due to the inertia of the light gas being used, there's still room upwards in the engineering envelope. Also, it's totally possible to accelerate things faster than the blast--famously the nuclear test "Operation Plumbbob" launched a steel plate at an estimated 66 km/s or shaped-charges like AlDante suggests $\endgroup$
    – Dragongeek
    Commented Dec 30, 2021 at 13:54
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At the risk of drifting even further away from OP's original intention, a nuclear explosion blew a 900kg steel cover (the "Manhole cover") into the atmosphere at a speed of 66 km $s^{-1}$.

The nuclear device was detonated at the bottom of a 150m borehole, covered by a steel cap. The principle is the same as in a gun. The explosion causes the gas, in this case air, to rapidly expand. Initial resistance by the cover allows pressure to build slightly, then release suddenly.

https://en.wikipedia.org/wiki/Operation_Plumbbob#Missing_steel_bore_cap

If this is too far off topic, let me know and I will delete the answer.

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  • $\begingroup$ Fascinating. If a 900kg steel cover can evaporate underways, maybe a bullet would ? When it's long range.. $\endgroup$
    – Goodies
    Commented Dec 31, 2021 at 1:04
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As Tech Inquisitor mentioned railguns, let's look at how they stack up. A projectile in an electromagnetic railgun will accelerate until the the induced back electromagnetic force is equal to the applied voltage. This is the theoretical limit; in practice the length of the rails determines how long a projectile can accelerate.

If we have arranged our railgun to operate in a vacuum, with no projectile resistance, then the muzzle velocity of the projectile is given by the following equation:

$$v_{muz} = \sqrt{\frac{2DF}{m}}=\sqrt{\frac{2DILB}{m}}=I\sqrt{\frac{2DL\mu_0}{m}}$$ where

  • $v_{muz}=$ Muzzle velocity (metres/second)

  • $D=$ Length of rails (metres)

  • $F=$ Force applied (Newtons)

  • $m=$ Mass of projectile (kilograms)

  • $I=$ Current through projectile (Ampères)

  • $L=$ Width between rails (metres)

  • $B=$ Magnetic field strength (Teslas)

  • $\mu_0=1.26 \times10^{-6}$ (The magnetic permeability of free space, Henries/metre)

(Equations taken from here)

So doubling the velocity requires four times the rail length, one quarter of the mass or twice the current.

Current railguns have a projectile mass in the region of 3kg and a muzzle velocity of about 3.5 km $s^{-1}$. Assuming nothing melts, reducing the mass to 5g (a factor of 600) would give a muzzle speed of the order of $3.5 \times \sqrt{600} \approx 86$ km $s^{-1}$.

As the Navy knows this, and chooses a larger projectile with a lower velocity, I think air friction would melt the smaller projectile.

Comparison of weapon systems:

System Energy
Naval Surface Warfare Center Dahlgren 8 MJ
University of Texas 9 MJ
BAE Systems 32 MJ
Dahlgren BAE Systems 33 MJ
Rheinmetall 120mm gun 9 MJ
BGM-109 Tomahawk 3000 MJ
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  • $\begingroup$ If anyone can explain why the table is fine in preview but not when published, I would be grateful. $\endgroup$
    – AlDante
    Commented Dec 30, 2021 at 10:09
  • $\begingroup$ In air, tungsten will burn before it melts. Same end result, though. $\endgroup$
    – Zeiss Ikon
    Commented Dec 30, 2021 at 15:11
  • $\begingroup$ @AlDante I solved the issue.. forward slashes are not allowed in these constructs with the old BBC, apparently they never solved that issue. I removed the / between Dahlgren and BAE. $\endgroup$
    – Goodies
    Commented Dec 31, 2021 at 1:02
  • $\begingroup$ @Goodies Thank you very much for finding and correcting this. $\endgroup$
    – AlDante
    Commented Dec 31, 2021 at 6:54

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