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As was discussed in the comments, I've crossposted this question to here, and am cross-linking them: https://worldbuilding.stackexchange.com/questions/234669/how-much-do-tunnels-extend-blast-waves-from-explosives

Someone taught me that explosive range is calculated with a simple inverse square law. If a blast is 10kpa at 1m in open air, then at 2m it should be 10kpa / 2m^2 = 2.5kpa? I've also heard it said it should be to ^3, but I think that's incorrect?

This is how I was taught it works, with blasts in tunnels:

Regardless, in a tunnel the blast gets focused, and I wondered how large the danger area is by comparison. What I was taught, is you just take the volume of a sphere where the radius is the blast radius in open air, and convert that to the volume of the tunnel, to get the approximate range. It was also recommended to halve the result, to roughly account for inefficiencies like the tunnel walls absorbing the blast.

Example: So if the blast radius is 20kpa at 10m in open air, and you blast it in a 4x4m straight tunnel of infinite length, the Volume of a 10m radius sphere is 4,188m^3, which would equal a 4x4x261.8m tunnel's volume. If you halved it, that'd suggest the blast pressure would be 20kpa 130.9m down the tunnel, and a lot more as you got closer.

Thoughts: Not sure how accurate that estimate really is even as a rule of thumb, though. I know blasts are more powerful in enclosed spaces, but turning a 10m radius into a 130m radius is pretty extreme. Maybe that would be the case with really hard rock?

Either way, was hoping to ask to learn a bit more about blasting radius underground. A friend wanted my help on a story with dwarves having tunnel wars with goblins, and range of explosives is something I'm not able to give him a good estimate for.

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    $\begingroup$ Hi user343708. Welcome to Phys.SE. Did you try to do a back-of-an-envelope-calculation? $\endgroup$
    – Qmechanic ♦
    Commented Aug 21, 2022 at 9:59
  • $\begingroup$ In view of the purpose you might find Worldbuilding more useful. Their rules on what is on-topic and off-topic are a bit less restrictive as homework and check-my-work type questions tend to be off-topic here. People also tend to post more expansive answers with alternative ideas. $\endgroup$
    – StephenG - Help Ukraine
    Commented Aug 21, 2022 at 10:10
  • $\begingroup$ @Qmechanic Hello. Thank you! I honestly am not sure how one does estimate this. The above is the best research I could muster on it, with an example of how it seems you might calculate it. But I strongly suspect that calculation might be very misleading/inaccurate. $\endgroup$
    – user343708
    Commented Aug 21, 2022 at 10:23
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    $\begingroup$ Honestly there are plenty of detailed answers on physics questions on Worldbuilding and indeed many physics members are also worldbuilding members. Subjects like this, which overlap engineering, also tend to get better responses on worldbuilding as we really tend to focus on very specific conceptual answers here. I'm not voting to close your question, I just think you'll find more use from worldbuilding for this purpose. $\endgroup$
    – StephenG - Help Ukraine
    Commented Aug 21, 2022 at 11:49
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    $\begingroup$ Cross posting is generally discouraged but not forbidden. Please do add links to the cross posted questions in your questions if you do this. $\endgroup$
    – StephenG - Help Ukraine
    Commented Aug 21, 2022 at 12:10

3 Answers 3

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There's an empirical rule given as equation 4 in "Field Tests on the Attenuation Characteristics of the Blast Air Waves in a Long Road Tunnel: A Case Study".

$\Delta p=\left( a\cdot\frac{m\cdot Q}{S\cdot x}+b\cdot\sqrt{\frac{m\cdot Q}{S\cdot x}} \right)e^{-n(x/d)}$

$\Delta p$ is the overpressure at a measurement point in Pa, $Q$ is the explosive charge in kg, $S$ is the cross sectional area of the tunnel in m$^2$, $x$ is the distance from the explosion to the measurement point in m, and $d$ is the equivalent diameter of the tunnel cross-section in m. The other variables are constants that have to be determined empirically. The paper suggests $a=2900000$, $b=730000$, $m=0.4$ $n=0.15$ as one set of parameters, but there are alternatives proposed in different sources.

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  • $\begingroup$ Thank you so much for joining just to answer my question, and with such a wonderful find! Have a couple of questions for you, if that's OK. AFAICS, Q is kgs of TNT (only explosive I could see referenced)? The walls of the tunnel seem to be assumed as bedrock or such? I don't notice a variable to calculate wall material. I played with the formula and have made a graph: desmos.com/calculator/9julpumgbp Apparently, it takes 2kg of TNT in a 2x2m tunnel to inflict lethal blast pressure at 2m!? That's a lot less than my calculations. Wondering if I did anything wrong with this formula. $\endgroup$
    – user343708
    Commented Aug 22, 2022 at 5:33
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A typical purpose of blasting is to apply energy to damage solid material, so it is usually not a sound wave (which propagates according to the elasticity and density of material), but a more complex shock wave (the elastic property assumes NON-damaging wave propagation).

So, a blast will expend significant energy locally, not just spread it as sound. Some PART of the blast energy becomes normal sound waves, and that part will obey the inverse square law in a uniform sound-conducting medium (rock, or air, or water...). The in-air residual sound created by a blast will indeed channel in a tunnel, not spreading out and remaining nearly constant as it travels.

But, only nearly constant; sound is an organized movement, and it decays by the processes that turn sound energy into heat; viscosity of air, scattering by reflections, and random collision-of-gas-molecules changes of direction of the motion are all removing sound energy and creating heat energy.

If there's dust disturbed in the passage of that sound wave, that is another loss of sound energy, soon to be just a tiny bit of heat. At a distance, only the lowest frequency parts of the original 'bang' will remain (a 'boom'), because the highest frequencies thermalize most rapidly. Most of the energy in a blasting event is at the highest frequencies, because that is most effective at doing the local damage required...

After a few corners and a few hundred meters, a BANG from a blasting operation in a tunnel will be a not-too-destructive BOOM, and a puff of wind.

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  • $\begingroup$ Thank you for the detailed answer, White. Nulius found a paper that gives a formula for tunnel explosions, and I made a calculator based off that, which I link in an answer I wrote. I found that the tunnel size makes a big difference due to the factors you mention, where there's an ideal size for any given explosive charge and distance. $\endgroup$
    – user343708
    Commented Aug 22, 2022 at 6:31
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With the help of this answer, I've made a calculator based off its formula, which I'm sharing here: https://www.desmos.com/calculator/yt9vuitqoh

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  • $\begingroup$ Welcome to Physics! In general, it's preferred to post such things as comments on the answer in question. As you stick around and gain rep, you'll be able to post comments on other users' answers. In the meantime, I've flagged this for moderator attention, and hopefully they'll convert it appropriately. $\endgroup$
    – Michael Seifert
    Commented Aug 22, 2022 at 14:45
  • $\begingroup$ Incidentally, if you calculate the volumes of the two tunnels it should be obvious why one requires so much more explosives to get a similar effect. $\endgroup$
    – J Thomas
    Commented Aug 23, 2022 at 13:14

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