How much do tunnels extend blast waves from explosives? 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.
 A: 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.
A: 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.
A: 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
