The blast wave that is formed as a result of an explosion has some import parameters. One of them is the peak pressure which is basically the pressure measured when the blast wave arrives at a given location. It is the case that this peak pressure decreases as one moves further and further from the center of the explosion. Ignoring the actual mathematical form, this is plausible.

Now the other parameter of interest is the decay time. It is the time it takes for the peak overpressure to decay to the atmospheric pressure at a fixed point in space. One could imagine an experimentalist putting a pressure gauge at some location. When the blast wave arrives, the pressure rises to the peak pressure and after some time $t_{d}$, it decays to atmospheric pressure. This $t_{d}$ is the decay time. From my readings so far, I don't seem to have an intuitive grasp of the decay time. The books I read suggest that the decay time increases as one moves away from the source. Could any body provide me with some simple explanation as to why this might happen?

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    $\begingroup$ Blast waves under what conditions? Are you talking about a spherical charge in atmospheric conditions? A planar charge in a channel or cylinder? $\endgroup$
    – tpg2114
    Commented Jul 18, 2013 at 0:28
  • $\begingroup$ More on overpressure of blast waves: Bulk transport of an aerosol by a blast wave $\endgroup$
    – user10851
    Commented Jul 18, 2013 at 1:19
  • $\begingroup$ @tpg2114 spherical charge in atmospheric conditions is what I am after. $\endgroup$
    – felasfa
    Commented Jul 18, 2013 at 3:02
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    $\begingroup$ From Needham's book: 'As a blast wave decays to very low overpressures, the signal takes on some of the characteristics of a sound wave. The positive duration of the pressure, density and velocity approach the same value. The magnitude of the peak positive pressure and the peak negative pressure approach the same value. The lengths of the positive and negative phases approach the same value and the material velocity approaches zero.' $\endgroup$ Commented Jul 18, 2013 at 3:45
  • $\begingroup$ Deer, thanks a lot. I think I now understand. I wrote up a solution based on the explanation you gave me. Let me know if I misunderstood something. $\endgroup$
    – felasfa
    Commented Jul 18, 2013 at 5:52

1 Answer 1


After a bit of looking around and thanks to comments on my question, I now have an answer to my question. I just put it here so others looking for an explanation can benefit. Here is it.

Accompanying any shock wave is a family of rarefaction waves.The change of any parameter(pressure,temperature,density) at a particular point in space corresponds to the arrival of a wave(shock or rarefaction). Here is the key idea to the problem. A rarefaction wave will decrease the speed of sound of the air it passes through. So consider point $A$ which is located at $X_{A}$ and point $B$ be located at $X_{B}$ where $X_{B}>>X_{A}$. Then at some time $t_{A}$, the shock wave arrives at point $A$. Then the pressure at that point changes as the different rarefaction wave pass through it till it attains the atmospheric pressure. But now the rarefaction wave heading to point $B$ are moving in air which has a lower sound speed so it takes them longer to reach the atmospheric pressure as compared to what happens at $A$. This implies that point $B$ will have a longer decay time.


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