Sciencemadness Discussion Board

Calculating blast radius

Dream of the iris - 28-9-2005 at 13:20

How do you calculate the blast radius of an HE? Safety zone/how far you'll be able to hear the explosive, etc.

12AX7 - 28-9-2005 at 13:37

Hm, at a glance, that's a kewlish question, but on second thought it's really a very good question. For sure it has to do with power (i.e., energy per time) of the explosive. Wish I could help, but now I'm as curious as you! ;)

Dream of the iris - 28-9-2005 at 16:48

I can see how one would think that, but trust me, I'm no kewl. Just ask the good people of totse BB and bbscience.org :)

vulture - 29-9-2005 at 05:06

ROFL, to prove you're not kewl you're providing a reference from a bunch of kewls, in casu totse. Nice going.

Anyways, be it kewl or not, this is practical discussion and thus outside the scope of this board.

Closed.

Ramiel - 29-9-2005 at 10:13

If I were to attempt to determine the 'blast radius' of my new explosive compound, how would I tell with limited apparati and on a small scale?

For those of us without Axt's blast-drop test equipment and wide open spaces - the only real way to test an explosives' power is to compare blast radi' (although the measurement may be very heuristic)

I am currently in deep experimental testing of novel organic peroxide polymer explosives. The only way I can test the power of resulting complexes is by comparing them to well known blast data (e.g. acetone peroxide [kewls laugh at your leisure] )

[Edited on 30-9-2005 by Ramiel]

Axt - 29-9-2005 at 11:39

If you can make sense of the screwy formated equations, your better then I. Would have been good if references were given. Check the link for <i>slightly</i> better formating.
Quote:
<a href="http://groups.google.com.au/group/alt.engr.explosives/tree/browse_frm/thread/c86beabac1bb8197/982d3ab5fa923bbe?rnum=1&hl=en&q=blast +calculate&_done=%2Fgroup%2Falt.engr.explosives%2Fbrowse_frm%2Fthread%2Fc86beabac1bb8197%2Fd1a92993c97fc769%3Fq%3Dblast+calculate%26rnum%3D1%26hl% 3Den%26#doc_982d3ab5fa923bbe">Originally posted by Francis Hermans</a>
The explosion will provoke an air shock wave which radiates in the air
and decrease with distance.
The intensity of the pressure wave depends from the quantity of explosive
that
detonate but also from it position or confinement.
Several formulae exist to calculate the air overpressure.
The two following are given for unconfined detonation at ground level
and are valid for a charge of TNT.
It is generally agreed that for commercial explosives the corresponding
charge can be reduce by 20 to 30 %.
1/3
P = 700 Q /R (mbar)
where
Q = kg
R = m

or by :
1/3 -1,2
Ps = 185 (R/Q )
where
Ps = KPa
If the same charge exploded freely underwater, then you could use the
Cole's formula to calculate the peak pressure of the shock wave.
1/3 1,13
Pm = 555 (Q /R)
where
Pm = peak pressure in bar
but here again, the formula is made for a TNT charge.
Francis

Try <a href="http://groups.google.com.au/group/alt.engr.explosives/tree/browse_frm/thread/f70e73227f8090ae/2a2754fb8f059b1c?rnum=1&hl=en&q=blast +calculate&_done=%2Fgroup%2Falt.engr.explosives%2Fbrowse_frm%2Fthread%2Ff70e73227f8090ae%2F25a275b534eac9b5%3Fq%3Dblast+calculate%26rnum%3D2%26hl% 3Den%26#doc_2a2754fb8f059b1c">here</a> as well.

This is the reference to the underwater blast calculations refered to in the quote above. I dont know where the air blast one come from.

R.H. Cole, “Underwater Explosions”, Dover Publications, NY (1948)

Thats for TNT, try relating that to some comparitive tables such as;

<center><img src="http://www.sciencemadness.org/scipics/axt/pressure-table.jpg"></center>


[Edited on 29-9-2005 by Axt]

vulture - 29-9-2005 at 12:53

Quote:

Regardless, if I were to attempt to determine the 'blast radius' of my new explosive compound, how would I tell?


Plate dent test? Lead block? Whole lot more practical than blast radius, because how on earth are you going to measure that reliably?

Furthermore, the energetics section of this board focusses on the chemistry of energetic materials, not the practical side.

Therefore this thread is closed, again.