malford
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Detonation Wave Initiation Points
Let us have a mass of explosive material with a velocity of detonation of 8km/s. Let this mass be cylindrical, one centimeter in diameter and one
meter in length.
If we initiate detonation at one end then the mass should complete detonation, in other words the detonation wave should arrive at the other end, in
1/8000th of one second.
Is not the power of an explosive derived from the volume of gases that is produced in a short period of time? In other words, generating x volume of
gas in time y yielding power z. If we divide y by two, would that double z?
This could be accomplished by initiating detonation at both ends of the mass. The detonation waves would meet in the middle at 1/16000th of one
second. The same volume of gas is generated in half the time. Would not this generate more power?
If so, one could take this concept to the extreme by, for example, spiraling an EBW through a cylindrical mass of explosive material so that no point
in the explosive mass is further than a very short distance from the initiating EBW, thusly detonating all of the material in a small fraction of the
amount of time it would normally take the detonate the entire cylinder.
Would this increase the power of a charge?
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franklyn
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A centimeter diameter column of a secondary high explosive such as
PETN , RDX , is a tad slim for high order detonation even in a steel container.
Depending on loading density it may go ~ 7000 m/s and may start slower
and speed up as it goes off.
Peak pressure is only available very close to the explosive at the point
in time the detonation wave passes by , a measure called standoff , a
mere centimeter away , about the same as the size of the explosive
charge and is proportional with the size.
The pressure applied at some given distance is the same whether it occurs
all in one short instant over the entire object affected or instead arrives
distributed over a little longer moment sweeping across the surface of the
affected object as the explosive wave passes by.
A special case is a shaped charge which contrives to converge the force
generated by a circular explosive wave to focus and direct the impact of
gas or liner creating greater transient pressure. In a similar manner having
all the explosive initiated simultaneously has somewhat more impact in
the immediate vicinity from the expanding gas. The effect is not very
pronounced and would not be readily apparent form cursory inspection ,
and only by careful measurement in a test setup.
If a mile long column of explosive detonates simultaneously , only that
in the immediate vicinity of the affected object really matters. The power
produced is great but irrelevant for your purposes. Having the column
detonate progressively along it's length similarly only matters close by.
Work = Force x Distance
Force = Mass x acceleration
acceleration = velocity x velocity
so _
Work = Mass x velocity x velocity x Distance
Work or the damage depends on the efficiency of coupling the energy
dissipated , which is constant as it is fixed by the quantity of explosive.
Proximity to the explosion determines how much of an effect there is.
I wrote on this here _
www.sciencemadness.org/talk/viewthread.php?tid=11195#pid1374...
The line plotted in the charts represents the sequence of instants in
time ticked left to right , showing the pressure felt ( the vertical scale )
at some fixed distance from the detonated explosive.
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The_Davster
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Look up "diamond shaped charge" a 2D diamond-shaped piece of plastic explosive is initiated from both ends by equal lengths of detcord going to the
same initiatory source. As the shockwaves meet in the middle, a more powerful shockwave is produced, giving a stronger cutting effect.
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malford
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Quote: Originally posted by franklyn  | The pressure applied at some given distance is the same whether it occurs
all in one short instant over the entire object affected or instead arrives
distributed over a little longer moment sweeping across the surface of the
affected object as the explosive wave passes by. |
Firstly, thank you for the immense amount of information in the link.
My use of the meter-long cylindrical charge was purely illustrative. I was trying to make most clear the thought experiment.
Now, I must, inevitably incorrectly, disagree with your above statement.
It seems to me that the work done at some given distance, rather than the pressure applied, would be the same whether detonation happened discretely
or continuously. Let's walk through a thought experiment to illustrate my argument.
Let's say we are attempting to destroy a wall and that the force required to dislodge the studs within the wall is 400 newtons. This is axiomatic.
Let's say the explosive charge has an impulse at the wall's location of 100 newton-seconds, an arbitrary figure, regardless of how it is detonated. In
other words, the charge contains an amount of potential energy, say 5,000 newton-seconds, divided by the area and distance of the wall or by its
steradians, to give us the maximum amount of energy that can be applied to that location. This is axiomatic.
If complete detonation of our meter-long charge, which is parallel to the wall, occurs continually over 0.5 seconds then the length of time between
when the front of the shock wave reaches the wall and when the end of the shock wave reaches the wall is 0.5 seconds. This would give us a force of
200 newtons for 0.5 seconds for a total of 100 newton-seconds. This is axiomatic.
If complete detonation occurs continually over 0.1 seconds then the length of time between when the front of the shock wave reaches the wall and when
the end of the shock wave reaches the wall is 0.1 seconds. This would give us a force of 1000 newtons for 0.1 seconds for a total of 100
newton-seconds. This is axiomatic.
Only in the second scenario would the wall be destroyed. This is axiomatic.
The premises necessitate the conclusion.
Also, someone of your knowledge may be able to answer our question in the RE thread hovering around the top of this sub forum.
[Edited on 13-7-2013 by malford]
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Metacelsus
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| Quote: Originally posted by malford |
Let's say we are attempting to destroy a wall and that the force required to dislodge the studs within the wall is 400 newtons. This is axiomatic.
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1) Pressure is what destroys walls. Force is too ambiguous (how is the force applied?).
| Quote: Originally posted by malford | In other words, the charge contains an amount of potential energy, say 5,000 newton-seconds
[Edited on 13-7-2013 by malford] |
2) 5000 N*s is a measure of impulse, not potential energy. The impulse of an explosive is not necessarily constant.
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franklyn
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What you postulate is correct but it is only academic. You will note
that all explosions have round fireballs , that is because the gas
pressure is equalized as it expands. In practical terms there is no
discernible distinction unless you are very close to the explosive.
Two inches or less for a charge diameter of two inches. The effect
from tamping is very much greater. The best you can achieve is
detonation from the center of a ball shape. A series of these in a
line detonated simultaneously can produce an elongated shape
blast at a short distance. Tamping along one side will increase the
blast effect away from it.
This is an integral in time , where the area under the pressure line
represents the total energy , called the impulse. It can be tall and
narrow , which is what you imagine from instantaneous explosion ,
or stretched out much lower overall for a longer period. The area
is constant and the same in both cases. The peak pressure will be
the same in both cases , it will remain high for a bit longer in the
first case but the impulse will act for a shorter time. Producing the
energy sooner means it dissipates quicker also.
The_Davster points out that when waves converge , the
resultant is the sum of both. Spherical implosion is the extreme
case and then only 1/3 the energy is directed inward but the peak
pressure is many multiples of the blast directed outward. The time
or length of the impulse is the same for both since it is the same
explosion. The area distribution of the impulse will be different for
each during a short initial period until the pressure is equalized.
________________________________________________
R.E. ( relative effectiveness ) is a terminology I only learned this week.
I know what TNT equivalence is. It is just a figure of merit to compare
a given weight of different explosives and so for a desired effect the
amount can be adjusted. I believe the sand crush test is the basis of
comparison. A TNT equivalence of 1.5 implies 1/1.5 = 2/3 as much
explosive is needed as one unit of TNT. It's simply a ratio.
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malford
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| Quote: Originally posted by franklyn | What you postulate is correct but it is only academic. You will note
that all explosions have round fireballs , that is because the gas
pressure is equalized as it expands. In practical terms there is no
discernible distinction unless you are very close to the explosive. |
Let us replace our cylindrical charge with round charge. Again, we will compare two detonation methods.
In the first, we will detonate the spherical charge from a single point central to its mass. If the charge is one meter in diameter and the explosive
used has 8km/s VoD, the completion of detonation will require 1/4000th of one second. For any point away from the charge which the shock wave passes
by, the amount of time from when the beginning of the shock wave reaches this point and when the end of the shock wave reaches this point is 1/4000.
The wave obviously would not be uniform in pressure over time, but I believe we can discount that.
Any waves, whether they be electromagnetic or pressure in a gas, have a wavelength. 0.5 meters is the length of our three dimensional wave.
In the second, we will detonate a spherical charge by spiraling an EBW through it which will be exploded by using a bank of electrolytic capacitors.
No point in the mass of the charge is further than 0.1 meters from a point along the EBW. If the charge is one meter in diameter and the explosive
used has 8km/s VoD, the completion of detonation will require 1/80000th of one second. For any point away from the charge which the shock wave passes
by, the amount of time from when the beginning of the shock wave reaches this point and when the end of the shock wave reaches this point is 1/80000.
| Quote: Originally posted by Cheddite Cheese |
1) Pressure is what destroys walls. Force is too ambiguous (how is the force applied?).
2) 5000 N*s is a measure of impulse, not potential energy. The impulse of an explosive is not necessarily constant. |
Pressure is measured in units of force (PSI.) The force is applied to the wall by increasing the pressure on one side of the wall, the side close to
the blast, more quickly than can the air equalize and travel to the lower pressure on the other side. This generates force. Probably with the help of
frank, we could exactly calculate what the force would be on a given surface area at a given distance using a given explosive, but our thought
experiments serve as well without spending that much time.
An amount of energy will be able to produce an amount of force for an amount of time. The strength of the force in this case would depend on the
distance from the charge. This is because, as frank points out, the "fireball" is round and as the distance from the charge increases, the surface
area of the shockwave increases. Since the force is being spread out over an increasingly larger area as the shockwave propagates, the force that is
applied to any point will quickly decrease to maintain the set amount of kinetic energy produced by the blast.
You are correct, the pressure is not constant, but the variables we discuss will still affect the peak pressure of an irregular shockwave as well as
they affect our theoretical simple shockwave.
Regarding Davster, I see how two waves converge to a more powerful wave, but wave synthesis is not the goal here. Rather, it is the generation of the
wave initially.
[Edited on 14-7-2013 by malford]
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franklyn
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Peak pressure at the surface of an explosive is determined by the explosive.
Whatever it's particular detonation pressure may be. Measured in Gigapascals
or Barometric multiples. This may be exceeded due to acceleration from the
expansion in close proximity to the surface or by ' shaping ' to converge blast
from a concave form.
The blast from Instantaneous detonation need only cover the distance from
the surface to the target. All the energy is present so the peak pressure is
present for longer as the impulse is collected into a shorter interval.
Point initiation at the center adds the radius to the time to target. Peak pressure
is more brief. Impulse is longer as the energy is released over a longer interval.
The effect will be the same beyond a minimal distance from the explosive's surface.
Because the pressure equalizes throughout.
For very large charges ( few hundred kilos ) the leading shockwave traveling
beyond the fireball , quickly loses amplitude ( pressure ) because compression
of the air as it travels rapidly dissipates energy.
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