waveguider
July 28th, 2008, 04:05 PM
I know I am fairly new to this forum but I've had a few questions bug me for some time and even though I searched for posts which might answer the question I only had limited luck, ( ''Al Koholics'' post is attached below).
However I digress, I am rather interested in the safe observation of conventional and enhanced blast explosions, I like getting as close as possible but still keeping the probability of death to a minimum. I NEVER use any dense or metal casings so shrapnel is not a real problem...reflection of the shock wave is not a problem either as all tests are carried out in a flat barren area.
On the page linked below there lies a equation ''to calculate the pressure of an explosion at a target zone Hyde (1965) gives the following''
p=(4120/z^3) - (105/z^2) + (39.5/z)
where
z=(R/W^[1/3])
R is the outward radial distance in feet from the target
W is explosive mass in lb (equivalent TNT).
Source:http://www.navweaps.com/index_tech/tech-048.htm
I wanted to test and see if this equation would help me in determining some factors so I plugged in the following hypothetical numbers.
A 5kg charge of TNT and I want to know the pressure at 25m from the site of detonation:
5kg= ~11lb
25m= ~82ft
z= (82/2.224) ==> 36.8705
p= (4120/50123)- (105/1359.4) + (39.5/36.8705)
p= (0.0822) - (0.07724) + (1.07132)
p= 1.07 PSI
which I think is enough to knock a person over and there is a possibility of eardrum rupture.
That's fine, now imagine the person was 100m away not 25m ant the charge mass stayed the same, we have:
100m=~328ft
z=328/2.224 ===> ~147.48
p= (4120/3207873.12) - (105/21750.94457) + (39.5/147.48)
p= (0.00128434) - (0.004827377) + (0.2678)
p= 0.264 PSI
Now as you can see the pressure at 25m according to this equation is ~ 4 times that at 100m. This seems inherently wrong to me as I would have expected it to be 64 times higher because of the well known inverse cube law.
Does this equation really work or am I in error? I would greatly appreciate any help in determining what I did wrong or if there is a equation out there for finding the overpressure produced..
My second question is related to my love feeling the shock wave (blast wave) hit me when explosives or fuel detonate I would like a measure at what PSI of overpressure this phenomenon occurs at?
In regard to ''the probability of a fatal injury'' equation on this page
http://www.boomershoot.org/general/Effects.htm
Since it never really goes to zero, what is an optimum probability for safe viewing but to also ''feel'' the explosion? I know that duration of overpressure plays an enormous role in FAE's so if there is something like a graph of injury which has overpressure on one axis and duration on the other that would be immensely useful.
Thank you.
As a way to show that I am not simply wanting to be ''spoon fed'' I am attaching a member of this communities post (without their permission) After an extensive search the closest thing I got to what I was seeking is the following post:
(1) The threshold level of overpressure which is estimated to cause lung damage is about 68.9 kPa for a simple unreinforced, unreflected blast wave. There will be considerable variation in this value with differing conditions of exposure.
(2) The threshold value for eardrum rupture is probably around 22 kPa (0.2 atm) and that overpressure associated with a 50% probability of eardrum rupture ranges from 90 to 130 kPa (0.9 to 1.2 atm).
Sooo....if by saying it's comforting to know that people can survive exposure to pressures many times in excess of 1000psi...
you mean that it is possible that a human being may live and function biologically in an stable pressurized environment...then ok. But in the context of a blast wave, no. To rupture your eardrums I need 3.19 psi. 22000 pa aint all that much. To cause lung damage to you, I need to apply a wave that generates just a hair under 10 psi overpressure. Im not saying death...but lung damage. The point at which fatal lung (or other organ (especially lung and gastrointestinal tract)) injuries occur is a little over this. So I would have to disagree that a human body, in the context of our conversation here, cannot withstand pressures many times in excess of 1000 psi. More like 10.
April 21st, 2002, 09:12 PM
Al Koholic
Praetor''
The poster, unfortunately, doesn't seem to be active here anymore.
However I digress, I am rather interested in the safe observation of conventional and enhanced blast explosions, I like getting as close as possible but still keeping the probability of death to a minimum. I NEVER use any dense or metal casings so shrapnel is not a real problem...reflection of the shock wave is not a problem either as all tests are carried out in a flat barren area.
On the page linked below there lies a equation ''to calculate the pressure of an explosion at a target zone Hyde (1965) gives the following''
p=(4120/z^3) - (105/z^2) + (39.5/z)
where
z=(R/W^[1/3])
R is the outward radial distance in feet from the target
W is explosive mass in lb (equivalent TNT).
Source:http://www.navweaps.com/index_tech/tech-048.htm
I wanted to test and see if this equation would help me in determining some factors so I plugged in the following hypothetical numbers.
A 5kg charge of TNT and I want to know the pressure at 25m from the site of detonation:
5kg= ~11lb
25m= ~82ft
z= (82/2.224) ==> 36.8705
p= (4120/50123)- (105/1359.4) + (39.5/36.8705)
p= (0.0822) - (0.07724) + (1.07132)
p= 1.07 PSI
which I think is enough to knock a person over and there is a possibility of eardrum rupture.
That's fine, now imagine the person was 100m away not 25m ant the charge mass stayed the same, we have:
100m=~328ft
z=328/2.224 ===> ~147.48
p= (4120/3207873.12) - (105/21750.94457) + (39.5/147.48)
p= (0.00128434) - (0.004827377) + (0.2678)
p= 0.264 PSI
Now as you can see the pressure at 25m according to this equation is ~ 4 times that at 100m. This seems inherently wrong to me as I would have expected it to be 64 times higher because of the well known inverse cube law.
Does this equation really work or am I in error? I would greatly appreciate any help in determining what I did wrong or if there is a equation out there for finding the overpressure produced..
My second question is related to my love feeling the shock wave (blast wave) hit me when explosives or fuel detonate I would like a measure at what PSI of overpressure this phenomenon occurs at?
In regard to ''the probability of a fatal injury'' equation on this page
http://www.boomershoot.org/general/Effects.htm
Since it never really goes to zero, what is an optimum probability for safe viewing but to also ''feel'' the explosion? I know that duration of overpressure plays an enormous role in FAE's so if there is something like a graph of injury which has overpressure on one axis and duration on the other that would be immensely useful.
Thank you.
As a way to show that I am not simply wanting to be ''spoon fed'' I am attaching a member of this communities post (without their permission) After an extensive search the closest thing I got to what I was seeking is the following post:
(1) The threshold level of overpressure which is estimated to cause lung damage is about 68.9 kPa for a simple unreinforced, unreflected blast wave. There will be considerable variation in this value with differing conditions of exposure.
(2) The threshold value for eardrum rupture is probably around 22 kPa (0.2 atm) and that overpressure associated with a 50% probability of eardrum rupture ranges from 90 to 130 kPa (0.9 to 1.2 atm).
Sooo....if by saying it's comforting to know that people can survive exposure to pressures many times in excess of 1000psi...
you mean that it is possible that a human being may live and function biologically in an stable pressurized environment...then ok. But in the context of a blast wave, no. To rupture your eardrums I need 3.19 psi. 22000 pa aint all that much. To cause lung damage to you, I need to apply a wave that generates just a hair under 10 psi overpressure. Im not saying death...but lung damage. The point at which fatal lung (or other organ (especially lung and gastrointestinal tract)) injuries occur is a little over this. So I would have to disagree that a human body, in the context of our conversation here, cannot withstand pressures many times in excess of 1000 psi. More like 10.
April 21st, 2002, 09:12 PM
Al Koholic
Praetor''
The poster, unfortunately, doesn't seem to be active here anymore.