Aces High Bulletin Board
General Forums => Aces High General Discussion => Topic started by: 2Slow on August 16, 2010, 10:45:57 PM
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What is the blast radius, in feet, for the 100, 250, 500, 1000, and 2000 pound bombs?
Heck, how about a chart for all the bomb ordinance's blast radius?
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What is the blast radius, in feet, for the 100, 250, 500, 1000, and 2000 pound bombs?
Heck, how about a chart for all the bomb ordinance's blast radius?
Looking forward for your work on it :)
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Someone get me the radii and I'll make up a chart, preferably the radius at wich it will destroy a town building. I'll even include dive bombing sights for the B-25, TBM, 110, and P-47/P-51.
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dive bomb sights=easy mode
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I have no idea but I know I killed 3 flakkers with one 1800kg bomb in a Stuka once.....then I LMAO all the way back to base. :lol
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I have no idea but I know I killed 3 flakkers with one 1800kg bomb in a Stuka once.....then I LMAO all the way back to base. :lol
:aok
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afaik splash damage is modelled correctly so there isnt a fixed blast radius, damage drops off exponentially with distance.
ie. blast damage = k/r2
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That is a good request that is needed. I have drop 1000 lbs bomb next to a GV and boom, then there is the 500 lbs bomb i have drop and no kill.
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preferably the radius at wich it will destroy a town building.
afaik splash damage is modelled correctly so there isnt a fixed blast radius, damage drops off exponentially with distance.
ie. blast damage = k/r2
Per Nemisis, that is what I am looking for. I understand the model as explained. So my question is, what is the blast radius of the different ord's for destroying a town building? There must be a point from the center of the blast from which the effect will not destroy a town building.
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you could try testing it in the TA, although I cant really see how this info would be useful :headscratch:
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IIRC,a delay of .54 will have 500# bombs overlap in spalsh damage.
Actually Rolex would have the exact delay but I think it's .54 for 500lbers.
Not sure it this helps but it will give you a starting point to experiment and find the answer yourself.
:salute
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you could try testing it in the TA, although I cant really see how this info would be useful :headscratch:
Yes, I know I can obtain this information by testing offline. I will, if need be. I just thought one might know or have access to this information in the real deal. Since I am pretending to be doing the real deal...
I could tell you why I want this information but virtual COMSEC, OPSEC, and Most Secret/Top Secret/NOFORN classification prevents this. :)
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IIRC,a delay of .54 will have 500# bombs overlap in spalsh damage.
Actually Rolex would have the exact delay but I think it's .54 for 500lbers.
Not sure it this helps but it will give you a starting point to experiment and find the answer yourself.
:salute
wouldnt it depend on the speed of the plane because if you are going 400mph in a 234 .54 seconds is quite a ways apart
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In the B-25, I use a delay of 0.17 when droping the '250s, and a delay of 0.27 with the 500lbers. For the Ju-88, I use a delay of 0.12 with the 50kg bombs. I've even found a method that doesn't require the use of the bombsight.
And I would test this in the TA if I could reliably determin the distance from the center of the blast. I have no measurments for the dimesions of the town buildings, so I can't even get a rough guess.
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I would test this in the TA if I could reliably determin the distance from the center of the blast. I have no measurments for the dimesions of the town buildings, so I can't even get a rough guess.
... which begs the question: if you have no way of reliably measuring it, how is knowing the exact distances in yards going to help?
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We have the exact length of a tank, and so you could convert the radius at which it will kill a building into 'tank lengths'. I agree that this wouldn't be some great tool that changes bombing in AH, but it may prove usefull in setting delays in bombers.
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Well the way I understand it is that the small bombs have small blast radius and as the bombs get bigger the blast radius grows exponentially. i.e. little bomb little blast. Big bomb big blast. :x
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I have no idea but I know I killed 3 flakkers with one 1800kg bomb in a Stuka once.....then I LMAO all the way back to base. :lol
I remember getting 18 GV kills when I dropped the 4k bombs from a set of lancs right when a ports VH popped. Best moment Iv ever had in AH. I just wish I could find the screen shot. :(
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lol, too bad you didn't drop 2 bombs from a 110C. Fighter perks are actually worth something.
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... which begs the question: if you have no way of reliably measuring it, how is knowing the exact distances in yards going to help?
It is I who was looking for the information. I do have a reliable method of measurements. For example, if one wants to know the measurements of a base hanger it takes some math (excel spreadsheet) and some recon. Fly the length of the bases main runway (or any runway). Go to bombsight, note e6b true speed and time the travel from one end to the other. Then do the math. For example, at 257mph one is moving 376.93 feet per second. Multiply this by the timed transit over the runway length. Now one knows how long the runway is. Print out the base map. Measure the runway in mm. Divide the runway length by the mm measurement. Now one has feet per mm. Measure the hanger with the ruler in mm. Do the math again. Multiply the hanger measurement in mm by the feet per mm. Now you have the size of the hanger.
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It is I who was looking for the information. I do have a reliable method of measurements. For example, if one wants to know the measurements of a base hanger it takes some math (excel spreadsheet) and some recon. Fly the length of the bases main runway (or any runway). Go to bombsight, note e6b true speed and time the travel from one end to the other. Then do the math. For example, at 257mph one is moving 376.93 feet per second. Multiply this by the timed transit over the runway length. Now one knows how long the runway is. Print out the base map. Measure the runway in mm. Divide the runway length by the mm measurement. Now one has feet per mm. Measure the hanger with the ruler in mm. Do the math again. Multiply the hanger measurement in mm by the feet per mm. Now you have the size of the hanger.
You have too much time on your hands sir!
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lol, too bad you didn't drop 2 bombs from a 110C. Fighter perks are actually worth something.
Bomber perks are worth a lot when you love flying the Ar234 as much as I do.
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It is I who was looking for the information. I do have a reliable method of measurements. For example, if one wants to know the measurements of a base hanger it takes some math (excel spreadsheet) and some recon. Fly the length of the bases main runway (or any runway). Go to bombsight, note e6b true speed and time the travel from one end to the other. Then do the math. For example, at 257mph one is moving 376.93 feet per second. Multiply this by the timed transit over the runway length. Now one knows how long the runway is. Print out the base map. Measure the runway in mm. Divide the runway length by the mm measurement. Now one has feet per mm. Measure the hanger with the ruler in mm. Do the math again. Multiply the hanger measurement in mm by the feet per mm. Now you have the size of the hanger.
Yikes! :O
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You have too much time on your hands sir!
Perhaps indeed.
My goal has been to conduct air raids in the same fashion as the Mighty 8th did in the real deal. Air Raid = 1 pass over target, one salvo. I think of it as bombing for effect and not for score. With old town, I and two others, could reduce the town to 90+% destroyed. With the new town, I and two others, can reduce the town to the same degree as we did with the old town. Just a bit more challenging. :)
Thus my question in regards to ordnance blast radius effect on town buildings. One can then plan bomb runs better.
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An exponential decay is y~e^-x rather than y~1/x^2. If you use an inverse square law to model blast damage you will have infinite bomb damage at zero radius (the whole dividing by zero thing).
For this case let our damage function be D(r)=D0 e^[-c r]: where D is the damage at radius r, D0 is the damage at radius 0 (the max damage of the bomb), c is some coefficient of decay that we have to determine from dropping a bomb and fitting the curve to our observations, and r is the radius.
The first thing that has to be done is to determine c. To do this we will take a 1000 lb bomb and drop it near something that takes a known amount of damage to destroy (town building = 250lbs if I remember right). So we make the drop and find that the maximum distance a 1000 lb bomb can land and take out a building is 100'. We now have all we need to know! We throw some algebra at it to give us c = -1/r Ln[D/D0], plug in the numbers, c = -1/100 Ln[250/1000] = 7.21*10^-3 (1/ft). Then we plug our new found c into our general formula and have a function of damage at a given radius for 1000 lb bombs. If things are modeled consistently in the game this function will hold for all bomb types. So for this example we would have
Damage = Bomb weight e^[(7.21*10^-3) r].
I made a nice plot of the function but I cant figure out how to post and image on here so the function and Idea will have to stand alone.
REMEMBER THIS IS AN EXAMPLE AND SOMEONE NEEDS TO MAKE A MEASUREMENT SO THE CURVE CAN BE PROPERLY FIT. So you just need a known bomb weight destroying an object of known hardness at a known maximum distance and then it can be fit.
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Perhaps indeed.
My goal has been to conduct air raids in the same fashion as the Mighty 8th did in the real deal.
You're gonna need a whole lot more planes :P
j/k, I get what you're saying. Bombing is more of a hobby to me than my main focus so it seems overly complicated to get into all that minutiae but if bombing is your thing, go for it.
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Ordtech-industries shows the following specs for modern general purpose bombs on their website:
pressure at 50 ft pressure at 100 ft armor penetration (distance not given)
Mk 82, 500lb 17 psi 5 psi 32 mm
Mk 83, 1000lb 26 psi 8 psi 46 mm
Mk 84, 2000lb 35 psi 11 psi 51 mm
I have not located any information like this for WWII bombs yet.
http://www.ordtech-industries.com/2products/Bomb_General/Bomb_General.html
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An exponential decay is y~e^-x rather than y~1/x^2. If you use an inverse square law to model blast damage you will have infinite bomb damage at zero radius (the whole dividing by zero thing).
For this case let our damage function be D(r)=D0 e^[-c r]: where D is the damage at radius r, D0 is the damage at radius 0 (the max damage of the bomb), c is some coefficient of decay that we have to determine from dropping a bomb and fitting the curve to our observations, and r is the radius.
The first thing that has to be done is to determine c. To do this we will take a 1000 lb bomb and drop it near something that takes a known amount of damage to destroy (town building = 250lbs if I remember right). So we make the drop and find that the maximum distance a 1000 lb bomb can land and take out a building is 100'. We now have all we need to know! We throw some algebra at it to give us c = -1/r Ln[D/D0], plug in the numbers, c = -1/100 Ln[250/1000] = 7.21*10^-3 (1/ft). Then we plug our new found c into our general formula and have a function of damage at a given radius for 1000 lb bombs. If things are modeled consistently in the game this function will hold for all bomb types. So for this example we would have
Damage = Bomb weight e^[(7.21*10^-3) r].
I made a nice plot of the function but I cant figure out how to post and image on here so the function and Idea will have to stand alone.
REMEMBER THIS IS AN EXAMPLE AND SOMEONE NEEDS TO MAKE A MEASUREMENT SO THE CURVE CAN BE PROPERLY FIT. So you just need a known bomb weight destroying an object of known hardness at a known maximum distance and then it can be fit.
An easy way to do this would be to drop a single 1000 lb bomb in the middle of the high building density area of a town. Fly back over and take a screen shot of the damage with something of known length in the frame (a tank as suggested earlier). Then the tank can be used as a scale and the radius at which a 1000lb bomb's damage decays to the 250 lb damage level by measuring the distance at which the last town building is destroyed.
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It is I who was looking for the information. I do have a reliable method of measurements. For example, if one wants to know the measurements of a base hanger it takes some math (excel spreadsheet) and some recon. Fly the length of the bases main runway (or any runway). Go to bombsight, note e6b true speed and time the travel from one end to the other. Then do the math. For example, at 257mph one is moving 376.93 feet per second. Multiply this by the timed transit over the runway length. Now one knows how long the runway is. Print out the base map. Measure the runway in mm. Divide the runway length by the mm measurement. Now one has feet per mm. Measure the hanger with the ruler in mm. Do the math again. Multiply the hanger measurement in mm by the feet per mm. Now you have the size of the hanger.
How can you know if the map is in correct scale? It might not be, I always thought it's just a way of locating the various buildings, I don't think it's in scale. Anyone knows about it?
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How can you know if the map is in correct scale? It might not be, I always thought it's just a way of locating the various buildings, I don't think it's in scale. Anyone knows about it?
Based on my work and results, it is scale.
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An exponential decay is y~e^-x rather than y~1/x^2. If you use an inverse square law to model blast damage you will have infinite bomb damage at zero radius (the whole dividing by zero thing).
sure about that? my gut says inverse square is a good, if simplified model. I'm not too bothered by infinite "damage" for r=0, although I'm too tired atm to work out exactly why (something to do with a finite amount of energy and a sphere with surface area of 0) :headscratch:
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An easy way to do this would be to drop a single 1000 lb bomb in the middle of the high building density area of a town. Fly back over and take a screen shot of the damage with something of known length in the frame (a tank as suggested earlier). Then the tank can be used as a scale and the radius at which a 1000lb bomb's damage decays to the 250 lb damage level by measuring the distance at which the last town building is destroyed.
Good idea! In case anyone is interested, 1 mm of measurement equals 33 scale feet. What I have done is insert the town.bmp into a MS Word document. I do not expand it or shrink. This is the map I work off of. Oh yeah, I am working off of a printed map document.
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sure about that? my gut says inverse square is a good, if simplified model. I'm not too bothered by infinite "damage" for r=0, although I'm too tired atm to work out exactly why (something to do with a finite amount of energy and a sphere with surface area of 0) :headscratch:
Energy falls off exponentially in a dampened medium in every case I have ever come across (bullets, electrons, electromagnetic radiation, water waves...) So if damage is related to energy it will also follow an exponential decay. We are essentially describing a shock-wave which would be modeled as a propagating dirac-delta function. (read:"moving high pressure front") In one of your earlier responses you called it an exponential decrease but then gave an inverse square relation for the equation. Now if you want to talk about the amount of shrapnel moving through the surface of some given sphere that would follow an inverse square law.
An argument against an inverse square law being appropriate is the zero radius problem, if you cant predict how much damage the bomb does exactly where it lands then how can you predict it anywhere else with any confidence? If you tried to do anything with it you would also find that working with an exponential decay mathematically is much easier than working with an inverse square law anyway. So if simplicity of the model alone were to be used as a measure of the usefulness of a model I would go with the exponential decay.
Finite energy in a sphere with zero surface area? Sounds like we are starting to talk about black holes now. How did you find out about the black hole bomb that the bish nation is developing?!?! :noid
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An exponential decay is y~e^-x rather than y~1/x^2. If you use an inverse square law to model blast damage you will have infinite bomb damage at zero radius (the whole dividing by zero thing).
For this case let our damage function be D(r)=D0 e^[-c r]: where D is the damage at radius r, D0 is the damage at radius 0 (the max damage of the bomb), c is some coefficient of decay that we have to determine from dropping a bomb and fitting the curve to our observations, and r is the radius.
The first thing that has to be done is to determine c. To do this we will take a 1000 lb bomb and drop it near something that takes a known amount of damage to destroy (town building = 250lbs if I remember right). So we make the drop and find that the maximum distance a 1000 lb bomb can land and take out a building is 100'. We now have all we need to know! We throw some algebra at it to give us c = -1/r Ln[D/D0], plug in the numbers, c = -1/100 Ln[250/1000] = 7.21*10^-3 (1/ft). Then we plug our new found c into our general formula and have a function of damage at a given radius for 1000 lb bombs. If things are modeled consistently in the game this function will hold for all bomb types. So for this example we would have
Damage = Bomb weight e^[(7.21*10^-3) r].
I made a nice plot of the function but I cant figure out how to post and image on here so the function and Idea will have to stand alone.
REMEMBER THIS IS AN EXAMPLE AND SOMEONE NEEDS TO MAKE A MEASUREMENT SO THE CURVE CAN BE PROPERLY FIT. So you just need a known bomb weight destroying an object of known hardness at a known maximum distance and then it can be fit.
I missed a set of parenthesis around the 1/100 when I ran the numbers for the "c" calculation through my calculator so the correct value for c is 0.0139 1/ft.
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Energy falls off exponentially in a dampened medium in every case I have ever come across (bullets, electrons, electromagnetic radiation, water waves...) So if damage is related to energy it will also follow an exponential decay.
decay/damping is certainly a factor but this is surely more of a propagation problem. I was thinking in terms of the amount of energy in the shock front hitting a fixed size object at varying distance, or equivalently the overpressure. I did forget that shock fronts arent elastic sound waves (whose energy does decrease at 1/r2) but are entirely different (single front, traveling in one direction and much faster than sound). a little sniffing around the net reveals that the shock wave energy starts to reduce with distance at 1/r3 (Ive woken up a little since dinner but why its a cubic deal is beyond me at this point) and the rate gradually, continuously drops to 1/r2 as the shock wave slows and eventually becomes a sound wave.
I still dont have a problem with r=0, expand the gas in a sphere with 0 volume and you will have infinite overpressure.
interesting problem :D
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:headscratch:
Think I just entered a foreign language class.
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:)
this page (http://www.makeitlouder.com/document_bombshockwaveestimation.html) has some good stuff on the physics of explosions.
there is an example there for a 1000lb bomb. the critical distance (ie. where the shock wave becomes a sound wave and the overpressure degenerates at 1/r2) is about 2000m.
so the overpressure degenerates at a rate of 1/r3 at r=0 down to a rate of 1/r2 at r=2000m. so for our town buildings splash, the short distance from the explosion means that 1/r3 is a decent approximation :aok
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In the real world over pressure may be the main mechanism for imparting damage to soft structures like non-reinforced town buildings. In the game the damage a structure can withstand is not measured in force/area of overpressure. It is measured in equivalent pounds of explosives. It takes 2750lbs of damage to take down a hangar not 14 psi of overpressure. If things in game were damaged considering over pressure alone, you could drop a 100lb bomb and destroy anything in the game just by hitting it directly since, as you say, it has infinite over pressure at r=0. In my experience a single 100 lb bomb does not kill much even with direct hits.
Until a r^-3 model can predict how much damage the bomb will inflict on a structure how useful is it?
If you can make it work so that Damage = (appropriate constants)/r^3 with appropriate units and all while satisfying the r=0 boundary condition I would be thoroughly impressed.
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:headscratch:
Think I just entered a foreign language class.
I just scrached my arse too.
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well the shock front propagates from the explosive/air boundary outwards, if you can find an existing bomb with r=0 I will be impressed :D
did you check the link?
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The bomb does not have radius equal to zero but r=0 is the origin for the explosion, so the casing air interface right before the casing ruptures and the shock wave begins propagating is r=0.
I read the link, it has a lot of good information about shock waves but no information about fragmentation or heat generation so it is missing a large portion of what causes damage when a bomb explodes.
The energy of a shock wave will dissipate much more rapidly than the energy of any bomb fragments so if we are looking for the single damaging effect that has the largest radius it would be the shrapnel anyway.
Q.E.D.
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Insert Quote
Quote from: tf15pin on August 18, 2010, 11:57:21 AM
An easy way to do this would be to drop a single 1000 lb bomb in the middle of the high building density area of a town. Fly back over and take a screen shot of the damage with something of known length in the frame (a tank as suggested earlier). Then the tank can be used as a scale and the radius at which a 1000lb bomb's damage decays to the 250 lb damage level by measuring the distance at which the last town building is destroyed.
Good idea! In case anyone is interested, 1 mm of measurement equals 33 scale feet. What I have done is insert the town.bmp into a MS Word document. I do not expand it or shrink. This is the map I work off of. Oh yeah, I am working off of a printed map document.
Based on my efforts at the bombing range (offline town bombing) I have a damage radius for 1000 pound bombs. If one measures from the inside wall to inside wall of destroyed buildings then I have it at 82.5feet. From outside edge to outside edge then the radius is 115.5 feet. 2.5mm or 3.5mm (length of 5 or 7mm divided by 2) measurements on that map as described above.
B17/B24 formation would have a destruction swath of 437' or 13.44mm on the map.
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1000lb town building at 41.25' nice :aok
you didnt happen to test 500lbers did you? :D
The bomb does not have radius equal to zero but r=0 is the origin for the explosion,
yup (in the centre of our spherical explosive charge).
... so the casing air interface right before the casing ruptures and the shock wave begins propagating is r=0.
nope. the explosive/air boundary at which the shock front starts to propagate through the air is at r=d/2, where d=diameter of the explosive charge.
overpressure instantly collapses buildings, not frag or heat. and I assume our bombs are GP bombs where the casing is just thick enough to deliver the explosive charge to target, designed for overpressure, not frag like a nade or cluster bomblet, or heat like an incendiary.
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1000lb town building at 41.25' nice :aok
you didnt happen to test 500lbers did you? :D
yup (in the centre of our spherical explosive charge).
nope. the explosive/air boundary at which the shock front starts to propagate through the air is at r=d/2, where d=diameter of the explosive charge.
overpressure instantly collapses buildings, not frag or heat. and I assume our bombs are GP bombs where the casing is just thick enough to deliver the explosive charge to target, designed for overpressure, not frag like a nade or cluster bomblet, or heat like an incendiary.
I confused some numbers and words. I have corrected below:
Based on my efforts at the bombing range (offline town bombing) I have a damage radius for 1000 pound bombs. If one measures from the inside wall to inside wall of destroyed buildings then I have blast diameter 165.5' (5mm on scaled map) when measured from the inside facing wall to the other inside facing wall. This gives one a radius of 2.5mm (83') for inside to inside. From outside edge to outside edge then the blast diameter is 231' (7mm on scaled map.) This gives one a radius of 3.5mm (115.5') for outside to outside.
B17/B24 formation would have a destruction swath of 437' or 13.44mm on the map.
I have not tested a 500 pound bomb sortie. Want to bet it turns out to be 50% of the 1000 lb. 250 lb would be 50% of the 500 lb. 100 lb would be a 10% of a 1000 lb bomb. Way to simple so probably not correct. ;)
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if the 1000lber can just drop a building at r=~25m (82'), a 500lber should do it at ~21m (68') using 1/r3. of course we dont know how the blast damage is actually modelled in AH so who knows. :)
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That is a good request that is needed. I have drop 1000 lbs bomb next to a GV and boom, then there is the 500 lbs bomb i have drop and no kill.
I had 8 500lbers dropped within a few hundred yards of me by F100s, didn't kill me, but I thought I was dyin. :lol
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:salute Red
After seeing the F100 at Thunder Over Michigan a few weeks ago, I've been reading about the Mistys using them as fast FAC over the trail. Amazing stuff in an extremely touchy plane from the sounds of it! :O
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I confused some numbers and words. I have corrected below:
Based on my efforts at the bombing range (offline town bombing) I have a damage radius for 1000 pound bombs. If one measures from the inside wall to inside wall of destroyed buildings then I have blast diameter 165.5' (5mm on scaled map) when measured from the inside facing wall to the other inside facing wall. This gives one a radius of 2.5mm (83') for inside to inside. From outside edge to outside edge then the blast diameter is 231' (7mm on scaled map.) This gives one a radius of 3.5mm (115.5') for outside to outside.
B17/B24 formation would have a destruction swath of 437' or 13.44mm on the map.
I have not tested a 500 pound bomb sortie. Want to bet it turns out to be 50% of the 1000 lb. 250 lb would be 50% of the 500 lb. 100 lb would be a 10% of a 1000 lb bomb. Way to simple so probably not correct. ;)
Good job collecting some data!
I recalculated the c value using your small radius to be on the safe side (c = 0.0169). So D=D0 e^(-c r) where D is the damage at radius r in lb, D0 is the damage of the bomb at radius=0 or the weight of the bomb, e is the natural exponent ~2.7, c is the damping constant, and r is the radius. So with this we can predict how much damage a given bomb will do at a given radius from the impact. If you want to calculate the radius at which a given bomb will destroy an object of given hardness you can do some algebra and get r = -(1/c) ln(D/D0) where the values are the same except for D you put in the object hardness. ln is the natural logarithm.
Some values we get from this when considering bombing hangers are: hit withing 22' with a 4000lb bomb, 5' with 3000 lbs of bombs, or 46' with 6000 lbs of bombs (i.e. .salvo 2 with 1000lb'ers in a 3 plane formation). These values will be on the low side since we used the low values from the data you collected when we determined c.
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actually none of us can predict the damage at r, because none of us knows how HT has modelled it, might even be a discrete function ... :uhoh
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There is a ceiling on the amount of destruction that a single weapon can produce, because the radius of destruction for any given level of blast increases approximately as the cube root of the energy output.
With 7 weapons dropped in a circle, at the midpoint between zero point distance, blast waves would collide at the overpressure points. On interior points of the circle, the blast waves, depending upon the angles at which they met, would be reflected or merge. The key feature of encirclement in terms of blast forces is that the energy on the interior of the circle would be concentrated in a comparatively small area instead of dissipating from the zero point at a 128° angle over an ever larger area [the angle formed at an intersections of a 7 sided polygon]. The result is greater blast destruction on the interior of the circle than is commonly suggested when contemplating single weapon effects.
Paul Cooper's text Explosives Engineering (1996) [p. 216]:
Take the amount of TNT - lets say 1,000 lbs. 4184 Joules = 0.001 Kton. Y^0.4), where Y is yield.
Approximate over pressures and distances for .001 Ktons as follows:
15 Psi - 0.02 miles
10 Psi - 0.02 miles
5 Psi - 0.03 miles
1 Psi - 0.01 miles
0.25 PSI - 0.19 miles
For the METRIC guys:
103 KPa - 0.02 Km
68.9 KPa - 0.03 Km
48.3 Kpa - 0.04...
You get the point. But keep in mind 2PSi is enough to destroy reinforced concrete buildings - but the blast radius falls off rather quickly. Tanks and other cans - at least now a days are sealed up pretty tight - so a close blast might be reflected out. If the tank were not sealed - the shock front would reflect off the interior of the tank walls - and amplify the blast effects to the crew. Hope that helps.
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Tell ya what once yall get this sorted out just tell me how high I gotta get to not blow my boston3s up while hittin a cv on the deck with 500lbers. :devil
Thanks. :aok