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When Should Volumetric Destruction Be Used?

Flashlight237

Flashlight237

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So here's something I thought of that recently concerned me. Hell, I even made that mistake myself. So, let's get down to the nitty-gritty of this topic.

I shall put up some examples that would involve things that qualify as structures...

Okay, so the first calc I'll bring up is one of my own: https://vsbattles.fandom.com/wiki/U...n:_Koshi-tan_et_al_Tanks_a_Gym_Roof_Explosion

So, as you can see, I've rated a gym explosion at 19 tons of TNT, which is City Block level. However, when I redid the same calc for the OBD using the blast pressure for sending people flying (25 psi), here's what I got:

Gymnasiums typically contain basketball courts: https://en.wikipedia.org/wiki/Basketball_court
We can see that there are two basketball courts laid along the side. Assuming two meters of space between each court, we can assume a radius of greater than or equal to 16.24 meters. Given people were sent flying, we would assume a blast pressure of 25 psi: https://apps.dtic.mil/sti/tr/pdf/AD0067365.pdf

This can be interpreted as either 10.3041 psi (25-14.6959) or 25 psi of overpressure. Let's do the math for both.

Reverse-engineering the United Nations' explosion formula, here's the math:

16.24^3*((27136*0.7104426862+8649)^(1/2)/13568-93/13568)^2=0.12780383124 tons of TNT

to

16.24^3*((27136*1.72369+8649)^(1/2)/13568-93/13568)^2=0.47192667517 tons of TNT

The next calc I'll bring up is a Senran Kagura calc rated at Low 7-C: https://vsbattles.fandom.com/wiki/U...Crimson_Homura_vs_Ultimate_Asuka_clash_redone

The calc is based on destroying a building that's a mere 46.4 meters tall. However, when I use the Explosion Yield Formula to calc the same feat, here's what I got.:

Building Height: 46.4 meters
Explosion Radius (assumed hemisphere based on height): 23.2 meters
Overpressure: 8 psi (maximum for complete destruction of wood buildings)

23.2^3*((27136*0.551581+8649)^(1/2)/13568-93/13568)^2=0.24973780471 tons of TNT

The third and last calc I'll put up is this Total Drama calc involving a Jumbo Jet: https://vsbattles.fandom.com/wiki/U...ra_Commits_Arson_And_Blows_Up_A_Plane_(S3E23)

This one is rated at 212 tons based on volumetric calculations. Keeping in mind that the size of the plane is an underestimate considering its earliest appearance in the cartoon it's from showed it to be much larger, I'll instead opt to use the size of a Boeing 747 since that's the first thing people think of when it comes to jumbo jets: https://en.wikipedia.org/wiki/Boeing_747#Specifications

Assuming the length of the plane as the explosion diameter here's what I got:

Plane Length: 70.7 meters
Explosion Radius: 35.35 meters
Overpressure: 5 psi (universal injuries, as Sierra was wheelchair-bound after the blast)

35.35^3*((27136*0.344738+8649)^(1/2)/13568-93/13568)^2=0.40688619338 tons of TNT

All three of the recalcs I've provided using the United Nations's Explosion Yield formula show that the volumetric calculations used largely overestimate the power needed to destroy structures, even in the last case where the size of the structure is very clearly underestimated. After all, buildings and jumbo jets are not monoliths. As such, I think it's high time we answered the question... "When should we use volumetric destruction?"
 
I’m under the case that we should NEVER use volumetric destruction, because our destruction values are incorrect, not only is it more complex than just using compressive strength and shear strength, but it’s also just flat out flawed, and we have accepted this, therefore using it should be the last thing we do. We should always use the explosion formula, if it’s an explosion.
 
As KLOL said, I'd prioritize the explosion formula. Volumetric destruction is a relatively new thing that has consistently returned higher yields. In the face of a calc method that I would deem more reliable, we should use the more reliable method. I think more or less all calcs should be approached with a descending order of reliability in mind- the most relevant, reliable calc method should be used where applicable.
 
I agree with KLOL, but I think the Inverse Square Law page would need to be changed in that case. The explanation should change to something that doesn't use volumetric destruction, and it should be noted that the Inverse Square Law shouldn't be used with volumetric destruction.

That seems to be what I'm feeling here.

A ground explosion with a radius of 5 meters has exactly enough energy so when it hits a brick with an area of 0.07116953508 square meters and a volume of 0.0010692559 cubic meters, it vaporizes it. How much energy does this explosion hold?

  • First we use the known value for vaporization onto the brick, which is 25700 joules per cubic centimeter. It requires 27.4798773 megajoules to vaporize said brick.
  • When dealing with ground explosions, one should use a hemisphere as a basis for the explosion’s shape. The area of a hemisphere with a radius of 5 meters is 157.08 square meters
  • 157.08 square meters/0.07116953508 square meters * 27.4798773 megajoules
  • If we plug in these numbers, it results in 60.651501 gigajoules, or 14.496 tons of tnt, which is City Block level
 
TheRustyOne said:
I agree with KLOL, but I think the Inverse Square Law page would need to be changed in that case. The explanation should change to something that doesn't use volumetric destruction, and it should be noted that the Inverse Square Law shouldn't be used with volumetric destruction.

That seems to be what I'm feeling here.

A ground explosion with a radius of 5 meters has exactly enough energy so when it hits a brick with an area of 0.07116953508 square meters and a volume of 0.0010692559 cubic meters, it vaporizes it. How much energy does this explosion hold?

  • First we use the known value for vaporization onto the brick, which is 25700 joules per cubic centimeter. It requires 27.4798773 megajoules to vaporize said brick.
  • When dealing with ground explosions, one should use a hemisphere as a basis for the explosion’s shape. The area of a hemisphere with a radius of 5 meters is 157.08 square meters
  • 157.08 square meters/0.07116953508 square meters * 27.4798773 megajoules
  • If we plug in these numbers, it results in 60.651501 gigajoules, or 14.496 tons of tnt, which is City Block level
Click to expand...
I typically don't use inverse-square law in these cases as those jack up the value by absurd amounts. So yeah, I'm fine with that recommendation.
 
Last edited:
I don't have anything to add... Rusty, KLOL and Bambu already said what I would've prolly said (ergo, I agree with them).
 
TheRustyOne said:
I agree with KLOL, but I think the Inverse Square Law page would need to be changed in that case. The explanation should change to something that doesn't use volumetric destruction, and it should be noted that the Inverse Square Law shouldn't be used with volumetric destruction.

That seems to be what I'm feeling here.

A ground explosion with a radius of 5 meters has exactly enough energy so when it hits a brick with an area of 0.07116953508 square meters and a volume of 0.0010692559 cubic meters, it vaporizes it. How much energy does this explosion hold?

  • First we use the known value for vaporization onto the brick, which is 25700 joules per cubic centimeter. It requires 27.4798773 megajoules to vaporize said brick.
  • When dealing with ground explosions, one should use a hemisphere as a basis for the explosion’s shape. The area of a hemisphere with a radius of 5 meters is 157.08 square meters
  • 157.08 square meters/0.07116953508 square meters * 27.4798773 megajoules
  • If we plug in these numbers, it results in 60.651501 gigajoules, or 14.496 tons of tnt, which is City Block level
Click to expand...

This is fine but isn't like for the most part exactly what we already do?
 
In this case, I think we need to add more PSI values. Some explosions vaporize certain structures and objects. And we don’t have any PSI values for this level of destruction.
 
I know I am late but I’d only use volumetric destruction when there is a clear shot of the destruction, like a crater or a hole in a wall.

Calculating the volume of a building is tricky and most of the times not worth the effort for such unreliable method.

If an explosion happened then use the explosion formula, if you see the crater then calc the crater as well. If the explosion happened on a building then use the explosion.
 
I feel in the case of blasting a hole through a wall or leaving a crater, shear strength and uniaxial compressive strength would probably be somewhat okay for that right?

I do agree that just using the explosion formulas is the safest and most reasonable bet though. Should a rule or some form of note be added to the calculations page stating that simply using the explosion formulas is preferable and more reliable than volumetric destruction in terms of accuracy?

This reddit thread also gives some slightly interesting insight on shear and compressive strength.
 
TheRustyOne said:
I agree with KLOL, but I think the Inverse Square Law page would need to be changed in that case. The explanation should change to something that doesn't use volumetric destruction, and it should be noted that the Inverse Square Law shouldn't be used with volumetric destruction.

That seems to be what I'm feeling here.

A ground explosion with a radius of 5 meters has exactly enough energy so when it hits a brick with an area of 0.07116953508 square meters and a volume of 0.0010692559 cubic meters, it vaporizes it. How much energy does this explosion hold?

  • First we use the known value for vaporization onto the brick, which is 25700 joules per cubic centimeter. It requires 27.4798773 megajoules to vaporize said brick.
  • When dealing with ground explosions, one should use a hemisphere as a basis for the explosion’s shape. The area of a hemisphere with a radius of 5 meters is 157.08 square meters
  • 157.08 square meters/0.07116953508 square meters * 27.4798773 megajoules
  • If we plug in these numbers, it results in 60.651501 gigajoules, or 14.496 tons of tnt, which is City Block level
Click to expand...
I feel like melting, vaporization, atomization, and subatomic destruction would likely be somewhat exempt from inaccuracy in volumetric destruction since they are literally the values for the methods of destruction they are stated to be outright, unlike fragmentation, violent fragmentation and pulverization, to my knowledge anyways.

So, they might actually be accurate for measuring an explosion (when they're actually applicable).

But I do suppose an example using volumetric destruction might be confusing to some.
 
Assuming the j/cc values are right, using j/cc x volume seems like the best / most simple method?

I would think it gives a good minimum for a value?

Why is that not right?
 
Ezaru said:
Assuming the j/cc values are right, using j/cc x volume seems like the best / most simple method?

I would think it gives a good minimum for a value?

Why is that not right?
Click to expand...
I mean, take a look at the feats measured using volumetric measurement, then take a look at the quick calcs I did in the OP determining the explosion yields needed to destroy those same exact things.
 
Flashlight237 said:
I mean, take a look at the feats measured using volumetric measurement, then take a look at the quick calcs I did in the OP determining the explosion yields needed to destroy those same exact things.
Click to expand...

Well, the first one seems to be calculating two different things, the destruction and the people being like, sent flying.

For the second one, are you sure that the standard for "complete destruction" and "pulverization" are comparable? If not I can see why one is higher, and does it account for the plaster and stone in the building mentioned in the calc?

For the third, are you sure a 5psi rating is fair for the explosion? What if it was higher?

It seems like destroying the edges of the plane would take more than injuring a person

But overall I am not asking if you are correct, but why that would be the case?

Like, you would think "It takes x to destroy y amount of z, this destroys 100y amount of z, therefore it is 100x" is the most simple logic and thus the most correct to use?

Idk I am only an amateur for calc stuff
 
Flashlight237 said:
I mean, take a look at the feats measured using volumetric measurement, then take a look at the quick calcs I did in the OP determining the explosion yields needed to destroy those same exact things.
Click to expand...
Can you explain how sending people flying is 25 psi?

Is destroying the plane (which seemingly wasn't even at the epicenter of the blast) really only 5 psi?
 
Moon_Crasher said:
Can you explain how sending people flying is 25 psi?

Is destroying the plane (which seemingly wasn't even at the epicenter of the blast) really only 5 psi?
Click to expand...
Though to be fair, even with 20 psi, destroying the plane is only around 3 tons of tnt (but the explosion did seem to be way bigger).
 
Moon_Crasher said:
Can you explain how sending people flying is 25 psi?
Click to expand...
Ah! That can be easily be explained. Page 16 of this military document PDF says that sending someone flying at 3 m/s is 25 psi of blast pressure, although it doesn't really specify if it meant the pressure itself or the overpressure:

Moon_Crasher said:
Is destroying the plane (which seemingly wasn't even at the epicenter of the blast) really only 5 psi?
Click to expand...
The overpressure rating chosen is based on the effects it had on people (in this case leaving Sierra wheelchair-bound for a time), which equates to "universal injuries." If I based it on the material used to build the thing, it would've been 4 psi, and that would've been pushing it since the plane wasn't exactly well-built.
Moon_Crasher said:
Though to be fair, even with 20 psi, destroying the plane is only around 3 tons of tnt (but the explosion did seem to be way bigger).
Click to expand...
I mean if you want the most accurate results from an explosion, you gotta look directly at what got destroyed.
 
Flashlight237 said:
Ah! That can be easily be explained. Page 16 of this military document PDF says that sending someone flying at 3 m/s is 25 psi of blast pressure, although it doesn't really specify if it meant the pressure itself or the overpressure:


The overpressure rating chosen is based on the effects it had on people (in this case leaving Sierra wheelchair-bound for a time), which equates to "universal injuries." If I based it on the material used to build the thing, it would've been 4 psi, and that would've been pushing it since the plane wasn't exactly well-built.

I mean if you want the most accurate results from an explosion, you gotta look directly at what got destroyed.
Click to expand...
How would I get overpressure from pressure?

Yeah, I do think 5 is probably the most accurate since it says most buildings collapse at that point and a commercial airplane is roughly building sized and it's not like the plane was reduced to unrecognizable rubble, so at least 4 to 5 psi is best.

(The blast does seem to be way bigger than just the plane's length but there doesn't seem to be anything to measure it against and the damage didn't seem to extend too far beyond the plane, so just using the plane's length as the diameter is probably for the best).
 
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