Energy Required to Destroy Stuff

This blog post is meant to be a reference for how much energy it takes to destroy materials such as rock, metals or some other material I haven’t thought of yet. There are various ways to go about destroying something.

A Note on how lots of people do this incorrectly…

I have seen some wildly incorrect ideas on how much energy it takes to destroy various materials on versus forums and the Internet in general, so I want to stamp out those wrong assumptions right now. At some point in time, people started believing that finding the shear strength of a material would lead to finding the total energy of, say, an explosive attack that destroyed some rock. Because you can actually convert pressure (Which the various strengths of materials are measured in) to j/cc (Joules per cubic centimeter). However, this is the pressure required to cut something, not destroy it. Shearing something is cutting it. That’s why we have a tool that, you guessed it! Cuts things! And they’re called shears. What we look at most of the time when it comes to destroying rocks is compressive strength. When a material like rock is being broken down into smaller pieces, it has been compressed, it’s as simple as that.

If you don’t believe me, read up.

Compressive Strength

Shear Strength

But I have seen people wildly misuse compressive strength as well. Some people seem to believe that using a rock’s compressive strength, for instance, you can see what it takes for rock to be ground up into a fine powder (AKA pulverization or spalling). If you read the link above, you’ll also find that this is incorrect. Compressive strength only tells us the maximum pressure a material can take before it fractures or fails in some way. It takes much, much more pressure and energy to convert a material to powder than it does to merely fracture it.

Finally, and most importantly, I want to note that things like compressive, shear, and tensile strength do not take into account impacts on materials. The tests to find these strengths slowly but surely apply the pressure on the material until it breaks. There is a huge difference in someone punching a piece of wood and trying to crush it without the help of an impact. Read up on impact loading for more info, but the short of it as that in most cases, knowing the compressive strength of a material won’t come in much handy. Not unless the feat we’re measuring is of a character slowly crushing something. The impact MUST be factored in, because it’s a lot tougher to crush something slowly than it is to do it with a high impact attack! Think of it this way. Which is harder? Damaging something by throwing a baseball at it, or damaging something by slowly pressing a baseball into it? The answer is obvious.

Fracturing/Fragmentation 

This produces visible breaks in the rock or whatever material is being destroyed and converting it into smaller substances. It always depends on the size of what’s being destroyed, but as long as it’s broken up into noticeable fragments, it can be considered fragmentation at the bare minimum.

Fragmentation of Rock is about 8 J/cc, per SD.net’s asteroid destruction calculator.

Fragmentation of iron has been found to be 5.83 Joules/in^3 or 96 J/cc.

From the book “The Engineering Record, Building Record and the Sanitary Engineer, Volume 56”:

iron

Ice has a fracture energy of .045 J/cc. Via this source:

ice1

ice2

1 * 1 * 5 = 5 cm

.226 / 5 = .045 J/cc

Explosive Fragmentation

Fragmentation merely breaks the rock into a bunch of pieces, but what about when the rock goes flying around?

Using this as a reference, we find that fracturing rock with explosives requires 15-20 joules/cc. I feel this is a good representation of most destructive anime/manga attacks.

Cratering/Violent Fragmentation

When a meteor hits the ground, it does several things at once: Vaporizes, melts, and fractures. The result is a huge mixed bag that none of the other definitions take into account. On some of the old forums I used to frequent, we called this “Violent Fragmentation”. Essentially rock is smashed so hard and with such energy that it flies very far, all over the place. Usually resulting in a large crater of sorts.

To figure out a number for this, I decided to use a real-life example. The Sedan Crater.

A 104 Kiloton Detonation in the ground created a violent explosion which combined everything mentioned above (melting, vaporizing, fracturing). The total amount of rock destroyed by the impact was around 5,000,000 m^3. Using these figures, we can get a j/cc amount easily.

5,000,000 m^3 = 5,000,000,000,000 cm^3

104 Kilotons of tnt = 435,136,000,000,000 J

J/cc is pretty explanatory!

435,136,000,000,000 / 5,000,000,000,000 = 87.027 J/cc

For the sake of simplicity, I’ll likely be rounding this to 87 J/cc when I use it for rock.

Cutting

Finding figures for cutting materials is tricky. Once again I must remind you not to fall into the trap of looking at a material’s shear strength, because that does not account for impacts. Impact cutting (Like a sword swing) accounts for most manga cutting feats, so we need to find something more specific. Here is what I have found so far.

Cutting of stone can be a range of different power requirements, but in most cases, at a minimum for something that instantly slices through stone, 727 J/cc is the bare minimum.

cutting

At this time, for non-rock/stone cutting that happens in an instant, I’m treating it as vaporization of that material on a small scale. This is because in real-life, vaporization of a material is the fastest way to cut it, and in these manga feats, most of the time it happens quickly.

Pulverizing

This is the act of crushing the material to the point where it is nothing but powder. If rocks, for instance, turn into nothing but a bunch of dust, this is pulverization. The powder/dust is important here. In anime and manga especially, I’ve never believed in just assuming something was pulverized because there’s no wreckage to be found from the damage. We’re dealing with magical explosions here, let’s not forget that. We are assuming they act exactly like explosions, or meteor impacts, or whatever, right? Well, it should be reflected, or at least stated in some way.

This one was incredibly hard to find sources on. As I mentioned before, others have incorrectly assumed that the compressive strength of a material leads to the results of what it takes to pulverize it. However, it turns out that’s only a small piece of the puzzle. In most anime/manga feats where converting material to dust is involved, it’s visible to the viewer that there’s some sort of haze. However, sometimes we’re told something has been pulverized and don’t see it. What I have recently found is that what the human eye can see depends on the size of the particle in microns. Anything below 10 microns in size will not be visible, while anything above will be. For this reason, I’ll be using the chart in the link above as my guide to how ground up a material has become.

Micron_Chart

The good news is that there’s the Kuznetsov Equation, which can determine the amount of energy required to break materials down into certain sizes. I’ve been looking for something like this for some time now.

The equation is as follows:

Kuznetsov1

k50= A(V/Q)^0.8 * Q^1/6

Kuznetsov2

Kuznetsov3

We can plug the numbers into this equation to find out how much kg of tnt is required for annihilating something into powder.

For example, let’s say a 5 m^3 rock was pulverized into the smallest we know of for limestone: 10 microns (As shown above) which is .001 cm. We’ll make A equal to 7 and assume this is a medium hardness rock.

.001 = 7(5/Q)^0.8 * Q^1/6

Q = 8,997,400 Kg of Tnt or almost 9 kilotons

This might seem extreme, but I also used the smallest possible micron level that the human eye can perceive on its own. The results vary greatly depending on the average size fragmented. For instance, if I did a still very tiny fragment size of 1 cm, the result becomes about 165 Kg of Tnt!

It seems to me that I’ll have to use close comparisons to real-life micron sizes in the case that I use pulverization, because it can really affect the results. Even still, I’m thankful to have found this formula.

Melting

Well this one is self-explanatory. If the material melts, it’s even more impressive than pulverizing something and takes lots of energy. It means the material has phased from a solid to a liquid.

Vaporization

Vaporization is, like melting, heat-based. But the difference here is that it takes things a step further in the phase conversion, making the material go from a solid to a gas. It takes a lot of heat energy to do this.

How will we get our figures?

A good question. In engineering and physics, materials are always given values (Based on extensive testing) for how much it can take before it fails, gets crushed, or has its’ phase converted.

specific energy

Explosions

I am keeping this blog post for future references that will be helpful for determining yields (Energy) of explosions. While it’s quite easy to find a calculator that will give us the energy of a nuclear bomb, finding ways to calculate explosions of a smaller variety is a lot more difficult. For instance, Sd.net’s handy nuke calculator  will not go any lower than 1 kiloton for a blast yield. In anime and manga, there are certainly explosive attacks below that range, and I feel it would be nice to have some light shed on those too.

Even more interesting is that, due to unique differences between types of explosions, we get different results depending on what kind of bomb we are comparing. For instance, according to the US government, stacking 500 tons of tnt and detonating it would produce similar results to a 1 kiloton nuke (Minus the radiation) in terms of shockwave and destruction.

nuke

If you’d like to read up more on that, DTRA History Series is where I got that snippet of info from. Just search for “Sailor Hat” in the document so you don’t have to read through hours of text. lol

It would seem though, that the size of the explosion depends on the material used and the speed at which the explosion happens. I am no expert on the subject (Why I’m posting it all here as a reminder), but I was a little surprised to find that the speed at which energy is released being slower would cause more destruction. On the other hand, Stacking a bunch of TnT vs. a much smaller and more convenient bomb that also spreads deadly radiation? From the perspective of a military, I suppose it still makes sense.

But because of the obvious differences in explosive types, I scowered the interwebs for some calculator or equation that could give us an explosive yield based on the size of the explosion. It has been surprisingly difficult to find anything (There are more minecraft articles about tnt than ACTUAL tnt for christ’s sake!), and I almost gave up. But after a while, I did find something that may just work.

It’s called Taylor’s Formula, named after a famous British fellow who was able to determine with surprising accuracy the yield of an atomic bomb just by getting a picture of it with some time frames.

taylor

Where E is energy (Duh), Po is density of the air, t is time after the explosion has formed, r is the radius of the explosion, and C is… erm… I’m just going to copy/paste this explanation:

“C is a dimensionless constant that is a function of the ratio of the specific heat of air at constant pressure to the specific heat of air at constant volume. The value of C is also affected by radiative losses, but for air, values of C of 1.00-1.10 generally give reasonable results.”

In short, it was previously thought that a generic specific heat ratio of 1.4 would be good to use, but Taylor showed through some complex formulas that this was not the case. Apparently the results of C are almost always between 1.00 and 1.10. It leaves margin for error, but if you want to take the easy route, it’s up to you. Figuring out that one constant alone is waaaay more difficult than the rest of the formula.

If you want to read up more on this, check out the wiki article, and if you really want to test your brain, have a look at Taylor himself explaining how he found the yield and the constant

But me? I’m actually going to use a variant of the formula discovered by Nigel B. Cook, a man who has put a lot of work into bomb research.

His blog details many of his findings, facts, and even corrections of mistakes we’ve made in regards to nukes, but specifically, I am looking at this article written by him.

nigel

All that has really changed here is that the constant has been replaced with extra figures in the formula. It looks rock solid to me, and is what I’ll be using to find explosive yields from here on out.

But if “It looks rock solid” isn’t good enough for you, how about a direct comparison? The Taylor Formula was derived from G.I. Taylor estimating the yield of the first ever nuclear bomb test (The Trinity Explosion). The yield of this nuke was later revealed to be around 20 Kilotons by the US government, but back in 1941, when the test was first performed, Taylor came up with a formula using dimensional analysis which, with only a time frame and the relative fireball radius during that time. Using these figures, Taylor estimated the yield of the nuke with surprising accuracy. It wasn’t 100% spot on, but short of doing some highly advanced physics based on direct testing of the nuke, this was by far the best estimate of its’ time. The US government was angered in 1951 when he released the formula and the yield of the bomb to the public. Back then, it was classified.

Anyway, Taylor was shown that the radius of the fireball was about 80 m within .006 seconds of the bomb going off.

As highlighted above, if we put his formula on paper with our given numbers…

Just to see how it affects the numbers, let’s make the Constant, C anywhere from the recommended .9-1.10. Let’s do one for .9, 1, and 1.10 respectively.

Taylor Formula (With .9 Constant)

(80/.9)^5 * 1.2 / (.0062^2)

5,549,324,256 * 1.2 / .00003844

6,659,189,107 / .00003844

173,235,928,000,000 J or 41.404 Kilotons

As we can see that’s way too high.

 

Taylor Formula (With 1 as Constant)

(80^5) * 1.2 / (.0062^2)

3,276,800,000 * 1.2 / .00003844

3,932,160,000 / .00003844

102,293,444,300,000 J or 24.448 Kilotons

Closer, but still a little high.

 

Taylor Formula (With 1.10 as Constant)

(80/1.10)^5 * 1.2 / (.0062^2)

2,034,596,842 * 1.2 / .00003844

2,441,516,211 / .00003844

63,514,989,870,000 J, or 15.18 Kilotons

And that sends the results too low. So obviously the correct constant is somewhere between 1 and 1.10, but the problem is that the constant is affected by a bunch of different variables that can change from explosion to explosion.

 

Taylor Formula (Nigel B. Cook Edit)

8*3.14*1.2*80^5 / [75(1.4-1).0062^2]

8*3.14*1.2*3276800000 / [75*.4*.00003844]

8*3.14*1.2*3276800000 / .0011532

98775859200 / .0011532

85,653,710,720,000 J or 20.471 Kilotons

This is remarkably close to the general figure that the US government, aided by advanced physics and first-hand testing, gave us. And is what I’ll be using going forward.

 

“But wait!“, I hear you cry.  Isn’t all of this used to determine the power of a nuke anyway? Why yes, yes it is. But it SHOULD apply to any explosion and here’s why: The time factor is what, as I mentioned and showed above, separates nuclear, high, and low explosives. How fast a bomb releases its’ energy can actually change the effects of it. It would be inconvenient, but according to the US government, detonating 500 tons of TNT would be similar (In blast effects only) to detonating a 1 kiloton nuke, and they specifically say it’s all to do with the speed at which that energy is released.

Now that brings us to how we find the value of time. And that all depends on what type of explosive we compare our calculated anime explosions to. Because there are figures given for the speed at which an explosion happens for each type. Therefore, with the size of the fireball and the speed, we can get time, no problem. It’s simply judging what type of explosion the calculated feat is closest to that matters here. And because of that, here are the standards I’ll be setting…

Comparing Explosion to a Nuke

The explosion would need to be large of course, with a shockwave that travels a long distance. If a large mushroom cloud is created by the explosion, that would also be grounds for comparing it to a nuke.

 

Comparing Explosion to a “High Explosive”

“High Explosive” being something like TNT. The key here is does the explosion create a shockwave? Does it make a large mushroom cloud? If it’s more of a smaller smoke plume, it’s more likely to be a high explosive.

 

Comparing Explosion to a “Low Explosive”

“Low Explosive” would be something like black powder. It’s actually not too hard to figure this one out. If no shockwave is given off, that’s indication that the explosion is on the slower side and most likely a “Low Explosive”.

burn rate.png

According to wikipedia (Referencing a book on explosives), Low Explosives like gunpowder and black powder can range anywhere from 171-631 m/s for their burn rate.

***More info to come***