Energy needed to vaporize soil/dirt?

[Neutralino]

I've been trying to calc this following feat where a large volume of earth is vaporized, forming the below crater. Unfortunately, the Calculations article doesn't list a vaporization value for soil, which is exactly the value I need. I've figured out the volume, which is 25,444,197 cm3 (assuming I've measured the depth of the crater properly; the angle makes it difficult to determine where to stop measuring), so I'm stuck here.

 

[Neutralino]

Is there a way to find the energy of the attack if it's treated as a standard explosion instead of vaporization?

 

[KLOL506]

Is there a way to find the energy of the attack if it's treated as a standard explosion instead of vaporization?

Yes.

Explosion Yield Calculations

First, you have to find the exact page, or scene where an explosion is shown as a feat. Depending on the environment where the explosion occurred, you may have to scale the size of a smaller object that is shown in the map. Let's do an example... An explosion of unknown size occurs in this...

vsbattles.fandom.com

 

[Neutralino]

Yes.

Explosion Yield Calculations

First, you have to find the exact page, or scene where an explosion is shown as a feat. Depending on the environment where the explosion occurred, you may have to scale the size of a smaller object that is shown in the map. Let's do an example... An explosion of unknown size occurs in this...

vsbattles.fandom.com

This formula?
W = R^3*((27136*P+8649)^(1/2)/13568-93/13568)^2
Is it valid? How was it mathematically determined? Why isn't the (1/2)/13568-93/13568)^2 just simplified to 0.00000000273? This part is entirely composed of constants without variables. The 13568-93 is especially unnecessary since it can just be 13475.

 

[KLOL506]

This formula?
W = R^3*((27136*P+8649)^(1/2)/13568-93/13568)^2
Is it valid?

Yes, it was taken from a UN paper on explosives.

How was it mathematically determined? Why isn't the (1/2)/13568-93/13568)^2 just simplified to 0.00000000273? This part is entirely composed of constants without variables. The 13568-93 is especially unnecessary since it can just be 13475.

@DontTalkDT should be able to explain more, he was the one who got the formula from the papers.

 

[DontTalkDT]

Why isn't the (1/2)/13568-93/13568)^2 just simplified to 0.00000000273? This part is entirely composed of constants without variables.

That (1/2) is an exponent for a start.
You could put the 93/13568 into a numerator, but that just gives you even more number and the same amount of operations.
The 93/13568 could be shortened. Then again, not sure if it's a number with finite digits after the comma, so it would potentially lower the precision.

The 13568-93 is especially unnecessary since it can just be 13475.

PEMDAS.

 

[Neutralino]

That (1/2) is an exponent for a start.
You could put the 93/13568 into a numerator, but that just gives you even more number and the same amount of operations.
The 93/13568 could be shortened. Then again, not sure if it's a number with finite digits after the comma, so it would potentially lower the precision.


PEMDAS.

Let me try to make it more readable.

W = R^3*((27136*P+8649)^(1/2)/13568-93/13568)^2

becomes

W = R³ x (27136P+8649)⁽⁽⁰.⁵÷¹³⁵⁶⁸⁾⁻⁽⁹³÷¹³⁵⁶⁸⁾⁾²

The exponent itself is squared, thus becoming 6.3803215449169652027872453212928e-14? Also, do you have a link to the UN paper where this formula can be verified?

 

[DontTalkDT]

Let me try to make it more readable.

W = R^3*((27136*P+8649)^(1/2)/13568-93/13568)^2

becomes

W = R³ x (27136P+8649)⁽⁽⁰.⁵÷¹³⁵⁶⁸⁾⁻⁽⁹³÷¹³⁵⁶⁸⁾⁾²

No, you still get the order of operations wrong. First parenthesis, then exponents and then division. Don't do division before exponents. (and edited exponents can't be copied into calculators, so I avoid them).

With Parenthesis for everything it's: R^3*((((((27136*P)+8649)^(1/2))/13568) - (93/13568))^2)

The exponent itself is squared, thus becoming 6.3803215449169652027872453212928e-14?

None of them can be multiplied out, cause there are + and - in the brackets that get exponentiated. (well, you could with the binomic formula, but then you end up with two P terms, which just makes things worse)

Also, do you have a link to the UN paper where this formula can be verified?

I'm not the one who actually got the formula. I think I can find the thread it was added with some work, but don't have the time right now.

 

[Neutralino]

No, you still get the order of operations wrong. First parenthesis, then exponents and then division. Don't do division before exponents. (and edited exponents can't be copied into calculators, so I avoid them).

With Parenthesis for everything it's: R^3*((((((27136*P)+8649)^(1/2))/13568) - (93/13568))^2)


None of them can be multiplied out, cause there are + and - in the brackets that get exponentiated. (well, you could with the binomic formula, but then you end up with two P terms, which just makes things worse)


I'm not the one who actually got the formula. I think I can find the thread it was added with some work, but don't have the time right now.

Thank you, all those parentheses seem messy at first but actually make the formula much easier to use.

VSB would benefit by sourcing important and potentially contentious information like this formula. I'm sure I'm not the only person who wondered how it was found, and an actual scientific paper would make it all the more credible.

 

[DontTalkDT]

I believe this is it.

http://publications.jrc.ec.europa.eu/repository/bitstream/JRC87200/lbna26456enn.pdf

 

[Neutralino]

I believe this is it.

http://publications.jrc.ec.europa.eu/repository/bitstream/JRC87200/lbna26456enn.pdf

Thank you very much. It must be the second formula on page 25? The formula VSB uses is the same formula but solved for W, or tons of TNT, instead of air pressure? I tried to solve it myself, but it's been years since I'd needed to do this kind of math...so I just plugged it into WolframAlpha and had it solve the formula for me. The result is



Is it equivalent to R^3*((((((27136*P)+8649)^(1/2))/13568) - (93/13568))^2)? Sorry, I've never been much of a math person...

 

[KLOL506]

I believe this is it.

http://publications.jrc.ec.europa.eu/repository/bitstream/JRC87200/lbna26456enn.pdf

Better add that to the reference section of the Explosion Formula page before it gets lost to time again.

 

[DontTalkDT]

Thank you very much. It must be the second formula on page 25? The formula VSB uses is the same formula but solved for W, or tons of TNT, instead of air pressure? I tried to solve it myself, but it's been years since I'd needed to do this kind of math...so I just plugged it into WolframAlpha and had it solve the formula for me. The result is



Is it equivalent to R^3*((((((27136*P)+8649)^(1/2))/13568) - (93/13568))^2)? Sorry, I've never been much of a math person...

I believe it's formula 10 solved for W.

 

[Neutralino]

I believe it's formula 10 solved for W.

Yes, that one. It's the second formula on page 25.

P=(6784W/R^3)+(93(√W/R^3)). I tried to solve it for W, but I couldn't figure out how to take the second W out of the radical.

Also, do you know where the air blast formula Y = ((x/0.28)^3)/1000 comes from?