- 6,014
- 2,122
Soo i have found another method for calculate the yield of an explosion but IDK if this is legit soo i need you for this
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This formula is applicable?
- Thread starter The_Causality
- Start date
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- Tags
- calculations
- 9,978
- 2,015
We already have two ways of calculate conventional explosion (although one of them isn't added to the guide), what's the principle behind this equation?
- 6,014
- 2,122
- Thread starter
- #3
it seems like it's the same thing except it's for the nuclear explosion (with a time frame)
- 9,978
- 2,015
Nuclear explosion is the same airburst formula, we don't have original equation, but we have the one for 20 psi, some online calculator use 4.7 - 5 psi too.
- 6,014
- 2,122
- Thread starter
- #5
I see, but I would just like to know if this method is usable (just out of curiosity)
- 9,978
- 2,015
If the principle behind it are bounds with nuclear explosions (yield and range) then it should work, however, you would need to elaborate in what means any of the variable of that formula.
- 6,014
- 2,122
- Thread starter
- #7
Okay, thanks you friend
