How to Calculate the Power of a Shockwave and Other Applications (Concept)

***A) INTRODUCTION***

There are countless feats in fiction that involve a shockwave generated, but it could be really difficult to calculate them because of the inner properties of a shockwave itself. In other cases, we could want to scale the durability of a character form an explosion, but they are outside of the blast radius and, like we saw here, the formulas for explosion won't work, so we have to use the shockwave generated by it. Whatever the reason is to calculate them, we need a good method and we will use Kingery's research.

***B) THE FORMULA***

This article gives us a formula along with several coefficients in function of what want to calculate. What we need is the incident pressure and we will calculate it in kPa. But first we will stop at the general formula:

# p

e

A
B
l
n
Z
C
(
l
n
Z

)
2

D
(
l
n
Z

)
3

E
(
l
n
Z

)
4

F
(
l
n
Z

)
5

G
(
l
n
Z

)
6

{\displaystyle p = e^{A + BlnZ + C(lnZ)^2+D(lnZ)^3+E(lnZ)^4+F(lnZ)^5+G(lnZ)^6}}

Where:

p: Pressure of the shockwave, in kPa.

A, B, C, D, E, F, and G: Dimensionless coefficients.

Z: Scaled distance [R/Y^(1/3)], in m/kg^(1/3).

* R: Distance from the epicenter to the target, m.
* Y: Yield of the explosion, kg of TNT.

Now we can simplify that formula, because it is really complex, while taking into account that for incident pressure we only have coefficients till E:

# p

e
A

⋅

e

B
l
n
Z

⋅

e

2
C
l
n
Z

⋅

e

3
D
l
n
Z

⋅

e

4
E
l
n
Z

{\displaystyle p = e^A\cdot e^{BlnZ}\cdot e^{2ClnZ} \cdot e^{3DlnZ} \cdot e^{4ElnZ}}

If we use properties form both exponentials and logarithms, we get:

# p

e
A

⋅

Z
B

⋅

Z

2
C

⋅

Z

3
D

⋅

Z

4
E

{\displaystyle p = e^A \cdot Z^B \cdot Z^{2C} \cdot Z^{3D}\cdot Z^{4E}}

And moreover:

# p

e
A

⋅

Z

B
2
C
3
D
4
E

{\displaystyle p = e^A \cdot Z^{B+2C+3D+4E}}

And that would be the simplified general formual to calculate the pressure from a shockwave generated from an explosion, but we will sustitute the coefficient's letters for some or other numbers depending on the value of Z. The article mentioned above has 3 cases:

**1. Z: 0.2-2.9 (not included)**

# p

1353.704246

Z

2.1436

{\displaystyle p = \frac{1353.704246}{Z^{2.1436}}}

**2. Z: 2.9-23.8 (not included)**

# p

1985.845407

Z

2.20514

{\displaystyle p = \frac{1985.845407}{Z^{2.20514}}}

**3. Z: 23.8-198.5 (included)**

# p

425.6425885

Z

1.4066

{\displaystyle p = \frac{425.6425885}{Z^{1.4066}}}

***C) CALCULATION METHOD***

Now that we have a formula, we need to know how to apply it properly. Here are the steps to follow:

1. Before anything, we need the yield of the explosion in kg of TNT. In order to calculate that, you have this article.
2. Then, we have to get the distance from the target to the epicenter of the explosion, in meters.
3. Now that we have those two values, we have to calculate Z by dividing what we got in point 2. by what we got at the point 1. to the one third (R/Y^1/3) and see between which two values is in, so we use one of the three cases mentioned above.
4. If all the steps are done correctly, we will get the pressure that the target received from the shockwave in kPa, so we need to multiply the result by 1000 to have it in Pa.

Extra: From this point, we probably don't want the result in pressure, since that is difficult to scale. In that case, we can get the Newtons of force that the target endured just by multiplying the result we got in Pa by the frontal surface area of the target.

***D) OTHER APPLICATIONS***

So far we've seen how to calculate the strength of a shockwave generated by an explosion in one point, but we can find some other types of feats that involve shockwaves that aren't caused by an explosion, per example: two characters' attacks collide. For those cases, we can make some changes on the formula, so we can get Y in function of R and p:

# p

e
A

⋅

(

R

Y

1

/

3

)

B
2
C
3
D
4
E

{\displaystyle p = e^A \cdot \left ( \frac{R}{Y^{1/3}} \right )^{B+2C+3D+4E}}

# p

e
A

⋅

(

R

B
2
C
3
D
4
E

Y

B
2
C
3
D
4
E

3

)

{\displaystyle p = e^A \cdot \left ( \frac{R^{B+2C+3D+4E}}{Y^{\frac{B+2C+3D+4E}{3}}} \right )}

Y

B
2
C
3
D
4
E

3

=

e
A

⋅

R

B
2
C
3
D
4
E

p

{\displaystyle Y^\frac{B+2C+3D+4E}{3}= \frac{e^A \cdot R^{B+2C+3D+4E}}{p}}

# Y

(

e
A

⋅

R

B
2
C
3
D
4
E

p

)

3

B
2
C
3
D
4
E

{\displaystyle Y = \left ( \frac{e^A \cdot R^{B+2C+3D+4E}}{p} \right )^{\frac{3}{B+2C+3D+4E}}}

And that is the general formula to calculate the energy of the impact (although it is made for explosions, we can equate them) in function of p and R.