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Brode's Method

Furudo_Erika

Furudo_Erika

1,276
328
Question: "Why isn't Brode's Method talked about on this site?"

Brode's Method is an actual formula used to calculate the energy from explosions:
Brode

Where:
E is the explosion energy in joules
P1 is starting or ambient pressure (1 atm; would typically be 101,325 pascals unless otherwise specified)
P2 is the burst pressure (We typically use 20 psi, so 137,895 pascals)
V is the volume of the gas (Would typically be a hemisphere)
γ is the heat capacity ratio of the gas in question. (1.4 for air)


One more question on the side:
For explosions on the ground we use the following formula:
W = R^3*((27136*P+8649)^(1/2)/13568-93/13568)^2

But, what exactly did this formula even come from? I cannot find any kind of reference to it at all, nor do I see any links or explanations to exactly what this formula even is.
 
Flashlight237 said:
Honestly, no idea. Even with the push for references, the wiki isn't exactly the best at citing stuff.
Click to expand...
Sadly, this is true.

Also. Brode's Method is more versatile due to the fact it uses ambient pressure in the equation...something the wiki's method cannot accomplish.
Which is nice for planets with differing atmospheric pressures, as unsurprisingly...atmospheric pressure, plays quite a big part in explosion calculation.
 
Second22 said:
vsbattles.fandom.com

Explosives and Shockwaves

vsbattles.fandom.com vsbattles.fandom.com
Click to expand...
Thanks.
Doesn't help with getting seizures looking at that mess of an equation tho, and a brain aneurysm just writing it down.

Brode's Method is way easier to read, and is actually useable for air-bursts as well (just use a full sphere instead of a hemisphere).
Which again...is something the ground explosion formula cannot accomplish, so a different formula is used for air-bursts.
 
Last edited:
Also, you multiply the result by 2 if it happens at ground level when using Brode's Method, as:
"When the explosion occurs at ground level, the calculated value of the energy is conventionally multiplied by 2 to take into consideration ground effects, like the reflection of the shock wave"

-The Italian Association of Chemical Engineering, 2013
 
Though, let's just see how different the two are mathematically.

I'm keeping the pressure the same: 20 PSI/137,895 Pa, and I'm using radii of 5m, 10m, 50m, 100m, 500m, 1,000m, and 1,609.34m/1 mile).
I'm also going to be using ground explosions for Brode's Method, as the other can only happen on the ground (so the 2x applies here).

I'm not gonna type everything, but you can do the calcs yourself for confirmation:

(5m)
Brode's: 0.01144 Tons (9-A)
Wiki's: 0.01005 Tons (9-A)

(10m)
Brode's: 0.09153 Tons (9-A)
Wiki's: 0.08037 Tons (9-A)

(50m)
Brode's: 11.44121 Tons (8-B)
Wiki's: 10.04605 Tons (High 8-B+)

(100m)
Brode's: 91.52967 Tons (8-B+)
Wiki's: 80.36844 Tons (8-B+)

(500m)
Brode's: 11.44121 Kilotons (7-C)
Wiki's: 10.04605 Kilotons (7-C)

(1,000m)
Brode's: 91.52967 Kilotons (7-C+)
Wiki's: 80.36844 Kilotons (7-C+)

(1,609.34m)
Brode's: 381.50947 Kilotons (High 7-C)
Wiki's: 334.98776 Kilotons (High 7-C)

Huh, all things considered, they are fairly consistent with each other.
 
I remember asking @DontTalkDT to give me the ground-based explosion formula and he gave it to me, but I forgot where. Guess I can ask him again.

EDIT: NVM I GOT BEAT.
 
BTW, does this work for underwater explosions, since the pressure there would actually make the explosions way more lethal by being able to transmit greater pressures at longer distances?
 
So a little googling for this method tells me this:
It is worth noting that the Brode equation is only a rough approximation of the reality, since it represents the energy required to compress an ideal gas, at constant volume, from P0 to P1 . Of course, this is not the case in a real explosion, where the gas reaches an equilibrium condition after expansion from an initial
volume at P1 to a final volume at P 0 . In addition, the expansion energy of the gas, even if properly calculated, still overestimates the actual explosion energy dissipated in the surrounding environment, because it would neglect several accompanying phenomena (like the energies required to rupture the vessel, to launch the fragments of the containment vessel, etc., which can amount up to 50 % of the total internal energy) as well as other aspects (such as the deformation of the vessel fragments, non-equilibrium effects, and so on).
Click to expand...
So, doesn't sound like a great method.

That aside, I don't doubt you could find a dozen explosion formulas, but there is no point. We have two that work and cover the relevant cases and when you start switching between formulas to get higher results that just becomes wank (and unscientific). Keeping it to one formula for each case is just more consistent. And might as well keep it to the established ones.

Furudo_Erika said:
Also, you multiply the result by 2 if it happens at ground level when using Brode's Method, as:
"When the explosion occurs at ground level, the calculated value of the energy is conventionally multiplied by 2 to take into consideration ground effects, like the reflection of the shock wave"

-The Italian Association of Chemical Engineering, 2013
Click to expand...
This is a rough estimation method that can be used for many explosion formulas. We don't use it because we have better alternatives. As in, a formula specifically designed for the specific ground / air case should be more accurate than such a simple geometrical adjustment.

KLOL506 said:
BTW, does this work for underwater explosions, since the pressure there would actually make the explosions way more lethal by being able to transmit greater pressures at longer distances?
Click to expand...
Seeing how that quote I cited above says something regarding ideal gas law I would guess that it doesn't work underwater.
Furudo_Erika said:
Also. Brode's Method is more versatile due to the fact it uses ambient pressure in the equation...something the wiki's method cannot accomplish.
Which is nice for planets with differing atmospheric pressures, as unsurprisingly...atmospheric pressure, plays quite a big part in explosion calculation.
Click to expand...
Actually, the pressure variables in our methods are overpressure, so I don't think that's necessarily true. Of course, the validity of either method to other planets is somewhat questionable, since I'm fairly sure neither method was designed for it.
(btw. since P2 is not the overpressure in Brode, in my understanding, 20psi is possibly not the right value to use?)
 
Think you might have made some mistakes here: Assuming this all hemispheric as alluded in the beginning. Also I'm not sure if you converted the psi/pascals to bars as I found it to be the value used in these examples here, here and here

Furudo_Erika said:
(5m)
Brode's: 0.01144 Tons (9-A)
Wiki's: 0.01005 Tons (9-A)
Click to expand...
V= 261.8m^3 (hemisphere of 5 radii)
p1= 101,325 pascals (1.01325 bar)
p2= 137,895 pascals (1.37895 bar)
γ= 1.4

E=(1.37895-1.01325)*261.8/(1.4-1)= 239.35065 Joules
239.35065x2= 478.7013 Joules (9-C)
Furudo_Erika said:
(10m)
Brode's: 0.09153 Tons (9-A)
Wiki's: 0.08037 Tons (9-A)
Click to expand...
V= 2094.4m^3 (hemisphere of 10 radii)
p1= 101,325 pascals (1.01325 bar)
p2= 137,895 pascals (1.37895 bar)
γ= 1.4

E=(1.37895-1.01325)*2094.4/(1.4-1)= 1914.8052 Joules
1914.8052x2= 3829.6104 Joules (9-C)
Furudo_Erika said:
(50m)
Brode's: 11.44121 Tons (8-B)
Wiki's: 10.04605 Tons (High 8-B+)
Click to expand...
V= 261799m^3 (hemisphere of 50 radii)
p1= 101,325 pascals (1.01325 bar)
p2= 137,895 pascals (1.37895 bar)
γ= 1.4

E=(1.37895-1.01325)*261799/(1.4-1)= 1914.8052 Joules
239349.73575x2= 478699.4715 Joules (9-B)
Furudo_Erika said:
(100m)
Brode's: 91.52967 Tons (8-B+)
Wiki's: 80.36844 Tons (8-B+)
Click to expand...
V= 2094395m^3 (hemisphere of 100 radii)
p1= 101,325 pascals (1.01325 bar)
p2= 137,895 pascals (1.37895 bar)
γ= 1.4

E=(1.37895-1.01325)*2094395/(1.4-1)= 1914800.62875 Joules
1914800.62875x2= 3829601.2575 Joules (9-B)
Furudo_Erika said:
(500m)
Brode's: 11.44121 Kilotons (7-C)
Wiki's: 10.04605 Kilotons (7-C)
Click to expand...
V= 261799388m^3 (hemisphere of 500 radii)
p1= 101,325 pascals (1.01325 bar)
p2= 137,895 pascals (1.37895 bar)
γ= 1.4

E=(1.37895-1.01325)*261799388/(1.4-1)= 239350090.479 Joules
239350090.479x2= 478700180.958 Joules (9-A)
Furudo_Erika said:
(1,000m)
Brode's: 91.52967 Kilotons (7-C+)
Wiki's: 80.36844 Kilotons (7-C+)
Click to expand...
V= 2094395102m^3 (hemisphere of 1000 radii)
p1= 101,325 pascals (1.01325 bar)
p2= 137,895 pascals (1.37895 bar)
γ= 1.4

E=(1.37895-1.01325)*2094395102/(1.4-1)= 1914800722 Joules
1914800722x2= 3829601444.01 Joules (8-C)
Furudo_Erika said:
(1,609.34m)
Brode's: 381.50947 Kilotons (High 7-C)
Wiki's: 334.98776 Kilotons (High 7-C)
Click to expand...
V= 8729754508m^3 (hemisphere of 1,609.34 radii)
p1= 101,325 pascals (1.01325 bar)
p2= 137,895 pascals (1.37895 bar)
γ= 1.4

E=(1.37895-1.01325)*8729754508/(1.4-1)= 7981178058.94 Joules
7981178058.94x2= 15962356117.9 Joules (High 8-C)

Maybe @KLOL506 or @DontTalkDT can double check if I followed the formula correctly and if it should be bars or pascals as I've seen only bars used. Also what about the other methods mentioned in the articles?
 
PowerToScale said:
Think you might have made some mistakes here: Assuming this all hemispheric as alluded in the beginning. Also I'm not sure if you converted the psi/pascals to bars as I found it to be the value used in these examples here, here and here

V= 261.8m^3 (hemisphere of 5 radii)
p1= 101,325 pascals (1.01325 bar)
p2= 137,895 pascals (1.37895 bar)
γ= 1.4

E=(1.37895-1.01325)*261.8/(1.4-1)= 239.35065 Joules
239.35065x2= 478.7013 Joules (9-C)

V= 2094.4m^3 (hemisphere of 10 radii)
p1= 101,325 pascals (1.01325 bar)
p2= 137,895 pascals (1.37895 bar)
γ= 1.4

E=(1.37895-1.01325)*2094.4/(1.4-1)= 1914.8052 Joules
1914.8052x2= 3829.6104 Joules (9-C)

V= 261799m^3 (hemisphere of 50 radii)
p1= 101,325 pascals (1.01325 bar)
p2= 137,895 pascals (1.37895 bar)
γ= 1.4

E=(1.37895-1.01325)*261799/(1.4-1)= 1914.8052 Joules
239349.73575x2= 478699.4715 Joules (9-B)

V= 2094395m^3 (hemisphere of 100 radii)
p1= 101,325 pascals (1.01325 bar)
p2= 137,895 pascals (1.37895 bar)
γ= 1.4

E=(1.37895-1.01325)*2094395/(1.4-1)= 1914800.62875 Joules
1914800.62875x2= 3829601.2575 Joules (9-B)

V= 261799388m^3 (hemisphere of 500 radii)
p1= 101,325 pascals (1.01325 bar)
p2= 137,895 pascals (1.37895 bar)
γ= 1.4

E=(1.37895-1.01325)*261799388/(1.4-1)= 239350090.479 Joules
239350090.479x2= 478700180.958 Joules (9-A)

V= 2094395102m^3 (hemisphere of 1000 radii)
p1= 101,325 pascals (1.01325 bar)
p2= 137,895 pascals (1.37895 bar)
γ= 1.4

E=(1.37895-1.01325)*2094395102/(1.4-1)= 1914800722 Joules
1914800722x2= 3829601444.01 Joules (8-C)

V= 8729754508m^3 (hemisphere of 1,609.34 radii)
p1= 101,325 pascals (1.01325 bar)
p2= 137,895 pascals (1.37895 bar)
γ= 1.4

E=(1.37895-1.01325)*8729754508/(1.4-1)= 7981178058.94 Joules
7981178058.94x2= 15962356117.9 Joules (High 8-C)

Maybe @KLOL506 or @DontTalkDT can double check if I followed the formula correctly and if it should be bars or pascals as I've seen only bars used. Also what about the other methods mentioned in the articles?
Click to expand...
Brode's Law uses pascals, not PSI or bars.
So, you use 137,895 - 101,325 Pa, not 1.37895 - 1.01325 bars
 
PowerToScale said:
Think you might have made some mistakes here: Assuming this all hemispheric as alluded in the beginning. Also I'm not sure if you converted the psi/pascals to bars as I found it to be the value used in these examples here, here and here

V= 261.8m^3 (hemisphere of 5 radii)
p1= 101,325 pascals (1.01325 bar)
p2= 137,895 pascals (1.37895 bar)
γ= 1.4

E=(1.37895-1.01325)*261.8/(1.4-1)= 239.35065 Joules
239.35065x2= 478.7013 Joules (9-C)

V= 2094.4m^3 (hemisphere of 10 radii)
p1= 101,325 pascals (1.01325 bar)
p2= 137,895 pascals (1.37895 bar)
γ= 1.4

E=(1.37895-1.01325)*2094.4/(1.4-1)= 1914.8052 Joules
1914.8052x2= 3829.6104 Joules (9-C)

V= 261799m^3 (hemisphere of 50 radii)
p1= 101,325 pascals (1.01325 bar)
p2= 137,895 pascals (1.37895 bar)
γ= 1.4

E=(1.37895-1.01325)*261799/(1.4-1)= 1914.8052 Joules
239349.73575x2= 478699.4715 Joules (9-B)

V= 2094395m^3 (hemisphere of 100 radii)
p1= 101,325 pascals (1.01325 bar)
p2= 137,895 pascals (1.37895 bar)
γ= 1.4

E=(1.37895-1.01325)*2094395/(1.4-1)= 1914800.62875 Joules
1914800.62875x2= 3829601.2575 Joules (9-B)

V= 261799388m^3 (hemisphere of 500 radii)
p1= 101,325 pascals (1.01325 bar)
p2= 137,895 pascals (1.37895 bar)
γ= 1.4

E=(1.37895-1.01325)*261799388/(1.4-1)= 239350090.479 Joules
239350090.479x2= 478700180.958 Joules (9-A)

V= 2094395102m^3 (hemisphere of 1000 radii)
p1= 101,325 pascals (1.01325 bar)
p2= 137,895 pascals (1.37895 bar)
γ= 1.4

E=(1.37895-1.01325)*2094395102/(1.4-1)= 1914800722 Joules
1914800722x2= 3829601444.01 Joules (8-C)

V= 8729754508m^3 (hemisphere of 1,609.34 radii)
p1= 101,325 pascals (1.01325 bar)
p2= 137,895 pascals (1.37895 bar)
γ= 1.4

E=(1.37895-1.01325)*8729754508/(1.4-1)= 7981178058.94 Joules
7981178058.94x2= 15962356117.9 Joules (High 8-C)

Maybe @KLOL506 or @DontTalkDT can double check if I followed the formula correctly and if it should be bars or pascals as I've seen only bars used. Also what about the other methods mentioned in the articles?
Click to expand...
Actually, let me rephrase.

That's in bar m³, which is equal to 100,000 j.

So that "478.7013 joules", is actually 478.7013 bar m³
Which is equal to 47,870,130 joules (9-A)
 
Last edited:
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