Cook's Edit to Taylor's Formula

[Arc7Kuroi]

If we're assuming an ideal gas, wouldn't E_k = 3/2*P*V be the case instead, if E_k refers to the kinetic energy of the molecules?

You bring up a good point here, I can't really ascertain much else from the PDF.

Talking about U, he asserts the following:
How does he prove the idea that it deceleration happens at a constant rate? I had the pleasure of looking into how the U.N.'s calculator for explosions calculates shockfront velocity for when I wrote the explosion dodging feats stuff. The formula they use is... complicated. Let's just say it involves a 14th degree polynomial and some other stuff. It was definitely not changing at a constant rate in any case.

Fair enough yeah. If I ever find any more in depth explanation of what Cook was doing, I'll come back to this, but if/until then I agree with your reason for doubt here.

 

[M3X_2.0]

Fair enough yeah. If I ever find any more in depth explanation of what Cook was doing, I'll come back to this, but if/until then I agree with your reason for doubt here.

 

[Arc7Kuroi]

I figured it's 2.35% error was enough to warrant validity but I guess not

 

[Arc7Kuroi]

If I ever find any more in depth explanation of what Cook was doing, I'll come back to this

Took me long enough.

If we're assuming an ideal gas, wouldn't E_k = 3/2*P*V be the case instead, if E_k refers to the kinetic energy of the molecules?

The reason he used P * V rather than classical energy of an ideal gas, is because the energy released by an expanding gas is the change in quantity (pressure) * (volume), i.e. pressure-volume work (work being forced dotted with distance, force being pressure times area, dotting pressure times area with distance gets you pressure times a volume element).

How does he prove the idea that it deceleration happens at a constant rate? I had the pleasure of looking into how the U.N.'s calculator for explosions calculates shockfront velocity for when I wrote the explosion dodging feats stuff. The formula they use is... complicated. Let's just say it involves a 14th degree polynomial and some other stuff. It was definitely not changing at a constant rate in any case.

The constant deceleration he mentions is for very early times, i.e. during the initial fireball formation. He isn't saying that the shock deceleration occurs at a constant rate across it's entire life time. However, for very early times (time =< time it takes fireball to fully form), it works to high accuracy. As denoted by the fact that Cook's edit only had a 2.35% error.

Just to recap:

E = 8 * pi * rho_0 * R^5 / [75 * (gamma - 1) t^2]
where:
rho_0 is the initial density of air (1.225 kg/m^3 usually near Earth's surface at 20 deg C)
R is the fireball radius in meters
gamma is the specific heat ratio (usually 1.4 near Earth's surface at 20 deg C)
t is the time it takes the fireball to reach radius R in seconds
E is the energy of an airburst explosion in joules

and this equation would only apply to explosions at least as massive as nukes that create rather large shocks.

 

[Dark-Carioca]

R is the fireball radius

t is the time it takes the fireball to reach radius R

In meters and seconds, I imagine?

How much would this new formula change a result?

 

[Arc7Kuroi]

In meters and seconds, I imagine?

Yes edited my post to make units clear.

How much would this new formula change a result?

Depends on the case I'd imagine. I haven't gone around applying it to any accepted airburst calcs.

 

[Arc7Kuroi]

Bump

 

[M3X_2.0]

@DontTalkDT

 

[DontTalkDT]

Took me long enough.


The reason he used P * V rather than classical energy of an ideal gas, is because the energy released by an expanding gas is the change in quantity (pressure) * (volume), i.e. pressure-volume work (work being forced dotted with distance, force being pressure times area, dotting pressure times area with distance gets you pressure times a volume element).

Wouldn't P then not have to be ambient air pressure instead of explosion overpressure? Volume work is ultimately expansion work against an outside force, no? In fact, why is volume work conflated with the expansion of a shockwave? Waves don't move matter (to any significant degree).

The constant deceleration he mentions is for very early times, i.e. during the initial fireball formation. He isn't saying that the shock deceleration occurs at a constant rate across it's entire life time. However, for very early times (time =< time it takes fireball to fully form), it works to high accuracy. As denoted by the fact that Cook's edit only had a 2.35% error.

Just to recap:

E = 8 * pi * rho_0 * R^5 / [75 * (gamma - 1) t^2]
where:
rho_0 is the initial density of air (1.225 kg/m^3 usually near Earth's surface at 20 deg C)
R is the fireball radius in meters
gamma is the specific heat ratio (usually 1.4 near Earth's surface at 20 deg C)
t is the time it takes the fireball to reach radius R in seconds
E is the energy of an airburst explosion in joules

and this equation would only apply to explosions at least as massive as nukes that create rather large shocks.

He uses said assumption of constant deceleration for his entire formula, though, and the UN formula is also for shortly after explosion, soooo....

Btw. where was the 2.35% error stated? Remember, we don't trust the author of this.

You also still need to address this:

And if this is supposed to be the equation for the kinetic energy of the molecules in an ideal gas, how is that equal to 1/2*M*U^2 with U being the outward blast velocity?

Well, the new version of this of: Why would pressure work be in any way equal to this.

 

[Arc7Kuroi]

He uses said assumption of constant deceleration for his entire formula

Which he derived for early times, aka his derivation is for small time scales.

I'll look into the rest later, it's past my bedtime.

 

[DontTalkDT]

Which he derived for early times, aka his derivation is for small time scales.

I'll look into the rest later, it's past my bedtime.

Early times for an explosion are something like 0.2 microseconds. The speed of an explosion does, in fact, decay very rapidly even at the first meter of expansion. (if the UN calculator is to be believed, which it is)
If this formula only holds for that extremely early times it is likely of no practical use.

 

[XSOULOFCINDERX]

Was this finished or am I bumping this for no reason?

 

[DontTalkDT]

Was this finished or am I bumping this for no reason?

Yes.

 

[XSOULOFCINDERX]

Yes.

....OK then.

 

[KLOL506]

Yes.

So uh... should we close this thread under the conclusion that the formula was rejected?

 

[DontTalkDT]

Well, if nobody can proof the formula, I guess we should do that.