Converting effects/Joules into yields

[The Grim Squeaker]

Having wanted to take a more active role in some of the easier calculation based yield debates I wanted to ask for a little primer from the fine physics minded members of the board:

First what is the energy required to :
A) turn an average 80 kg human male into ash (62 MJ?)
B) the energy needed to turn 1 kg of rock into dust

Secondly how would I convert the yield in Joules of a weapon into K/M/G tons?

Also is the following calculation based on a Halo Super Mac correct?:
v= 0.4c = 0.4*299 792 458= 119,916,983 m/s
m= 3,000 tons= 2,721,554.22 kg

Ek (kinetic energy in joules)= m*v^2=! 3.91*10^22 Joules, so how many MT or Gigatons is this?

Thank you very much, I tried asking my physics teacher but he didn't know any of it.

[phongn]

1 kT = 4.186e12 J = 4.186 TJ.

[The Grim Squeaker]

What exactly is e? Does is stand for 10^x?.

Also does it then follows the standard k=1,000, Mega=1,000,000. Giga= 1,000,000,000. Peta= 1,000,000,000,000. Tera=....?

Sorry if the questions are dumb, but we dont use e unless it's part of a set equation and I haven't used it for ages and can't remmember half of what I learned last year in Phy,Math,electronics.

[The Grim Squeaker]

If I understood the conversion correctly then the energy released by lets say the aforementiones Super MAC is:

3.91 × 10^22 J/4.186*10^12=9.34065934 × 10^9KT= 9.34pt?

That can't possibly be right , it should be in the low giga ton range,
I should divide the sum by 1,000=10^3 to go up from Kilo ton to mega ton, right?

[Surlethe]

What exactly is e? Does is stand for 10^x?.

Yes. It's shorthand for 10^x -- for example, 5e13 = 5*10^13

Also does it then follows the standard k=1,000, Mega=1,000,000. Giga= 1,000,000,000. Peta= 1,000,000,000,000. Tera=....?

I don't know how trustworthy this site is, but it sounds right, and gives:

kilo = 1e3
mega = 1e6
giga = 1e9
tera = 1e12
peta = 1e15
exa = 1e18
zeta = 1e21
yota = 1e24

Also is the following calculation based on a Halo Super Mac correct?:
v= 0.4c = 0.4*299 792 458= 119,916,983 m/s
m= 3,000 tons= 2,721,554.22 kg

Ek (kinetic energy in joules)= m*v^2=! 3.91*10^22 Joules, so how many MT or Gigatons is this?

If you're dealing with velocities which are a significant percentage of c, you may want to use the relativistic KE formula, which is
E = {(1/(1 - v^2/c^2) - 1}mc^2.

[Sea Skimmer]

1 kT = 4.186e12 J = 4.186 TJ.

Note that it is in fact highly inappropriate to use kilo or megatons to refer to the yield of anything but a nuclear weapon.

[Mobiboros]

1 kT = 4.186e12 J = 4.186 TJ.

Note that it is in fact highly inappropriate to use kilo or megatons to refer to the yield of anything but a nuclear weapon.

I'll probably sound dumb for asking, but why?

[Lord Zentei]

1 kT = 4.186e12 J = 4.186 TJ.

Note that it is in fact highly inappropriate to use kilo or megatons to refer to the yield of anything but a nuclear weapon.

I'll probably sound dumb for asking, but why?

Try measuring the size of your funiture in miles.

[phongn]

Try measuring the size of your funiture in miles.

Well, for large-scale weapons the energy release may still be technically correct but MT/kT/etc have certain connotations that may not be wanted.

[Sea Skimmer]

I'll probably sound dumb for asking, but why?

Its because nuclear weapons release energy in a variety of ways at once which cause a specific set of effects, a combination of hard radiation, thermal radiation and light along with blast. Another weapons might release the same amount of energy but with different effects. More of its energy may be blast for example, or hard radiation making a supposed 200kt Explosive 'X' weapons cause more or less damage then a 200kt nuclear device.

[CmdrWilkens]

I'll probably sound dumb for asking, but why?

Its because nuclear weapons release energy in a variety of ways at once which cause a specific set of effects, a combination of hard radiation, thermal radiation and light along with blast. Another weapons might release the same amount of energy but with different effects. More of its energy may be blast for example, or hard radiation making a supposed 200kt Explosive 'X' weapons cause more or less damage then a 200kt nuclear device.

The problem is that kT and MT are not arbitrary measurements used to describe nuclear weapons in paticular but are exact and specific energy amounts set in stone. Using them for non-nuclear events simply means you are using a different scale than Joules in order to get a smaller number of digits in your answer if still being equal in its result. Again kT and such are still derived from the fact that TNT was such a stable and consistent measure that pund for pound you could get consistent energy yields out of it and those yields became standardized to the point that it is NOT an arbitrary number or even one which is solely in use for nuclear weapons. It came to be associated with nuclear weapons but it is every bit as precise as the rest of the metric system and there should be no stigma attached (as a quick showing of this I'd point everyone to Mike's own Planet Killer's essay page.)

[Mobiboros]

Well, for large-scale weapons the energy release may still be technically correct but MT/kT/etc have certain connotations that may not be wanted.

The 'connotations' aspect seems to make the most sense of why not to use it even if it's a valid way to measure the energy released.

Thanks!