# Calculating weapon yield of antimatter, was Re: antihydrogen atoms

From: Michael M. Butler (butler@comp-lib.org)
Date: Sun Feb 24 2002 - 16:19:32 MST

Richard Steven Hack wrote:
> Well. we have 20-odd megatons and 43 kilotons. Are we all using the same
> physics here?
>
> Jeeez....

Slipped decimal points were even more frequent back in the days of slide
rules.

There are power-of-ten differences and factor-of-two differences in some of
the posted numbers. The factor-of-two difference is due to different starting
assumptions about taking the total converted mass as one gram vs. taking just
the antimatter mass as one gram.

Some errors in other folks' calculations could be in part attributable to an
incorrect/inconsistent value for c. You have to stick with one system (I used
SI, which is kilograms-meters-seconds) or you'll get powers-of-ten errors. In
particular, using a centimeters/sec value for c when one should be using
meters/sec will put you a factor of 10,000 off when you square c.

As a sanity check, we can recall that c is roughly 186,000 miles/sec in
English units; this makes it about 300,000 _kilometers_/sec, or 300,000,000
_meters_/sec: 3 * 10^8, just about the number I used.

Eliezer and I agree. 1 g of antimatter + 1 g of matter yields 43 kt if we have
100% efficient conversion.

I'm pretty sure my numbers are right. 21.5 megatons per _kilogram_ of mass
totally converted, or 21.5 kilotons per _gram_ of mass totally converted. I
showed all my work in my earlier post.

```--
butler a t comp - lib . o r g
I am not here to have an argument. I am here as part of a civilization.
Sometimes I forget.
```

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