50.95 divided by the speed of light squared? So, for all
practical purposes - if that's right - it's zero. Why not
just say so?

That depends entirely on the units you're using.

If you measure the velocity of C in typical human-scale units (e.g.
metres per second) then C^2 is, indeed, a numerically-large value, and
50.95/C^2 is numerically small (and will be in units of
watts seconds-squared over metres-squared).

If, on the other hand, you measure velocities on another scale, things
look very different. Specifically, let's use a scale which represents
each velocity as a fraction of the maximum possible velocity... let's
call this unit of velocity "skedaddles". Measured this way, C is
precisely 1.0 skedaddle, C^2 is precisely 1.0 skedaddle squared, and
50.95/C^2 equals 50.95 (watts per square skedaddle).

Same result... only the units of measurement are different. Neither
result is zero. "Numerically small" is not equivalent to "zero for
all practical purposes".

(obLinguistic: "skedaddle" is a somewhat quirky American term of
uncertain heritage, which means "leave in a hurry, scram, escape", and
seems a reasonable term for a scale of zerch up to as-fast-as-
possible. I believe that the equivalent British term was defined as
"runawayrunaway" by King Arthur, as cited in "Monty Python and the
Holy Grail").

--
Dave Platt AE6EO
Friends of Jade Warrior home page: http://www.radagast.org/jade-warrior
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