### Re: Meier Pictures

From: **Bruce Maccabee <brumac@compuserve.com>**
Date: Sat, 25 Jul 1998 15:40:35 -0400
Fwd Date: Sun, 26 Jul 1998 00:20:01 -0400
Subject: Re: Meier Pictures
>Date: Fri, 24 Jul 1998 12:31:33 -0700 (Pacific Daylight Time)
>From: Nick Balaskas <nikolaos@YorkU.CA>
>To: UFO UpDates - Toronto <updates@globalserve.net>
>Subject: Re: UFO UpDate: Re: Meier Pictures
>>Date: Thu, 23 Jul 1998 21:03:29 -0400
>>From: Bruce Maccabee <brumac@compuserve.com>
>>Subject: UFO UpDate: Kinetic Energy Bullets
>>To: updates@globalserve.net
<snip>
>>Don't forget that the definition of "kinetic energy" is the
>>product of the mass times the velocity squared. If I knew the
>>speed and mass of the .22 WNR I could calculate its K.E. and
>>compare with the 45 (234 grain, 870 ft/sec; KE = 234 x 870^2 =
>>177,114,600 in the appropriate units.
>Bruce, looks like you also forgot the definition of kinetic
>energy (the energy of moving objects). It is equal to the mass
>times the velocity squared divided by 2. As a result, your
>value above is twice as large as it should be.
>Of course, if a given mass is totally converted into energy,
>then the energy obtained is the product of the mass time the
>velocity of light squared (Einstein's mass-energy formula).
I agree. I left the 1/2 off and erroneously described the
"remainder", mv^2 , as the "kinetic energy"., (I should have
said twice the kinetic energy). I did not include 1/2 because
what I was thinking about was the COMPARISON between the KE of
the two bullets, and, in particular, the RATIO of the KE of one
to the KE of th other, and in this ratio the (1/2) factor would
cancel.
On the other hand, if these were relativistic bullets, then the
total energy would be mc^2 where m is the relativistic mass, m =
m'/(1-v^2/c^2) where m' is the rest mass). Clever mathematicians
can "expand" this expression in powers of v^2/c^2 and find that
in the low velocity limit m = (1/2)m'v^2/c^2.