Note #1: This is based on the assumption that an average key is 2 inches
long. I think that's pretty close to true, and I even measured my own keys
to make sure.
Note #2: This is, of course, based on all the keys, though we probably
won't have to go through all of them.
Lastly, I think I did the math right, but please correct me if I'm wrong.
Orbit of earth
====================
9.39951 x 10^13 cm
OR
3.70059448819 x 10^ 13 inches
Number of keys
==================
2^56 keys * (2 inches per key) = 2^57 inches
Calculation
===================
2^57 / (3.70059448819 x 10^13) = 3894 times around the sun
Alex
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mailto:abischof@vt.edu http://www.vt.edu:10021/A/abischof/
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| export-a-crypto-system-sig RSA-3-lines-PERL |
| #!/bin/perl -sp0777i<X+d*lMLa^*lN%0]dsXx++lMlN/dsM0<j]dsj |
| $/=unpack('H*',$_);$_=`echo 16dio\U$k"SK$/SM$n\EsN0p[lN*1 |
| lK[d2%Sa2/d0$^Ixp"|dc`;s/\W//g;$_=pack('H*',/((..)*)$/) |
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