Author: Uri Blass
Date: 22:51:29 09/02/01
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On September 02, 2001 at 14:27:18, Chessfun wrote: >On September 02, 2001 at 12:17:24, Torsten Schoop wrote: > >>Hi Uri, >> >> >>>If it is 5% slower than it means that it seems to be 5 elo >>>and my impression based on posts of amir in the past is that 5% >>>is more realistic. >> >> >>IMO it is not possible to measure a speed different of even 5% in ELO at 40/120 >>on 1 GHz machines (maybe 1 ELO?). For that reason it makes no difference to use >>Junior 7 or Deep Junior 7. >> >>Torsten > >On my TB945 mhz from the start position >Junior 7 655 kN/s >Deep Junior 7 604 kN/s > >Though I don't see how 5% translates into only 5 elo. > >Sarah. you can use christhope formula 70*log(1.05)/log(2) in order to get 5 elo difference from 5% in your case 655/604=1.08... so christophe found 8 elo improvement. This is enough to explain the 5 elo and the rest of my post is to explain the way that I got the 5 elo without using a calculator because I was too lazy to use a calculator(maybe it was better to use a calculator in order to avoid this explanation) ). I thought in the following way: Being 100% faster means 70 elo improvement but being 1% faster means 1 elo improvent and the reason is that I remember that 1.01^70 is close to 2. Being 5% faster means 5 elo improvement because for small changes the improvement in elo is close to be proprtional to the improvement in speed. being 5% faster is almost the same as 5 times being 1% faster but being twice faster is not almost the same as 100 times being 1% faster. In numbers: 1.01^5 is close to 1.05 but 1.01^100 is not close to 2. Uri
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