trying prelim approach for distance times equivalence from physics alone
based on this :
http://www.letsrun.com/forum/flat_read.php?thread=3415815&page=6
the key is
E = s * v^3 = s * ( s/t )^3 = s^4/t^3
the exponent is 4/3
the key step is that considering other distances, the "energy equivalent" performance is the inverse of the exponent ( not sure about technicalities of this yet - we can discuss, as exponent is 4 for s & 3 for t ) :
ratio of distances ^ (3/4)
using it, we get some tentative figures : lets try 1'45
E = 800 * (800/105)^3 = 353827 u
for 1500m
E1 = 353827 * (1500/800)^(3/4) = 566947 u
v1 = (566947/1500)^(1/3) = 7.23 m/s ->3'27.46
for 3000m
E2 = 353827 * (3000/800)^(3/4) = 953486 u
v2 = (953486/3000)^(1/3) = 6.82 m/s ->7'19.60
for 5000m
E3 = 353827 * (5000/800)^(3/4) = 1398624 u
v3 = (1398624/5000)^(1/3) = 6.53 m/s ->12'44.53
for 10000m
E4 = 353827 * (10000/800)^(3/4) = 2352195 u
v4 = (2352195/10000)^(1/3) = 6.17 m/s ->26'59.98
for 1/2 M
E5 = 353827 * (21097.5/800)^(3/4) = 4117624 u
v5 = (4117624/21097.5)^(1/3) = 5.80 m/s ->60'38
for M
E6 = 353827 * (42195/800)^(3/4) = 566947 u
v6 = (566947/42195)^(1/3) = 5.47 m/s ->2"08'27
strong bias to 800 & 10k & longer
i'll throw this out for discussion...