r = player's speed / opponent's speed
If my hypothesis is right, the player should be K*log2(r) points stronger than his opponent. According to the Elo formula, this rating difference should also be -400*log10(1/s-1), where s is the player's expected score (see my earlier article Rating Points Revisited).
K*log2(r) = -400*log(1/s-1)
-K*log(1/r) / log(2) = -400*log(1/s-1)
Let k = K / (400*log(2))
k*log(1/r) = log(1/s - 1)
r-k = 1/s -1
s = 1 / (1 + r-k)
The graph below plots the player’s expected score s against his speed ratio r for K=100, K=150 and K=200 Elo points:
If the player’s speed is equal to that of his opponent, his score is 0.5. If he is infinitely slow, his score is 0. If he is infinitely fast, his score is 1. This graph can also be interpreted as showing how much the player’s score improves if he is given more time on the clock than an equal opponent. The equation can be rewritten as:
rk = s / (1 - s)
The right hand side of this equation is the player's score divided by his opponent's score. Interestingly, when k = 1, which occurs when K = 400*log(2) = 120, the speed ratio is equal to the score ratio; so if you are twice as fast, you score twice as many points.
Suppose that a player takes time ts to achieve a score s, when an equal opponent has time t. This is equivalent to both players having equal time, but with the speed ratio:
r = player's speed / opponent's speed = ts / t
(ts / t)k = s / (1 - s)
t0.5 / t = 0.5 / (1 - 0.5) = 1
t0.5 = t
(ts / t0.5)k = s / (1 - s)
t0.5 = ts * [s / (1 - s)]-1/k
Given the player’s time and his score, this equation gives the time that he would need to score 0.5. Having calculated t0.5, we can calculate the score for any time ts using:
s = 1 / (1 + (ts / t0.5)-k)
N.B. In this and the earlier article, I have used the logistic distribution (used by the USCF and others) rather than the normal distribution (used by FIDE). Also see my later articles More Rating , Time and Score and Rating, Time and Score Revisited.