Then the results are as follows:
| t | b Observed |
b Calculated |
δ |
| 20 | 58.2 | 57.0 | +1.2 |
| 64 | 44.2 | 46.7 | -2.5 |
| 526 | 35.8 | 34.5 | +1.3 |
| 1440 | 33.7 | 30.4 | +3.3 |
| 2×1440 | 27.8 | 28.1 | -0.3 |
| 6×1440 | 25.4 | 24.9 | +0.5 |
| 31×1440 | 21.1 | 21.2 | -0.1 |
The deviations of the calculated values from the observed values surpass the probable limits of error only at the second and fourth values. With regard to the latter I have already expressed the conjecture that the test might have given here too large a value; the second suffers from an uncertainty concerning the correction made. By the determination made for t, the formula has the advantage that it is valid for the moment in which the learning ceases and that it gives correctly b=100. In the moment when the series can just be recited, the relearning, of course, requires no time, so that the saving is equal to the work expended.
Solving the formula for k we have
This expression, 100—b the complement of the work saved, is nothing other than the work required for relearning, the equivalent of the amount forgotten from the first learning. Calling this, v, the following simple relation results:
To express it in words: when nonsense series of 13 syllables each were memorised and relearned after different intervals, the quotients of the work saved and the work required were about inversely proportional to a small power of the logarithm of those intervals of time. To express it more briefly and less accurately: the quotients of the amounts retained and the amounts forgotten were inversely as the logarithms of the times.
