I find it interesting to predict one’s time in an important running race based on a test run, as to know what a good pace is to keep up during that race. I always used a self-conceived rule, which is quite different from the generally accepted rule according to research engineer Peter Riegel, although it took some math to figure that out (see below).
Self-conceived rule
I have experienced for myself early on in my running career that my pace increases by a factor of 1.07 when doubling the running distance, under ideal circumstances. For example, a 10 km race tended to be 7 percent slower than a 5 km race. I generalized this idea into a formula.
1: p2 ÷ p1 = 1.07 n
2: d2 ÷ d1 = 2 n
1 & 2 ⇒ n = ²log(d2 ÷ d1)
p2 ÷ p1 = 1.07 ²log(d2 ÷ d1)
p2 = p1 × 1.07 ²log(d2 ÷ d1)
p1 = t1 ÷ d1; p2 = t2 ÷ d2
⇒ t2 ÷ d2 = (t1 ÷ d1) × 1.07 ²log(d2 ÷ d1)
t2 = t1 × (d2 ÷ d1) x 1.07 ²log(d2 ÷ d1)
Factors for 1.07 ²log(d2 ÷ d1)
The table below shows some favorite distances from road races. In the rows are the distances from which the pace is known and in the columns the distances from which the corresponding factors can be read as numbers in the table. Multiplying a known pace for a particular distance with this factor gives the corresponding unknown pace for another distance. For example, for a 15 km race (third row in the table) multiply this race’s pace by a factor of 1.1062 to calculate the ideal pace for a marathon (eighth column in the table). A sub-3 hour marathon would require a sub-36 minutes 10 km, according to this table.
| 5 km | 10 km | 15 km | 10 mi | 20 km | ½ mar | 30 km | mar | |
|---|---|---|---|---|---|---|---|---|
| 5 km | 1.0000 | 1.0700 | 1.1132 | 1.1208 | 1.1449 | 1.1509 | 1.1911 | 1.2314 |
| 10 km | 0.9346 | 1.0000 | 1.0404 | 1.0475 | 1.0700 | 1.0756 | 1.1132 | 1.1509 |
| 15 km | 0.8983 | 0.9612 | 1.0000 | 1.0069 | 1.0285 | 1.0339 | 1.0700 | 1.1062 |
| 10 mi | 0.8922 | 0.9546 | 0.9932 | 1.0000 | 1.0215 | 1.0268 | 1.0627 | 1.0987 |
| 20 km | 0.8734 | 0.9346 | 0.9723 | 0.9790 | 1.0000 | 1.0052 | 1.0404 | 1.0756 |
| ½ mar | 0.8689 | 0.9297 | 0.9673 | 0.9739 | 0.9948 | 1.0000 | 1.0350 | 1.0700 |
| 30 km | 0.8395 | 0.8983 | 0.9346 | 0.9351 | 0.9612 | 0.9662 | 1.0000 | 1.0339 |
| mar | 0.8121 | 0.8689 | 0.9040 | 0.9102 | 0.9298 | 0.9346 | 0.9673 | 1.0000 |
Rule according to Riegel
According to Peter Riegel the increase in pace for a longer distance is radically different:
t2 = t1 × (d2 ÷ d1) 1.06
For times between about 3 and 230 minutes.
t2 = t1 × d2 1.06 ÷ d1 1.06
t2 ÷ d2 = t1 × (d2 1.06 ÷ d2) ÷ d1 1.06
t2 ÷ d2 = t1 × d2 1.06 - 1 ÷ d1 1.06
t2 ÷ d2 = t1 × d2 0.06 ÷ (d1 × (d1 1.06 ÷ d1))
t2 ÷ d2 = t1 × d2 0.06 ÷ (d1 × (d1 1.06 - 1))
t2 ÷ d2 = (d2 0.06 ÷ d1 0.06) × (t1 ÷ d1)
t2 ÷ d2 = (d2 ÷ d1) 0.06 × (t1 ÷ d1)
p1 = t1 ÷ d1; p2 = t2 ÷ d2
⇒ p2 = (d2 ÷ d1) 0.06 × p1
If the distance doubles, the pace increases by a factor of 2 0.06, or 1.0425. That would mean that as the distance increases, the pace increases drastically less than with the aforementioned calculation method of my own.
A similar table is shown below, containing factors to be multiplied with known paces to calculate unknown paces, depending on the distances.
Factors for the Riegel calculation
The table below shows some favorite distances from road races. In the rows are the distances from which the pace is known and in the columns the distances from which the corresponding factors can be read as numbers in the table. Multiplying a known pace for a particular distance with this factor gives the corresponding unknown pace for another distance. For example, for a 15 km race (third row in the table) you can multiply the pace by a factor of 1.0640 to calculate the ideal pace on a marathon (eighth column in the table). For a sub-3 hour marathon, a time of sub-39 minutes on the 10 km is fast enough, according to this table.
| 5 km | 10 km | 15 km | 10 mi | 20 km | ½ mar | 30 km | mar | |
|---|---|---|---|---|---|---|---|---|
| 5 km | 1.0000 | 1.0425 | 1.0681 | 1.0726 | 1.0867 | 1.0902 | 1.1135 | 1.1365 |
| 10 km | 0.9593 | 1.0000 | 1.0246 | 1.0289 | 1.0425 | 1.0458 | 1.0681 | 1.0902 |
| 15 km | 0.9362 | 0.9760 | 1.0000 | 1.0042 | 1.0174 | 1.0207 | 1.0425 | 1.0640 |
| 10 mi | 0.9323 | 0.9719 | 0.9958 | 1.0000 | 1.0131 | 1.0164 | 1.0381 | 1.0596 |
| 20 km | 0.9202 | 0.9593 | 0.9829 | 0.9870 | 1.0000 | 1.0032 | 1.0246 | 1.0458 |
| ½ mar | 0.9172 | 0.9562 | 0.9797 | 0.9839 | 0.9968 | 1.0000 | 1.0213 | 1.0425 |
| 30 km | 0.8981 | 0.9362 | 0.9593 | 0.9633 | 0.9760 | 0.9791 | 1.0000 | 1.0207 |
| mar | 0.8799 | 0.9172 | 0.9398 | 0.9438 | 0.9562 | 0.9593 | 0.9797 | 1.0000 |
What to use?
It seems rather obvious to me that a generally accepted rule makes more sense than something you came up with yourself based on your own race results. Other than that, there are strict limits to when the rule according to Riegel applies. For slow runners (e.g. marathon in 4½ hours) or extremely short distances (e.g. 400 m) the rule does not apply, or, at least, the prediction value will be rather low.
An acquaintance of mine ran a time of 1h01m36s in a “15 km” race (14.87 km according to his GPS). According to my calculation, that would be enough for 3h15m39s on a marathon with certified course (1% longer than 42.195 km). According to Riegel, it would be enough for 3h08m03s on the same certified marathon course. In both cases, the result of the “15 km” race was too slow for an “official marathon” (i.e. with certified course) under 3 hours. A sub-3 hour marathon would have required a time of 56m40s or 58m57s for the “15 km” race (my rule, Riegels rule respectively). This acquaintance has to get faster somehow, either by training and/or good fueling during the race. Ideal circumstances would help too.
Anyway, I think I can put my home-grown calculation method to rest and start using the much better tested method according to Peter Riegel.