Rowers train indoors on the Concept 2 ergonometer, which measures the time to row a given distance. These numbers are generally used to assess fitness.

Frequently the question comes up, given an existing performance like 3:20min over 1000m, what performance can an athlete expect for another distance like 2000m or 5000m?

Paul's Law, a folklore whose origin in rowing is not clear, provides a rule of thumb:

When the total distance doubles, the time over 500m increases by 5 seconds.

In rowing the time over 500m is called the split and is the most common way to refer to speed. Given 3:20min over 1000m implies 1:40 over 500m. For 2000m the expected time would be $1:45 \times 4 = 7:00$ minutes.

Paul's Law expressed as a formular when $t_0$ over distance $d_0$ is known and we are looking for the predicted time $t_1$ over distance $d_1$:

$$ t_1 = d_1 \left( \frac{t_0}{d_0} + \frac{1}{100}\mathrm{log}_2 \frac{d_1}{d_0}\right) $$

Concept 2 maintains an online ranking for performance on the ergonometer, the Logbook. Rowers can create an account and upload training data. When this is done using Concept 2's ErgData app, the result is marked as verified since it takes the data directly from the rowing machine.

==> pairs.csv <==
year,gender,name,age,dist1,time1,dist2,time2
2025,f,CA3C50104C50E18DF97B5FDB6DD61A51,41,1000,194,5000,1074.3
2025,f,30E7F8121E7BB974935271278BDE71CB,38,1000,195,2000,419.9
2025,f,30E7F8121E7BB974935271278BDE71CB,38,1000,195,5000,1126.9
2025,f,5B641684246CBD871E90A6BF1CDD8DE2,30,1000,196.6,2000,426.9
2025,f,6FB4C1A2D71A158778AD2EDB85E0EE32,40,1000,197.6,2000,417.8
2025,f,6FB4C1A2D71A158778AD2EDB85E0EE32,40,1000,197.6,5000,1109.7
2025,f,CE825D590582EEFBD5C529E6A12E0250,32,1000,201.6,2000,424.1
2025,f,CE825D590582EEFBD5C529E6A12E0250,32,1000,201.6,5000,1115
2025,f,4AE5B7B3679C3ECD9C4F8E858788F458,39,1000,208,2000,433.2

File pairs.csv contains 2000+ data points from the 2025 ranking that only includes verified results.

• gender (male, female)
• MD5 hash of the athlete's name
• age
• distance 1 in meters and time 1 in seconds
• distance 2 in meters and time 2 in seconds

So each line captures two performances of the same athlete over two distances. The same athlete may have more entries in this file:

2025,m,22D90C393F79EC2180C6427A5F6FA0BE,45,1000,179.7,2000,391.8
2025,m,22D90C393F79EC2180C6427A5F6FA0BE,45,1000,179.7,5000,1047
2025,m,22D90C393F79EC2180C6427A5F6FA0BE,45,2000,391.8,5000,1047

We compute in pauls-law.r for each line the time change in split time per double the distance that predicts the second performance from the first. If Paul's Law is accurate, we expect this constant to be around 5s.

Below is the distribution of the Paul Constant as we found it:

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
-29.500   3.721   5.035   5.029   6.335  40.725

Paul's Law appears to be quite accurate on average: An increase of 5.0 seconds per 500m per double the distance predicts the seconds performance based on the first. At the same time, the distribution reveals data that suggest not both efforts were maximal:

• A constant close to zero suggests the same split over both distances
• A negative constant indicates a faster performance over the longer distance.

We could recalculate the distribution by defining some criterion that eliminates questionable efforts.

An alternative way to predict performace in endurace sports is using a power law described in

Drake, J.P., Finke, A. & Ferguson, R.A. Modelling human endurance: power laws vs critical power. Eur J Appl Physiol 124, 507–526 (2024). https://doi.org/10.1007/s00421-023-05274-5

The power law describes an expected finishing time $t_1$ for distance $d_1$ based on an earlier performance of $t_0$ over distance $d_0$:

$$ t_1 = t_0 \times (d_1/d_0)^\alpha $$

with $\alpha \approx 1.064$.

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
 0.6607  1.0464  1.0650  1.0643  1.0828  1.5240

Below is the distribution of $\alpha$ for the same (slightly flawed) dataset.

If we use $\alpha=1.64$, what does this mean for our initial example of 1000m in 3:20min?

$$ 3:20 \times 2^{1.064} = 06:58.14 $$

It would predict a 2000m time of 06:58.14, slighty faster than the prediction by Paul's Law of 7:00.