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There are various approaches to predicting Comrades Marathon finishing times. Lindsey Parry, for example, suggests that you use two and a half times your recent marathon time. Sports Digest provides a calculator which predicts finishing time using recent times over three distances. I understand that this calculator is based on the work of Norrie Williamson.

Let’s give them a test. I finished the 2013 Comrades Marathon in 09:41. Based on my marathon time from February that year, which was 03:38, Parry’s formula suggests that I should have finished at around 09:07. Throwing in my Two Oceans time for that year, 04:59, and a 21.1 km time of 01:58 a few weeks before Comrades, the Sports Digest calculator gives a projected finish time of 08:59. Clearly, relative to both of those predictions, I under-performed that year! Either that or the predictions were way off the mark.

It seems to me that, given the volume of data we gather on our runs, we should be able to generate better predictions. If the thought of maths or data makes you want to doze off, feel free to jump ahead, otherwise read on.

## Riegel’s Formula

In 1977 Peter Riegel published a formula for predicting running times, which has become popular due to its simplicity. The formula itself looks like this:

which allows you to predict the time it will take you to run distance , given that you know it takes you time to run distance .

## Reverse-Engineering Riegel’s Model

Riegel’s formula is an empirical model: it’s based on data. In order to reverse engineer the model we are going to need some data too. Unfortunately I do not have access to data for a cohort of elite runners. However, I do have ample data for one particular runner: me. Since I come from the diametrically opposite end of the running spectrum (I believe the technical term would be “bog standard runner”), I think these data are probably more relevant to most runners anyway.

I compiled my data for the last three years based on the records kept by my trusty Garmin 910XT. A plot of time versus distance is given below.

At first glance it looks like you could fit a straight line through those points. And you can, indeed, make a pretty decent linear fit.

> fit <- lm(TimeHours ~ Distance, data = training) > > summary(fit) Call: lm(formula = TimeHours ~ Distance, data = training) Residuals: Min 1Q Median 3Q Max -0.64254 -0.04592 -0.00618 0.02361 1.24900 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) -0.1029964 0.0107648 -9.568 <2e-16 *** Distance 0.1012847 0.0008664 116.902 <2e-16 *** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 0.1394 on 442 degrees of freedom Multiple R-squared: 0.9687, Adjusted R-squared: 0.9686 F-statistic: 1.367e+04 on 1 and 442 DF, p-value: < 2.2e-16

However, applying a logarithmic transform to both axes gives a more uniform distribution of the data, which also now looks more linear.

If we were to fit a straight line to the data on logarithmic axes then the relationship we’d be contemplating would have the form

or, equivalently,

which is a power law relating elapsed time to distance. It’s pretty easy to get Riegel’s formula from this. Taking two particular points on the power law, and , and eliminating gives

which is Riegel’s formula with an unspecified value for the exponent. I am going to call the exponent the “fatigue factor” since it determines the degree to which a runner slows down as distance increases.

How do we get a value for the fatigue factor? Well, by fitting the data, of course!

> fit <- lm(log(TimeHours) ~ log(Distance), data = training) > > summary(fit) Call: lm(formula = log(TimeHours) ~ log(Distance), data = training) Residuals: Min 1Q Median 3Q Max -0.27095 -0.04809 -0.01843 0.01552 0.80351 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) -2.522669 0.018111 -139.3 <2e-16 *** log(Distance) 1.045468 0.008307 125.9 <2e-16 *** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 0.09424 on 442 degrees of freedom Multiple R-squared: 0.9729, Adjusted R-squared: 0.9728 F-statistic: 1.584e+04 on 1 and 442 DF, p-value: < 2.2e-16

The fitted value for the exponent is 1.05 (rounding up), which is pretty close to the value in Riegel’s formula. The fitted model is included in the logarithmic plot above as the solid line, with the 95% prediction confidence interval indicated by the coloured ribbon. The linear plot below shows the data points used for training the model, the fit and confidence interval as well as dotted lines for constant paces ranging from 04:00 per km to 07:00 per km.

Our model also provides us with an indication of the uncertainty in the fitted exponent: the 95% confidence interval extends from 1.029 to 1.062.

> confint(fit) 2.5 % 97.5 % (Intercept) -2.558264 -2.487074 log(Distance) 1.029142 1.061794

## Estimating the Fatigue Factor

A fatigue factor of 1 would correspond to a straight line, which implies a constant pace regardless of distance. A value less than 1 implies faster pace at larger distances (rather unlikely in practice). Finally, a value larger than 1 implies progressively slower pace at larger distances.

The problem with the fatigue factor estimate above, which was based on a single model fit to *all* of the data, is that it’s probably biased by the fact that most of the data are for runs of around 10 km. In this regime the relationship between time and distance is approximately linear, so that the resulting estimate of the fatigue factor is probably too small.

To get around this problem, I applied a bagging analysis, creating a large number of subsets from the data. In each subset I weighted the samples to ensure that there was a more even distribution of distances in the mix. I calculated the fatigue factor for each subset, resulting in a range of estimates. Their distribution is plotted below.

According to this analysis, my personal fatigue factor is around 1.07 (the median value indicated by the dashed line in the plot above). The Shapiro-Wilk test suggests that the data is sufficiently non-Normal to justify a non-parameteric estimate of the 95% confidence interval for the fatigue factor, which runs from 1.03 to 1.11.

> shapiro.test(fatigue.factor) Shapiro-Wilk normality test data: fatigue.factor W = 0.9824, p-value = 1.435e-12 > median(fatigue.factor) [1] 1.072006 > quantile(fatigue.factor, c(0.025, 0.975)) 2.5% 97.5% 1.030617 1.107044

## Generating Predictions

The rules mentioned above for predicting finishing times are generally applied to data for a single race (or a selection of three races). But, again, given that we have so much data on hand, would it not make sense to generate a larger set of predictions?

The distributions above indicate the predictions for this year’s Comrades Marathon (which is apparently going to be 89 km) based on all of my training data this year and using both the default (1.06) and personalised (1.07) values for the fatigue factor. The distributions are interesting, but what we are really interested in is the *expected* finish times, which are 08:59 and 09:18 depending on what value you use for the fatigue factor. I have a little more confidence in my personalised value, so I am going to be aiming for 09:18 this year.

Comrades is a long day and a variety of factors can affect your finish time. It’s good to have a ball-park idea though. If you would like me to generate a set of personalised predictions for you, just get in touch via the replies below.

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