Monday, 28 January 2019

Treadmill and road running - equivalent pace estimations

The effort expended running on a treadmill differs from that on road for a number of reasons, the main one being the lack of wind resistance. It is often stated that a gradient, which is easily set on most treadmills, of between 0.5-1.5% makes treadmill paces equivalent to running on a flat road. Of course, a single value of gradient cannot work for all paces since fast runners will experience a greater wind resistance on the road. Fast runners must therefore require a greater gradient on a treadmill to create an equivalent effort to that experienced when moving through air quickly on road. Equally, we might expect a zero gradient to be required for slow runners as they experience very little wind resistance.

Creating equivalent pace tables, for comparison between running on the road and treadmill, is not only useful for setting the right gradient to mimic a road-based effort - such tables might allow runners to trade treadmill-speed with gradient. This has some important implications, not least making high intensity efforts on a treadmill safer. If one could use gradient increments, in a semi-quantitative way, to replace speed increments then near maximal efforts become possible without recourse to harnesses and large amounts of padding.

If you Google for such tables what pops-up are mostly straightforward pace and speed conversion tools. There does, however, appear to be one table that might provide some better data and that is from HillRunner. That table is in miles per hour and ranges from gradients of 0%-10% and speeds from 5-12mph. In the FAQ Ryan Hill mentions that the data came from some students who were making oxygen measurements both on a track and treadmill, however, no references exist. This is not 'published' work in any formal sense. But, in all likelihood, it is a decent starting point.

In the FAQ Ryan Hill states that the data does not 'nicely' fit a formula and that he cannot see how a 'meaningful' calculator can be produced. I found this a slightly odd statement. My experience is that physiology is often well-modelled by simple maths, and running on a treadmill should be relatively simple to model. Looking at the table of data I wondered if using pace data was what made the maths appear to be 'complicated'. So, I attempted to produce a model. My first step was to convert the pace values to something likely to be easily modelled. We know that oxygen consumption values tend to scale linearly with speed, so that is where I started. I choose meters per second (m/s), although any distance and time unit would have done.

I then plotted the relationship between treadmill speed and the equivalent speed on road, given by the table, for each gradient. What was apparent was that for any one gradient there was a very good linear fit between treadmill speed and road speed (r2>0.99). That surprised me - I was expecting some kind of curve due to wind resistance. But, the straight-line nature of the relationship made the model easy to construct since it was now a simple matter of deriving an equation for the parameters that determine each straight-line fit for a given gradient. Plotting the straight-line fit coefficients (gradient and offset: y=mx+c) against the treadmill gradient revealed that they could be modelled well by second order polynomials. In fact plotting my modelled coefficients versus those calculated from the linear fit showed a linear correlation of r2>0.99999 - so, the model was able to predict the real values very well indeed. In most cases the model calculation of equivalent pace was within 0.5s per km of the table value - and at worst 1.5s away for a few values. The difference between this model and the table was 0.02 seconds per km with an SD of 0.33 seconds per km. That is a pretty good fit.

Here is the formula (metric) where treadmill speed is in kph and grade in percent (i.e. 1% is 0.01) which returns equivalent road pace in mm:ss per km as fractions of 24 hours (which is Excel's built in time unit).

Equivalent Road Pace =0.01157/(Speed*(Grade^2*0.72+Grade*0.0528+0.266)-Grade^2*29.27+Grade*13.656+0.006)

For those who want a 'turn-key' solution or to see the fitting, here is a link to the spreadsheet:
(This link will become inactive after 28th May, 2019 - after which you will need to email: for a new link)

If you just want to convert a treadmill speed (kph) and gradient to the equivalent road speed then this is the formula you might want to use:

Road speed (kph) =Treadmill Speed*(Grade^2*2.59+Grade*0.19+0.958)-Grade^2*105+Grade*49.2+0.0216

If you would like a bespoke table (either metric or imperial) over a set of gradients, let me know and I will endeavour to produce one (if it makes sense to do so - i.e. it isn't a vast extrapolation).


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