A friend of mine pedaled for an hour on a cycle ergometer, a stationary bicycle in a gym.

At the end of the ride the cycle computer reported a distance of $40\:\mathrm{km}$ and an average power of $250\:\mathrm{W}$.

I would like to know whether the reported data are meaningful or comparable with a real ride on road.

The average power can be reasonable: see for example a ride of a professional cyclist with an average power of $\approx200\:\mathrm{W}$.

I have some doubts about the distance: I have no idea how the cycle ergometer computes it but I would like to do anyway a meaningfulness check.

The problem is that I can think only to a very simple model: the mean speed is $v=40\:\mathrm{\frac{km}{h}}$ and the energy she spent is $E=1\:\mathrm{h} \times 250\:\mathrm{W}=0.25\:\mathrm{kWh}$; if I assume that, in an hypothetical road ride equivalent to her gym exercise, all the energy she spent is converted into kinetic energy $E=\frac{1}{2}mv^2$ then I get for the total mass (her mass plus the bicycle mass) $m=\frac{2E}{v^2}=\frac{0.5\:\mathrm{kWh}}{1600\:\mathrm{\frac{km^2}{h^2}}}=14600\:\mathrm{kg}$ that is obviously wrong because it is too high.

If my model was good enough then I would state that the distance is not plausible and so perhaps the cycle ergometer need to be calibrated or its algorithm is not working well.

But I suspect that my model is not a good one and so what is a more adequate model to help to assess the plausibility of the reported distance?


1 Answer 1


The prediction made by the cycle are quite accurate. In a real situation because of Air drag, rolling resistance and slope it takes a power output of around 300kWh for a 90kg cycle+rider. So at 250kWh, a less than 90kg rider+cycle can travel almost 40kms.

The model you made is wrong because then energy spent is not equal to kinetic energy, it's rather equal to the energy required for acceleration + the retarding factors I mentioned above.

See: https://en.wikipedia.org/wiki/Bicycle_performance#Power_required


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.