# How much more efficient is a road bike than a mountain bike? [closed]

What would be a good estimate on the difference of efficiency between a road bike and a mountain bike?

A number of links cite all the usual reasons: thinner tires, better aerodynamics...

But I'm not able to find one where they quantify this efficiency

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## closed as off-topic by Kyle Kanos, ACuriousMind, Brandon Enright, John Rennie, KostyaJul 30 '14 at 11:11

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Define efficient. – Kyle Kanos Jul 29 '14 at 16:31
Related: bicycles.stackexchange.com/q/1505 , bicycles.stackexchange.com/q/12947 and links therein. – Qmechanic Jul 29 '14 at 19:52

I often wondered about these things - then I came up with a simple experiment that works for me because I have a simple bike computer (thing with a magnetic pickup on the spokes that updates my speed every second).

I find a flat piece of road, and ride at a certain speed (say 20 mph on my road bike, or 15 mph on my mountain bike). I then stop pedaling at a specific point, and take note of the speed dropping (it conveniently updates every second: use iphone or other voice recorder and just read out the numbers as you see them: 20.0, 19.5, 19.1, 18.7, 18.2, 17.8, 17.4, etc).

Now comes the fun part: turning this into power needed to keep a certain speed going.

You should have a pretty good idea of your mass plus that of your bike. At a given speed, this gives a certain kinetic energy ($\frac12mv^2$). The drop in your speed means your kinetic energy is being dissipated (road friction, air drag, slope of the road...). To be accurate, you need to take account of the fact that your wheels have rotational kinetic energy - almost all the mass is at a certain radius $r$ (typically 35 cm for a road bike, variable for a mountain bike). For a wheel of mass $m$ rolling at velocity $v$, the total energy is

\begin{align} KE&=\frac12mv^2 + \frac12I\omega ^2\\ &=\frac12mv^2 + \frac12mr^2\omega ^2\\ &=mv^2\end{align} so exactly double what it would be if you had not taken the rotational energy into account. The simplest way to account for this is just to double the mass of the wheels, then use $\frac12mv^2$. The correction factor is quite small, given that you are probably much heavier than your wheels.

Now you create a table (Excel works well for this) with columns for time (sec) and speed (mph) - these are the data columns. You then compute speed (m/s), KE (J), change in KE (J) in the next three columns. Now you can create a plot of power needed at a given speed. Using the numbers above, I came up with the following:

which shows that maintaining 20 mph on my road bike on that day (gentle tail wind) required about 225W of sustained power - which is quite comfortable. According to the website calculator at http://www.tribology-abc.com/calculators/cycling.htm I should have expected about 275W with no wind when going 32 km/h; this is certainly in the right ball park. The same calculator shows that the power needed drops to 141 W at 15 mph (24 km/h) - again, quite close to what my simple experiment gave.

Another look at the breakdown of the bike calculator shows that the rolling resistance is independent of speed, and that the factor that changes quickly is the wind resistance. This tells me a few things:

• at low speeds (below 12 mph) the rolling resistance is critical: this is where pumping up the tires of your mountain bike can really help
• At higher speeds, the wind resistance dominates the power dissipation: a good posture helps to streamline your body. This is where the recumbent and TT bikes excel, and where the mountain bike really loses out.

I don't have the same data for a mountain bike as I collected for my road bike, but I'm sure you could do the experiment yourself - and it's more fun...

EDIT a bit more about rolling resistance.

Rolling resistance is poorly understood by many people. There are different factors that come into play:

• Tire dimensions (radius, width, curvature)
• Tire pressure

For example- if the road is rough, a small tire keeps "having to ride up hill" while a larger tire will "glide over the bumps". A soft surface (like sand) creates a dip, and again the tire keeps hiving to "climb out of the hole". This climbing is felt as rolling friction. A mountain bike tire, being wider, digs less of a hole - and so fat tires are best on soft surfaces.

But the really interesting thing is friction on a smooth road. Here, the key factor is the shape and size of the contact patch - specifically, the length of the contact patch. There is a nice diagram (from http://velonews.competitor.com/2012/03/bikes-and-tech/technical-faq/tech-faq-seriously-wider-tires-have-lower-rolling-resistance-than-their-narrower-brethren_209268) that helps to show this:

The thing that matters most is the difference between the length of the contact patch, and the corresponding arc of the tire that is touching this patch. For a patch length $l$ and a wheel radius $r$, the angle $\theta$ (from start of contact to end of contact, as measured about the axle) is given by

$$tan\frac{\theta}{2}= \frac{l}{2r}$$

This means that the amount of rubber that is confined along the length $l$ is in fact a little bit bigger - the excess amount of rubber is

$$e = r\theta - 2r\ sin\frac{\theta}{2}$$

Small angle expansion ($sin\theta = \theta - \frac{\theta^3}{6} + ...$) tells us that for small $\theta$ this difference is roughly

$$e = \frac{r \theta^3}{3}$$

Further, if we assume an elliptical patch with a constant aspect ratio (this is approximately true for a given tire dimension), then the length will scale with the inverse square root of the pressure (since $F = P \cdot A$, force is product of pressure and area); and since theta is approximately linear with length, you see that the motion of the rubber (and thus the energy dissipation) will go down with pressure.

Now comes the "knobby" mountain bike tire. Because much of the tire is not touching the road, the "effective pressure" is lower than you think it is - more specifically, the tire will start to touch the ground earlier, and leave later, for a much higher effective contact length, and thus $\theta$. And that means much higher friction.

How much higher? I have no measurements, but here is an estimation:

The pressure on a mountain bike tire is typically in the 30's of PSI - let's say 1/4 of the pressure in a road bike tire. But it's also much wider - say 3x wider than a road bike tire - which will shorten the contact length for a given pressure. Finally, with the knobbly nature of the profile, it might have an "effective contact length" that is 20% longer than it would have been for a smooth tire (because the stiffness of the tire will offset some of the knobbly nature of the tire).

With all those assumptions, you get a contact length ratio (vs road bike) of $(4/3)*1.2 = 1.6$. Now we computed earlier that rubber friction goes as the third power of the contact length, or 4x greater.

Rolling friction on a road bike is around 5N (see link above). Four times more rolling friction corresponds to an additional 15N, which at 15 mph is about 90 W. That's a lot of power - by the plot I derived above, it would drop your speed by about 3 mph for the same power. That is quite similar to the value quoted by Lubos. Note that at 15 mph, your wind resistance is quickly dropping, and that the position of the body (upright vs dropped) doesn't really have a huge impact (although more so if you are riding into a stiff headwind).

This just goes to show that you really need to pay attention to your tires - you pay a price for having tires that cannot sustain a high pressure (and you pay even more for not inflating the tires appropriately for the road surface...)

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+1 for the experiment, but really this would only answer the question if you had actually done it with both a road- and mountain bike. – leftaroundabout Jul 29 '14 at 17:08
@leftaroundabout - thanks, and you are right. I am supplementing Lubos's answer: he says "the best thing to do is to do the experiment", and I'm saying "here is a good way to do the experiment". There is a ton of variability, so there is no "one answer" other than the usual "it depends". – Floris Jul 29 '14 at 17:10

The general rule-of-thumb

I would guess that at no wind, the resistance of the air (drag) is a dominant relationship between wattage and the equilibrium speed. Because the drag scales like $v^2$, a 15-20 percent increase of the equilibrium speed corresponds to something closer to 30-40 percent increase of the efficiency.
This figure looks very large and it is probably an overestimate, indeed, because some other effects slowing the bike down increase less quickly than $v^2$ with the speed.