Why is riding in mountains slower than on flat ground?

It may be obvious to most of you, but it wasn't to me. After all, there are always two sides of a mountain. While you are slower going up, you are much faster going down. The average should be equal to riding on flat. Hmmm, this calls to some math.

Let's say we are riding up and down a symetrical hill of road length 2S. We cycle up with average velocity v1 and go down with average velocity v2. The up and down parts have equal length S. What is our overall average speed v?

The enticing answer is v=(v1+v2)/2, but, as you may presume, the world is not that simple. Combining the formulae: v1=S/t1, v2=S/t2, t=t1+t2, v=2S/t, we have: v=2S/t=2S/(t1+t2)=2S/(S/v1+S/v2)=2/(v2/v1v2+v1/v1v2)=2v1v2/(v1+v2).

So the answer is: v=2v1v2/(v1+v2) and it's independant on length S. For example, if you climb with 10 km/h and descend with neck breaking 70 km/h, your overall speed is not 40 km/h as you might have wished, but only 17,5 km/h, as you will find out looking incredibly at your computer at the end of the downhill. The only reasonable way to increase the overall speed is to climb faster. If you climbed just 50% faster, with 15 km/h instead of 10 km/h, then the overall speed increases to respectable 24,7 km/h. If, instead of climbing half faster you descended twice faster, i.e. 100% faster, at the world record breaking 140 km/h, your overall speed would still be miserable 18,7 km/h.

The point of this story: instead of risking your neck at speeds where there is no margin of error, beter have an ultralight pack which will enable you to climb a little faster.

We'll make some more use of mathemathics. We first express climbing speed as a percentage f of descending speed, v1=f v2. The average overall speed is then v=v2 2f/(1+f), as compared to intuitive (and wrong) average speed (v1+v2)/2=v2(1+f)/2. The difference, together with the ratios of climbing/descending speeds from the above example (10/70, 15/70 and 10/140), is indicated in the graph.