Cycling Maths

Why do hills slow cyclists?

Cycling up hill we slow down but a simple look at the situation suggest we get that speed back going down the other side. Unfortunately as we slow down below the average on the way up the hill, it is not a simple matter of speeding up over the average the same amount down the hill to maintain our average: it is the amount of time spent at a speed that determines its weight in the average calculation and the hill (at the same gradient) will have run out by then!

So at an average A if we spend 10 minutes cycling up hill at A-5 mph we need to spend 10 minutes cycling at A+5 mph which will be (A+5)/(A-5) times further. If our average was 15 that is twice as far downhill at 20mph as we cycled up hill. Doesn't seem fair but that's the maths.

On a symmetrical hill (with equal gradient up as down and the same distance up as down) the following equation tells us how fast we must shift downhill (sd) in terms of the Average (A) and speed uphill (su):

sd = A * su / (2*su - A) (1.0)

or better if su = A * p [p being the percentage drop in speed]

sd = A * [ p / (2p - 1) ] (2.0)

Unfortunately if the speed uphill falls to half, or below half, the Average, then there is not enough distance downhill to recover the average without having to maintain speed over the average after the hill.

Hills thus sap the cyclists average speed and this is why they slow cyclists.