# How to rate a hill

I’m not much of a cyclist. I have a bike, and I occasionally take it out for a longish ride, but generally I struggle to tell the difference between a peloton and a catamaran.

However, a question from @tzed250 on Twitter caught my attention: is there a way to rate a climb?

### The problem

If you’re cycling along, is it harder to get up a short, steep slope or a long, gradual one? Is there a way to assign a rating so you can compare climbs?

### My thoughts

For me, the obvious way to rate a climb is: how much work do you have to do? How much energy does it take to get to the top?

There are three major forces you need to work against:

**Gravity**. This is the main one: without it, there wouldn’t be a hill.**Air resistance**. This is only significant if you’re going at a fair speed – that’s obviously never the case for me, certainly not uphill, but it might affect a more competent cyclist.**Friction**((I originally had the wrong model for this, but @robjlow put me on the right track)). You have to overcome a certain amount of resistance just to get your bike to move. This is typically small.

The energy you require to just barely get up the slope, crawling along at pretty much zero metres per second, is $mgh$. Assuming you can get moving at all, it takes effectively the same energy to climb 100m at a 3º angle as at a 15º angle. But objectively, those are not equally difficult slopes.

### Power

The key is to think about the amount of *power* a cyclist generates. I found this extremely helpful site, which came at it from a slightly different angle.

The power required to climb a hill in still air turns out to be ((I’ve simplified their equation, but only slightly)):

- $P = Mgv \sin(\theta) + K_a v^3 + MK_r v$

The three terms correspond to the three forces listed above, with:

- $M ~ 90$ as the mass of the cyclist and bike in kilograms
- $g \approx 9.8$ as the gravitational constant
- $\theta$ the angle of the slope (the arctangent of the gradient)
- $K_a \approx 0.368$ is some constant relating to air resistance
- $v$ is your speed along the road
- $K_r \approx 0.005$ is some constant relating to the rolling resistance of your bike.

Given $P$ and $\theta$, that gives a cubic equation in $v$, which can be solved (exactly if you’d like to, numerically if you’re sensible).

Once you have $v$, and you know how far you have to travel ($d$m) up the hill, you can work out the time required ($t = \frac{d}{v}$).

Once you have $t$, you can work out the work you need to do as $E = Pt$.

Or, you can get the computer to do it. This gives energy ratings and time for cyclists developing power of 100W, 200W and 300W, any one of which you could use to grade the hill.

### The upshot

So, what does that tell us about different hills? Suppose we gain 40m over a kilometre, a 4% gradient. The machine says that’s 38kJ for a 100W rider, 43kJ at 200W and 49kJ at 300W.

If we gained the same height over 500m, the ratings would be 36, 37 and 38 kJ, respectively. However, it’s not a fair comparison: you’re cycling further on the shallow gradient. To match it up, we need to add in 500m on the flat, which costs 8, 12 and 16kJ for a total of 44, 49 and 54kJ.

I’m not certain this is the best way to compare – part of me wonders if dividing the energy rating by the distance (which gives something in units of force) would make for a better comparison. I’m genuinely unsure.

Have you got a better suggestion?