# Effect of hills on cycling effort

Cyclists usually quantify their cycling by distance: century rides, 200k Audax, 150 miles per week, and so on. Distance is an easy measure to understand, and many of us use a cycling computer with an odometer and trip mileage to record distance travelled. However, the effort required (or fitness benefit) on a hilly 100-miler is obviously very different to one on the flat. During our recent trip to Holland the effect of flat terrain was noticeable for the ease with which you can cycle all day and consequently, the number of people cycling. It’s not the same in hilly England. How do you judge the physical difficulty of long cycle rides: Chris’s 152 mile coast to coast for example, or Mary’s Audax rides, or my own Rivington 100?

A nonsensical cycling effort formula

Since my last one broke some time ago I’ve stopped using a bike computer. I now measure rides by time, not distance, on the basis that my level of effort is fairly constant regardless of whether I’m cyling uphill or horizontally. Time spent cycling seems more meaningful (although my watch also broke recently).

I thought it would be interesting to try and calculate the effect of hilly terrain, to compare the effort with that required on a flat course of similar distance. Various mathematical formulas and tables can be found on the web, and there’s also an online calculator for the aerodynamic drag and propulsive power of a cyclist at various degrees of gradient. Powermeters are available but they’re expensive, or you can compare calories burned using a Garmin GPS – it would give an indication, but a bit of amateur mathematics seems more fun.

So I’d like to compare the physical demands of two 100-mile bicycle rides: one on the flat and the other with 5,000 feet of climbing. I’m no good at maths, physics, or mechanics so this may well be full of errors and false logic. For the time being I’m also ignoring the effect of freewheeling downhill.

5,000 feet = 0.946969697 miles. Call it 0.947 miles. A 100 mile ride that climbs 0.947 miles therefore has an average gradient of 0.947%. There are various ways to express road gradients. A traditional gradient of 1 in 10 (1:10) is 10% and 1 in 1 is 100% – a 45° angle. 100% is not a vertical wall.

Velocity chart (from Validation of a Mathematical Model for Road Cycling Power)

The chart illustrates that the effect of increasing a gradient is *linear* in that a hill which gets steeper and steeper slows you down at a constant rate: about 11% for every 1% change in steepness. So a slope twice as steep as another slows you twice as much, or requires twice the effort to maintain the same speed. More slope doesn’t add effort exponentially.

In other words, it makes no difference whether you climb the 5,000 feet over a short distance or spread out over 100 miles. The effort required to climb it is the same on both cases. But this is purely the climbing, I think. Adding in the ‘moving along anyway’ factor and allowing for freewheeling downhill complicates things. Freewheeling first.

Assuming the 100 mile ‘climbing’ bike ride ends where it began there will obviously be as much down hill as up. I don’t know the minimum gradient for freewheeling along with never needing to pedal but if it’s minus 2% the easiest course will have the steepest possible climbs and the longest possible minus 2% gradients, with all the climbing at the start. This course probably doesn’t exist, so looking at Chris’s Morecambe to Mappleton gradient profile I’ll assume the freewheeling downhills are 10% of the course, during which time no effort is required.

To take account of the *moving along anyway* factor Cycling Performance Tips suggests calculating calories required for a horizontal component added to those for a vertical component to arrive at the total energy requirement for cycling a course with uphill gradients. As stated, it doesn’t matter whether the gradients are steep or not, nor how many there are. It’s total elevation gain that matters. So using the formulas from Cycling Performance Tips, here is my calculation:

**10 mph over level ground for 100 miles:**

- Pw = 4.4704 m/sec [3.509 + {0.2581 (4.4704)(4.4704)}]
- = 4.4704 [3.509 + 5.158]
- = 38.74 watts
- Pc = 38.74/4186.8 = 0.0093 Cal/sec
- T = 100/10 = 10 hrs
- = 10 x 3600 sec/hr
- = 36000 sec
- Ce = 0.0093 x 36000 sec = 334.8 Cal
- Ci = 334.8 Cal/0.25 =
**1339.2**Cal =**Eh** - (plus approx 50 Cal/hr x 10 hrs for basal metabolism Calories)
- Eh =
**1839.2**Cal

**Allow for 5,000 feet of climbing:**

- W = 85 kg x 1524 metres
- = 129540 kg-m
- Ce = 129540/418 = 309.9 Cal
- Ci = 309.9/0.25 =
**1239.6**Cal =**Ev**

This shows how climbing 5,000 feet increases the effort needed over a 100 mile course: 1839.2 Cal for the horizontal element and an extra 1239.6 for the climb. With no climbing, you could cycle 150 miles for the same effort, assuming calories equate to effort. I climbed 7,311 feet on the Rivington 100, equating to about 75 more miles than if the course had been perfectly flat. That’s probably not quite true. I returned to where I started so the freewheeling down hills effect has to be factored in. Assuming 10% freewheeling with no pedalling, the energy required for the climbing component is not **1239.6** Cal but rather less at **1116**.

In the real world other things come into play: wind speed and direction, cycling slower up hills, going faster than 10 mph on the flat, length of downhills and the momentum effect, and so on, but the calculation gives a fair idea on the effect of hills on cycling effort, or how much easier it is to cycle a flat course. If you’ve climbed 10,000 feet on a 100 mile ride in England you could just as easily have cycled 200 miles in Holland!