# Cycling in Brighton - Can You Climb Ditchling Beacon?

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Brighton is a great city for cycling. There is a good network of cycle lanes and easy access to the South Downs National Park so it's a great way to keep fit. There are also plenty of challenging hills! A good target for any amateur cyclist in the area is to try and cycle up Ditchling Beacon. In this post, we will use some simple ideas from GCSE level maths and physics to help understand the various forces that need to be overcome when out on the bike, and how we might train for hill climbing.

First consider cycling in a straight line at a constant speed. Newton's first law says that there is no net force on a body that moves with a constant velocity, so to keep going at the same speed, the force provided by the cyclist has to match the forces acting on the cyclist and the bike. These come in two general classes corresponding to two things that all cyclists find hard... riding into the wind and riding up hills.

The first type of force is due to air resistance. The different contributions from these so-called drag forces can be thought to act on the bike and rider as a whole, and also on the front and back wheels. They are all proportional to the air density $\rho$, a typical cross-sectional area, and most importantly to the square of the speed, $v$. The forces can be written as:

In these equations, the subscript $air$ refers to general air resistance on the bike and cyclist and subscripts $fw$ and $bw$ refer to air resistance on the front and back wheel. The $C$ are empirical coefficients, $A$ is the cross sectional area of bike and rider, $r$ is the radius of the wheel, and $s$ is a fraction between 0 and 1 to account for rear wheel shielding by the rest of the bike. Typical values are $A=0.5m^2$, $\rho=1.2kg/m^3$, $r=\frac{1}{3}m$, $C_{air}=\frac{1}{2}$, $C_{fw}=C_{bw}=\frac{1}{20}$ and $s=\frac{1}{4}$. The relative size of $C_{air}$ compared to $C_{fw}$ shows that the air resistance term will be at least 10 times more important than the wheel resistance terms.

The velocity $v$ in the formulae is typically measured in $ms^{-1}$. Because $10knots = 5.14ms^{-1}$ and $10kph = 2.78ms^{-1}$, and the appropriate velocity when riding into the wind is a sum of the speed of the rider and the wind speed, one can make the rough approximation

With the parameter values given above, the total contribution from wind resistance can be roughly approximated as

The second type of forces are due to gravity and rolling resistance. These are given by:

where $M$ is the combined mass of bike and rider, $g\approx 10ms^{-2}$ is the local acceleration due to gravity, $C_{rr}$ is another empirical coefficient to account for frictional rolling resistance, and $\theta$ is the angle of the hill measured against the horizontal. Typical values are $M=100kg$ and $C_{rr}=\frac{1}{100}$. The gradient of a slope is defined as 'rise over run', or the change in height $\Delta y$ divided by the change in horizontal distance $\Delta x$. $\Delta y$ is opposite the slope angle $\theta$, and $\Delta x$ is adjacent, so the gradient is just $\tan\theta$. The gradient is more usually expressed in terms of the grade of the slope which is given in percent, so that $\text{gradient}=\text{grade}/100$. For small slope angles (typically any grade less than 15%), one can make the approximation $\sin\theta = \tan\theta \approx\theta$, so that the sum of the gravity and rolling resistance becomes:

The total force on the bike and rider from wind resistance and gravity can now be approximated by the formula

The assumptions contained in this approximate formula are detailed above. The main one that might need to be adjusted is the combined mass of bike and rider, which has been approximated here as $M=100kg$. To recap, this formula gives the overall force that the rider has to generate to maintain a constant speed when riding up a slope of a given grade and riding into the wind with a particular constant wind speed.

Let's use this formula to think about various aspects of cycling. If we assume a modest cycling speed of 20kph, the power generated by the cyclist is given by $F\times v$ (with $v$ in $ms^{-1}$), so the power expended when cycling along the flat with no headwind is calculated by our formula as approximately 70W. With a headwind of 10knots, the power expended is 130W, and with a headwind of 20knots, the power expended is 230W. So riding into a headwind of 10 or 20 knots is equivalent to riding up a constant gradient of 1.5% or 3.5%.

To get a feel for different hill gradients, let's look at some places in Brighton. The figure below shows the profile of Dyke Road Drive and Bear Road. Dyke Road Drive is about 1.2km long, with an elevation gain of 57m. This gives an average grade of 4.8%. Bear Road is about 1.4km long with an elevation gain of 115m, so its average grade is 8%. Coldean Lane is a longer hill at 1.75km with elevation gain of 100m, so the average grade is 5.7%. The biggest hill in the area is Ditchling Beacon, with a length of 1.3km and an elevation gain of 153m giving an average grade of 11.8%. A lesser known hill almost as steep as the Beacon is Steyning Round Hill, with an elevation gain of 92m over a distance of 0.8km, giving an average grade of 11.5%. The shorter overall length of this hill makes it a good training hill for someone working up to tackling the Beacon.

Another possible way to train for a steep hill is to find a milder hill and add more weight to the bike. So for example, if we wanted to train for Ditchling Beacon by riding up Bear Road, we can figure out how much extra weight to add to the bike to make the effort comparable. If we assume no headwind and equate the gravity forces due to the original mass at the steep gradient with the augmented mass at the lower gradient, this will give the extra mass needed. So

where $m$ is the extra mass, and $g_h$ and $g_l$ are the grades of the steeper and shallower hills. A little algebra leads to

If we plug in $g_l=8$ and $g_h=11.8$, corresponding to Bear Road and Ditchling Beacon, we find $m=0.42M$, so would need to add an extra 42% of the combined bike and rider weight to make Bear Road feel like Ditchling Beacon.

Now let's consider what strategies would help to minimize the forces working against the cyclist. It really all depends on the conditions on the day. If the riding is mainly flat then wind drag becomes the most important factor and the best way to reduce this is by becoming more aerodynamic through a good choice of clothing. As the speed of the rider increases, the power expended goes up as speed cubed. This means that someone who is cycling past at twice the speed of another is actually working 8 times as hard.

If the terrain is more hill-based, then weight of bike and rider becomes the main factor as the gravity forces will dominate the drag forces on hills. Carbon fibre bikes and traveling as light as possible is the best strategy in this case.

Finally, let's consider what kind of power an amateur cyclist can realistically expect to be able to generate. The following figure shows estimates of the gradient along Bear Road and also my speed based on GPS readings on a bike ride I did just before Christmas 2017. One can see how the grade of the hill fluctuates around the average. Based on the GPS readings shown here, the average grade is 9.5%, and my mean speed up the hill is 3.1$ms^{-1}$, which corresponds to around 12kph. The combined mass of me and my bike is approximately 90kg, which gives the power expended as 305W. At an approximate mass of 80kg, my power to weight ratio is around 3.8W/kg of body mass, which is about right for an amateur cyclist working over a time of around 5mins according to this article. A typical professional cyclist should be able to generate closer to 7W/kg of body mass over a 5 minute interval.