Pure power vs Watts / Kg in cycling

Zwift racing has taken off during this lockdown period and I thought I would enter into one of the races myself. At the time of entering the race I had an FTP of 230W and weighed 55Kg, giving me a Watts / Kg of 4.2W/Kg. This placed me in category A:

Zwift categories

Given it was my first race and I knew that my relatively low Watts yet high Watts/Kg would not favour me on a flat course, I decided to enter a category B race yet it quickly became apparent that I was in over my head! That race taught me in the most brutal way that watts per kg isn’t everything. So when is watts/kg king, and when does just simple raw power count the most?

Keen cyclists obsess over two numbers: their functional threshold power or FTP, and their power-to-weight ratio. FTP is the highest average power output a cyclist can sustain for an hour, and is measured in Watts. In a time trial over a flat course, the rider with the higher FTP will be able to sustain a faster average velocity and hence complete the course in the shortest time. But on a climb gravity starts to come into the equation, and the riders’ power-to-weight ratio (measured in Watts/kg) is increasingly important. Intuitively, the steeper the gradient, the more the advantage to the lighter rider.

Typically a lighter rider also has a lower FTP (if a rider is both lighter and more powerful, they are always going to win!). How steep does a climb have to be before the lighter rider has the advantage over a heavier and more powerful rider?

Some starting assumptions

  • The riders put out constant power for the duration of the course, at their FTP.

  • The course is a constant gradient.

  • Given constant power output and a constant gradient, riders will therefore be riding on the course at a constant velocity

Where does all the power go?

To turn the power produced by the cyclist into speed on the road, a cyclist must overcome three forces: road friction, wind resistance, and gravity. The Climbing Cyclist blog provides us with an equation showing the power output P that a cyclist needs to produce to travel at a certain speed.

Informally, P = power required to overcome rolling resistance + power required to overcome wind resistance + power required to overcome gravity.

P = P_{rolling} + P_{wind} + P_{gravity}

Let’s look at each of these in turn.

Overcoming rolling resistance

The power required to overcome rolling resistance, P_{rolling}, is given by:

P_{rolling} = k_r M g s

(Note, this is a revision of the formula given in The Climbing Cyclist blog, which failed to include the gravity based term, see Understanding Rolling Resistance for more details).

where

  • k_r is the rolling resistance coefficient, around 0.005 for a road bike on the road
  • M is the combined mass of bike and rider,
  • s is the speed of the bike on the road.
  • g is the gravitational constant (9.8m/s^2)

As an example, for a 70kg rider + bike combination going uphill at 10km/h (2.78m/s), we have:

P_{rolling} = 0.005 \times 70 \times 9.8 \times 10

or about 34W to overcome rolling resistance.

Overcoming wind resistance

The power required to overcome wind resistance, P_{wind}, is given by:

P_{wind} = k_a A s v^2 d

where

  • k_a is the wind resistance coefficient, which we will set at 0.5 representing neither a headwind nor a tailwind
  • A is the frontal area of the bike and rider. Typically a lighter rider will have a small frontal area, but here we’ll make the simplifying assumption that all of our riders have a frontal area of 0.6m^2.
  • s is the speed of the bike on the road
  • v is the speed of the bike through the air, which with no wind will be equal to s
  • d is the density of air, which at sea-level is roughly 1.226kg/m^3.

Using these assumptions we have:

P_{wind} = 0.5 \times 0.6 \times 1.226 \times s^3

And if we are climbing at say 10km/h (2.78m/s), then P_{wind} is approximately 8W.

Overcoming gravity

The power required to overcome gravity, P_{gravity}, is given by:

P_{gravity} = giMs

where

  • g is the gravitational constant (9.8m/s^2)
  • i is the gradient of the climb (e.g. 0.05 for a 5% climb)
  • M is the combined mass of rider and bike
  • s is the speed of the bike on the road

At 10km/h up a 5% climb a 60kg rider needs to produce 9.8 \times 0.05 \times 60 \times 10 = 294W , and a 90kg rider needs to produce 441W!

A tale of two riders

Armed with this mathematical model, we can compare two riders, rider A, with mass M_A and FTP P_A, and rider B with mass M_B and FTP P_B. We want to know when rider A has an advantage over rider B, which given we have assumed the riders are going at a constant speed, is the same thing as asking when s_A, the speed of rider A, is greater than s_B, the speed of rider B, on a climb of gradient i.

Who’s best at overcoming gravity?

P_{gravity} = giMs

So s_A = \frac{P_A}{giM_A} and s_B = \frac{P_B}{giM_B}.

We’re interested in when s_A > s_B, i.e., \frac{P_A}{giM_A} > \frac{P_B}{giM_B}.

We’re considering positive gradients only, and gi is a constant term on both sides so we can remove it, and the condition becomes:

\frac{P_A}{M_A} > \frac{P_B}{M_B}

Note that P/M is just the power-to-weight ratio, so what we’ve discovered is that power-to-weight ratio is the only thing that matters when it comes to overcoming gravity!

Who’s best at overcoming rolling resistance?

For rolling resistance we have

P_{rolling} = k_r M g s

So s_A = \frac{P_A}{k_r g M_A} and s_B = \frac{P_B}{k_r g M_B}.

We’re interested in when s_A > s_B, i.e. \frac{P_A}{k_r g M_A} > \frac{P_B}{k_r g M_B}.

Once again k_r g is a constant on both sides and we’re left with:

\frac{P_A}{M_A} > \frac{P_B}{M_B}

So once again, power-to-weight ratio is the only thing that matters when it comes to overcoming rolling resistance.

Who’s best at overcoming wind resistance?

For overcoming wind resistance we have

P_{wind} = k_a A s^3 d

So s_A = \sqrt[3]{\frac{P_A}{k_aAd}} and s_B = \sqrt[3]{\frac{P_A}{k_aAd}}.

We’re interested in when s_A > s_B, i.e. when \sqrt[3]{\frac{P_A}{k_aAd}} > \sqrt[3]{\frac{P_B}{k_aAd}}.

k_a A d is a constant term, so we’re left with:

\sqrt[3]{P_A} > \sqrt[3]{P_B}, or simply P_A > P_B.

In other words, for overcoming wind resistance, pure power is the only thing that matters.

Putting it all together

When it comes to climbing and rolling along the road, power-to-weight ratio is king. But for overcoming wind resistance absolute power is the only thing that matters. For longer duration time-trial efforts of the kind we’ve been considering therefore, it’s not the grade of the climb per se that determines whether power-to-weight or pure power is more important, but the speed the cyclist is travelling at. Power-to-weight is most important on climbs because speeds tend to be lower. On a flat course, speeds tend to be higher, and the power output required to overcome wind resistant is proportional to the cube of the speed.

For example, we saw that the power required to overcome wind resistance when climbing at 10km/h was just 8W. But racing on a flat course at 30km/h the power requirement goes up to 213W, and at 40km/h it requires just over 500W! (Our cyclist is in an inefficient upright position, getting down in the drops to reduce A, the frontal area of the bike and rider makes a big difference in real riding).

Another way of putting this is that pure power matters more than power-to-weight ratio when wind resistance is the dominant factor, i.e.

P_{wind} > (P_{rolling} + P_{gravity})

k_a A s^3 d > (k_r M g s + g i M s)

(0.5 \times 0.6 \times 1.266 \times s^3) > ((0.005 \times 9.8 Ms) + (9.8Mis))

0.38s^3 > (0.049Ms + 9.8Mis)

0.38s^3 > Ms(0.049 + 9.8i)

0.38s^2 > M(0.049 + 9.8i)

Or \frac{0.38s^2}{M(0.049 + 9.8i)} > 1.

For M = 75kg, we can explore the space formed by gradients from 0 to 15%, and speed from 1 to 30km/h (accepting no-one is likely to cycle up a 15% gradient at 30km/h!) to find the frontier where pure power has the advantage.

So there you go, if you are a lighter, less powerful rider you may want to consider dropping down a category (or two!) before your first Zwift race!

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