I am curious how to estimate drag for a bicycle. This has two different framings:

  • Find the amount of power required to ride at a certain speed. That is, when one is traveling at a constant speed, the power added to the system (with my effort and/or coasting on a descent) equals the power subtracted by aerodynamic drag, rolling resistance, ascending, deflecting objects thrown by spectators, etc.

  • Find the amount of pushing force required to maintain a given speed (again, the pushing forward equals the pushing backward at constant speed). The most obvious example is aerodynamic drag, which one can feel “pushing” back when one is riding at speed.

Online calculators and formulas exist, but they assume coefficients of rolling resistance or aerodynamic drag, or assume that I can provide these coefficients. How do they make these assumptions, and/or how can I make those estimates myself?

(Credit goes to R. Chung for urging me to ask, in the commentary of this question.)


Your question is simple but a full answer is complex. The simplest answer is to point to Part 2 (especially chapter 4) of Wilson and Papadopoulos (2004), or the recent review by Debraux et al. (2011), or the paper by Martin et al. (1998). However, even these papers do not cover approaches that take better advantage of the data available from modern bicycle computers and GPS units. Some background on the power-drag equation will help you understand why there are so many different ways (with accordingly different levels of accuracy, precision, difficulty, and cost) of estimating drag.

The equation to convert speed to power is well-understood. Total power demanded has four parts:

Total power = power needed to overcome rolling resistance + 
              power needed to overcome aerodynamic resistance + 
              power needed to overcome changes in speed (kinetic energy) + 
              power needed to overcome changes in elevation (potential energy)

Of these, the simplest piece is the power needed to overcome changes in elevation. The power needed to account for the change in potential energy, and to overcome changes in speed are straightforward:

watts(PE) = slope * speed in meters/sec * total mass * 9.8 m/sec^2
watts(KE) = total mass * speed in meters/sec * acceleration

There is a small part of the KE component due to moment of inertia in the wheels but for bicycles that tends to be small and we often ignore it. However, the equations needed to describe the rolling resistance and aerodynamic resistance are a bit more complicated. The article by Martin et al., cited above, gives more detail but if we can ignore wind then the aerodynamic component simplifies to

watts(aero) = 0.5 * rho * CdA * (speed in m/s)^3

where rho is the air density in kg/m^3 and CdA is the drag area ("A" is the frontal area and "Cd" is the coefficient of drag; CdA is their product and can be thought of as the "equivalent" area of a cube held perpendicular to the direction of the wind with a face of area A). If there is non-zero wind, the aerodynamic component is approximately

watts(aero) = 0.5 * rho * CdA * (groundspeed in m/s) * (airspeed in m/s)^2

When there is no wind, airspeed and groundspeed are the same so the equation simplifies to the one above. A headwind increases airspeed above groundspeed while a tailwind decreases airspeed below groundspeed. I say "approximately" because these are approximations when the wind comes from directly in front or behind (or is zero); when the wind is off-axis, it affects CdA.

Finally, the power needed to overcome rolling resistance (which includes tires, tubes, and bearing friction) is

watts(RR) = Crr * total mass * 9.8 m/sec^2 * speed in m/s

Crr is the coefficient of rolling resistance.

Now, if you go to an online calculator like the one at Analyticcycling.com you'll see that you must provide values for rho, Crr, Cd, and A; then, given a particular value of speed and slope, it will calculate power. It's easy to find online calculators for the air density, rho, but much harder to find estimates of Crr and CdA (or separately, Cd and A).

The easiest (but most expensive) way to estimate CdA is in a wind tunnel. There, an object is mounted on a scale (basically, a very precise and accurate bathroom scale), wind at a known speed is applied, the air density is measured, and the total force on the object is measured by the scale. Watts are force (in Newtons) * speed (in meters/sec) so force (in Newtons) = watts/air speed = 0.5 * rho * CdA * (airspeed^2). The tunnel operator knows rho, knows airspeed, and the expensive bathroom scale measures the force so you can calculate CdA. Wind tunnel estimates of CdA are considered the gold standard: when performed in a good tunnel with experienced operators, the measurements are precise and repeatable. In practice, if you want to know the Cd separately, you'd measure the frontal area A with a digital camera and compare it to a digital photograph of an object (like a flat square) of known area. As an historical aside, nearly 100 years ago Dubois and Dubois measured frontal area by taking photographs of a person and a reference object, cutting out the photos along the outlines of the object, and then weighing the cut-outs on sensitive scales.

However, the resistance in tires, tubes, or bearings are not affected by air speed, so one cannot estimate Crr from wind tunnel data. Tire manufacturers have measured rolling resistance of their tires on large rotating drums but they cannot measure aerodynamic drag. In order to measure both Crr and CdA, you need to find a method that measures both and allows you to differentiate between the two. These methods are indirect field estimation methods and they vary a great deal in their accuracy and precision.

Until the last 20 years or so, the most common indirect field method was to coast down a hill of known slope and to measure either maximum speed (also known as terminal velocity) or else the speed when passing a fixed point on the hill. Terminal velocity doesn't let you differentiate between Crr and CdA; however, if one measured speed at a given point and were able to control the "entry" speed at the top of the hill, you could then test at different entry speeds and get enough equations to solve for the two unknowns, Crr and CdA. As you might expect, this method was tedious and liable to poor precision. Nonetheless, many ingenious alternatives were explored, including coasting down wind-free corridors or inside large airplane hangars, and measuring speed to relatively high precision using "electric eyes" or timing strips. Some of these methods are described in the review by Debraux et al., cited above.

With the advent of on-bike power meters, new opportunities emerged to measure aerodynamic and rolling drag. In short, if you could find a flat wind-sheltered road, you would ride at a constant speed or power on the road; then, repeat at a different speed or power. The requirement of "flat and wind-sheltered at constant speed" meant that you could ignore the PE and KE components of power and only had to deal with the rolling resistance and aerodynamic components so the overall power equation simplifies to

Watts = Crr * kg * g * v + 0.5 * rho * CdA * v^3; or 
Watts/v = Crr * kg * g + 0.5 * rho * CdA * v^2

where g is the acceleration due to gravity, 9.8 m/sec^2.

The latter formula can be easily estimated by linear reqression where the slope of the equation is related to CdA and the intercept is related to Crr. This is what Martin et al. did; they used an airplane runway, averaged the runs in both directions, and measured barometric pressure, temperature, and humidity to calculate rho, and measured and corrected for wind speed and direction. They found that the CdA estimated by this method agreed to within 1% of the CdA measured in wind tunnels.

However, this method requires that the road be flat and that speed (or power) be constant over the length of the test run.

A new method for estimating CdA and Crr has been developed that exploits the recording capability of many modern bike computers and bicycle power meters. If one has moment-by-moment recording of speed (and optionally, power) then you can directly measure changes in speed so the KE component of power can be estimated. In addition, if you ride around in a loop, the road need not be flat since you know that upon returning to the start point of the loop the net elevation change will be zero so the net PE component will be zero. This method can be and has been applied to coasting down hills of known net elevation change (that is, you don't need to have constant slope, and if coasting you know the power is zero). Examples of this approach can be found here and here and, when performed carefully have been shown to agree with wind tunnel estimates of CdA to well within 1%. A short video presentation on the method can be found starting at about the 28:00 mark here. A short video of the method in use on a velodrome can be found here

  • I guessed this was a R.Chung answer by line 2.... – Criggie Oct 31 '17 at 2:58
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    You just got mentioned on a GCN show youtube.com/watch?v=mJrzRDqQ5vQ at around 14 min 25 sec. – Criggie Apr 19 '19 at 12:19
  • That mention was apparently enough to kill the rest of the show. – R. Chung Apr 19 '19 at 15:03
  • Crosswinds increase air drag as well, not just the drag coefficient. This is a direct consequence of the force being proportional to the square of the relative air speed. For example, if you compare riding at 10m/s without wind to riding the same speed with a 10m/s cross wind, the relative air speed rises to sqrt(2) * 10m/s = 14.1m/s. This leads to a force that is twice as big and at a 45° angle. So, the force component in the direction of travel, and thus the required watts(aero) will be increased by a factor of sqrt(2) = 1.41. – cmaster - reinstate monica Feb 8 at 18:05
  • This unintuitive result can be explained by the fact that the rider interacts with more air in the crosswind scenario. The mass of air that the cyclist interacts with grows with the same sqrt(2) factor as the watts(aero) does. – cmaster - reinstate monica Feb 8 at 18:09

If you could find several long hills of different but relatively constant (and not too steep) slope, then determine the slope and your terminal velocity on each hill (assuming that velocity is below some safe speed), you should be able to do the math to determine aerodynamic drag (working on the reasonably valid assumption that rolling resistance is negligible a higher speeds).

Or, with very careful observation, you could determine how rapidly you decelerate on a level road.

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    One could also, using a long rope (to avoid "drafting" effects), tow the bike and cyclist at a constant speed on level ground, with a spring scale between rope and cycle to measure applied force (which would equal drag). A little hazardous, but probably not incredibly unsafe if reasonable precautions are taken (including having a quick/easy way for the cyclist to release the tow rope). – Daniel R Hicks Jun 18 '12 at 18:52
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    The "tow rope" method is discussed in the Debraux et al. article linked elsewhere. It doesn't have good precision. The deceleration method works well if you have a way to record moment-by-moment speed, such as with one of the increasingly popular Garmin bike computers. A method for doing this is discussed at forum.slowtwitch.com/cgi-bin/gforum.cgi?post=3590389#3590389 and, when done on a calm day without passing cars or other traffic, it has produced results in agreement with wind tunnel estimates. – R. Chung Jun 18 '12 at 21:06
  • Yeah, the deceleration method would work well with an accurate GPS or other time/position logger. And one could combine it with the tow to get up to the higher speeds that would normally (with a non-superhuman cyclist) require a downhill to attain. WRT the basic tow rope (with force gauge) technique I suspect the most difficult part is measuring the force accurately, and probably some modern electronic signal processing techniques could be applied to assist that. – Daniel R Hicks Jun 18 '12 at 22:40
  • I buy this explanation! +1 (specially considering that drag is, in practice, independent of power). – heltonbiker Jun 18 '12 at 23:10
  • Heltonbiker, the problem is that terminal velocity not only has poor precision but also only gives you ordinal ranking for comparisons (that is, under the best conditions you can tell that A has lower drag than B but not by how much) which means you don't get an estimate of CdA. Similarly, people have tried rollout distance. Moment-by-moment modeling works much better. – R. Chung Jun 18 '12 at 23:45

Jan Heine & crew at Bicycle Quarterly recently reported the results of their wind tunnel research. A summary is available online, but the full results are only available in the printed journal.

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    Sadly, that article focuses only on one component of drag that cyclists experience (viz., aerodynamic drag) and answers the question "how can one estimate drag?" with "in a wind tunnel." – R. Chung Jun 18 '12 at 13:25

Oh boy. Aerodynamics on a bicycle. I want to show you a picture of the back end of a triathlete as he walks beside his bicycle. Except I can't find it.

Okay, how's this for an analogy then. Find a brick. Find a pencil. Stand the pencil up on its end and tape the brick on top of it. Put this contraption in a wind tunnel. Measure the drag of this contraption.

Now, take away the pencil. Measure the drag again.

You are the brick. The pencil is your bicycle.

The next time you are tempted to spend money on bicycle parts to reduce the drag in this operation, you should think very carefully about this analogy. Especially considering that it's been found that the wrinkles in your jersey contribute more to the drag of your aerodynamic shape than both aero bars and an aero helmet combined.

In other words, your money is better spent on a skinsuit, or sunscreen. And sunscreen has less drag.

  • Actually, a properly-fitting skinsuit has less drag than sunscreen on naked skin. We know this because we've measured the drag of riders with bare arms and covered arms, and with shorts that cover more and less of the thigh. Skin turns out to be faster than loose clothing but not as fast as the right skinsuit. – R. Chung Nov 2 '17 at 13:25
  • Well I'll be damned. – Ernie Nov 2 '17 at 15:54
  • Of course, that requires that the skinsuit fit exactly right. :) – Ernie Nov 2 '17 at 15:55
  • It's kinda surprising what one learns when one can actually measure drag. It turns out the eyeball is okay for spotting big changes in drag but not so great for distinguishing between small changes -- and if you race, even small changes can be consequential. A similar thing has been found for Olympic swim suits: they're faster than swimming nude. – R. Chung Nov 2 '17 at 16:20
  • Be that as it may, the drag induced by the person on the bike is many times higher than the drag induced by the bike itself - even in the worst possible examples of bike aerodynamics. Which was my point. Make thine own self aerodynamic first, before spending loads of money on doing the same with the bike. An entire industry thrives on this lack of understanding. – Ernie Nov 2 '17 at 20:09

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