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Bicycle performance

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A bicycle's performance, in both biological and mechanical terms, is extraordinarily efficient. In terms of the amount of energy a person must expend to travel a given distance, investigators have calculated it to be the most efficient self-powered means of transportation.1 From a mechanical viewpoint, up to 99% of the energy delivered by the rider into the pedals is transmitted to the wheels, although the use of gearing mechanisms may reduce this by 10-15% 2 3. In terms of the ratio of cargo weight a bicycle can carry to total weight, it is also a most efficient means of cargo transportation.

Energy efficiencyEdit

A human being traveling on a bicycle at low to medium speeds of around 10-15 mph (16-24 km/h), using only the power required to walk, is the most energy-efficient means of transport generally available. Air drag, which increases roughly with the square of speed,[1] requires increasingly higher power outputs relative to speed. A bicycle in which the rider lies in a supine position is referred to as a recumbent bicycle or, if covered in an aerodynamic fairing to achieve very low air drag, as a streamliner.


On firm, flat, ground, a 70 kg man requires about 100 watts to walk at 5 km/h. That same man on a bicycle, on the same ground, with the same power output, can average 25 km/h, so energy expenditure in terms of kcal/kg/km is roughly one-fifth as much. Generally used figures are

  • 1.62 kJ/(km∙kg) or 0.28 kcal/(mile∙lb) for cycling,
  • 3.78 kJ/(km∙kg) or 0.653 kcal/(mile∙lb) for walking/running,
  • 16.96 kJ/(km∙kg) or 2.93 kcal/(mile∙lb) for swimming.

The average "in-shape" person can produce about 3 watts/kg for more than an hour (e.g., around 200 watts for a 70 kg rider), with top amateurs producing 5 watts/kg and elite athletes achieving 6 watts/kg for similar lengths of time. Elite track sprinters are able to attain an instantaneous maximum output of around 2,000 watts, or in excess of 25 watts/kg; elite road cyclists may produce 1,600 to 1,700 watts as an instantaneous maximum in their burst to the finish line at the end of a five-hour long road race. Even at moderate speeds, most cycling energy is spent in overcoming aerodynamic drag, which increases with the square of speed; since the power required is the speed times the force (here drag), the power needs increase approximately with the cube of speed.

Typical speedsEdit

Typical speeds for bicycles are 15 to 30 km/h (10 to 20 mph). On a fast racing bicycle, a reasonably fit rider can ride at 50 km/h (30 mph) on flat ground for short periods. The highest speed officially recorded for any human-powered vehicle (HPV) on level ground and with calm winds and without external aids (such as motor pacing and wind-blocks) is 132.449 km/h (82.3 mph). That record was set in 2008 by Sam Whittingham in the Varna. In the 1989 Race Across America, a group of HPV's crossed the United States in just 6 days. [2] [3][4][5] The highest speed officially recorded for a bicycle ridden in a conventional upright position under fully faired conditions was 82.52 km/h (51.29 mph) over 200m [6][7]. That record was set in 1986 by Jim Glover on a Moulton AM7 at the 3rd international HPV scientific symposium at Vancouver.

Weight vs powerEdit

There has been major corporate competition to lower the weight of racing bikes through the use of advanced materials and components. Additionally, advanced wheels are available with low-friction bearings and other features to lower resistance, however in measured tests these components have almost no effect on cycling performance. For instance, lowering a bike's weight by 1 lb, a major effort considering they may weigh less than 15 lb to start with, will have the same effect over a 40 km time trial as removing a protrusion into the air the size of a pencil. For this reason more recent designs have concentrated on lowering wind resistance, using aerodynamically shaped tubing, flat spokes on the wheels, and handlebars that allow the rider to bend over into the wind. These changes can impact performance dramatically, cutting minutes off a time trial.

Kinetic energyEdit

Consider the kinetic energy and "rotating mass" of a bicycle in order to examine the energy impacts of rotating versus non-rotating mass.

The translational kinetic energy of an object in motion is:[8]

E = \tfrac{1}{2}mv^2,

Where E is energy in joules, m is mass in kg, and v is velocity in meters per second. For a rotating mass (such as a wheel), the rotational kinetic energy is given by

E = \tfrac{1}{2}I \omega^2,

where I is the moment of inertia, \omega is the angular velocity in radians per second. For a wheel with all its mass at the outer edge (a fair approximation for a bicycle wheel), the moment of inertia is

I = m r^2.

Where r is the radius in meters

The angular velocity is related to the translational velocity and the radius of the tire. As long as there is no slipping,

\omega = \frac{v}{r}.

When a rotating mass is moving down the road, its total kinetic energy is the sum of its translational kinetic energy and its rotational kinetic energy:

E = \tfrac{1}{2}mv^2 + \tfrac{1}{2}I\omega^2

Substituting for I and \omega, we get

E = \tfrac{1}{2}mv^2 + \tfrac{1}{2}mr^2 \cdot \frac{v^2}{r^2}

The r^2 terms cancel, and we finally get

E = \tfrac{1}{2}mv^2 + \tfrac{1}{2}mv^2 = mv^2.

In other words, a mass on the tire has twice the kinetic energy of a non-rotating mass on the bike. There is a kernel of truth in the old saying that "A pound off the wheels = 2 pounds off the frame."[9]

This all depends, of course, on how well a thin hoop approximates the bicycle wheel. In reality, all the mass cannot be at the radius. For comparison, the opposite extreme might be a disk wheel where the mass is distributed evenly throughout the interior. In this case I = \tfrac{1}{2}m r^2 and so the resulting total kinetic energy becomes E = \tfrac{1}{2}mv^2 + \tfrac{1}{4}mv^2 = \tfrac{3}{4}mv^2. A pound off the disk wheels = only 1.5 pounds off the frame. Most real bicycle wheels will be somewhere between these two extremes.

One other interesting point from this equation is that for a bicycle wheel that is not slipping, the kinetic energy is independent of wheel radius. In other words, the advantage of 650C or other smaller wheels is due to their lower weight (less material in a smaller circumference) rather than their smaller diameter, as is often stated. The KE for other rotating masses on the bike is tiny compared to that of the wheels. For example, pedals turn at about \tfrac{1}{5} the speed of wheels, so their KE is about \tfrac{1}{25} (per unit weight) that of a spinning wheel.

Convert to kilocaloriesEdit

Assuming that a rotating wheel can be treated as the mass of rim and tire and 2/3 of the mass of the spokes, all at the center of the rim/tire. For a 180 lb rider on an 18 lb bike (90 kg total) at 25 mph (11.2 m/s), the KE is 5625 joules for the bike/rider plus 94 joules for a rotating wheel (combined 1.5 kg of rims/tires/spokes). Converting joules to kilocalories (multiply by 0.0002389) gives 1.4 kilocalories (large calories, nutritional calories, or Calories with a capital "C").

That 1.4 kilocalories is the energy necessary to accelerate from a standstill, or the heat to be dissipated by the brakes to stop the bike. These are kilocalories, so 1.4 kilocalories will heat 1 kg of water 1.4 degrees Celsius. Since aluminum's heat capacity is 21% of water, this amount of energy would heat 800 g of alloy rims 8 °C (15 °F) in a rapid stop. Rims do not get very hot from stopping on flat ground. To get the rider's energy expenditure, consider the 24% efficiency factor to get 5.8 kilocalories—accelerating a bike/rider to 25 mph (40 km/h) requires about 0.5% of the energy required to ride at 25 mph (40 km/h) for an hour. This energy expenditure would take place in about 15 seconds, at a rate of roughly 0.4 kilocalories per second, while steady state riding at 25 mph (40 km/h) requires 0.3 kilocalories per second.


The advantage of light bikes, and particularly light wheels, from a KE standpoint is that KE only comes into play when speed changes, and there are certainly two cases where lighter wheels should have an advantage: sprints, and corner jumps in a criterium.[10]

In a 250 m sprint from 36 to 47 km/h to (22 to 29 mph), a 90 kg bike/rider with 1.75 kg of rims/tires/spokes increases KE by 6,360 joules (6.4 kilocalories burned). Shaving 500 g from the rims/tires/spokes reduces this KE by 35 joules (1 kilocalorie = 1.163 watt-hour). The impact of this weight saving on speed or distance is rather difficult to calculate, and requires assumptions about rider power output and sprint distance. The Analytic Cycling web site allows this calculation, and gives a time/distance advantage of 0.16 s/ 188 cm for a sprinter who shaves 500 g off their wheels. If that weight went to make an aero wheel that was worth 0.03 mph (0.05 km/h) at 25 mph (40 km/h), the weight savings would be canceled by the aerodynamic advantage. For reference, the best aero bicycle wheels are worth about 0.4 mph (0.6 km/h) at 25, and so in this sprint would handily beat a set of wheels weighing 500 g less.

In a criterium race, a rider is often jumping out of every corner. If the rider has to brake entering each corner (no coasting to slow down), then the KE that is added in each jump is wasted as heat in braking. For a flat crit at 40 km/h, 1 km circuit, 4 corners per lap, 10 km/h speed loss at each corner, one hour duration, 80 kg rider/6.5 kg bike/1.75 kg rims/tires/spokes, there would be 160 corner jumps. This effort adds 387 kilocalories to the 1100 kilocalories required for the same ride at steady speed. Removing 500 g from the wheels, reduces the total body energy requirement by 4.4 kilocalories. If the extra 500 g in the wheels had resulted in a 0.3% reduction in aerodynamic drag factor (worth a 0.02 mph (0.03 km/h) speed increase at 25 mph), the caloric cost of the added weight effect would be canceled by the reduced work to overcome the wind.

Another place where light wheels are claimed to have great advantage is in climbing. Though one may hear expressions such as "these wheels were worth 1-2 mph", etc. The formula for power suggests that 1 lb. saved is worth 0.06 mph (0.1 km/h) on a 7% grade, and even a 4 lb saving is worth only 0.25 mph (0.4 km/h) for a light rider. So, where is the big savings in wheel weight reduction coming from? One argument is that there is no such improvement; that it is "placebo effect". But it has been proposed that the speed variation with each pedal stroke when riding up a hill explains such an advantage. However the energy of speed variation is conserved; during the power phase of pedaling the bike speeds up slightly, which stores KE, and in the "dead spot" at the top of the pedal stroke the bike slows down, which recovers that KE. Thus increased rotating mass may slightly reduce speed variations, but it does not add energy requirement beyond that of the same non-rotating mass.

Lighter bikes are easier to get up hills, but the cost of "rotating mass" is only an issue during a rapid acceleration, and it is small even then.


Possible technical explanations for the widely claimed benefits of light components in general, and light wheels in particular, is as follows:

  1. Light weight wins races with significant climbing because the heavier bike can't make up the gap on descents or on the flats: the rider on the lighter bike just drafts. Alternatively, if the identical riders of heavier and lighter bikes simultaneously reach the bottom of a climb to the finish, all of the advantage goes to the lighter bike. This is not the case in a hilly time trials (or riding solo), where the advantage of heavier, but more aerodynamic wheels would easily make up the distance lost in climbs. In climbing, lighter wheels offer no particular advantage vs. a lighter frame, because there is no net loss of KE.
  2. Light weight wins sprints because it accelerates more easily. But note that heavier aerodynamic wheels gain significant advantage as speed increases, and for a good part of a sprint a rider is doing little accelerating but is working hard against a high-speed wind. So many sprint situations may favor heavier but more aerodynamic wheels.
  3. Light weight wins in criteriums because of the constant acceleration out of every corner. Heavier but more aerodynamic wheels offer little advantage because the riders are in a group most of the time. The energy savings from lighter wheels is minimal, but it may be more significant that the leg muscles have to put out just that bit of extra effort at each jam.

There are two "non-technical" explanations for the effects of light weight. First is the placebo effect. Since the rider feels that they are on better (lighter) equipment, they push themselves harder and therefore go faster. It's not the equipment that increases speed so much as the rider's belief and resulting higher power output. The second non-technical explanation is the triumph of hope over experience—the rider is not much faster due to lightweight equipment but thinks they are faster. Sometimes this is due to lack of real data, as when a rider took two hours to do a climb on their old bike and on their new bike did it in 1:50. No accounting for how fit the rider was during these two climbs, how hot or windy it was, which way the wind was blowing, how the rider felt that day, etc.

Another explanation, of course, may be marketing benefits associated with selling weight reductions.

In the end, the "incremental muscle power requirement" argument is the only one that can support the claimed advantages of light wheels in "jump" situations. This argument would state that: if the rider is already at the limit on each jump or each stroke of the pedals, then the small amount of extra power required for the extra weight would be a significant physiologic burden. Whether this is true is not clear, but it is the only explanation for the claimed advantage of wheel weight savings (compared to saving weight from the rest of the bike). For these accelerations, it makes no difference whether 1 lb is taken off the wheels or 2 lb off the bike/rider. The miracle of light wheels (compared to saving weight anywhere else in the bike/rider system) is hard to see.

Aerodynamics vs powerEdit

Heated debates over the relative importance of weight savings and aerodynamics are a fixture in cycling. This is an attempt to at least get the equation-based parts of the debate clarified. There will always be those who argue that "experience trumps mathematics" on this issue, so this will attempt to highlight those areas where experience might disagree with the math. From this, perhaps further discussion can focus on the topics of dispute rather than questioning known physics. To be as clear as possible, this will cover 1) the power requirements for moving a bike/rider 2) the energy cost of acceleration, and then 3) why experience and the math might disagree.

Power requiredEdit

There is a well known equation that gives the power required to push a bike/rider through the air and to overcome the friction of the drive train:

P = g m V_g (K_1+s) + K_2 \times V_a^2 V_g

Where P is in watts, g is Earth's gravity, V_g is ground speed (m/s), m is bike/rider mass in kg, s is the grade (m/m), and V_a is the rider's speed through the air (m/s). K_1 is a lumped constant for all frictional losses (tires, bearings, chain), and is generally reported with a value of 0.0053. K_2 is a lumped constant for aerodynamic drag and is generally reported with a value of 0.185 kg/m [11].

Note that the power required to overcome friction and gravity is proportional only to rider weight and ground speed. The aerodynamic drag is roughly proportional to the square of the relative velocity of the air and the bike. Being that the total power requirement to propel the bike forward is a sum of these two variables mulitplied by speed, the degree of proportionality between power requirement and speed varies according to their relative magnitude, in an interval between square and cube: at higher speeds (e.g., downhill) power required will be close to being a cube function of speed, at lower speeds (uphill) it will be close to being a square function of speed.

The human body runs at about 24% efficiency for a relatively fit athlete, so for every kJ delivered to the pedals the body consumes a kCal (4.2kJ) of food energy.

Obviously, both of the lumped constants in this equation depend on many variables, including drive train efficiency, the rider's position and drag area, aerodynamic equipment, tire pressure, and road surface. Also, recognize that air speed is not constant in speed or direction or easily measured. It is certainly reasonable that the aerodynamic lumped constant would be different in cross winds or tail winds than in direct head winds, as the profile the bike/rider presents to the wind is different in each situation. Also, wind speed as seen by the bike/rider is not uniform except in zero wind conditions. Weather report wind speed is measured at some distance above the ground in free air with no obstructing trees or buildings nearby. Yet, by definition, the wind speed is always zero right at the road surface. Assuming a single wind velocity and a single lumped drag constant are just two of the simplifying assumptions of this equation. Computational Fluid Dynamicists have looked at this bicycle modeling problem and found it hard to model well. In layman's terms, this means that much more sophisticated models can be developed, but they will still have simplifying assumptions.

Given this simplified equation, however, one can calculate some values of interest. For example, assuming no wind, one gets the following results for kilocalories required and power delivered to the pedals (watts):

  • 175W for a 90 kg bike + rider to go 9m/s (20 mph or 32 km/h) on the flats (76% of effort to overcome aerodynamic drag), or 2.6m/s (5.8 mph or 9.4 km/h) on a 7% grade (21% of effort to overcome aerodynamic drag).
  • 300W for a 90 kg bike + rider at 11m/s (25 mph or 40 km/h) on the flats (83% of effort to overcome aerodynamic drag) or 4.3m/s (9.5 mph or 15 km/h) on a 7% grade (42% of effort to overcome aerodynamic drag).
  • 165W for a 65 kg bike + rider to go 9m/s (20 mph or 32 km/h) on the flats (82% of effort to overcome aerodynamic drag), or 3.3m/s (7.4 mph or 12 km/h) on a 7% grade (37% of effort to overcome aerodynamic drag).
  • 285W for a 65 kg bike + rider at 11m/s (25 mph or 40 km/h) on the flats (87% of effort to overcome aerodynamic drag) or 5.3m/s (12 mph or 19 km/h) on a 7% grade (61% of effort to overcome aerodynamic drag).

Shaving 1 kg off the weight of the bike/rider would save 0.01 m/s at 9m/s on the flats (1 second in a 25 mph (40 km/h), 25 mile (40 km) TT). Losing 1 kg on a 7% grade would be worth 0.04m/s (90 kg bike + rider) to 0.07m/s (65 kg bike + rider). If one climbed for 1 hour, saving 1 lb. would gain between 225 and 350 feet (107 m) - less effect for the heavier bike + rider combination (e.g., 0.04 mph (0.06 km/h) * 1 hr * 5,280 ft (1,609 m)/mile = 225 ft). For reference, the big climbs in the Tour de France have the following average grades:

The equation can be separated into level ground power

P_{level} = g m V_g K_1 + K_2 V_a^2 V_g ,

and vertical climbing power given by

P_{climbing} = mg(h/t) \approx gm (V_g \times s).

Energy cost of accelerationEdit


Why experience and the math might disagreeEdit


See alsoEdit



  1. Wilson, David Gordon; Jim Papadopoulos (2004). Bicycling Science (Third Edition ed.). The MIT Press. p. 126. ISBN 0-262-73154-1. 
  2. Template:Cite web
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  8. Ruina, Andy; Rudra Pratap (2002) (PDF). Introduction to Statics and Dynamics. Oxford University Press. p. 397. Retrieved 2006-08-04. 
  9. Template:Cite web
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  11. Corresponding to a surface area of 0.4m^2 with a drag coefficient of 0.7: Drag (physics)#Power
  12. Sastre wins the 2008 L'Alpe d'Huez stage. July 23, 2008. Retrieved 2009-01-14. 

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