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. 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.
- 1 Mechanical efficiency
- 2 Energy efficiency
- 3 Typical speeds
- 4 Reduction of weight vs reduction of rotating mass
- 5 Aerodynamics vs friction and grade
- 6 See also
- 7 References
- 8 External links
From a mechanical viewpoint, up to 99% of the energy delivered by the rider into the pedals is transmitted to the wheels (clean, lubricated new chain at 400W), although the use of gearing mechanisms reduces this by 1-7% (clean, well-lubricated derailleurs), 4-12% (chain with 3-speed hubs), or 10-20% (shaft drive with 3-speed hubs). The higher efficiencies in each range are achieved at higher power levels and in direct drive (hub gears) or with large driven cogs (derailleurs).  
A human being traveling on a bicycle at 16–24 km/h (10–15 mph), using only the power required to walk, is the most energy-efficient means of human transport generally available. Air drag, which increases with the square of speed, requires increasingly higher power outputs relative to speed, power increasing with the cube of speed as power equals force times velocity. 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 (150 lb) person requires about 60 watts  to walk at 5 km/h (3.1 mph). That same person on a bicycle, on the same ground, with the same power output, can travel at 15 km/h (9.3 mph) using an ordinary bicycle, so in these conditions the energy expenditure of cycling is one-third of walking.
Active humans can produce between 1.5 W/kg (untrained women for longer periods) and 24 W/kg (top-class male athletes during 5 s). 5 W/kg is about the level reachable by ordinary male athletes for longer periods. Maximum power levels during one hour range from about 250 W ("healthy men") to 500 W (exceptional men athletes) 
The energy input to the human body is in the form of food, usually quantified in kilo-calories [kcal] or kilo-Joules [kJ=Ws]. This can be related to a certain distance travelled and to body weight, giving units such as kJ/(km∙kg). The rate of food consumption, i.e. the amount consumed during a certain period ot time, is the input power. This can be measured in kcal/day or in J/s = W (1000 kcal/d ~ 48.5 W).
This input power can be determined by measuring oxygen uptake, or in the long term food consumption, assuming no change of weight. This includes the power needed just for living, called the basal metabolic rate BMR or roughly the resting metabolic rate.
The required food can also be calculated by dividing the output power by the muscle efficiency. This is 18-26%. From the example above, if a 70 kg person is cycling at 15 km/h by expending 60 W and a muscular efficiency of 20% is assumed, roughly 1 kJ/(km∙kg) extra food is required. For calculating the total food required during the trip, the BMR must first be added to the input power. If the 70 kg person is an old, short woman, her BMR could be 60 W, in all other cases a bit higher.  Viewed this way the efficiency in this example is effectively halved and roughly 2 kJ/(km∙kg) total food is required.
Although this shows a large relative increase in food required for low power cycling, in practice it is hardly noticed, as the extra energy cost of an hour's cycling can be covered with 50 g nuts or chocolate. With long and fast or uphill cycling, the extra food requirement however becomes evident.
To complete the efficiency calculation, the type of food consumed determines the overall efficiency. For this the energy needed to produce, distribute and cook the food must be considered.
In utility cycling there is a large variation; an elderly person on an upright roadster might do less than 10 km/h (6.2 mph) while a fitter or younger person could easily do twice that on the same bicycle. For cyclists in Copenhagen, the average cycling speed is 15.5 km/h (9.6 mph).
Cycling speed records
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, but including a defined amount of gravity assist) is 133.78 km/h (83.13 mph) set in 2013 by Sebastiaan Bowier in the VeloX3, a streamlined recumbent bicycle. In the 1989 Race Across America, a group of HPVs crossed the United States in just 5 days. The highest speed officially recorded for a bicycle ridden in a conventional upright position under fully faired conditions was 82.52 km/h (51.28 mph) over 200m. That record was set in 1986 by Jim Glover on a Moulton AM7 at the Human Powered Speed Championships during Expo86 World Fair in Vancouver.
Reduction of weight vs reduction of rotating mass
There has been major corporate competition to lower the weight of racing bikes. The UCI sets a limit on the minimum weight of bicycles to be used in sanctioned races, in order to discourage making structures so thin that they become unsafe. Although less weight results in time savings on uphill terrain, less rotating mass is doubly effective during acceleration, as shown below.
Kinetic energy of a rotating wheel
where is the moment of inertia, (pronunciation: 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
Where 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,
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:
Substituting for and , we get
The terms cancel, and we finally get
In other words, a mass on the tire has twice the kinetic energy of a non-rotating mass on the bike. 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 and so the resulting total kinetic energy becomes . 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 the speed of wheels, so their KE is about (per unit weight) that of a spinning wheel. As their center of mass turns on a smaller radius, this is further reduced.
Convert to kilocalories
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 (82 kg) rider on an 18 lb (8 kg) bike (90 kg total) at 25 mph (40 km/h ; 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 (nutritional calories).
Those 1.4 kilocalories are 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.
Advantages of light wheels
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.
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.
Aerodynamics vs friction and grade
Heated debates over the relative importance of weight savings and aerodynamic optimization are a fixture in cycling. By calculating the power requirements for moving a bike and rider, the relative energy costs of air resistance, rolling resistance, slope resistance and acceleration can be evaluated.
There is a well-known equation that gives the power required to push a bike and rider through the air and to overcome the friction of the drive train:
where is in watts, is bike + rider mass in kg, is Earth's gravity, is ground speed (m/s), is the grade (m/m), and is the rider's speed through the air (m/s). is a lumped constant for all frictional losses (tires, bearings, chain), and is generally reported with a value of 0.0053. is a lumped constant for aerodynamic drag and is generally reported with a value of 0.185 kg/m. If there is no wind, and the result simplifies to:
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 multiplied by speed, the degree of proportionality between power requirement and speed varies according to their relative magnitude, in an interval between the linear and cube: at higher speeds (riding fast on a flat road) power required will be close to being a cube function of speed, at lower speeds (climbing a steep hill) it will be close to being a linear function of speed.
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, air speed is not constant in speed or direction if there is wind. 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.
Given this simplified equation, one can calculate some values of interest. For example, assuming no wind, one gets the following results for kilocalories required (assuming 24% muscular efficiemcy) and power delivered to the pedals (watts):
- 175 W for a 90 kg bike + rider to go 9 m/s (32 km/h or 20 mph) on the flat (76% of effort to overcome aerodynamic drag), or 2.6 m/s (9.4 km/h or 5.8 mph) on a 7% grade (2.1% of effort to overcome aerodynamic drag).
- 300 W for a 90 kg bike + rider at 11 m/s (40 km/h or 25 mph) on the flat (83% of effort to overcome aerodynamic drag) or 4.3 m/s (15 km/h or 9.5 mph) on a 7% grade (4.2% of effort to overcome aerodynamic drag).
- 165 W for a 65 kg bike + rider to go 9 m/s (32 km/h or 20 mph) on the flat (82% of effort to overcome aerodynamic drag), or 3.3 m/s (12 km/h or 7.4 mph) on a 7% grade (3.7% of effort to overcome aerodynamic drag).
- 285 W for a 65 kg bike + rider at 11 m/s (40 km/h or 25 mph) on the flat (87% of effort to overcome aerodynamic drag) or 5.3 m/s (19 km/h or 12 mph) on a 7% grade (6.1% of effort to overcome aerodynamic drag).
Reducing the weight of the bike + rider by 1 kg would increase speed by 0.01 m/s at 9 m/s on the flat (5 seconds in a 32 km/h (20 mph), 40-kilometre (25 mile) TT). The same reduction on a 7% grade would be worth 0.04 m/s (90 kg bike + rider) to 0.07 m/s (65 kg bike + rider). If one climbed for 1 hour, saving 1 lb would gain between 69 metres (225 ft) and 110 m (350 ft) – less effect for the heavier bike + rider combination (e.g., 0.06 km/h (0.04 mph) * 1 h * 1,600 m (5,200 ft)/mi = 69 m (226 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
and vertical climbing power given by
This approximation approaches the real solution for small, i.e. normal grades. For extremely steep grades such as 0.35 the approximation gives an overestimation of about 6%.
Power under acceleration
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