Cycling Power Calculations

| May 2, 2010 12:52 pm

by Franz Kelsch

In another post I wrote about the many cycling power meters that are available to measure in real time the power a cyclists is applying to the pedals. This article provides some of the science behind cycling power and formulas that are being used on the Ultra Cycling website to estimate power for those riders who do not have a power meter. If you are one of the many cyclists who mistakenly say that the effort goes up the the square of the speed, you might want to read this article.

Work, Energy and Power

These terms all mean something different, but are indeed related.  A basic understanding is needed before we move on to discussing Power in cycling.

Work refers to an activity of a force being applied and movement over a distance in the direction of the force.  If you cycle up a hill you are doing Work. The typical unit of the force being applied is newtons.  One newton is equal to the amount of force required to accelerate a mass of one kilogram at a rate of one meter per second per second.

Energy is the capacity to do Work.   The source of energy came come from potential energy, as when descending, or energy produced by your body. The typical unit of measurement is the joule. One joule is the energy exerted by the force of one newton acting to move an object through a distance of one meter. The calorie is a pre-SI metric unit of energy. It is still used for food energy, referred to as Calorie (capital C).  This is called a kg calorie, or 1,000 gram calories. One gram calorie equals approximately 4.2 joules so one Calorie (as in food) equals 4,200 joules, or 4.2 kj.  The body is not that efficient in using the energy in food and only about only 18 to 26 percent of the energy available from respiration is converted into mechanical energy. Considering this efficiency, 1 Calorie of food consumed (4.2 kj energy) can produce 1 kj of energy to the pedals.

Power is the rate of using Energy.  If Energy were money in your pocket, then Power would be how fast you are spending the money and what you bought with your money would be the Work accomplished. The typical unit of Power is the watt.  One watt is equal to 1 joule of energy per second. When we are taking about Power output when cycling, as measured in watts, we are taking about the rate we are expending Energy to moving the bicycle and rider forward (Work).  If the cyclists is applying a power of 100 Watts to the pedals, that means 100 joules per second, or 360 kj per hour. To replace that energy, the cyclists would need to consume about 360 Calories of food per hour.

Forces in Cycling

There are certain forces opposing motion of the bicycle that the rider needs to provide energy to overcome. These forces are:

  • Rolling Resistance. This  is friction from contact with the road. It is affected by the bike quality, tire, road surface, tire pressure and weight of of the ride and bicycle.  At very low speeds, on a flat surface, this is the main force.
  • Air and Wind Resistance.  Air is a fluid (although one with low density) and any object moving through the air will encounter friction. It is a function of the speed of the bike plus the wind speed, the area and shape of the cyclists and bike, and the speed being traveled.
  • Gravity. When climbing the rider needs to put in sufficient energy to “lift” their own body weight plus the weight of the bike. It is is a function of the grade and speed.  As the elevation increases, the potential energy increases.  This potential energy can provide energy back whenever the cyclists descends.

Power is the work required per unit of time to overcoming the net forces acting on the rider and bicycle. If you add each of the above forces and multiple by the speed, the result is the power required. The power is applied by the pedals and equals the force applied to the pedals times the velocity of the pedal movement.

Estimating Power Output

Those interested in the math can read further on how to estimate the power required to overcome each of the forces on the cyclists. These are simplified formulas dealing primarily with static forces and do not take into account all items that affect the forces such as wind, impact of turbulence, mechanical fiction in the drive train, etc.

Rolling Resistance


  • Frl – Force, in newtons, caused by rolling resistance
  • Prr – Power, in watts, to overcome Frl
  • Crr –  coefficient of rolling resistance – typically 0.004 but can be as high as 0.008 for bad asphalt or as low as 0.001 for a wooden track.
  • g – acceleration due to gravity – 9.8 m/s2
  • Wkg – mass of the ride plus bicycle in kg
  • Vmps – Veloicty in meters/sec


  • Frl = Wkg x  g x  Crr
  • Prr = Frl x Vmps


Take a rider and bike combined weight of 165 lbs (75 kg) traveling at traveling at 20  mph ( 8.92 meters per second), using Cff of 0.004 and with g being 9.8 meters/sec/sec.  The force would be:

  • Frl = 75kg x 9.8 m/s2 x 0.004 = 2.94 newtons.
  • Prl = 8.92 m/s x 2.94 newtons = 26 watts

Since the power is proportional to speed, the same rider traveling at 5 mph would require 6.5 watts to overcome rolling resistance.

Air and Wind Resistance:


  • Fw – Force on rider and bicycle due to wind drag
  • Cw – drag coefficient, typically 0.5
  • Rho – air density in kg/m .  Depends on temperature and  barometric pressure. Some typical values are sea level: 1.226, 1500m: 1.056 and 3000m: 0.905
  • Vmps – Speed in meters/sec
  • A – effective frontal area of the rider and bicycle in m^2.  Typical value is 0.5.


  • Fw =  1/2 A Cw Rho Vmps^2
  • Pw = Fw Vmps


Take a rider and bike combined weight of 165 lbs (75 kg) traveling at traveling at 20  mph ( 8.92 meters per second), with no headwind, using Cw of 0.5, Rho of 1.226 and front area of 0.5. The force due to wind drag would be:

  • Fw = 1/2 x 0.5 x 0.5 x 1.226 x 8.92 x 8.92 = 12.19 newtons
  • Pw = 12.19 newtons  x 8.92 m/s  = 108 watts.

If you at traveling at 5 mph, instead of 20 mph then:

  • Pw = (1/2 x 0.5 x 0.5 x 1.226 x 2.23 x 2.23) x 2.23 = 1.7 watts



  • Fsl – Force in newtons due to the pull of the rider and bicycle down the slope
  • Psl – Power in watts required to overcome the force of Fsl
  • Wkg – Combined weight of the rider and bicycle in kg
  • g – Acceleration due to gravity, 9.8 m/s^2
  • GradHill – gradient of the hill, in decimal, the ratio of the rise to the horizontal run.


  • Fsl = Wkg x g x GradHill
  • Psl = Fsl x Vmps


Take a rider and bike combined weight of 165 lbs (75 kg) traveling at traveling at 5  mph ( 2.23 meters per second), climbing a hill with a grade of 12% (GradHill = 0.12).  The force due to gravity would be:

  • Fsl = 75 x 9.8 x 0.12 = 88.2 newtons
  • Psl = 88.2 x 2.23 = 196 watts.

Combined Forces


  • Total Power = Prl + Pw + Psl   or Total Power = (Frl + Fw + Fsl) x Vmps


Using the values we already calculated in the above examples:

  • Flat Road, 20 mph:  Total Power = 26 +108 = 134 watts.  Here most of the power is used to overcome air drag
  • Flat Road, 5 mph: Total Power = 6.5 + 1.7 = 8.2 watts.  Here most of the power is used to overcome rolling resistance
  • Climb, 5 mph, 12% grade: Total Power = 6.5 + 1.7 + 196 = 204 watts.  Here most of the power is used to overcome gravity.

Some cyclists mistakenly say the power needed goes up by the square of the speed. Although the force due to air drag goes up by the square of the speed, the power required due to air drag goes up by the cube the cube of the speed.  Speed has a linear impact on rolling resistance force and no impact on gravitational forces.  Remember to get power we multiple the force by the speed.  Air drag forces already have the square of the speed in the formula so to get power your multiple by speed once again.

Real World

The above calculations are based on a simplistic model and exclude the effect of wind and some other dynamic forces.   Wind is very seldom zero and even on a circular course there is net loss of power due to wind, assuming wind is constant. There are also factors influenced by aerodynamics of the type of clothing being worn, the type of helmet, the biking position, turbulence caused as the air flows past the rider.  Except in the case of a tail wind, all these other factors will increase the power required. For climbing significant grades these additional factors are small compared with gravity and can be ignored.  However for flatter terrain at high speeds, some additional watts or power output will usually be measured.


3 Responses to “Cycling Power Calculations”

John W. wrote a comment on July 15, 2010

Thanks a lot. Ron @ Cozy Beehive also does a solid review of the math. He considers both wind and density change with altitude.

Richard Vanderplas wrote a comment on October 20, 2013

Thanks Franz. I’ve Looking online for three days for an article like this. I needed equations and examples for my C language simulation of a biker pedaling up and down hills and shifting gears etc. I do this for my own amusement in retirement. Thanks again.

Mahmoud Elghoneimy wrote a comment on October 21, 2015

Hello Sir,

Your blog is very interesting, and I have a question; why you used the rolling resistance coefficient in calculations instead of static coefficient ?