Power and forces

In the triangle formed by VS, V F and parallel to V W the cosinus-rule can be used ; NB.: cos (180-αw) = cos αw   VS²  = VF² + FW² + 2 * VF * Vw * cos αw. So we can derive VS . The cosinus-rule, used in the triangle with αS enables us to determine cos αS;  cos αS = (VS² + VF² - FW²) / (2 * V F * VS )                          

 The strength of the apparent wind FS = 0,5 * ρ air * C w * A * VS².  The resistance of the wind head-on is :  R wind = Fs * cos αS . The power needed for air + wind resistance, i.e. the energy required to overcome the air resistance + wind,  is:  Pwind = R wind * VF / 0,95. The wind direction in the computer must be entered in degrees, 0 against wind and 180 wind right behind. This is the wind direction at a standstill!

 (2) The rolling resistance R rol = m * g * Cr. The power consumption of rolling resistance, can be calculated if we multiply Rrol with velocity v in m/s, and divide it by the efficiency of the drivetrain.

The rolling coëfficiënt Cr for the following tyres on smooth concrete (eg a gym or racetrack) is: 

Clement Seta tubular with 8 bar = 0.002   ;   Racing tyre 23 mm 7 bar = 0.003   ;    Touring tyre 32 mm with 5 bar = 0.005   ;   ATB tyre rough profile 47 mm 3 bar = 0.009   ;    Anti-puncture tyre 37 mm with 5 bar = 0.007    ;     Anti-puncture tyre 37 mm 3 bar = 0.01

  NB: especially in a tough anti-puncture tyre, the tyre-pressure is hard to estimate by squeezing it with your hand!  If we do not use smooth concrete, but rough asphalt or poor road surface, rolling resistance will be much higher), on good pavement the value for 23mm racing tire is 0.005 (so + 0.002) .

 (3) The acceleration resistance R accel = m * a is mass (m) multiplied by acceleration (a). This variable (3) is actually only included for completeness. If we ride at a constant speed, this value is zero. If there is more power produced, then used at the moment, the cyclist will accelerate. At a higher speed restistance increases (especially the airdrag). When the available, and used power are equal, we riding at a constant speed again.

 Rotating parts use extra energy as we accelerate, especially if the radius of the movement is large, wich is the case with wheels and tires. (This extra energy, about 2%, is neglected  in the calculations below).

  (4) The slope resistance  is: R slope = m * g * "%". Here, m is the mass in kg, g gravity 9.81 N / kg, and  "%" is the percentage slope. The use of the inclination is not entirely correct. This is because we should use the tangent of the angle of inclination. Actually, we are using the sinus, but the difference will only be noticeable above 21% (tan12 = 0.213 and sin12 = 0.208 : so both 21%). A negative inclination is downhill!

  The calculation: This program can be run if you have a spreadsheet like Excel on your computer! You can download power&speed.xlsx at the end of this page . A free spreadsheet program is available through Open Office; there is one in Google apps too. The effect of the wind on our power-out (P total) can be compared wind-less weather (Pcalm). NB. At the bottom of the calculation is a field in which your poweroutput can be converted into an expected speed; it's a cube root. This calculation is complicated, see J.Lieh at “Info on the web”. The wind and acceleration are not included in this calculation.

 An example: a rider and his bike weigh 75 kg; they climb the Mont Ventoux. The climb is 18 km long and is completed in one hour (5 m / s); the inclination is 9%. The required power for climbing is about 331 W. But our hero has also got an air resistance of 31W and a rolling resistance of 18W to overcome. In total about 380W. Unfortunately, there is still 5% loss in the drivetrain, so he must deliver 400W.

  If his gearing used was 42-21 and the tire had a circumference of 2.1m, one revolution of the crank is (42/ 21) * 2.1 = 4.2 m; this is 238 revolutions per kilometer and about 71 rpm. When he is sixtie and 25 kilos heavier, he want to do the climb again, but he only delivers 150W. How long will it take? What gearing should he use, when  he wants to ride at 60 rpm (1 per second) ? 

  If the power output is low, almost all power is used for climbing. Roughly Pslope is about 135W = m * g * "%" * V / η  Now we can calculate his speed V; this  is 1.5 m / s = 5.4 km / h. The climb will take approximately 3:20 h. This is 200 minutes or 12,000 revolutions. The distance traveled per revolution is 18000m: 12000 = 1,50 m. If we take an ATB-crankset, with a small sprocket 24, used in combination with a 34 as largest sprocket on the cassette, we find  a development of 1.48 m per revolution of the crank. So the gearing 24-34 would be a good choice.

 What forces do cyclists apply? The main forces are exerted on the pedals. In formula: the average effective force is the average power, divided by the pedaling speed (rotationspeed in m/s) : Favg = Pavg / vpedal 

If Pavg is 300W at 60 rpm, and the cranklengte 175mm (d = 0.35 m),  vpedal is: pi * 0.35 M * 1/s = 1.1 m/s ;  Favg is: 300 Nm/s / 1.1 m/s = 273N

Here we see actual forces exerted by a real cyclist on the pedal. The direction of the force is more or less vertical. We will decompose the force into two forces: one component perpendicular to the crank and a force in the longitudinal direction. The portion that is converted into labor, is the component perpendicular to the crank arm. This is converted through the axle into chaintension. The component in the longitudinal direction is absorbed by the crank. This is similar to the reaction force of the floor. There is no work and no loss of power!  

  As you can see, the rider looses power, because he lets his pushing leg, bring up the other leg. So the actual force on the pedal, is higher than the effective force. At the dead spots ( 6h clock), no power is developed!  In all steep slopes (> 15%) you will notice this effect. It is also felt very well on exercise bikes without flywheel. Normally a good circular motion is maintained on the basis of the kinetic energy in the system.

  Trained cyclists use their strength better than this rider. They will push later and stop earlier. In addition, they pull their clipless pedals in the upward movement; it is small contribution, a few percent, but it assists and does not cost energy.

 The peak force on the pedals will be higher than the average effective force, about 1.3-1.5 as high. In a sprint even 2-3  times as high. The rider can no longer remain seated, because he exerts forces greater than its weight. He will get out of the saddle and combine rotation, muscle strength and weight. This is called "jacking". It is similar to the run of a high jumper, wherein energy forward, is converted into speed in the vertical direction. Powerpeaks above 2000 N have been measured!

 In efforts ± 250W, the force will be around 20% of body weight (this is inter alia dependent on the resistance). Our riding force is not a pure torque, because we only use one leg at a time to push. Twice per revolution there is a pulse. 

  If the top position of the pedal is called 0 degrees and the lower position 180, we generate about 90% of our power between 50 and 140 degrees. Moment (or torque) is force multiplied by lever-arm: M = F * r

 With a locked shoe we can excert a pulling action (max.5%) on the lifting pedal. The torque of trained cyclists is never really zero, but the minimum is not more than a few percent.

  Now what is the tension in the chain? Say a touringcyclist with luggage rides a 20% hill; the maximum force exerted is  500N, the cranklengte is 175mm and it is used for a triple crankset with a 28  sprocket (a radius of 57mm). The maximum time is 500N * 0,175m = 87.5 Nm; it also passes through the gear: F = M / r = 87.5Nm: 0,057m = 1535N. This is the peak of the tension in the chain.

  At a gearratio of 42 teeth in front and 21 teeth behind, the speed of the rear axle doubled compared with the cranks and the torque halved!  If the wind is head-on, we need to shift down; as pedaling and step rate will remain the same, take speed and speed of the rear wheel off. The torque is increased to compensate for the increased resistance. The power remains the same: it is linked times crankshaft speed.

  The pedaling frequencies for tourists are usually between 60-80 rpm. For racers 90-120 rpm is common ( good trackracers, even more than 200 rpm). High speeds are not really efficient; the rider goes for maximum power, not maximum efficiency.

  What force F does a cyclist need to ride a long 17% slope? If the total weight of rider + bike 1000N, then:   F = 1000N * 0.17 = 170N.

  This force should be delivered to the street by a wheel with 670mm diameter; the torque to be delivered : M = 170N * 0,335m = 57Nm.

  This torque also passes through spokes and hubs. Set our flange has a radius of 0.025m; then the force in the hub flange: 57Nm / 0.025m = 2280N. If there are 36 spokes present, it is at least 63N per spoke.

  This is the case when the spoke is perpendicular to the axis of the flange ; at 3x crossed spoke-pattern the angle to the axis is 60° , so we must divide by sin 60° ; this gives a value of 73N per spoke.

  Is this rider able to climb that hill, if he has a defect rearderailleur, and just one gear 42-14? Say the maximum average pedaling force he can devellop is 500N and the cranks are 170mm. Then his maximum torque will be (0,170m * 500N) / (42/ 14) = 28.3 Nm! Not at all, he needs twice as much: 57 Nm ; he should use 42- 28 at least!

  Low pedaling frequencies (<40) can harm the knee joints. How much power he will  have to produce, if we keep 40 as a minimum?

  If he does 40 rpm (= 0.67 per sec) the wheel circumference = pi * D = 2, 1m; the gear ratio is 42 / 28 = 1,5.

 The speed (velocity) v : 2,1m * 0,67 1/s * 1.5 = 2m/s (= 7,2km /h). The required climbing power P slope = 1000N * 0.17 * 2 m/s = 340W. If we estimate 40W for rolling resistance , air resistance, and efficiency- losses, we need 380W total output.