# do not edit here, go to Cycle Alberta Community Discussions page A Little Bike Physics

Drag Force on a Bike

Ever wonder why your cycling buddy can coast down those hills so much faster than you? It can be really quite puzzling since each bike has roughly the same parameters. Could it be due to the rider’s bike or weight….

Many people attribute the maximum downhill velocity of a rider to the weight of the individual which is not correct since all objects in a vacuum will fall at the same velocity. The atmospheric Drag forces have the most impact on the speed of the bike.

The Drag force on a biker is mainly due to the frictional force created as the biker moves through the atmosphere. Only about 3% of the drag forces experienced by a biker are caused by friction within the mechanism of the bike itself (ie bearings, gear, and chain friction). It is also important to remember that weight refers to the force experienced by an object due to gravity.

Assumptions:
1. The biker is not peddling downhill.

2. The wind speed is zero.

Drag is dependent on the air density, velocity squared, the air’s viscosity and compressibility, the size and shape of the body and the bodies inclination to flow.
Please note that the dependence of the body shape and inclination, air viscosity, and compressibility is very complex. The drag coefficient (Cd) represents the effects of form drag, skin friction drag, wave drag and induced drag and would be determined experimentally in a wind tunnel . For the purposes of this calculation, the average Cd for a person on a bike is 0.63.

Drag = (Cd * rho(air) * (velocity)**2)/2) * reference area
rho(air) = 0.63  kg/m**3 (ave)
reference area = 0.5 m**2 (ave)
V**2 = m**2/s**2
Drag force (F drag) = (0.63 *0.81*0.5*V**2)/2 = 0.13*V**2 kgm/s**2 = 0.13 * V**2 Newtons
Force downhill due to gravity (F dh) = mass (m)* acceleration due to gravity = m*9.81 *sine(slope of hill or grade)

Now calculate the bike velocity when F dh = F drag   Assume the grade is 8.4.
A. Use Mass =11.72 Kg therefore weight = m* a = 11.72 kg (9.81) = 115 N
115 *sine (8.4) = 0.13 V**2
V = 11.52 m/s = 11.52/.447 = 25.8 mph
B. Use Mass= 20.39 Kg therefore weight =m*a = 20.39*9.81 = 200 N
200* sine(8.4) = 0.13V**2
V = 15.19 m/s = 15.19/.447 = 34 mph

Conclusion:   Assuming all other factors are equal, an individual with a larger mass will achieve a higher speed (stall speed) before having to pedal to overcome the drag forces. Of course then their momentum (mv) will also be higher on the run out at the bottom of the slope.