JAPANic said:You think they go fast on the flats?
You should see them going downhill in a straight line...
Our club trains every Sunday on a flat stretch and depending on the wind the heavier guys have an advantage. Especially if there is a tail wind. They are disadvantaged in a head wind though as the momentum is fighting the wind. Assuming a heavier rider is bigger they also have more surface area/drag.
Well my explanation works also for downhill.
The heavier rider contains more potential energy at the top of a climb than the lighter rider. As any object (rider plus bike) descends, this potential energy is converted into kinetic energy and is also used to overcome wind and rolling resistance. At terminal velocity, almost all of the 'power' produced from this effect goes to overcoming wind resistance.
You can calculate the 'power' produced by descending by this simple equation.
P= V*Gradient* MG
Thus a rider and bike weighing 100 kg, which is descending at 20 meters second, or 72 kph, on a 1 in ten (.1) slope, is converting ~2000 joules of KE per second into what is essentially free 'power'. As Delta E/T = power then it is 2000 watts of 'power' from gravity.
Now, assume that this rider can also pitch in 500 watts from their muscles, and you have a whopping 2500 watts to overcome wind resistance.
Now, imagine a 50 kg combo of anorexic climber and bike, who can produce, say 300 watts. They only get 1000 watts from gravity, making a total of 1300 watts.
In this case, the 1300 watts may not be sufficient to overcome the wind resistance produced from going downhill at 72 kph, and the rider will slow down. Also, the 2500 watts from our 100kg rider/bike combo may be more than the wind resistance produced by smashing thought he air at 72 kph, and the rider will speed up.
At terminal velocity, the total power will ~ equal wind resistance. We can see that unless lighter riders have bigger power outputs than heavier riders, or greatly superior aerodynamics, heavier rider will tend to reach a greater terminal velocity. In this case, the lighter rider would have to have almost half the CdA of the heavier rider.
As Docspoc has demonstrated thriugh his example of cubes, this is just not the case.
To find terminal velocity going downhill, just solve the equation below
(V*Gradient* MG) + Rider power = CdA* V^3