### Why does weight matter?

#### By terryh on Jun 30, 2005

Why do (some) cyclists obsess about weight? Why does it matter? Well, if we all rode at a constant speed on level ground, and had perfect bearings in our bikes, it wouldn't. But in the real world, the story is not a very simple one. The human body is not a very capable power plant. I can sustain about 200 watts, and even Lance doesn't do much more than 600 watts except in very short bursts. So the game is to optimize the power budget of the system to make the most of this puny output. Where does the power go? To a first approximation, there are three items:

- Rolling resistance
- Aerodynamic drag
- Gravity (or to be more precise: rate of change of gravitational potential energy)

Rolling resistance is typically small - bearings and tires are very good these days. If we consider riding on the flat, then it is clear that there is no contribution from gravity. So most of our power output goes to overcoming aerodynamic drag. It is somewhat difficult to characterize the drag of a cyclist. Even wind-tunnel modelling is not fully adequate. The reason for this is that in the real world, the apparent wind "seen" by the cyclist is the vector sum of their speed and the prevailing surface wind. This means that the effective wind might not come from head-on. This means that both the effective area and the coefficient of drag are not constants, as they are usually considered to be. Be this as it may, we can still say in general terms that aerodynamic drag goes like the square of the speed relative to the air. From basic physics, power is the rate of doing work, and work is force times distance. So the power is drag times speed, in other words it goes like speed **cubed**.
So if you want to go twice as fast, you need eight times the power output, or put another way, twice the power output will take
you about 26% faster.

Now how about hills? The power lost (stored, actually) due to climbing is simply your mass times the local gravitational accelleration, times your vertical speed. If you have a steep hill, and you know the altitude at the bottom and at the top (from a map, or from measuring it), this leads to a simple way to calculate your power output. Provided the hill is steep enough that you are going quite slowly (less than 10 mph (16 kph) but the slower the better), there shouldn't be too much aerodynamic drag to mess up the calculation. Weigh yourself, your bike, and all your gear that you ride with. We need to convert to S.I. units so that the answer will come out in watts. Say it was 150 lbs + 20 lbs + 5 lbs. Divide by 2.2 to get Kg mass - almost exactly 80 Kg. Now multiply by little-g, the earth's gravitational accelleration, to get weight in Newtons. 80x9.8 is about 780 N. Now suppose the climb is 1900' and you do it in 49 minutes. 1900 feet is abuot 580 m, and 49 minutes is 2940 s. So the climb rate is 0.197 m/s. Power is weightxclimb rate, or in this case about 154 watts.

What can we say in general? well for a given hill, power is proportional to speed. It is also proportional to weight. So if you save a couple of pounds, by whatever means, you potentially go around 1% faster. What if we hold the speed constant and vary other parameters? Power goes like the sine of the steepness of the hill. But for the kind of angles we're talking about, sin(x) is approximately equal to x. So if the hill gets a bit steeper, the power output to maintain speed goes up close to proportionally. Power is proportional to weight in the constant speed case, just like in the constant hill case.

So the general conclusion we can draw here, is that riding on the flat (in a time trial, where there's no drafting) is driven by power to drag ratio, whereas riding up hills is driven by power to weight ratio. That's why you don't see the average time trialist get to the top of a mountain first - and why the ultra-elite like Lance and co. are so special

Tag: Cycling

Posted by

fat cycliston June 30, 2005 at 10:20 AM PDT #Posted by

vishwason June 30, 2005 at 06:42 PM PDT #