great post. Inspired me to play with a few numbers.
Suppose you wanted a significant advantage from a motor. I think 25W would be enough to make a difference. To put it in perspective, an extra 25W would be the difference between about 6 W/kg, and 6.4 W/kg in the average sized racer. From the specific power calculations that I've seen, this sounds like the difference between missing the podium, and winning the race.
Suppose you only needed it in the business end of the race. I think a 1 hr duration would be enough to make a difference.
Suppose the above motor could be constructed to fit in the space envelope of the rear (or front?) hub. I am not a motor guy, but it sounds plausible to me.
Suppose this motor weighed .3 kg. If anything, this sounds generous, and I suspect it could be done with less mass.
From what I've read recently, some people think the above sounds reasonable, but think you would be devastated by the weight or size of batteries. Is that true? Let's have a look:
How heavy, and how big, would the battery be to power the above motor?
Suppose the efficiency of the motor is 75%. Suppose the efficiency of the battery is also 75%.
Target Power (P): 25 W
Target Duration (t): 1 hr = 3600s
Efficiency Motor (n_m): .75
Efficiency Battery (n_b): .75
Required Energy (E):
E = P * t / n_m * n_b
E = (25 W / 1 000 000 W / MW) * 3600 s / .75 * .75
E = 0.16 MJ
So, with 0.16 MJ of Energy, we can power our proposed motor for a sufficient amount of time.
From wikipedia (
http://en.wikipedia.org/wiki/Energy_density) a lithium battery has the highest specific energy of any readily available, and practical, battery chemistry. In particular, the numbers are as follows:
Lithium Battery Specific Energy: 1.8 MJ/kg
Lithium Battery Energy Density: 4.32 MJ/L
So,
mass battery (m_b):
m_b = 0.16 MJ / 1.8 MJ / kg
m_b = 0.09 kg
volume battery (V_b):
V_b = 0.16 MJ / 4.32 MJ / L
V_b = 0.037 L
To put that into context, the battery would weigh about 90 g, and take up approximately 40 mL of space, or (allowing for a bit of packaging, etc) about 4 cm3. Much too large to fit inside a hub, but why would you want to put it in the hub anyway? Easily would fit in down tube, among other places.
The weight of our proposed system is only 90 + 300 ~ 400 g. We can round up to half a kg....