I have been riding S40 regularly with my church group this year, after training by myself last year as I was just too afraid to ride on the road. I am now able to keep up and go faster on downhills and flats but lag behind in hills. Part of that is just my training, no doubt. I think part of that is also the weight penalty. I started thinking about that and tried to pull the analysis together to see what the weight penalty would equate to additional wind speed.
Weight penalty = 15 lb (31 lb vs 16 lb)
On a 6 deg hill, sine(6 deg) = 0.1
Additional "pull"force due to hill = 15 x 0.1 = 1.5 lbf = 6.67N
Drag equation
F= CdA x air density x velocity squared / 2
Using CdA = 0.32 for a generic bike front area in drop position and generic Cd. Assume air density at 1.225
6.67 = 0.32 x 1.225 x V squared / 2
V squared = 34
V = 5.8 m/s = 13 mph
That tells me the 15 lb weight penalty is equivalent to 13 mph wind on a 6 deg incline hill! I know the actual calculation is more complicated than that. However, for simple approximation, does the calculation make sense? I guess I need to lose more weight!! :-(
Thanks,
Michael
Weight penalty = 15 lb (31 lb vs 16 lb)
On a 6 deg hill, sine(6 deg) = 0.1
Additional "pull"force due to hill = 15 x 0.1 = 1.5 lbf = 6.67N
Drag equation
F= CdA x air density x velocity squared / 2
Using CdA = 0.32 for a generic bike front area in drop position and generic Cd. Assume air density at 1.225
6.67 = 0.32 x 1.225 x V squared / 2
V squared = 34
V = 5.8 m/s = 13 mph
That tells me the 15 lb weight penalty is equivalent to 13 mph wind on a 6 deg incline hill! I know the actual calculation is more complicated than that. However, for simple approximation, does the calculation make sense? I guess I need to lose more weight!! :-(
Thanks,
Michael