I felt it was important to know how heavy my various sets of wheels actually are: I have a set of low-drag high-mass wheels and a set of high-drag low-mass wheels, and quantifying what all those various ‘high’ and ‘low’ terms are is at least interesting and hopefully useful.
|Where||Hub||Skewer||Rim||Tyre||Tube||Mass (kg)||Drag (N)|
|Front||Shimano Ultegra HB-6800||Flo 30||Veloflex Corsa 23||Vittoria Latex 19/23||1.35||0.736|
|Rear||Shimano Ultegra FH-6800||Flo 30||Veloflex Corsa 23||Vittoria Latex 19/23||1.50||0.736|
|Both||Shimano Ultegra 6800||Flo 30||Veloflex Corsa 23||Vittoria Latex 19/23||2.85||1.47|
|Front||Shimano Dura-Ace HB-9000||Mavic Reflex||Vittoria Corsa SC 23||1.05||2.45|
|Rear||Shimano Dura-Ace FH-9000||Mavic Reflex||Vittoria Corsa SC 23||1.25||2.45|
|Both||Shimano Dura-Ace 9000||Mavic Reflex||Vittoria Corsa SC 23||2.30||4.90|
The mass column is taken from my (inaccurate) kitchen scales. The drag column is taken from the Flo Cycling Aero Data graph at 12.5º yaw and assuming that my Reflex rims have similar drag characteristics to the Open Pro they measured: it is actually likely to be worse as the Reflex is a noticeably shallower rim than the Open Pro. I have converted the stupid, nonexistent, gram-force units they use into SI newtons.
As of today the combined weight of myself and the bike with the kit I usually carry is about 90kg, so the absolute weight saving of 0.55kg is a relative saving of only 6‰.
I do not have a facility to determine my own drag, but a random page I found vie google suggest that the average drag force on a cyclist is 46.2N. Using this for the sake of making a calculation the absolute drag difference of 3.43N is a relative saving of 7.4%. So there is an order-of-magnitude difference in scale between the drag-benefit of the low-drag wheels and the mass-benefit of the low-mass wheels: roughly the low-drag wheels are 10 times less draggy than the low-weight wheels are low-weight. It must be noted that the drag and drag-reduction are calculated at 48kph, a speed much higher than that at which I typically ride.
Can I use these numbers to calculate the gradient at which I am better-off using the lighter wheels? Sort of, but not really. I can calculate the road angle at which the two cross-over assuming that I ride at 48kph: it’s 2.73º which is about 4.8%. This isn’t very useful because for me to ride my bicycle at 48kph up a 4.8% hill requires something in the region of 1000W which is a little unrealistic. I can maybe do 1000W for one or two seconds, only enough to get up a 25-metre long hill.
I climb hills at around 250W on a good day, so I will try and re-calculate the crossover angle for that power. For the sake of simplicity I will calculate the motion in a single second, during which by definition I will output 250J. This energy input must be balanced by energy losses. The energy lost as a result of the weight of the bike is (87.5 + m) * 9.81 * o wherein m is the mass of the wheels in kg and o is the vertical height increase in metres. The energy lost as a result of aerodynamic drag scales by the square of the airspeed which we can calculate as (h/13.4)2 * (41.3 + d) wherein h is the distance ridden in metres (same number as speed in metres per second) and d is the drag of the wheels in newtons at 48kph. The variables h and o are related by the equation h*sin(α) = o wherein α is the angle of the road to the horizontal.
For each bike configuration we can construct a quadratic equation in terms of h and α. With the low-drag wheels we have:
0 = (1/13.4)2 * 42.77 * h2 + 90.35 * 9.81 * sin(α) * h – 250
0 = (1/180) * 42.77 * h2 + 886 * sin(α) * h – 250
0 = 0.238 * h2 + 886 * sin(α) * h – 250
And with the low-weight wheels we have:
0 = (1/13.4)2 * 42.77 * h2 + 89.8 * 9.81 * sin(α) * h – 250
0 = (1/180) * 46.2 * h2 + 881 * sin(α) * h – 250
0 = 0.257 * h2 + 881 * sin(α) * h – 250
These formulae are correctly derived, but of course they do not take into account other energy losses which it turns out are significant. At 250W on a flat course (α = 0) in zero wind my speed is about 8.9m/s (32kph, 20 mph) but these formulae as they are predict a speed of over 30m/s (110kph, 65mph)! I wish! I need to estimate the effect of these various forces and it seems reasonable to me to assume that they scale linearly with speed. I therefore add another term of 26 * h to both formulae in order to make the predicted speeds on a flat course approximately equal to reality. This adjustment gives me speed numbers which look reasonable for flat ground and uphill gradients, in that they correlate reasonably with my Strava data, but they seem to be optimistic for the downhill sections.
So, at what angle α of road do these two formulae predict the same value of speed h? 0.0208 radians. Until road signs and maps start to give gradients in radians it’s probably more useful to give this as a gradient percentage: 2.1%. If I can train and increase my power to the Tony Martin or Bradley Wiggins level of 500W the cross-over moves out to 3.3%.
This is actually less steep than I was expecting. However it is also worth pointing out that the relative difference in speeds is not that great at any point, never more than 0.6% in either direction in the range from flat to 25% upward incline.
So which wheels should I use for my 10-mile time trial at RAF Weston on the Green? My GPS records the incline once per second so I can run those inclines through the formulae to predict the speed for each 1-second interval and then compute the average of those. I can then compute the time required to cover 10 miles at those speeds. This process predicts that the low drag wheels will give me a time of 28 minutes 26 seconds and the low weight wheels a time of 28 minutes 40 seconds.