Denavelo wrote:Shallower rims spin up faster than say a 50mm/60mm combo. Climbing with a set of enve 2.2 and using the same 4.5 with the same weights. Pretty sure the 2.2 will spin up faster based on rim height. 404's and rims of that depth regardless of weight lose momentum on tight steep switchbacks. At least in my experience. Every switchback is like starting over again on the climb. Haha
Says the guy who just bought a set of Aeolus 5.0 tubular wheels.
The speed differences are minute. The weights are minute. It's not only not noticeable, the myth of micro-acceleration has been obliterated before.
Indeed, considering a higher rim is more aero, it will spin up faster. Here's an example how an even heavier setup destroys a lighter, but less aerodynamic setup from a standstill. Note the weight difference is even bigger than between the wheels we are discussing^^.
http://alex-cycle.blogspot.com.au/2014/ ... ts-ii.html
Clearly your wheels a re faulty or your perception is wrong. The higher rim will be faster (with the same weight).
Now onto the climbing/rotating mass myth. To quote MarkMcM who did this excercise with kilo's:. Now considering the amount he was working with, the idea that a few gram matter is beyond ridiculous.
viewtopic.php?t=6394&start=45#p60824
I can't believe that people keep arguing that rotating mass climbs slower than non-rotating mass under the same power. When you are working against gravity, mass is mass, it doesn't matter if it rotates or not. The idea that micro-accelerations due to pedal force fluctations make a difference in the overall picture is a strawman. During pedal force fluctuations, accelerations are decelerations cancel out. All that really matters is average power output vs. gravity.
Since Ras11 complained that no math has been offered, I decided to set up a model to simulate the accelerations/decelerations due to pedal fluctuations. The equations and variable values were taken from the Analytic Cycling web page.
Pedaling force: The propulsion force (from pedaling) was modeled as a sinusoidal. Since it is assumed average power is constant, the nomimal drive force will vary inversely with velocity. So, the propulsion force is modeled as:
Fp = (P/V)(1+Sine(2RT))
Fp = Propulsion force (pedaling)
P = Average power
V = Velocity
R = Pedaling revolution rate
T = Time
(Note: The angle in the sine term is double the pedal revolution rate, since there are two power strokes per revolution)
The drag forces on the rider are aerodynamic drag, rolling resistance, and gravity. These three terms together are:
Fd = (1/2)CdRhoAV^2 + MgCrrCosine(S) + MgSin(S)
Fd = drag force
Cd = Coefficient of aerodynamic drag
Rho = Density of air
A = Frontal area
M = total mass of bike and rider
Crr= Coefficient of Rolling Resistance
g = Acceleration of gravity
S = Slope of road
The total force is thus:
F = Fp - Fd
From Newton's second law, the equation of motion is:
dV/dt = F/I
I = Inertia
Because there is both rotating and non-rotating mass, total mass and total inertial will not be the same. Because mass at the periphery of the wheel as twice the inertia as non-rotating weight, the total mass and inertia of a bike are:
M = Ms + Mr
I = Ms + 2Mr
Ms = Static mass
Mr = Rotating mass
The complete equation of motion is thus:
dV/dt = {(P/V)(1+sin(2RT)) - [ (1/2)CdRhoAV^2 + (Ms+Mr)gCrrCosine(S) + (Ms+Mr)gSine(S) ] } / (Ms + 2Mr)
This equation is non-linear, so I solved it numerically with a 4th order Runge-Kutta numerical differentiation.
Borrowing the default values in the Analytic Cycling web page for "Speed given Power" page, the values used are:
P = 250 Watts, Cd = 0.5, Rho = 1.226 Kg/m^3, A = 0.5 m^2, Crr = 0.004, g = 9.806 m/s^2, S = 3% (= 1.718 deg.)
(http://www.analyticcycling.com/ForcesSpeed_Page.html)
For this simulation, the pedal revolution rate was selected as 540 deg/sec. (90 rpm cadence)
To solve this equation, a 4th order Runge-Kutta numerical differentiation was set up using an Excel spread sheet. Step size was selected at 0.01 sec., and the initial Velocity was 1 m/sec. The solution was calculated for 3 cases of equal total mass, but different distributions of static and rotating mass, calculated over a 200 second period, by which time each case had reached steady state. As expected, the velocity oscillated with the pedal strokes. The average, maximum, and minimum velocities during the oscillilations during stead state were:
Case 1:
Ms = 75 kg, Mr = 0 kg (0% rotating mass)
Average Velocity: 7.457831059 m/s
Maximum Velocity: 7.481487113 m/s
Minimum Velocity: 7.434183890 m/s
Speed fluctuation: 0.047303224 m/s
Case 2:
Ms = 70 kg, Mr = 5 kg (5.33% rotating mass)
Average Velocity: 7.457834727 m/s
Maximum Velocity: 7.480016980 m/s
Minimum Velocity: 7.435662980 m/s
Speed fluctuation: 0.044354000 m/s
Case 3:
Ms = 65 kg, Mr = 10 kg (10.67% rotating mass)
Average Velocity: 7.457837584 m/s
Maximum Velocity: 7.478718985 m/s
Minimum Velocity: 7.436967847 m/s
Speed fluctuation: 0.041751139 m/s
These results agree very strongly with the solution on the Analytic Cycling web page, which predicted an average speed with constant power of 7.46 m/s (16.7 mph)
The results show that as expected, the smaller the percentage of rotating mass, the greater the magnitude of the velocity oscillations (which are quite small). But a more interesting result is in the average speed. As the amount of rotating mass decreased, the more the average velocity _decreased_, not increased (at steady stage). This result is actually not unexpected. The drag forces are not constant, but vary with velocity, especially aerodynamic drage (Because aerodynamic drag increases with the square of velocity, power losses are increase out of proportion with speeds - so, for example, aerodynamic losses at 20 mph are 4 times as much as they would be at 10 mph). Because speed fluctuates as the propulsion force oscillations, in the cases of the low rotating mass, the maximum peak speeds reached are higher than for the cases with the high rotating mass. This means that when a lower percentage of rotating mass there will be greater losses during the speed peaks. Because of the total drag losses will be greater over the long run, the greater momentary accelerations with lower rotating mass actually results in a lower average speed.
To see what happens at a steeper slope, which will have a lower speed (and presumably larger speed oscillattions), I ran the model again with a 10% (5.7 deg.) slope. Here are the results:
Case 1:
Ms = 75 kg, Mr = 0 kg (0% rotating mass)
Average Velocity: 3.217606390 m/s
Maximum Velocity: 3.272312291 m/s
Minimum Velocity: 3.162540662 m/s
Speed fluctuation: 0.109771630 m/s
Case 2:
Ms = 70 kg, Mr = 5 kg (5.33% rotating mass)
Average Velocity: 3.217613139 m/s
Maximum Velocity: 3.268918539 m/s
Minimum Velocity: 3.165997726 m/s
Speed fluctuation: 0.102920813 m/s
Case 3:
Ms = 65 kg, Mr = 10 kg (10.67% rotating mass)
Average Velocity: 3.217618914 m/s
Maximum Velocity: 3.265921742 m/s
Minimum Velocity: 3.169047012 m/s
Speed fluctuation: 0.096874730 m/s
This data follows the same pattern as above. The speed oscillations (micro-accelerations) are greater with the lower rotating mass, but the average speed is also slightly lower with lower rotating mass. So next time you want to claim that lower rotating mass allows faster accelerations, remember too that the greater speed fluctuations (due to greater accelerations) will also results in greater energy losses due to drag forces.
But, in reality, the differences in speed fluctions and average speeds are really very small between all these cases. For all practical purposes, when climbing, it is only total mass that matters, not how it is distributed.
I'd be happy to send the Excel spreadsheet to anyone that is interested.
