On Bicycles, and.... what else is there?

I looked at how grade variability affected average power when climbing a hill. Honestly I thought the result was going to be larger, but the reality was it was a relatively minor effect. When the hill is very gradual, for example 1%, variations in grade of a certain fraction have little effect on speed. When the hill is very steep variations in grade are more significant, but since they increase power only via wind resistance, and wind resistance is relatively unimportant (assuming still air), again variations in grade have little effect. It's only important in the middle ground where speeds are high enough that wind resistance is relatively important but where grade variations have a relatively large influence on speed. A virtually equivalent logic applies to rolling resistance variation. A variation in rolling resistance about an average value (averaged over distance) will have the same effect as a variation in grade by the same absolute amount. So the effect of variabilit...

I've looked at this matter before, but one factor which I've seen continually neglected in all of the climbing power analysis estimates is the effect of grade variability. Road grade on climbs is almost never constant: it varies about a mean in some fashion. Yet the estimates are almost always done assuming constant speed, constant power. Now these estimates end up remarkably accurate anyway. Why? Because the grade variability effect is negligible? Well, no. It's because you're canceling one mistake with another. For example, you neglect grade variability, which always increases power, but you also neglect drafting, which always decreases power. How does grade variability increase power? It's because grade variability typically results in speed variability and speed variability yields variability results in variability in wind resistance and wind resistance, by virtue of being superlinear, is increased more by increases in speed than it is decreased by d...

I couldn't figure out how the hummingbird feeder worked. Why didn't it overflow? Forces need to balance, of course. Neglecting surface tension, there liquid level is higher in the inner reservoir than in the feeding chamber, so there must be a corresponding pressure difference. Suppose the pressure in the inner chamber were zero. Then the column height difference would need to be atmospheric pressure / (density of liquid × gravity). But this is over 9 meters! Obviously the height difference is only approximately 1% of this. So the pressure difference inside versus outside is only approximately 1%. The inside is only slightly below atmosphere. So air is getting in. How? Does it diffuse through the liquid? If this were the dominant mechanism, it wouldn't take long for the pressure inside to go from 99% to 99.3%, for example, which should be plenty to push the column of liquid down in the inside chamber and thus push liquid out through the holes. It would over...

I was looking at an excellent web page in how to make your own wall-mounted bike rack ( link here ). But one aspect of it was bugging me... the calculation of the force acting on the hook in the wall. I encourage you to look at that site -- it's a bit annoying in that it requires you to click through 5-separate ad-laden pages to see it all, but I suppose that pays the bills. Here's my diagram of how the rack would look on my wall... either in the recommended arrangement of hanging the bikes wheel up, or in a way a bike shop friend of mine recommended, wheel down. Either or a mix would be compatible with the 2-tier arrangement of hooks. With only a single tier, you'd need to alternate up-down. Here's a separate example of such a rack with bikes hanging from the front wheel: Going back to the original web page where the bikes are also hung from the front wheel, the part I'm most interested in here is on page 3, which described the physics. Here's the ...

The last time I considered the constant cadence approximation as it applies to circular chainrings. Recall the principal issue is that the constant cadence approximation makes the following assumption for each pedal stroke: = × where ω is the angular velocity of the pedals, τ is the propulsive torque, and brackets signify a time-average. The error from this approximation is obviously: × − which is fairly trivially shown to equal: − ) × (ω − )> which is proportional to the correlation coefficient between torque and angular frequency, where a positive correlation results in an underestimation of power. Angular frequency is proportional to what I refer to as "instantaneous cadence": the rate at which the crank arms are revolving. So the issue comes down to whether the instantaneous cadence is correlated with applied torque, or similarly, if it's correlated with applied power (assuming applied power fluctuations are due more to torque changes than cadence c...

A big issue with power accuracy is to not only get the force versus time accurate but also to get an accurate cadence versus time. Power is the instantaneous product of propulsive force times pedal velocity, pedal velocity being proportional to cadence multiplied by crank length, and therefore errors in cadence translate directly to errors in power. It is typical in the power meter business that cadence is approximated as constant over a full or perhaps half pedal stroke. Indeed, Garmin has announced they are using this approximation on the Vector. This is of course technically incorrect: cadence varies over a given pedal stroke just as it varies from one pedal stroke to the next. Ideally cadence would be sampled at a sufficient rate to get multiple points within a half-pedal-stroke, so the variation in pedal speed between the strong and weak portions of the pedal stroke would be captured. I have previously looked at this issue and I concluded the power error would be proportion...

The last time I compared DC Rainmaker's total power numbers from Vector, Powertap, and Quarq on a typical ride near DC ( Zip file here ). The result was excellent agreement by my standards, considering the meters are measuring different points in the power transmission path. That leaves the Stages , which he also used. The Stages is not a total power meter; it's a left-leg power meter. It produces a derived number for total power by doubling left-leg power, but if you even glance at any Vector data, you realize that left leg and right leg power differ. Vector is also not a total power meter: it's two power meters. It's a left-leg meter and it's a right leg meter. These are distinct and are calibrated separately. You can derive total power by adding the two together, and the assumption here is total power is the sum of the left leg power and the right leg power. This is an excellent assumption as long as I'm not pushing with my hand on the crank arm. So...

DC Rainmaker has posted another dataset to his blog ( Zip file here ). His early installation issues behind him, these data serve as a valuable comparison between the power meters he's using: the spanking new Garmin Vector, the Quarq Elsa, Powertap, and Stages. The point of this post is to quantitatively compare the powers. But first some discussion... Each of these power meters is measuring power at a different point in the transmission path. Vector gets first shot at it, picking up power transmitted through the pedal axle. Stages is next, measuring power in the left crank arm. Next, Quarq measures it in the crank spider. Finally Powertap picks up the power which manages to make it to the rear hub. The largest losses are expected between the Quarq and hub, as mechanical losses in the chain and in the rear derailleur pulleys converts mechanical power into heat before the Powertap sees it. Between the Vector and the Quarq, it takes more imagination. Brim Brothers, not yet...

Garmin Vector vs Quarq Elsa: L-R balance comparison The Garmin Vector is clearly the most anticipated power meter to come onto the market. There's a few reasons for this. One is the freedom of choice it affords in selecting components. With a Powertap, you need essentually a new Powertap for every wheel. With crank-based systems, restrictions are more limtited, but swapping pedals is generally considered easier than swapping cranks (this is debatable, however: my Lightning crank is super-easy to take off and on). But perhaps more than this is the ability to measure independently the left and right pedal. This is power measurement at essentially the point of contact. It's directly measuring the forces applied by the rider. That was the inspiration for the name: "Vector". It's measuring the force vector applied by the rider's feet. This is somewhat of an academic point in comparison to the crank spider, since it's generally considered to be the ...

This Tour de France has experienced truly epic increases in people calculating how much power riders generate on climbs, typically towards assessing whether Chris Froome is practicing illicit performance enhancement. One of the more popular practioners of this assessment is VeloClinic, for example in his Haiku-esque Tumblr page . In previous years, there has been a good deal of discussion about VAM, Ferrari's statistic of rate of vertical ascent. Since climbing primarily involves mass overcoming gravity, VAM is related to power/mass, but but it's only a crude instrument, since the more gradual the climb, the greater the fraction of power going into wind resistance and rolling resistance. Additionally, it is possible to sustain higher VAM for shorter climbs compared to longer ones: on shorter climbs you can use your anaerobic energy reserves more rapidly, increasing the energy per unit time, which is power. Furthermore, at higher altitude oxygen concentrations in the atmos...

One issue with wheelsize is angular momentum. Jan Heine claims that 622 mm rims (700C) are best for up to 32 mm tire width, then 650B are good up to 42 mm, then "26-inch" are best for larger size. The idea is the wider tires yield larger mass and also yield a larger rolling radius relative to narrower tires at the same rim radius. This results in more angular momentum. Turning requires changing angular momentum, so more angular momentum creates more stability: more reluctance of the bike to change direction. It's been commonly asserted that trail is what controls bike stability, not angular momentum of the wheels (the "gyroscopic effect"). Trail's important, for sure, and you can make bikes where trail is the only contributor to stability (for example, ski-bikes), but the detailed analysis by Andy Ruina at Cornell has shown that angular momentum and trail both contribute, as well as center of mass. Bikes are a complex dynamic system. Angular momentu...

I already showed my "simple" model for how an arbitrary wind affects speed. This was based in part on a numerical fit to an implicit power-speed calculation, where I found to decent approximation the logarithm of speed as a function of tailwind/headwind was well fit by a parabola. The formula I used, combining the effect of a tailwind/headwind with the effect of a crosswind, was: s' = exp[ −(s w ' / 3) 2 ] exp[ 2 s wx '/ 3 ], where v 0 ' is the ratio of flat-road speed with the wind to flat-road speed without the wind, s w ' is the ratio of wind speed to flat-road speed without wind, and s wx ' is the component of that in the direction of rider travel. I already showed an analytic derivation of the cross-wind term, which is proportional to the square of s w '. I also did a first-order dependence of s' on s wx '. However, to justify this full model other than numerically requires a second-order dependence of s' on s wx '. ...

Last time I gave up trying to solve this integral. I wanted to watch Cyclocross Worlds so got lazy and did a numerical solution and fit. But I realized while out running after the racing that I could have done much better. So I'm back for more. (1 / 2π) ∫ dφ exp[ (s w ' / 3) 2 ] exp[ −2 s wx ' / 3 ], where the integral is over the full circle and φ is the angle of the wind relative to the rider (0 = pure tail wind). The key here is to recognize that this can be well-approximated by a Gaussian for s w ' to at least 1. This isn't the solution of the integral, but it's a good approximation, so once I recognize the analytic form of the solution I can get away just matching derivatives with respect to s w ' = the ratio of the wind speed to the zero-wind rider speed. So my solution will be the following, where I must solve for K: exp[ (s w ' / K) 2 ]. I recognize that for every value of positive s wx ', there is a corresponding negative value...

Last time, I discussed a simple approximation which comes fairly close to predicting the speed at constant power in an arbitrary wind of reasonable magnitude. The formula can be written as a normalized speed s' (normalized to no wind) and a normalized wind s w ' (normalized to the same scale) as follows: s' = exp[ −(s w ' / 3) 2 ] exp[ 2 s wx '/ 3 ], where s wx ' is the normalized speed of the wind in the direction of the rider (positive for tailwind, negative for headwind). For a typical closed circuit, the rider will head in random directions relative to the wind. If the wind is constant than the average distance-weighted direction is zero: the rider starts where he begins. Last time I calculated a result assuming a square course aligned with the wind. That's an over-simplification, even if it's probably representative. But I'd prefer to consider the more generally applicable case of all directions equally likely. So I assume the rider ...

The motivation for the preceding analysis was to add wind effects to my heuristic speed model. The philosophy of the heuristic speed model is to not rely on a constant power approximation, but try to model cyclist behavior directly. So how do riders behave when faced with a wind? When I first started riding with a heart rate monitor, an early lesson was my heart rate dropped when I was riding into the wind. This was obviously psychological: the wind was defeating me and I lost the motivation to pedal hard. Then I moved from the San Francisco Bay area to Austin where I lived for three years and winds were a predictable part of every ride, while extended hills were essentially gone. Instead of challenging myself on long hills, I learned to use treat the headwinds as a challenge rather than bad luck. As a result, I began increasing my heartrate, rather than decreasing it, when I encountered headwinds. With groups it's different: the group may be motivated to hammer into the ...

Back in July 2009 I did a series of first-order calculations on the effect of various parameters on riding speed. Calculating speed from power is difficult to do explicitly, but to first order the calculations become straightforward. First-order analysis is where much of the intuition is, anyway. One of these calculations was the effect of wind resistance on riding speed . Then in November 2009 I extended that analysis to the effect of wind speed on riding speed . To my dismay, I found an error in that result. I'd even rationalized the wrong result with incorrect arguments. I had to track the consequences of that error through the following two blog posts. I think I fixed everything. The corrected result was: d s / d s w = 2f / [ 2f + 1 ‒ s w / s ] where s is the rider speed, s w is the tailwind speed, and f is the initial fraction of retarding force due to wind resistance. There appears to be a singularity issue for strong tail-winds (see the denominator) but then ...