Friday, October 30, 2009

Tour of California 2010: stage 3?

Steve Hill has proposed a preliminary route for stage 3 of the 2010 Tour of California. Check out his map here.

However, this prediction is, well, a bit dull. Basically repeating what was done in 2009 with the exception of the crossing of the Golden Gate Bridge from Sausalito, We already know that there will be an extra climb on this year's route. This one actually has one less climb, omitting the climb out of Sausalito.

So I propose an alternate. It adds only 8 km to Steve's route, from 194 km to 202 km. But it's more a quality thing than a quantity thing.

Here's my map.

Steve's route has riders turning right at the top of Tunitas Creek Road, climbing briefly on Skyline Blvd. before descending to Highway 84, then descending 84 and continuing the final 10 km of that road to the coast. Yawn....

Instead I propose they cross Skyline at the top of Tunitas, descend technical Kings Mountain Road, turn right on Highway 84, left on Portola Valley Road, right at the intersection with Sand Hill Road, then right on Old La Honda Road.


the top of Old La Honda Road on the day of the Low-Key Hillclimb this year (Josh Hadley photo)

Local weekend warriers have been speculating for decades about what the big boys would do on Old La Honda. For example, I suggested Contador could ride it in 12:35 assuming his same effort he demonstrated at Verbier in the 2009 Tour de France. Well, sorry to disappoint, but at this point in the stage nobody's going to be making a maximal effort up Old La Honda. Indeed, I doubt the present record will fall, which rumor has it is 13:45 by Eric Wohlberg in training. Still, it would be amazing to see the pros riding the road so many of us love.

The only question: will the residents of Woodside feel similarly?

Anyway, back to the stage. After climbing Old La Honda to Skyline, I have them descending the narrow, scenic opposite side to Highway 84. Then they continue from there to the coast and southward, following last year's route.

But then I add an additional twist, one I admit is unlikely. Before climbing Bonny Doon, I have the riders detouring over Swanton Road north of Davenport. Sure, there's a lot of scenic detour options along the route, and the organizers probably prefer to keep it simple. But I couldn't resist on this one: Swanton is so cool. A nice little climb, one which Pro Tour legs will hardly notice, but still some very nice spectator options here. I virtually never pass Swanton by when I'm riding down there.

I don't know when Tour officials will commit to route details. It's a long time before the May date for the race. But if they're listening: give Old La Honda the go. It won't affect the final result, but will be extremely popular among local riders.

Metrigear Vector Pt 8: Pedal Speed (2)

In the last post I discussed some data Metrigear presented on their blog showing accelerations measured at the pedal spindle, decomposed into radial and tangential components. I wanted to step back and discuss the derivation of power from pedal speed. Okay, to be honest I'd prepared this blog post and skipped past it to post the one responding to Metrigear's data. So I'm a bit out of sequence: this one was mostly written before the Metrigear post.

Mechanical power is the component of force in a particular direction times the speed in that direction at which the power is being applied. We've already worked out the force components, and determined the direction of propulsive power, so all that remains is to determine the speed along that direction. So all that remains is to determine the speed along that direction.

We'll assume again the coordinates of the right pedal:


Axes for right pedal

The propulsive force is applied in the x direction, which is a fixed direction in the coordinates of the spindle. So we need to determine the rate at which the pedal is rotating in that direction.

If the bike is in a zero-gravity inertial frame, for example at rest on a trainer in a space station, then the solution is easy: there is only one component of acceleration, which is that associated with the pedal rotating. From the schematic, the pedal is moving in the +z direction, orbiting about an axis of rotation in the x direction, and is therefore accelerating in the -y direction. A way to remember this relationship is that all planets accelerate toward the sun, which attracts them gravity, and yet their trajectory is roughly circular about the sun, so a circular motion has an acceleration toward the center of rotation. In this case, the magnitude of that acceleration is:

aω = ω² L,

where ω is the rate of rotation (radians per unit time) and L is the crank length. Note aω differs slightly from the aorbital used in the the last post as they differ in the reference. aω is referenced to the bicycle frame, where aorbital was referenced to inertial space. More on that in a bit...

Of course the Vector isn't designed just for stationary trainers in space stations. There's other forces at play. There's gravity, for example. Then there's the acceleration of the bike moving forward. For simplicity, I'll make what seems the reasonable assumption that the spindle is always perpendicular to the direction of forward motion. To consider more components of bicycle acceleration, more coordinates are needed. These are defined in the following diagram:

angular coordinates
Angular coordinates

Two coordinates are defined here: φ, which is the angle of the crank arm relative to the bike frame horizontal, and θ, which is the "roll angle" of the bike relative to the pull of gravity.

An important point: why is φ relative to the bike frame and not the pull of gravity? Propulsion is propulsion is proportional to the rate of change of φ times applied torque. Now suppose that φ is changing at a steady rate, but the effective grade of the road is also changing. Let's define the angle of the road (and therefore of the bike) relative to the horizontal as defined as the plane perpendicular to the pull of gravity as φbike. Then does the applied propulsion depend on whether φbike is increasing or decreasing? The answer is clearly not, since what matters is the speed with which the chain is driving the rear cassette, and the relevant chain speed is the rate of rotation of the crank relative to the bicycle frame. If the whole bike is rotating because the road grade is changing, and a force is applied to the pedal, but the pedal is locked into position somehow, no work is done even though there is a torque being applied and being moved through an angle relative to the direction of gravity. The key is the pedal is not being moved through an angle relative to the the bike frame. Of course the laws of physics are generally defined with respect to inertial reference frames. But the key again is Newton's Third Law, which suggests in the locked pedal example the measured torque is being balanced by an unmeasured one. That unmeasured torque in this case is through the wheels: the wheels are being pushed by the earth in a way which cancels out the propulsive torque of the pedals moving through an angle in inertial space. The component of torque not canceled by the wheels is that due to motion of the crank relative to the bike frame, as we've defined.

So with that long-winded discussion, we can define applied power:

Applied power ≡ P = Fz ω,

ω = dφ / dt.

All that remains is to measure ω for real-world situations. If we can derive the rotational acceleration aω we can derive

ω = √(aω / L)

We then need to extract aω.

I discussed last time how to perhaps estimate ω by averaging radial accelerations for the left and right pedal spindle. A caveat there is my discussion about the relevant rotation being relative to the bike frame. Spin the entire bike, and the pedals will spin with the bike, and yet no propulsion is being applied to the chain. Nevertheless, this is a relatively small error, since the pedals typically change their angle a lot faster than the bicycle does. Were one to extract rate of crank rotation from an optical sensor between the crank arms and the bottom bracket, the rotation would be relative to the bicycle frame and not due to inertial space. This would be better, but of course adding an additional cadence sensor adds mass, cost, and complexity for dubious benefit.

Next I'll try to guess how to do it for one pedal spindle. It's not obvious, however, I admit. Maybe I'll stall for time by returning to a discussion of climbing power, or maybe discuss the joys of endoscopy. But that's off topic.

Thursday, October 29, 2009

Metrigear Vector Pt 8: Pedal Speed (1)

The last post in my Metrigear series I conjectured how the Metrigear determined which direction was propulsive. Since the spindle spins with the pedal stroke, but since acceleration due to orbital motion is always in the same direction, averaging over time indicates which is the direction to the center of the orbit, and the propulsive direction is orthogonal to this in a well-defined way depending on if a pedal is a right pedal or a left pedal. So all that's left is to determine how fast the pedal is moving.

Given a crank length L, each revolution of the pedal the pedal moves a distance 2πL, or using the definition of a radian, each time the pedal moves though an angle of one radian it moves an arc length L. So if we know the angular rate at which the pedal is rotating, and we know the crank arm length, then we know the rate at which the pedal is sweeping out its orbit: its orbital speed which we multiply by the force in the orbital direction to derive propulsive force.

I was going to conjecture about how this is done, but then Metrigear published a blog post in which they showed how two accelerometers per pedal can be used for just this purpose. The two accelerometers in the pedal spindle measure accelerations along each of two directions orthogonal to the axis of the pedal spindle. Of course, the orientation of the spindle, and therefore of these orthogonal accelerometers, cannot be guaranteed relative to the pedal direction due to differences in the thread orientation in different crank arms. So a linear combination of these accelerations is needed to derive the acceleration along the direction to the center of the orbit, and similarly along the pedal's orbital trajectory. The linear transformations to derive these components use directions perhaps derived in a fashion similar to the one I described: over time, accelerations along the pedal's arc of motion diminish with averaging.

I extracted data from the Metrigear plots with g3data, a handy application for this purpose, and did a regression to sinusoids as shown in the following plot:

Reproduction of data from the Metrigear Blog, with fits

These data were recorded from a bike on a trainer, where any accelerations of the bike are insignificant. So the only contributions to the acceleration signal are gravity (which Einstein taught us is indistinguishable from acceleration) and the rotation of the pedal. Gravity manifests itself as an oscillating component with a rotation rate proportional to the cadence. The fit to the data shows a rotation rate of 12.06 radians/second, or a cadence of 115 rpm. This is a bit lower than Metrigear derived from the same data, but they measured the spacing between zero crossings, while I did a regression to the whole data set, which may be more robust.

There are two components to the acceleration: the tangential one (along the direction of pedal motion) and the radial one (which includes the orbital acceleration).

The tangential acceleration signal oscillates with an amplitude of 1 g, which is a standard gravitational acceleration (although effective gravity varies from point to point on the Earth, and the effective gravity will be reduced if the pedal spindles are not perfectly horizontal with respect to the pull of gravity, either of these variations are expected to be small in comparison to the resolution of these data).

The radial component also oscillates. I did regressions on both, each set yielding a similar oscillation frequency. Then I fixed the oscillation frequency at the average of these fits and refit the curves, adjusting the phase and, in the case of the radial curve, the offset from zero. This offset provides the average orbital acceleration rate in these data, which represent smooth steady pedaling. The derived value for this offset was 2.60 g.

Metrigear states the crank length can be derived from these data, and it's true. Using my derived rate of pedal rotation: ω = 12.06 radians/sec, and a value of gravity g = 9.807 m/sec², the crank length L is calculated L = aorbital / ω², where a is the measured acceleration. Running the numbers yields 175 mm. So I think this is the crank length.

However, this approach used a regression of data to derive the rate of oscillation of the gravity component. Better would be to measure the rate of pedal rotation at every discrete point in time, independent of other points in time, so that if the pedal rate changes then we have rapid and accurate estimation of the new rate. For this purpose, if the orbital acceleration can be extracted from the radial acceleration signal, stripping out the offset from the components due to gravity and bike acceleration, then the rate of pedal rotation can be derived:

ω² = aorbital / L

From this we can get the pedal's orbital speed:

ω L = ( aorbital L )

So how do we derive this orbital component? Well, one way is to look at data from th e left pedal and the right pedal together. Assuming the pedals are 180 degrees apart (not always true with Power Cranks!), and that the spindles are parallel (crank arm and pedal spindle flex are within acceptable limits), then the acceleration of gravity and of the bike's accelerations are in the opposite direction relative to the left spindle and relative to the right spindle, with the direction from each spindle to its center of rotation taken as a reference. So taking the average of these radial acceleration components does the job:

ω L = [ aradial,L + aradial,R ) L / 2 ]

Does Metrigear do this? Well, maybe. But they claim the Vector works with just one Vectorized pedal, and I'm sure PowerCrank users will be upset if it doesn't work with their set-up. So there needs to be a back-up scheme.

Hmm....

Tuesday, October 27, 2009

Soda Springs power: error estimate

Last post I made an estimate of my power up Soda Springs road on Saturday. It's worth doing an estimate of possible errors with this:
  • body mass: around 1 lb
  • equipment mass: let's say 1 lb
  • rolling resistance coefficient: 20% seems reasonable. This was really seat-of-the-pants.
  • wind drag coefficient: 10% seems about right
  • distance 0.1 miles. Exact starting location, the placement of the finish, and line through the corners are all factors
  • altitude gained: 15 feet sees consistent with individual reports
  • wind speed: Zero was assumed. Given variable winds reported at Redwood Estates, a conservative estimate of error of this estimate is probably 1 mph
  • transmission coefficient: based on differences in chainline, I'll say 1%. This doesn't affect the derived Powertap-equivalend power, but it does affect the significance of comparisons in power from one ride to the next.
  • time: 2 seconds seems good: 1 second with recording the finish time, and 1 second for a delay in starting from when the horn was honked
  • acceleration: I may have crossed the finish as fast as 15 mph
Let's run the numbers, then, on how each of these errors would affect the result. Fearlessly diving in:
  • body mass & equipment mass: I'm assuming each affects power the same. These affect climbing power, rolling resistance power, and acceleration power. The resulting changes are:
    1. climbing power: ΔP = 1.7 W
    2. rolling resistance watts: ΔP = 0.08 W
    3. acceleration power: ΔP = 0.02 W.
  • rolling resistance coefficient: changes rolling resistance power in proportion to error: ΔP = 2.2 W.
  • wind resistance coefficient: changes wind resistance power in proportion to error: ΔP = 1.8 W
  • distance: this affects wind resistance, rolling resistance, and acceleration power. First, a 0.1 mile difference out of 5.35 miles is a 1.87% difference in speed. From this:
    1. Wind resistance increases by approximately 3 × 1.87%, +5.71%: ΔP = 1.0 W.
    2. rolling resistance increases by 1.87%: ΔP = 0.21 W.
    3. Acceleration power increases by approximately 2 × 1.87%, ΔP = 0.01 W
  • altitude gained: weight × height / time: ΔP = 1.5 W
  • wind speed: wind resistance changed proportional to twice the ratio of wind speed error to bike speed: ΔP = 3.6 W.
  • transmission coefficient: error in power = error in transmission coefficient × total power: ΔP = 2.6 W
  • time: the time error was 0.10%. This affects climbing and rolling resistance power by the same fraction. Wind resistance power is affected by 3 times this proportion. Acceleration power is affected by twice this proportion (an insignificant contribution). In total: ΔP = 0.31 W
  • acceleration: changes acceleration power by up to +125%: ΔP = 0.4 W
Whew! Okay, so you might be tempted to add up all of these error terms as "worst case" estimation (in circuit design this is called "designing to the worst-case corners"), but for uncorrelated errors (for example, an error in distance doesn't imply anything about the error in time or weight) the better approach is to sum the squares of the error terms, then take the square root.

Josh Hadley photo

Okay.... drum roll....
ΔP = 6.1 W

So my estimate of 263 W could in effect be anywhere from 257 W up to 269 W. Given the 8 W difference from OLH distance to Soda Springs distance, that would be an effective OLH power of as high as 277 W powertap-equivalent. Were I to have extracted that value, given my present fitness, I'd not be concerned, especially since there's similar error bars on that 290 W estimate from my Old La Honda climb this summer.

So the end result is: don't place too much importance in these sorts of physics-based power estimates. There's too many variables, too many sources of error, even for something as straightforward as a steep hillclimb.

Perhaps surprising in the strong influence of wind resistance on the error estimate: that term alone increased the uncertainty by 1.2 W. And we made a seemingly trivial estimate of the uncertainty in the wind speed: only 1 mph, essentially impossible to detect without instruments.

So don't sell that power meter yet.

Monday, October 26, 2009

Soda Springs power estimate

The Soda Springs Low-Key went rather well in most regards. As a participant,I was pleased with my finishing place. I usually train with a PowerTap, but for most races (especially hillclimbs!) I leave the powertap wheel home and instead use a carbon race-only tubular (either my Reynolds MV-32T or my Mt Washington).

On hillclimbs, though, I can generally estimate the power. The key parameters are total weight, rolling resistance coefficient, wind drag coefficient, and air density. For PowerTap-equivalent power I do not need drivetrain efficiency, since drivetrain losses aren't measured by the PowerTap, anyway. Given these, it's simple enough to calculate power from speed.

So, some numbers:
  • weight: I was 125 lb in the morning upon wakening. Given input - output, I was probably close to this at the start. Curiously, I'd been almost 3 lb lighter the last time I weighed myself two days before. A lot depends on what I've been eating and how hydrated I am. Additionally, my bike weighed out at 11 lb 6 ounces before the start. This was one ounce more than I expected, but I'd accidentally mounted the wrong tire on my front wheel. Hard to comprehend how I could have gone through an entire tire-glueing procedure without noticing this.... I only noticed it, in fact, when I was looking at a photo of myself after the race! Add in the weight of shoes, clothing, and my water bottle and I figure around 4 extra pounds is pretty close: should be within 1% of actual total weight.

The pre-climb weigh-in (Rich Hill photo)

  • Rolling Resistance: My rear tire has a value reported by Al Morrison to be 0.234%. I accidentally had a Veloflex Carbon on my front wheel, with a value of 0.312%. Given the slope, the rear wheel should be responsible for around 2/3 of the total, the front wheel around 1/3, putting the average around 0.26%. The road on the climb is rough in sections, so I'll assume an actual value of around 0.4%.
  • Air density: At Redwood Estates in the hills over Los Gatos (1690 ft elevation), air pressure was 29.98 inches of Hg, while the temperature was 67.2 F with 70% humidity at 10:32 pm, around the time of the climb. These values yield an air density of 1.13 kg/meter³.
  • Wind drag coefficient: I'll use the value I extracted from Old La Honda data, CdA = 0.36 meters².

Okay, then. I climbed 2368 feet over 5.35 miles in 32:04. For these sorts of calculations, with mixed imperial and metric units, the Unix "units" program is great, as it handles the unit conversions.
  1. Climbing power = weight × feet gained / time = 234 watts
  2. Rolling resistance power = weight × Crr × speed = 11 watts
  3. Wind resistance power = ½ ρ CdA × speed³ = 18 watts
  4. Acceleration power: I went from 0 to 10 mph over 32:04, which works out to 0.3 watts, or zero with the 1 W precision used here. I never braked during the climb, so there were no other accelerations which were not canceled by other decelerations.
  5. Total = 263 watts
If I wanted to determine SRM or Quarq Cinqo watts (or MetriGear Vector), I'd divide by an efficiency term, for example 0.97, and thereby estimate maybe 271 watts.

I was probably around 290 watts (PowerTap equivalent) when I set my Old La Honda PR in July. Sure, Soda was almost twice the duration, but that's only responsible for a few %. Specifically, assuming a critical power model where AWC = CP × 60 seconds, which generally fits my data, I'd expect a 2.7% loss in power going from OLH to SS: around 8 watts. This at least brings my effective OLH power above 270 watts PT-equivalent, not bad for the Noon Ride, but not up to competition standards.

I think it shows a difference between off-season (lifting, running, generally not pushing things too hard) and in-season fitness. The same applies for everyone in the Low-Keys, however. None of the top guys are in mid-summer fitness.

D'oh! (Rich Hill photo)

So what was the penalty of my bad tire choice? Let's use Al's numbers of 0.08% difference for the wheel, blow that up to 0.11% for the rougher road, and assume 1/3 of weight on the wheel, yielding 1.0 watts lost to propulsion out of 263. Using a formula I derived earlier, with 88% of the power mass-proportional, this yields a speed penalty of 0.30%, or 5.9 seconds. There's also a mass penalty, as the Record is slightly heavier, around 1 ounce. That penalty is an extra 0.6 seconds, bringing the total to 6.5 seconds. I was 22 seconds down on Eric, however, so this mistake didn't affect my placing.

For error analysis, a large source of uncertainty is the total climbing. For every 10 feet of error in the total climbing feet that's a difference of 1.0 watt. But the climbing feet agree with those recorded by two other riders on their barometric altimeters, each recording slightly less, not more climbing than the Garmin data I used as a reference. I'll stick with the Garmin numbers, since they're less susceptible to barometic drift due to the GPS signal used as a calibration correction to the barometric signal, the latter better for detail.

Sunday, October 25, 2009

Metrigear Vector Pt 7: Direction

Saturday was another Low-Key Hillclimb, up Soda Springs Road, and once again I got to see the Vector in action as Clark Foy rode his prototype up Soda Springs Road. I'm sure we'll see data from that test on the MetriGear Blog.

For now, I'll move my discussion of how I view this really cool power meter away from force and towards acceleration and velocity. The Vector's motto is "Direct your Forces": power is the product of a velocity (in this case the pedal's velocity around its circle) and the force in that direction. I've already talked about how the Vector might measure a variety of force components. The key is to convert those force measurements into a measure of propulsive power.

The first problem to be solved on generating power is to determine in which direction the pedal is driving the bike. As I've already discussed, the relevant velocity is the velocity is the rotational valocity of the pedal rotating in its orbit The translational velocity of the bike doesn't matter. Imagine the thought experiment of pushing forward with both pedals while the bike coasts. I'm applying a force in the direction of the motion, and typically a force in the direction of motion provides power. Indeed, were someone running next to the bike and pushing the bike by the pedals while the bike coasted, the bike would accelerate. But more typically a rider would be pushing forward on the pedals, but by Newton's Third Law, would be pulling back with an equal and opposite force on the handlebars and/or saddle. So the net force in the direction of motion would be zero.

Pedal force diagram, shamelessly borrowed from the MetriGear Blog

Rather, the way the bike is typically propelled is by pushing the pedals through the arc of their trajectory about the crank. So it's important to isolate the component of force in this direction. Fortunately, the pedal spindle rotates with the crank, and the electronics are encapsulated within the spindle, so this force is always in the same direction within the spindle. Well, not quite always... the pedal may tighten or loosen during riding, or if the pedal is moved from one crank to another the phasing of the threads may change, for example. So the unit needs to be able to determine which is the direction of rotational motion, but can afford to be a bit lazy about it, relying on relatively long-term observations, rather than rederiving orientation each sample.

As a bike moves, it may accelerate in a wide variety of ways. But the two dominent acceleration sources are the orbital acceleration of the pedal's rotation and the acceleration of gravity. Einstein took as a postulate of Special Relativity that gravity and acceleration were indistinguishable. Therefore an accelerometer has no way of knowing whether it is in a gravitational field, or is accelerating in the absence of a gravitational field along the same axis. So from the perspective of the Vector, a gravitational field is just another component of acceleration.

So how can the Vector determine its orientation? Remember the Vector is rotating as the pedal rotates. Therefore the acceleration associated with its circular motion (in the frame of the bicycle; imagine the bicycle is at rest on a stationary trainer) always points in the same direction, as long as the pedal is fixed within the crank. There is an additional acceleration which rotates over time, the acceleration associated with gravity. On top of these, if the bike is not on a trainer, are acceleration components associated with the bike speeding up or slowing, the bike turning, the bike "rolling" (tilting right or left), the bike bouncing (or otherwise changing rate of ascent), the bike changing grade, and the cadence speeding up or slowing. But average these accelerations over time, and the one which remains is the acceleration associated with the pedal rotating. Note this acceleration is the same direction whether the pedals are rotating forward or backward. So just spin the pedals for awhile and it will eventually become evident how the spindle is oriented.

But there's more. There's also gravity. Gravity provides a signal of how the spindle is oriented in space, of the rate of rotation. For uniform orbital motion, the acceleration associated with the circular orbit is equal to ω²L, where ω is the rate of angular change in radians per unit time, and L is the crank length. So if we can determine ω from the rate of change of the direction of gravity, and we can measure the acceleration, we can determine L.

But the Vector does not determine L, instead it requests that it be provided by the user. I was initially puzzled by this. But then I realized, why not ask for L? it's just one less thing to derive, reducing the opportuinity for error, as long as it is well known from the crank supplier. Side note: the Ruegamer Vf, a really cool crank, has an adjustable crank arm length, separately for each crank arm. So for something like this, you'd need to measure the crank length yourself.

The Ruegamer Vf, which has an adjustable length

So anyway, the Vector can find the direction of dominant orbital acceleration. The direction of propulsion is then perpendicular to this direction. Rather simple. All that remains is to determine which direction is pedaling forward, and which direction is pedaling backward. And unless I've rigged my bike with a cross-over chain such that I backpedal to move the bike forwards, this is fairly simple, as long as I know which pedal I am (right or left). And fortunately left and right pedals have opposite threads, and so are not interchangable, even with Speedplays for which the pedal bodies could be swapped between the left and right. (I could fool it by reverse-threading my cranks, in which case the pedals likely won't stay tight.) So the Vector knows if it's in a right or left pedal, and therefore knows in which direction it is supposed to spin.

So there it is: with two acceleration components, one along each of the two directions orthogonal to the spindle, the Vector should be able to determine the direction to its center of rotation. Then based on whether it is a right or left pedal, it should be able to pick the one of two orthogonal directions in the rotation plane which corresponds to pedaling forward. Then it can determine which force component is moving the bike forward, or in the case of a fixed gear or tandem captain crank, possibly slowing it down.

Friday, October 23, 2009

Metrigear Vector Pt 6: Forces and Bending (3)

The previous analysis described how local curviture of the pedal spindle might be used to extract out moments of force applied to the body: the total downward force and the twisting torque. The sensors would measure the distortion of the pedal spindle, which with an appropriate calibration factor, would yield the values of the bending moment, from which the force and torque on the pedal body can be derived. Of course, this is total speculation on my part, as I really do not know how the Vector really works: it is just a plausibility argument that (1) the problem is non-trivial, and (2) the solution may be tractible.

Given the coordinate axes defined for the pedals (right pedal shown here):

Axes for right pedal

it is clear that this analysis applies to the force in the y and z directions, and to torques applied about these axes. For a force in the x direction, the spindle does not bend, but rather stretches. In this direction, it should behave like a linear spring: fractional stretch in proportion to the applied force. This could be directly measured with a strain sensor in the spindle, with the associated spring constant calibrated after assembly.

For the associated torque, associated with the spindle being twisted, additional sensors could be used, again calibrated to the mechanical properties of the spindle. As I noted early, the derived twisting torque, when multiplied by the rate of pedal rotation, would indicate the amount of power being lost in the pedal bearings. Even if this is an insignificant fraction of total power, it could be an indication of the need to service a pedal, for example.

So, to get the full set of three forces (for example, x, y, and z) and three torques (for example about these three axes) applied to a pedal one needs:
  1. two measurements along the spindle along each of two orthogonal axes of the bending of the spindle,
  2. a measurement of the stretch of the spindle,
  3. a measurement of the twist distortion of the spindle.

Parameters which require calibration include:
  1. the stiffness of the spindle to bending at each of the two points, yielding the proportionality factor relating spindle curvature to bending moment (torque), perhaps separately along each of the two orthogonal axes,
  2. the spring constant assocated with elongation of the spindle,
  3. the spring constant assocated with twisting the spindle.
  4. For extra accuracy, the mass of the pedal body and the portion of the spindle from the center of the sensors outward, to calculate the power going into changing the rotation rate of these mass components. This kinetic energy component is directly between the shoe and the pedal, not transmitted through the pedal spindle, and therefore wouldn't otherwise be measured by Vector.
A quick comment on this last one: the shoe mass is also spinning, and acts as a reservoir of kinetic energy, as does the leg. But I assume the Vector is only reporting power delivered to the bike, and the bike includes the pedals but not the shoes or legs. In addition to kinetic energy of the pedals, there's also potential energy, but unless the left and right crank don't weigh the same or aren't oriented 180 degrees from each other, the potential energy changes of one side are canceled by those of the other.

In addition to these calibration factors, since the piezoelectric effect has a strong temperature coefficient, a temperature sensor is needed to compensate for this.

So that's it: enough on forces! I could be completely off base, and Metrigear does things completely different. But at least to the extent of my freshman physics simplicity, I think I've made my plausibility argument that the thing may actually work.

The next issue is how to use this force information to generate power. That's where the accelerometers come in.

Thursday, October 22, 2009

Metrigear Vector Pt 6: Forces and Bending (2)

Back to measuring forces on the pedal: a reminder of the force diagram:


Force diagram of pedal


Assume the sensors measure the torque (the bending moment) on the pedal spindle at each of the associated points, x1 and x1. Forces F1 and F2 are applied at the points x3 and x4. The goal is to determine F3 and F4 from the measurements τ1 and τ2.

Simply enough: two equations, two unknowns (F1, F2):

τ1 = F1 ( x3x1 ) + F2 ( x4x1 ),
τ2 = F1 ( x3x2 ) + F2 ( x4x2 ).

Note the second equation is the same as the first with the x1's replaced with x2's, and the τ1 replaced with τ2. Multiply the top equation by ( x3x2 ), and the bottom by ( x3x1 ):

( x3x2 ) τ1 = F1 ( x3x1 ) ( x3x2 ) + F2 ( x4x1 ) ( x3x2 ),
( x3x1 ) τ2 = F1 ( x3x2 ) ( x3x1 ) + F2 ( x4x2 ) ( x3x1 ),

and substract the two and combine terms:

( x3x2 ) τ1 ‒ ( x3x1 ) τ2 = F2 [ ( x2x1 ) ( x4x3 ) ],

yielding the solution

F2 = [ ( x3x2 ) τ1 ‒ ( x3x1 ) τ2 ] / [ ( x2x1 ) ( x4x3 ) ].

Now for the other solution, swap F1 and F2, and swap x3 and x4, then move terms around to get rid of excess negatives:

F1 = [ ( x4x1 ) τ2 ‒ ( x4x2 ) τ1 ] / [ ( x2x1 ) ( x4x3 ) ].

Of primary interest is the total force, trivially adding the previous two results, and simplifying:

F1 + F2 = (τ1τ2) / ( x2x1 ).

Wow. That's embarrassingly simple.

Interestingly, the only position data you need to know here is the separation of the two torque sensors. This is known, of course, assuming the electronics are inserted straight into the hole, and the hole is itself straight within the spindle. Both of these are likely good approximations within the 1.5% net error budget of the Vector, since errors in orientation only contribute second order to errors in separation along the spindle axis. You don't need to know how far the electronics are from the crank arm, nor how far the pedal body is from the sensors.

Also of interest is the torque on the pedal body:

( F2F1 ) ( x4x3 ) / 2 =
[ ( [ x4 + x3 ] / 2 ‒ x2 ) τ1 ‒ ( [ x4 + x3 ] / 2 ‒ x1 ) τ2 ] / ( x2x1 ).

Wow -- I hope I got all of that correct (I rarely do). Anyway, here you need an additional bit of information, which is the separation of the center of the pedal body from the sensor. This is a parameter which would need to be determined by a calibration step. Fortunately, accuracy on this torque number probably isn't important. It's power we really want to nail down, and that depends on propulsive force, not how hard the pedal is being twisted.

Luckily you do not need to know how far anything is from the crank arm. If you did, for example inserting a spacer under the pedal to change the Q-factor would mess up the result. Perhaps more significantly, bending of the crank arm may yield an effective pivot point beyond the point where the pedal is screwed into the crank, and the position of this virtual pivot may depend on pedal force. So since the result doesn't depend on the location of the pivot point, it may be relatively insensitive to crank flex.

So what do I get out of all of this analysis? Basically that with good bending data taken at two points along the spindle, you can get the average force on the pedal body without knowing the distance to the pedal body. That's very nice. Additionally, you can derive the torque exerted on the pedal body about its center if you know the distance from the bending sensors to the center of the pedal body. This could be calibrated for maximum accuracy, or it could be more crudely estimated given the lack of relative importance of this value.

Metrigear Vector Pt 6: Forces and Bending (1)

As I remain busy with the Low-Key Hillclimbs, I am trying to squeeze in some time to continue with my thoughts on the MetriGear, which I think will be the most exciting power meter to hit the market yet.

As I've emphasized in previous blog posts, Vector's motto of "Direct your Forces" makes it clear the ephasis is on force, not just power. Power is a single scalar quantity for each pedal, but force contains many more components. Indeed, since the pedal rotation varies around the pedal stroke, and since the orientation of the pedal inspindle into which the Vector is installed cannot predicted, the Vector must measure multiple force components and extract the one which contributes to propulsion. Providing users with not just this propulsive force, but non-propulsive forces as well, may be beneficial to pedal stroke optimization or bike fit, as I've discussed.

Key to all of this is being able to measure these forces and torques. Recall each pedal has three axes, and force components can be applied to the pedal along each of these axes, and additionall torque components can be applied which attempt to rotate the pedal about each of these three axes. So is measuring each of these components possible?

The Metrgear is a module inserted into the spindle of the pedal, with piezoelectric sensors which measure the bending of the spindle. The bending moment on the spindle, relative to a given point on the spindle, is an applied torque: a force times a distance. Before continuing, consider the following schematic of a pedal:


Force diagram of pedal


Starting from the left, the threads of the spindle are considered to be screwed into a crank arm, which is considered rigid. Real cranks are not rigid, nor are the bicycles to which they attached. However, for the purpose of measuring forces on the pedal, that should not be important: as long as the pedal is prevented from accelerating, for a given set of applied forces, the bending of the spindle will be the same whether the crank or rest of the bicycle is rigid or compliant.

Then I assume two idealized torque sensors. These measure the bending moment on the pedal from their respective positions, which are within the spindle distances x1 and x2 from the crank arm. The force moments they measure are labeled τ1 and τ2.

Next are the applied forces, which are assumed on the pedal body. Of course forces may be applied all along the pedal body, since the cleat may be in contact with the entire pedal body surface. However, it simplifies things if it is assumed force is applied at only two discrete points. These points are assumed to be at distances x3 and x4 from the crank arm. The magnitude of these forces are labeled F1 and F2.

There are also forces applied to the threads. Consider the case where the spindle is not accelerating. Then if there were no force on the threads, by Newton's second law, the spindle would accelerate or undergo a state change like cracking or heating. So the crank must induce forces opposite to the forces applied to the pedal body, with a torque which cancels that applied to the pedal body. If the pedal were not attached to the bike, for example, but only to the cleat, pushing on the pedal will just accelerate the pedal.

Of course, the spindle is accelerating: it is rotating, which implies continuous acceleration, and the bike may be changing speed or orientation or direction, all form of acceleration which apply as well to the spindle. So these forces do not really balance. Since the spindle has non-zero mass, it takes net force to accelerate the spindle. Were the pedal and spindle zero mass, things would be simpler. Then the forces and torques would need to balance, even with acceleration.

Further with the static loading assumption is that the bending of the spindle is changing only slowly. For example, whack the pedal without a rider touching it and the pedal will oscillate back and forth for awhile, like a pendulum attached to springs. Obviously in this case there is no rider power -- the rider isn't even touching the pedals. Yet the spindle is bending. Typically this sort of thing can be neglected if the analysis is averaged over times sufficient that the pedal would oscillate frequently over the averaging time. That assumption seems valid in this case.

So many assumptions. Typically in the analysis of bending cantilevers (see this example, after Bernouilli) the forces at the constrained end (in this case, the threads) are not considered in detail. Rather they are treated as an idealized constraint. In his case, that the position and orieniation of the spindle are fixed. For example, in the diagram here, that the coordinates and slope of the spindle are both held at zero at the threads.

Well, that's already too much for one blog post. To be continued.

Sunday, October 18, 2009

Low-Key Tunitas Creek

Despite fears of the pumpkin-seeking-hoard wrecklessly hurtling their oversized vehicles down narrow Tunitas Creek Road, yesterday's Low-Key Hillclimb was a wonderful success.

First, Bike Hut. Bill, who runs Bike Hut and Potrero Nuevo Farms, met us and was super-friendly, extending us wonderful hospitality in allowing us to start our ride there. It made me wish we could start every climb at Bike Hut. I assure you as long as Low-Key and Bike Hut continue to co-exist, we'll be back. It's only too bad the ride options from that location are a bit limited: Tunitas is basically the only option. Lobitas Creek Road is an alternative start to Tunitas Creek Road, but it adds a dangerous, narrow descent on what's a bidirectional road.

Then the weather: early morning fog retreated to reveal blue skies and warm temperatures. Perfect!

Then the volunteers: what a great group we had! Registration went smoothly, results went smoothly, the course marshals were great, and we had fantastic photos from Josh and Christine. Josh overcame the challenging finish line lighting with is external flash: his images really capture the exact moment of riders crossing the line.

"One!" I shout my number as I cross the line, finally able to ride a climb in this year's series. Josh Hadley photo.

And the climb: I love this one! Tunitas Creek climbs gradually, then steeper, then gradual again on its run to Skyline Road. But the Star Hill - Swett detour, rarely ridden by most racing types, gives it all the more character. The detour avoids the section near the top of Tunitas Creek which was omitted from the county-specific repaving job done this past February, adding instead a narrow road with more climbing and a character-building little wall at the intersection of the two roads. There was a bit muck at the Tunitas-to-Star Hill turn which had me slightly nervous, and that little 19% wall can be dangerous if passing cars don't exercise caution (Christine Holmes had the amazing misfortune to have cars coming in both directions as she climbed it: she prudently bailed out and walked). But overall it worked very well.

Well, except for those living in the house at the finish line. I'm afraid we disturbed them with all of our number shouting. Apologies for that...

So a great day. Low-Key does take a lot of time. But the payback is so fantastic.

One example of the payback was when little Emma Stahl, racing in the under-5 junior category, reached the top on the tail end of her father Geoff's bike. It was her first bike race ever, and she clearly appreciated the heart-felt cheers as she approached and crossed the finish. That's what I love about Low-Key: Emma's out there working hard for close to two hours and she's as much a winner as those of us finishing in the 33-34 minute range. Really nice.

I didn't miss the chance to look at the pumpkins.

Tuesday, October 13, 2009

Pumpkins from Hell

It was clearly an act of inspired brilliance, not a brilliance of genius, but a brilliance which transcends genius in its impression: a brilliance of anti-genius.

I scheduled the Tunitas Low-Key Hillclimb for the weekend of the Half-Moon Bay Pumpkin Festival.


Okay, I try to be open-minded, and I'm sure there's more to it than this... but why are people willing to jam Highway 92 into a quagmire of creeping traffic to look at squash? Okay, the squash might tend to be largish, exciting I'm sure, but I still fail to embrace the thrill. Squash is good for soup.

In any case, it's too late to change plans now, so on we go to Tunitas Creek, and hope the squash seekers have better sense than to toss their over-sized vehicles onto the narrow, winding descent of Tunitas Creek Road.

Sunday, October 11, 2009

Low-Key Old La Honda

Wow -- Old La Honda road has worried me every since I decided to put in on the schedule. The locals don't, I'm sorry to say, have a reputation of being happy about cyclists riding up their beautiful road. And so many do -- at any given time on a weekend day there's probably around 10 of them on the 3.2 mile climb. It's perfect in almost every way: not too long, but not short; not too steep, but steep enough; decent pavement but not smooth enough to be without character; a steady enough grade to stay in a single gear all the way, but varied enough to shift if you want to; and shaded from the wind and the sun. Combined with the relatively low car traffic (maybe 2-3 cars pass on a given climb up), and it's proximity to the densely crowded peninsula, it's as close to cyclist heaven as you can find in our car-obsessed culture.

But despite a few hic-cups, everything went amazingly well. This was due to the excellent work of coordinator Doug Simpkinson and all the great volunteers. I really am grateful to all of them. And Clark Foy had his MetriGear Vector prototype there. I think that thing is cooler than cool, as my blog posts here have made obvious.

The biggest change we need to make is in timing. Two weeks in a row timing has caused issues: time go get beyond the "recording times on paper" dinosaur stage. I found this ap, which I downloaded onto my girlfriend's iPhone:



I love the name: Pencil Busters. Art Walker, former Stanford cycling coach, once said "no number should ever be written more than once". I agree, except even better is to never write a number at all. With this app, we can record times, annotate them with the iPhone (for example, invalid numbers, or even note rider numbers), then email them to my account. This will substantially facilitate results, as long as the iPhone battery holds out....

Anyway, we'll give it a try at Tunitas Creek this weekend.

Tuesday, October 6, 2009

Metrigear Vector Pt 5: Interpretation of Torques

Now for torques....

There's three torque modes for each pedal, or six for the pair, which I've described by using sums and differences of the corresponding torques for each pedal. I'll focus here on the torques for each pedal.

First, there's Tx. This is the torque driving the pedal spindle to screw into the crank arm. I say screw into because the handedness of my coordinate system is the opposite of the handedness of the threads: positive torque is counterclock for the right pedal and counter-clockwise for the left pedal, looking from the outward side of the pedal. I hope I got that right ... Since the pedal is spinning relative to the spindle, and this induces a force, this torque should generally be somewhat positive. How positive depends on how effective the pedal bearings are.

Work can be calculated as the integral of torque with respect to angle. So <Txl + Txr> × cadence, where <> denotes an average with respect to rotation angle (not time), is proportional to the power being lost to bearing friction, at least assuming pedal deflection is small (if the pedal deflection were large, it becomes too complicated for my simple level of analysis). In other words, it assume the pedal and crank are mostly rigid.

A further complication is that the average of the torque with respect to the angle can't be determined. The relevant angle is the angle of the pedal body with respect to the spindle (and therefore with respect to the crank arm). We can determine the angle of the spindle (and crankarm) from the accelerometers, but we then need to assume an orientation for the pedal body relative to space. For example, assume the pedal body is always at a fixed angle, for example always horizontal. If the pedal body is at an angle which changes throughout the pedal motion, as it always will be, then its bearings will sometimes be spinning faster, sometimes slower, than the crank arm. This will introduce an error.

Moving onward: there's two other torque components for each pedal: Ty and Tz. With the right pedal at 3 o'clock and with the pedal body horizontal, Tz is similar to Tx in that it describes rotation about an axis which is nominally free. As the foot rotates about the pedal body, on an X-series Speedplay there should be only a small restoring force (the "pedaling on ice" feeling people love or hate about these pedals). On a Zero, there should be a restoring force dependent on the tension setting.

However, it gets more complicated. One is the Vector measures a torque, but not the angular motion, and you need both to calculate a power. It has no idea what the shoe is doing relative to the pedal body.

But another issue is the assumption that the pedal body is horizonal. Of course, it could be at any angle. Again, the vector has no idea. So I'm not sure how useful Tz turns out to be.

Again with the horizontal pedal body at 3 o'clock, Ty would be a measure of how uniform the propulsive force is between the inside and the outside of the pedal body, or at least how uniformly this applied force is transmitted to the inside and outside bearings at the spindle. Interesting? I don't know. But things become more complicated as the shoe rotates and as the pedal body rotates. Whoa -- all too much for my small brain!

So that's the torques. About the only interesting torque mode to me is Txl + Txr, which may be able to estimate power loss in the pedal bearings. That would be really cool.

Sunday, October 4, 2009

Metrigear Vector Pt 4: Interpretation of Forces

Okay, so back to the Metrigear Vector for a bit.

What I discussed before was how the MetriGear Vector is trying to move the emphasis from power to force. Get force right, get velocity right, and power is automatically right.

I mentioned how one can derive twelve components of force. One of these components, Frz-Flz, drives the chain, and therefore provides propulsive force. It is assumed none of the other force components are propulsive.

I need to make a disclaimer here: I don't work for Metrigear, so can't be sure they make the same assumptions I do. I'm just running with the idea and saying what assumptions I might make.


Axes for pedals at 3 and 9'oclock

So, other than the propulsive mode, I think the next most interesting mode is Frz+Flz. This describes right-left imbalance. Obviously at any point in the pedal stroke, the right will be applying more or less torque to the crank than the left. But averaged over the pedal stroke, the difference should, if the pedal stroke is perfectly balanced, average to zero. If this value is positive, then the right foot is doing more work than the left. If negative, the left is working more. Obviously it will never average to exactly zero. What it it's not zero? That's a good question. But I can easily imagine a rider might want to focus on engaging the less active side more.

Here I'm discussing force on the pedal. But what drives the chain is torque on the crank. The relationship between the two, for the moments consisting of Frz and Flz, is a factor equal to the crank length. Now, Metrigear says the Vector needs to be updated with the present crank length. This is interesting, but basically it uses crank length to convert force into torque. Torque becomes power by multiplying by angular velocity, which is equivalent (for pedals moving in a circle) to cadence.

There's other forces, however, which could be measured by Vector. For example, Fx (on each pedal) is a force applied along the spindle direction. This is equivalent to pushing out with a show (for a positive value) or pushing inward with a shoe (for a negative value). This may be very useful. For example, when cornering, it's conventional practice to push down with the outside foot, using that pressure to help control the bike. There will then be a component of Fx to the outward pedal.

However, for straight-line pedaling, simple-minded thinking suggests Fxl and Fxr should be small. If there were a generally positive value of Fxl and Fxr, that might suggest the cleats were mounted too far out, it seems. Or a negative value might imply the cleats were mounted too far in. In each case the foot might be struggling to find a more comfortable lateral position. Improper cleat placement can lead to torquing the knee, which leads to knee pain. I think one of the reasons some riders claim Speedplays lead to knee pain is that Speedplays have no lateral float (only rotational float) and thus proper setting of lateral cleat position is critical. Not everyone is going to get it right. An alternative to moving the cleat is to move the pedal. For example, spacers may be used at the crankset, or a crankset of a different "Q-factor" could be used, or Speedplay sells spindles in different lengths. I'm not sure you'd be happy about swapping spindles if you'd just dropped the better part of $1k on Metrigear Vector spindles in the wrong length, though.

Then there's Fy. I've already discussed Fyl + Fyr, pushing forward on the pedals. FylFyr would be a measure of pushing forward with one pedal and backward with the other. I'm not sure how to interpret these forces. Some would say any force not involved with direct propulsion should be minimized. But I don't believe there's any evidence to support this simplistic view. So it will be interesting to see how numbers like these get used.

One more comment: I'm being sloppy with coordinates, since I'm assuming pedals in the 3 o'clock and 9 o'clock position. But pedals move in circles in the frame of the bike. In this frame, the x-direction as I define it stays x. I'll assume the z-direction always points in the direction of propulsive force. Then the y-direction, if it is to remain orthogonal to x and z, will not always point forward. In my discussion of the forward force, then, a measure of the force pointing forward would need to be a linear combination of the x and y forces, with coefficients determined by a sine and cosine of the crank orientation in the frame of the bicycle (although in the frame of the spindle, forward is always the same direction) The actual Fy values may be less meaninful than parameters like these derived in part from Fy.

Next time: the torques.

Low-Key Montebello

Well, we had some hiccups with results.... a malfunction on the timer causing times to be deleted, then some mis-shouted numbers at the finish. The combination made for a tough time untangling the results, with heavy reliance on photos. Ron Brunner saved the day by recovering results from the video.

I really want to have a back-up position for double-checking numbers of finishers. Someone to check the number with the stem sticker to make sure everything is good. Good numbers are by far the most important thing in good results, especially on popular climbs where other riders are passing by, making photos tough.

We've done helmet stickers to help with this before. Maybe time to break out that option again. Or a ribbon tied around the handlebars.

Anyway, week 1 results are posted. Still preliminary: two riders still unidentified.

Next week may be a challenge. We'll be starting in small groups to mitigate traffic impact. Turn-out might be on the high side, so we'll really need a few extra volunteers for this one.

In any case, it was a fun time, and it's great to see the series thriving! Thanks to everyone who has helped give the series such a positive start for 2009!

Friday, October 2, 2009

MetriGear Vector Pt 3: Forces and Torques

A single point mass can aquire three independent momentum coordinates, for example along principal axes x, y, and z. The macroscopic rigid body (i.e. not compressable or deformable) can additionally aquire momentum associated with rotation about these axes. With each momentum component, there are associated force (for translation) or torque (for rotation) components.

For modeling purposes, we can consider each pedal a rigid body. There are then two times six equals twelve total degrees of freedom, constituting six force degrees of freedom and six torque degrees of freedom.

For one pedal, the forces and torques associated with these degrees of freedom might be listed:
  1. Fx
  2. Fy
  3. Fz
  4. Tx
  5. Ty
  6. Tz
where Fx is the force through the center-of-mass along the x-axis, Tx is the torque about the x-axis passing through the center of mass, and similarly for the axes y and z.

For two pedals, one can then define separate forces and torques for the left and right pedals:
  1. Flx
  2. Fly
  3. Flz
  4. Tlx
  5. Tly
  6. Tlz
  7. Frx
  8. Fry
  9. Frz
  10. Trx
  11. Try
  12. Trz
where Flx is the force through the center-of-mass along the x-axis of the left pedal, where Frx is the force through the center-of-mass along the x-axis of the right pedal (note this is a different x-axis; it points in the opposite direction), and similar for other axes and for the torque values.

A side comment: it's not trivial that two pedals would have twelve degrees of freedom. Treating the single pedal as a simple rigid body is an approximation. But the pedals aren't independent of each other; they are constrained in their motion by the crankset. Additionally, while the rigid body may have six degrees of freedom of momentum, it does not imply only six force components are possible. Higher order force components may also be applied. For example, you could apply a positive-z force to the center of a pedal body and a negative-z force along the body perimeter, net force zero. But these higher-order force moments are perhaps of less interest than the moments listed here, and in any case may be unmeasurable by piezoelectric sensors in the spindle. On the other hand, I believe it's plausible (but I don't know) that the Vector would be able to report the force components described here.

So these may be a reasonable set of "measurables". However, rather than use these values, the degrees of freedom can also be quantified using any orthogonal linear combination of these values. For example:
  1. Frx + Flx
  2. FrxFlx
  3. Fry + Fly
  4. FryFly
  5. Frz + Flz
  6. FrzFlz
  7. Trx + Tlx
  8. TrxTlx
  9. Try + Tly
  10. TryTly
  11. Trz + Tlz
  12. TrzTlz

The advantage of this partitioning of the twelve degrees of freedom of the two pedals is that it makes clearer that one and only one degree of freedom is contributing to propulsion. Recall that it's being assumed the right pedal is at 3 o'clock, the left pedal at 9 o'clock. So in this instance, a positive-z force on the right pedal with a negative-z force on the left pedal rotates the pedals and drives the bike forward. This mode is described by the component Frz-Flz.

Since the bike moves in the positive-y direction, you might think a positive-y force on both pedals (Fry+Fly) would also drive the bike forward. However, I already mentioned it's being assumed here that Newton's Third Law implies a counter-force is being applied to the saddle and/or handlebars, and thus in the net the bike does not experience propulsion from such a force.

Okay, so there's the propulsive force. More on the other forces and torques later.

Thursday, October 1, 2009

Low-Key Hillclimbs begin on Saturday!

Okay, a bit of a change in topic....

The Low-Key Hillclimbs begin this Saturday! Very exciting. Week one is Montebello Road, an excellent road for a hillclimb: basically dead-end with a bit of steep (but not too much) and some gentler slopes in the middle to give beginners some recovery but test the pacing strategy of the stronger riders:



I won't be able to ride any of them myself until week 3 (I'm working the first two weeks), but it's always great to be out there any enjoying the day, generating data.

If you want to join in on the fun, but don't want to climb, please go to the website and volunteer to help! We can always use help at registration on week 1 of the series.