Alberto Contador On His Final Angliru : Climbing Speed & Power to Weight Ratio

This post is modeled on the calculation method shown in a past post where I calculated Contador's VAM and power to weight ratio on the Angliru during the 2008 Vuelta a'Espana.

Today, Contador won the final mountain stage of the Vuelta, and again on the Alto d'Angliru.

My preliminary calculation suggests that Contador climbed 950 height meters in a time of roughly 35 minutes. I clocked his climbing time from the 9.3K to go mark.

The estimation is therefore :

The Ferrari method to estimate power to watt ratio is therefore :

His VAM today is less than my estimation for his 2008 VAM (done for the last 4K and quite high due to road steepness in the last sections), his power to weight ratio is also less than my estimation for his 2008 power to weight ratio. 

However, reductions seem reasonable for a man at the twilight of his career and do not make room for suspicion. The power to weight ratio displayed at the end of a grand tour is remarkable nevertheless.

This is a clean performance unless further data instructed otherwise. 

The calculations are based on the following raw data.

Wind & Altitude : Effects On Power To Weight Ratio

The following, divided into 3 parts, is a simple physics exploration of wind and altitude's independent effects on power to weight ratio, a subject that has not been treated by any cycling book so far.  Independent effects are important to understand in order to get a feel for how they affect power to weight ratio when acting in combination. The climb chosen is the famous Col du Tourmalet which is featured in this year's Tour. Assumptions in the analysis, for simplicity, include constant speed, constant grade and a formula for frontal area developed by Bassett & Kyle. Validation is done against calculations done by Alex Simmons (see conclusions) for wind vs power to wt results. Power models from Tom Compton's "Analytic Cycling" site and data from prominent researchers are compared with for altitude vs power results.

Courtesy : New Scientist

We have explored VAM calculations  here and here, However, climbing rate by itself does not provide an understanding of the effect of transient wind and altitude on performance. The wind and altitude can make a big difference. On the other hand, one could also argue that the effect of switch-backs and spectators along the road may effectively cancel out the contributions from head and tail winds. I'm not sure how much that contributes but that can be a topic in the comments section.

 
The equation explored in a previous post to understand climbing rate explored only the power needed to increase a rider's potential energy. Lets call this A. If W is total weight of bike and rider in Newtons, Vg ground speed in m/sec :

A = W.V_g.sin[arctan(grade/100)]  Watts

While this is a big piece of the power pie for climbing, there are  couple of other elements to the power equation. One deals with rolling resistance at low speeds. Let's call that B. If Crr is the co-efficient of rolling resistance (typically 0.004 or so for a 100psi tire)

B = W.V_g.Crr.cos[arctan(grade/100)]  Watts

Both A and B are constant if grade and ground velocity are steady.

There's a third element that comes to play which is the effect of wind and density. We can call it C. This is composed of frontal area A and associated drag force. If Cd is the co-efficient of drag (0.9 typical for cyclist on hoods) and this multiplied with area gives drag area, then :

 C = 1/2.C_d.Area

Say very little change happens with the cyclists' drag area (Cd.A where Cd is coefficient of drag and A = area), velocity V and say, density altitude is fixed as p. Let's also assume speed to be steady and gradient of slope to be an average. If  Vw is the wind speed (negative for tailwind) and Vg the ground speed, then not accounting for the tiny drive-train losses in your super efficient bicycle, the power to climb will then be a function of relative velocity as follows :

Power to climb, P  = A + B + C.p.Vg.(Vg + Vw)²  Watts

Let's work with just the third term, calling it Term. Its interesting to see what happens when you normalize Term by the the cyclist's ground speed on the climb. Dividing and multiplying throughout by Vg to change its form but preserve its meaning, we get : 

Term  = C.p.Vg³(1+ Vw/Vg )²

  => K.p.(1+ Vw/Vg )²

where K is defined to be K = C.Vg ³

Vw/Vg is called velocity ratio. Now you can see that when there is no headwind, Vw/Vg =0 and power becomes a function of K and p only. 

PART 1 : Power Variation With Velocity Ratio While Keeping Density Constant

Let's look at how that wind term varies with a change in velocity ratio. The density of the air on Tourmalet is kept uniform for this exercise, at a value of 1.06 kg/m³

When a headwind picks up, the velocity ratio increases from its baseline of  0. The following geometric growth is seen in Term :

• -When headwind is 1/10th of cyclist's forward velocity, Term is an extra 21% compared to no wind condition. 
• -When headwind is 1/4th of cyclist's forward velocity, Term is an extra 56% from no wind condition.  
• -Similarly, when it is half of cyclist's forward velocity, Term is an extra 125% from no wind condition and you must supply 44% more power from case 3).  
• -At an extreme case (assuming you're traveling atleast faster than 5mph), (1+ Vw/Vg) could be highest when the headwind is same as your forward velocity. When that happens, 1+1 = 2 and 2 raised to itself = 4 meaning that all things kept constant, Term is more than 4 times its original value, a 300% increase.  

But this component of power, Term, is added to A and B to produce power, P. Hence, percentage increases shown above don't mean the sum power will also increase by same percentage. So for the last case above, P does not increase by 300%. When one does the math properly (unlike my poorly done first edition of this article), the increase in P from a no-wind condition is actually  :

300% x [ Kp/(A+B+Kp) ]  

Or generally, if all other things are kept the same, 

% increase in power, P%. = [(1+ Vw/Vg )²-1].[Kp/(A+B+Kp)] 

increase in absolute power = P%.(A+B+Kp)

 

We know how to calculate the first part.  The second term, Kp/(A+B+Kp) can be called the Correction Factor. Once this puzzle is solved, we can solve for the proper required increase in power.

Without going into too many details, here are graphs I produced. Maximum height and weight were extracted from the Navy Seals' website. These are some of the fittest people in the world and they were a good choice because it makes no sense to work with data of unhealthy, overweight people. (Note, their dimensions are not like Tour contending emaciated cyclists but it should be close). Frontal area was then calculated using Bassett and colleague's formula that was retweeted by Jonathan Vaughters.  The speed is kept constant at 10mph.  p is kept constant at the average of 1.06 kg/m3. The gradient was uniform at 7.5%, with no change. 

FIG 1 & 2 (above) : First and second plots show the wattage expended to sustain 10mph on the Tourmalet for men and women. The numbers are shown for potential energy and rolling resistance components. Sanity check is to note that wattage needed to counter rolling resistance, for a constant velocity but varying rider weight (shown in red line), is pretty small compared to the wattage needed to fight the grade which is true.

FIG 3 & 4 (above) : Frontal area across riders vary with height and weight. The above plot shows how C varies for men and women according to their dimensions. These dimensions are the maximum accepted in the Navy Seals, who are among the fittest people. Frontal area was calculated using a formula developed by scientist Bessett & Kyle. The area is an approximation ofcourse, and does not account for changing positions on the bike. 

 FIG 5 : The last plot is the power correction factor plot, the puzzle I was seeking to unravel before. Does it make sense? Below, I describe how to use this plot to calculate percentage increase in the total power requirement. 

Going back to the question we were asking earlier : How much increase in power is required to combat a 10 mph headwind (velocity ratio of 1) for a 70 kg male Navy Seal cycling at 10 mph (4.4704 m/sec)? 

Here's how to solve this question. Looking at the figures above, the corresponding numbers for his weight are:

A = 253 W (power due to grade)
B =  15 W (power due to rolling resistance)
C = 0.14 m²
Corresponding correction factor for his weight is ~0.047.
Kp = C.Vg ³p = 0.14(4.4704³).1.06 = 13.25 kg.m²s^-3

Solving using the equations shown earlier in the post,
   

 % increase in power= [(2)²-1].[0.047]~ 0.141 (14.1%)
 increase in absolute power= 0.141(253+15+13.25)~ 40W
 increase in power to weight ratio = 40/70 = 0.57 W/kg

Similarly, we can quantify how much extra boost one would need per kilogram of body weight for different velocity ratios. 

Here's a case scenario for a Tour de France contender on the Tourmalet :
Weight = 65 kg
Height = 1.78m
Power to weight 6 W/kg riding
Desired speed = 10mph.

The plots are shown below : 

 FIG 6 : The operating zone of a Tour contender can, for example, be 6.0 W/kg. This example is for a rider weighing 65 kg. Other parameters assumed are shown in red. In an ideal case, the rider traces a straight horizontal line on the 0 point (y-axis) when there's no wind. Any additional wind will force him to trace horizontal lines as he pushes his body harder. Horizontal trace will move vertically up on the plot with wind change.

The above plot perhaps proves why it pays to prepare the extra mile for July. If 6.4 Watts/kg is what a person needs to win the Tour, perhaps only 6.2 is really needed. The additional 0.2 W/kg is used as insurance, against attacks, accelerations and misfortune winds. Very rarely are riders gifted as such to keep doing this over the course of 3 weeks.

Note : You could also have a tailwind of similar proportions, except that now the (1+ Vw/Vg) term is modified to (1- Vw/Vg). A tailwind of the same proportion as your forward velocity means 1-1 = 0 which multiplied to K effectively cancels your need to combat any wind. It is free speed which you can devote to the terms A and B.  The funny part is that a tailwind never gives you back the "same amount" of speed as the headwind will take away from you. Its a law of nature.

PART 2 : Power Variation With Density Change As Velocity Ratio Remains Constant

Previously, we looked at change in power to weight ratio with change in wind speed. 

Fact of the matter is, air density does change appreciably with altitude as you ascend.  So this term varies with density as climbing progresses :

Term = K.p.(1+ Vw/Vg )²

where K is defined to be K = C.Vg ³ 

The Standard Atmosphere chart, which I pulled out of a book and approximated by a polynomial fit in Excel, tells me the following :

1) At an altitude of 500m (1640 ft), the density is 95% of what it is at sea level. That's a 5% decrease.
2) At an altitude of 1000m (3281 ft), density has decreased by 9.2%. 
3) At an altitude of 1500m (4921 ft), density has decreased by 13.5%. 
4) At an altitude of 2000m (6562ft) , density is now only 82% of its sea level value, a decrease of 18%.

Air density is gently geometrical for the first 1 or 2 miles of the earth's atmosphere. Here is how it behaves :

Like before, I derive the % decrease in power requirements due to altitude with zero headwind (Vw/Vg=0). Let L be the amount representing % of sea level density at any given altitude. If po is sea level density,

 % decrease in power =

[1- (L/100)] / [1+ (A+B)/(Kpo(1+Vw/Vg)² ]

The effect altitude and hence air density has on relieving a rider is very gently geometrical, and not linear. This is proven below. The linear trend-line as you can see doesn't fit 100%.

FIG 7: The number on the y-axis shows how much power to weight ratio one can potentially save while climbing at higher altitude. This is because the higher the altitude, the lesser is the pressure and density of air. This is also why World Hour Record attemps are often done in stadiums that are located high above sea level. Beyond 2500 m ofcourse, advantages in power savings are countered by the thinning of the air. 

FIG 8 :  This diagram simply shows to what degree headwinds and air density contribute to power requirements.

Part 3 : Conclusion & Validation

If wind conditions changed as climbing progresses, density decreases. Whereas wind exponentially increases the power requirements, the density moderately decreases it since you require lesser effort to propel in a less dense medium. But wind is the biggest power soaker.

Here is an interesting chart showing the effects of wind on power to weight ratio given finishing times for Alpe d'Huez. This legendary climb is much more shorter and steeper than Tourmalet but located at lower altitude. It was compiled by Alex Simmons, a cycling coach from Australia.  Perhaps my charts can supplement his or vice versa? It is really well made. 

Simmons' plot can be a sanity check in that power to weight ratio needed with wind on Alpe d'Huez closely correspond with mine for the Tourmalet. I must however add that I have not considered the losses due to drivetrain in my model. For example, a +2.5 mph headwind is a 0.25 velocity ratio. Corresponding additional power to weight ratio (Fig 6) is about 0.13 W/kg from the no wind condition. Assuming a bike is 95% efficient, 0.13/0.95 = 0.14 W/kg. This is my required increase.

Simmons' model for the steeper Alpe d'Huez, on the other hand, says that it should be +0.5 W/kg.  Is this +0.5 W/kg an overestimation, considering that a 2.5 mph wind is classified as just "Light Air" in the Beufort Wind Scale?

If 70 kg rider is producing 6W/kg, he is riding at 420 Watts. A +0.5 W/kg increase means he has to propel himself now at 455 Watts! Phew!

I'm not surprised it is higher than mine since the grade is steeper and my rider weight is 65 kg, although my choice for bike weight is same as his. Also, an important thing to note is that Alpe d'Huez is at lower altitude than the Tourmalet. Hence, it is more than likely that a rider has to put forth more watts on the Alpe than on the Tourmalet. Remember the mildly geometrical reduction effect of density decrease?

Another interesting bit to notice also from the plot above is that a cyclist climbing Alpe d'Huez doesn't get back the same relief from power to weight ratio with a tailwind as the extra amount he has to spend in a headwind. Its funny how that works. Its like climbing hills. You don't get back the time you spent going uphill by going fast downhill.

How about verification of the altitude's effect? First, Analytic Cycling's "Forces on Rider" model shows the following power requirements at the bottom, mid point, and top of Tourmalet. Speed was kept constant at 10 mph and all other parameters were made to match my choices before.

You can notice a 1 Watt power demand reduction just by riding from the bottom of the Tourmalet to approximately its midpoint (a 600m elevation change). Not much really in the overall scheme of things since that's a 1W/65kg = 0.003 W/kg decrease. My model (Fig 7) says 0.016 W/kg is the reduction so they're pretty close. From 1400 to 2200m, the reduction is again 1 W, or 0.003 W/kg for the Tom's model. My model says 0.02 W/kg.

Secondly, although data is severely lacking with regards to the effect of altitude on power to weight ratio, a couple of other researchers in the exercise physiology field looked at how aerobic power varied  for 4 groups of runners, not cyclists. They estimated that aerobic power as a percentage of that at sea level, signified by "y", drops off wrt to elevation by the following relationship :

Below, I have plotted those 3 estimations along with mine (65 kg rider producing 6W/kg) upto an altitude of 2200m (the max height of the Tourmalet).

What we see above is apples and oranges. Mine calculates the relief in effort to maintain the same speed due to decrease in altitude density while the others have calculated the decline in power output due to a decline in VO2 max output. They really don't co-relate, do they (advantage vs degradation) ? But is it fair to say that the net effect of both of these phenomena is a curve in-between them?


I must conclude by stating that Alex's theoretical calculations for the wind effect matches closely with mine but this is not the same case for the altitude effect (see above). However, it remains to be said that all our numbers remain to be validated from real powermeter data and CFD simulations for cycling. I'm curious to see how both wind, and high altitude, affect a cyclist in the real world and I'll bet the changes are highly non-linear in nature.  Discussions ongoing at Cycling Forums.

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The Rate Of Climbing Uphill Explained

I've noticed that some cyclists who have been introduced to the concept of the rate of climbing uphill falsely think that Michele Ferrari pulled this out of his own pocket, as in, he invented it or something of that nature. Negative. He only went ahead and popularized the idea, putting a confusing trademark name to it such as "VAM" and developing some of his own methods to look at it with respect to cycling performance. But the idea itself is rooted in centuries old elementary geometry and Newtonian physics. Click here for the history of vectors.

Let's have some *serious* fun with climbing rate. You'll start to see simple relationships that may finally make some sense to you as a cyclist. If you love analysis, this may be for you.

I'll show 5 different ways to study vertical speed, or rate of climbing. 3 of them are using math. The 4th involves using some field work with math. I'll show you what Tom Danielson's climbing rate was for the record ascent up Mount Washington and I'll also show you an error analysis done on it with an example, which is pretty important on anyone's estimations. Finally, the 5th method shows how using some devices do this all for you instead of number crunching.

Let's begin. Enjoy over a cup of tea. If you see something fishy, I welcome you to call out any errors in the proceedings.

METHOD I - A Power Perspective

The overwhelming portion of total propulsion power in Watts needed to just climb uphill on a bicycle is given by the following relation :


9.81 = acceleration due to gravity on earth's surface (meters per second squared)
M = total mass of a bicycle and rider (kilograms)
Vg= ground speed (meters per second).
G = grade of hill, expressed as a fraction = (Rise/Run). To see a graphic of what rise and run mean, click here.

This is only power for climbing and does not include power to overcome wind and rolling resistance, drive-train losses or power for acceleration.

Note 1 : For steep grades, instead of simply (Rise/Run), G should be replaced with :


To put into perspective, at a steady 10% gradient, error % between the small angle approximation of (Rise/Run) and the real formula above is about 0.5%. At a steep 20% grade, it is 1.9%. At a near to impossible 30% grade, it is 4.2%. At a vertical asphalt of 45% grade, the error is way off at 9%.


Rate Of Climbing Broken Down

Now if I were to take the above power equation and break it down into 2 little packages, here is what each package would mean in a practical sense.


In other words, power equals the product of a force (total weight acting downwards) and vertical speed, which in other words is the rate at which you cover vertical distance.

Let's take the second package, the climbing rate.

This is it. Its a beautiful equation. Its units are in meter/second and you can convert it to meter/hour. 1 meter/second = 3600 meters/hours. Its also called VAM by Ferrari, which may be a pretty confusing term to people.


Practical And Theoretical Limits Of Climbing Rate

The equation for climbing rate, with G substituted, is :


If we chuck out the sine function using an overhead crane and plot it, it looks like the following :

Fig 1 : The Climbing Rate Curve for a ground speed of 1 m/s. Cyclists climbing on human power alone can only use a tiny portion of this curve above y=0, ranging from 0% grade to 40-45% grade.

In the linear yellow region, any point on the curve can be given by the equation (rise/run) with negligible error. I have chosen about 10-12% for the upper limit of the linear portion. For any hill grade above this, its better to represent G in the power equation with "Vg.sin[arctan(rise/run)]" than just "Vg.(rise/run)" to avoid errors.

Also notice the point on the curve that signifies 45 % gradient. It is impossible to ride efficiently above 40%. After 45% is the curve signifying landslide possibility, which cannot be attempted by any cyclist. This is the upper limit for practical climbing rate.

Observe the upper and lower limits of the curve in blue. They are called asymptotes because the curve tends to approach towards them but never reaches -1 or +1.

The effect of the ground speed term Vg (other than 1m/s), when multiplied to this sine function f(x) will be to stretch out f(x) like a rubber band and extend the upper and lower limits of the asymptote to -Vg and +Vg. What this is telling us is that if a cyclist's ground speed is, say, 5 m/s (11.2 mph, 18 kph), his climbing rate cannot go below -5 m/s (-18,000 m/hour) or above +5m/s (18,000 m/hour). These are the absolute upper limits of theoretical climbing rate for his given ground speed of 5m/s. Its the heaven and hell of climbing. Both are impossible. (Whether hell is above or below is upto you to decide. I think in climbing, hell should be above!)

Note 2 : If grade kept increasing and increasing, do you think gravity will actually allow you to keep climbing on your own power? In other words, would rate of climbing keep on increasing with step increases in % gradient, as Ferrari's website may have you believe? Yes, it does. But there's a limit where we can't go further and vertical speed drops to zero. As you ascend uphill your muscles have to supply the power to increase your potential energy. It doesn't come from thin air. Eventually, you will become tired. Your mechanical gearing advantage will decrease as you have a finite set of gears. Your speed will decrease exponentially and at some critically steep grade possibly 40% or more, your velocity will be reduced to near zero and you can roll backwards or fall. It doesn't become efficient to propel yourself anymore. See Fig 2 below. This is why I argued in the past that on very steep grades, it is more practical energy wise to get off your bike and walk. Why? Because the speeds are more or less the same cycling or walking!

Note 3 : Human muscular power is also very different from electrical motor power. The capability of a human to deliver bicycle propulsion is a function of time before becoming exhausted, also called endurance time represented by the symbol tau (T). Every individual has a power curve (Watts vs Time) that generally curves downwards from left to right. It tells us that high power can be sustained for lesser time than lower (but slower) power output. It is very unlikely that a high climbing rate can be sustained for a very long time through human power alone (unless you're some freak). This can be seen in the graph below.

Fig 2 : Pick a certain W/lb and while staying on the horizontal line, notice how your ground velocity Vg decreases drastically as you move towards the sloping lines representing higher grades.

Relationship Of Climbing Rate To Grade Inside And Outside Linear Curve

In the linear portion of Fig 1, a 10% relative increase in grade in the linear region (say from 7% to 7.7%) should theoretically result in a 10% increase in climbing rate. In other words, climbing rate is directly proportional to the vertical ascent and inversely proportional to the horizontal length of climb. So if we kept ground speed and run constant, a higher rise leads to higher climbing rate and vice versa. Say we doubled the ascent, then climbing rate is double the initial rate. Conversely, if we kept ground speed and rise constant, a longer length of climb will decrease climbing rate or vice versa. Say we doubled the length of the climb, then the climbing rate is halved from the initial rate. If we halved the length, the rate is doubled.

Outside of this linear curve after about 10% gradient, a 10% relative increase in grade (say from 17% to 18.8%) should only result in a 3.8% increase in climbing rate. So even though climbing rate increases, it doesn't increase as much due to the nature of the curve above. You can see how its trying to level off.


An Example : Andorre Arcalis 10.6 km, avg. 7.1 %, catégorie HC

We can take the profile information of the ascent to the ski resort of Arcalis and compute our climbing gradient for each kilometer using kilometer specific gradient and kilometer specific ground speed. The grade isn't steep, hence you can use G = (rise/run).

Do this for all 10.6 kms, add them up and take their average. You should end up with an approximation of your total climbing rate for the mountain. Here is first example done for Kilometer 1 :

Fig 3 : Km by Km Calculation Of Climbing Rate. Click to zoom in.

METHOD II - A Geometric/Vector Perspective

If you remember some vector theory, you know that any vector can be resolved into component vectors. In our 3-D world, the velocity vector of a cyclist uphill can be resolved into an x-component, y-component and z-component velocities. For the sake of simplicity, if we looked at it in 2-D, it would look like the following. The upward velocity component is the climbing rate or vertical speed. Refer to the equations on the right side to solve for these Cartesian component vectors.

Fig 4 : Resolving Ground Speed. Click to zoom in.

Try to visualize this picture of blue, red and green arrows in your mind as you imagine the cycling moving uphill. The length of each vector signifies the magnitude of the speed. As the cyclist changes speed or accelerates, the length of the ground velocity vector changes in real-time and so will its components since they're all connected. Once the cyclist approaches the downhill section over the other side, the direction of ground velocity vector points downward and its length increases because of the action of gravity and the quickening of pace (until terminal velocity is reached or above). The vertical component will then point directly down and its length signifies the descending rate.

Note 4 : In the real world, since we have a third vector for the lateral swaying/zigzagging motion as you climb a hill, the climbing rate calculated from 2-D resolution of vectors would be some percentage off from the real value. This is the error resulting from simplification.

METHOD III - A Time Perspective

Suppose you don't care for any of the above nonsense to calculate your climbing rate, you can still find it out using your time. Go out to a climb like Mt. Washington in the United States and time yourself using a stopwatch. For example, back in 2002, Tom Danielson set an unbeaten record of 49'24" for the 1432m ascent at 11.9% ave.


Plug in what we know for Mt. Washington. Ascent = 1432 height meters. Climb time = 49.4 minutes.

Tom's VAM estimate from time comes out to 1739 meters / hour. If you want to break his record, you need to aim for round about this vertical speed. Else... you can sit at home and eat ice cream.


Climbing Rate Error Analysis

Now we all are human beings. Our stop watches are not very precise instruments. And when someone tells you that the ascent in meters of a mountain is "x", what is the error in it and how does it propagate into the final climbing rate calculation based on Method III?

Well, that is simple. Here the error equation :

where :

δClimbing Rate = Error in climbing rate (meters/hour)
δAscent = Error in ascent (meters)
δTime = Error in time measurement on a device.

Lets put this into perspective. Lets just suppose that the person who measured Tom Danielson's climbing rate was 50 meters off in his estimate of the total vertical climb and there was a time measurement uncertainty of 0.2 seconds or .003 minutes (20% of least count of a watch of 1 sec is a good rule of thumb). If we plug this into the error equation, we get :


What this is telling us is that because of the uncertainties in the vertical rise of the climb and time taken in our example, the uncertainty in Danielson's climbing rate (or VAM) is 103.8 m/hour. The relative error is then a 6% error in the initially calculated figure of 1739 m/hour.

This is just an example, mind you. I did not declare that someone estimated the ascent 50 meters off the real value. Yet, this shows that its important to do an error analysis on anyone's climbing rate measured out on the field. Michele Ferrari does not show an error analysis on his data presented on his website concerning this year's Tour de France. People are bound to look at inflated values and declare, "HE DOPED, HE DOPED!! CATCH HIM!"

METHOD IV. An Analytic Perspective

If you still want to dig deeper into some relations using the power of partial differential equations, see Dan Connelly's page on an analytic treatment of climbing rate. You get more mathematical ideas and can do more cool things with it. Dan is a semiconductor engineer from California and like all engineers, loves his number crunching.

METHOD V - A Device Perspective

If you think math and science are beyond you, then spare some coin and try a nifty little device called a variometer, or vario for short. Paragliders use wrist-mounted variometers to check their vertical speed and altitude. This is an expensive but lightweight gear. It uses electronic pressure sensors to sense the change in pressure around them. Since there is a correlation between pressure and altitude, it is possible to calculate instantaneous rate of ascent or descent. For example, between the foot and the peak of the Arcalis climb, there is approximately 0.08 kg/cm2 of pressure difference.

The one shown in the picture, made by a company called Ascent, displays vertical speed with adjustable averaging (m/s or ft/min). Its resolution is 0.1 meters/s or 360 meters/hour. It can record data for 200 'flights' and has a rechargeable lithium-polymer battery with life of upto 10 hours or 4 hours standby time. Because a cyclist's vertical speed range is so small in comparison to a paraglider (the speed range of a paraglider is typically 20,000–60,000 metres per hour), such a device may even be impractical. Here's a video of how it works. I have not used one so it'll be interesting to find out whether they work in any useful manner for a cyclist.

Fig 5 : Ascent Vario, $250-300

Another option is the Avocet Vertech II Alpin which is built for mountain hikers. It is claimed it can show you current climbing rate of range 0 to 28,000 feet per hour in with 100 ft/hour resolution or 30 meters/hr. See other features here.

Fig 6 : Avocet Vertech II Alpin, $250

A reader also commented that Mavic's Wintec Ultimate cyclocomputer that costs about $200 can also show you vertical climbing rate, but I cannot confirm this.





ADDITIONAL RESOURCES :

Power To Weight Ratio
Air Pressure And Altitude Above Sea Level Reference Table
The Standard Rule For Climbing Bragging Rights
Contador's Climbing Rate & Power to Weight Ratio on L'Alto de Angliru (Spain)
Error Propagation

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Old Willunga Hill

Our hearts may be full of love and warmth for Oz, yet our wallets may be obscenely empty.

For those of you who can't afford a decent visit to "Your Chance To See The Old Man", hop on the Beehive teleportation system. We'll try to tour the biggest hill of the TDU : Old Willunga Hill Road.

To get to Down Under, we'll probably have to first get the hell out of 'Up Over'. How will we do that? Simple. Give a spank on that globe, let it dance around and voila! Embrace the Southern Hemisphere!

Please, do not try to jump into Australia. This is not drawn to scale.

Patience now. Lets zoom into the wonderful land of Kangaroos.

Calm down. I see some of you are getting excited...

We're now approx. 100,000 feet above Willunga, in the sunny state of South Australia.

Please calm down, we'll get there. Do not eject or do a Chesley Sullenberger either...

Our Chinese digital zoom frankly sucks. Switching to European opticals.........

.....aha, there's Willunga!

Do you know Willunga is known for its almonds? Sadly, the Aboriginal word for Willunga does not mean "Land of Almonds". Gotcha there. See how I tricked you? Infact, the word actually means "Black Duck" - that's more uneventful than almonds.

The road shown with the red arrow is our destination. Its the Old Willunga Road.


It is featured in Stage 5, the longest stage of this scorching Tour. Riders ride two times over it and it could perhaps be the most strategic spot for attacks.




Old Willunga is basically a 1000 foot climb, little over 6K in length at almost 5-6% in ave gradient.

The ancient Aboriginals never rode bikes, but it is said that they could jump so damn high, the could reach the top of this road with little effort.

Marvel at the beauty of the ups and downs of terrain. For essential cows, beautiful sunflowers and a McDonalds, please use your wild imaginations. There are only bushes and bush fires here.

We have come to the end of our trip. Thank you and please boast to all your friends that you just went to Willunga... in a little over , let's see....one fifth of a second. Thanks for joining.

Safety Moment : Crashing On Your Nose

Fellow blogger Will Davies becomes the surprise subject of my first safety moment this year. After a coffee stop, Will was fiddling for change in his jersey pocket with one hand, and the other on his brakes as he was descending the wicked Col de la Croisette. It was a little too late when he realized he had picked up an uncomfortable amount of speed, and so squeezed the brakes not 'expecting' much braking force. His expectation turned out to be wrong instead as he found himself flying over the handlebars James Bond style and dipping into the road nose first. After some x-ray scans, his favorite Swiss clinic on the other side of the mountains told him he had a broken nose and a strained shoulder.

Will Davies could put most of us cyclists to shame any day. He's a classic mountain goat for starters, setting ridiculous climbing challenges every year and then going out and logging the miles on the French mountains through thick or thin. In 2008, Will climbed a total of 160,000 meters of vertical ascent. That translates to 525,000 feet or roughly one foot for every minute of the day of the year. He loves epic adventures, hairpin turns, and never misses a single Col sign for the photo ops.

Will, a tiny speck in this pic, climbs the Val Thorens - one of the highest ski resorts in Europe

For 2009, he declares his "Challenge" is to complete 100 mountain passes. For what? "To lose a few pounds," he snaps.

Holy crap. Now that's flahute.

You can read his adventures and see breathtaking photos on his blog Cycling Challenge. Its one of the best there is.

Alberto Contador On The Angliru : Climbing Speed & Power to Weight Ratio

One can only sit back in astonishment and marvel at the speeds at which the breakaway flew up Angliru. Once again, Contador teaches his rivals how to tame mountains.









Anyone want to take a guess at his VAM for this climb? I'm pretty curious...

Okay, lets see if I can do it.

The EUROSPORT recording above kicks off right from the 4K to go banner. Assuming it is live like it says and there wasn't any cut and paste job by the uploader, Contador took about 15.5 minutes to get to the top at 1600m (5250 feet), as he crosses the banner there and proceeds towards the downhill. Referring to slope information for the mountain, the altitude difference between 4k to go and the top is roughly 500m. 15.5 minutes to climb 2 miles? Think about how steep that road was. VAM will be proportional to steepness. In this case, you can pretty much guess that it'll be in the high range.

So for the given altitude change (not for the whole climb) and from around the point he accelerated from Valverde and Rodriguez :



Ofcourse, there'll be an error percentage + or - something, but c'mon, it can't be huge. Nevertheless, a VAM of anything close to this number is bloody frickin remarkable!!!! Anyone disagree?

For a perspective, just remember - when Marco Pantani climbed the Alpe d'Heuz at record speed in 37 minutes & 35 seconds back in 1997 , he produced a VAM of around 1791!! And the climb was only at 7.9% average as opposed to Angliru's 10.13%, but oh well....he must have taken the pill too.

Finally, according to the evil Dr. Ferrari's method (and I have no idea where he digs this from), one can get an estimate of Contador's power to weight ratio (PWR) by dividing his VAM by 300. For the sake of calculation, lets take 1930 as his VAM.

That'll be just sufficient to be one the top climbers in history, eh?

Multiply that with his recorded racing weight of 62 kg, and you can see that he roughly produced around 400 Watts or close to it during those 15 minutes.

Don't grill me because of accuracy issues. I'm just giving you the numbers for the idea.

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