Climbing Power Formula
I came across an interesting formula yesterday that may be used to predict how much power is necessary to climb a given hill on a bike. It's from Allen Lim, PhD. Allen has a long list of accomplishments related to power-based cycling the most notable of which is consulting with Floyd Landis. Floyd credits Allen with determining what he would have to do powerwise to win the now infamous stage 17 of the 2006 Tour de France. I haven't talked with Allen in some time and no longer have his email address after a computer crash this past fall so I haven't been able to confirm what I'm about to show you. But it seems to work so I suspect it's right.
Dr. Lim's formula to estimate the power necessary to climb a hill:
bike + rider weight (kg) x 9.8 x elevation gain (meters) / time (seconds) = power (watts). Add 10% for rolling and air resistance.
So yesterday I tested it. I climbed a known one-mile hill and captured the data on my power meter. Here are my metrics from that climb:
Weight of my Cervelo Soloist with 1 water bottle - 8.18kg
Body weight with winter clothes, shoes, etc - 74.55kg
Elevation gain (est based on average 5% grade) - 78.87m
Time to climb hill - 312 seconds
Plugging all of these into the formula and adding 10% predicts that it would take me 225 watts to climb this hill. The actual was 218. Remarkably close, especially when you consider that I am estimating the elevation gain and there is some variation between power meter readings.
Of course this formula could be used to figure out how long it would take you to climb a given hill in a race at a certain power output by simply rearranging the formula components. You could also use it to figure out how much faster you would climb a hill if you reduced your total system weight by a given amount but kept power the same.
Interesting stuff. If Allen sees this I hope he comments.
posted by Joe Friel @ 6:42 AM
16 Comments
16 Comments:
That seems to make sense. I just put in some numbers for Alpe d'Huez - rider 75KG + bike of 7.5KG - 1071m elevation gain.
This gives 264W required to climb the Alpe in an hour - sounds about right to me.
Joe,
Great stuff. My comments here
Keebler
Hello Joe
I had a vo2max Q? What Percent of max heart rate does vo2max occur at? thank you
john
Anon--It will be quite close to MHR but varies with the indivldual and the test protocol. For some fatigue will set in at about the time VO2max is achieved but well short of MHR. Some will be in such poor condition that in a test with long stages they will hit a "peak" VO2 which is well short of their MHR.
Of course this formula works. It is just basic physics 101 :) Power = Force / Time. Force = Mass * Acceleration (gravity, 9.8 m/s^2). Thus Power = mass * acceleration / time. All that is left is to get the units correct. It turns out that a Watt is defined as a kg*(m/s^2)/s or kg*m/s^3. So just enter mass as kilograms, gravity as 9.8 m/s^2 and time in seconds. Wind resistance is a very large factor on the flats, but while climbing a hill you are slowed down and fighting gravity more than wind resistance. Therefore only a small correction of around 10% is necessary.
Virtual trainers emulate the resistance using those equations... except that they do a resistance test to calibrate some values.
Question: Do you also know a climbing formula for runner? I've tried to find one on the web, but i haven't found something useful.
Thanks
serge--No, I've never come across anything like this for runners. Sorry.
Oops, I left out distance in my post above. I meant to say Power = Work/Time. And Work = Force * Distance, where distance is the vertical gain. This makes P = mass*acceleration*distance/time. So a Watt is actually kg*m^2/s^3.
For any climbing method, whether it be by bike or by foot, the power equation is basically the same. Like poindexter correctly pointed out, it's basic physics. So the formula still holds: Power = (mass of runner)*9.8*(distance of elevation)/(time to climb elevation). The only other factor I would change is the additional 10% that you add for wind resistance and frictional forces. For an efficient runner, the factor one would add to the power would most likely be less (e.g. ~7%) since a runner encounters less wind resistance than a cyclist. The factor can easily be determined by calculating the theoretical power in a frictionless, windless environment and dividing by the actual power used to climb the distance in the theoretical calculation.
well, Isaac Newton could have calculated the same 2 centuries ago, even us at the high school when studying basic physics. I can imagine Dr. Lim could perform a more accurate calculation without making it too complicated, like including other parameters.Best regards
I created an online calculator and posted it on my blog http://oncycling.blogspot.com/2008/01/climbing-calculator.html
I plugged this formula into the spreadhseet that I record my climbs on, and over more than 100 climbs there is an 87% correlation so it seems to hold water.
Is there a formula for working out approximate power values for a specific distance in a specified time e.g a 40km TT
Hi Gianni--Not that I've ever seen. But if you go to www.analyticcycling.com you can play with the variables and come up with some estimates.
Hey are you a professional journalist? This article is very well written, as compared to most other blogs i saw today….
anyhow thanks for the good read!
anon--No, I'm a coach. I wish you had been around when I was in college to talk with my Creative Writing prof. She gave me the lowest grade I ever got in school. Thanks for the kind comment.
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