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Suppose I ride at a cadence of 50RPM for 10 minutes with a 39x23 gear ratio on a hill with 10% grade. Is there a simple formula to calculate the power output required?

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You need to know what torque is being applied, in addition to the RPM. Or you could simply calculate the power required to climb a given grade at a given speed. (Personally, on a 10% hill, I'm just thankful to be able to keep moving.) –  Daniel R Hicks Jan 20 '12 at 4:08
    
How can I know my torque ? –  Rick Ant Jan 20 '12 at 5:45
    
Attach torque wrenches in place of your crank arms and read them out several times a revolution. (Which is to say that that's the hard problem.) –  Daniel R Hicks Jan 20 '12 at 12:36
    
can I calculate power from calories I burned divide by time I spent ? –  Rick Ant Jan 24 '12 at 3:10
    
Rick: Yes, but you also need to account for your gross metabolic efficiency in converting between calories burnt and power. In general, GME ranges between about 20% and 25%. As it happens, 1 calorie = 4.184 joules, or 1 joule = .239 calories, so a common rule of thumb is to assume a GME of 23.9%, in which case the number of food calories (= Calorie, = 1000 calories = 1 kilocalorie) is roughly equivalent to 1 kilojoule. Typically it's easier to estimate work in joules (or kilojoules) than it is to estimate Calorie expenditure so the conversion is usually in the opposite direction: power -> Cal. –  R. Chung Jan 24 '12 at 3:55

4 Answers 4

up vote 6 down vote accepted

You don't quite supply enough information in your specific question (that is, "50RPM for 10 minutes with 39x23 with 10% hill") to provide a full answer in absolute terms but, if we assume you're riding a standard sized 700c bike there's enough information to make a good estimate in relative terms.

First I'll give a short answer, then a rule of thumb that's easy to calculate and will put you within about 10%, then a longer more detailed answer.

The short answer to your question, in relative terms, is ~ 3 watts/kg of total mass. To convert that to total absolute watts, you just multiply 3 watts/kg * total mass (in kg) for you, your bike, and all the equipment you're carrying. For example, if you weigh 70 kg and your bike and all its equipment together weigh an additional 10 kg, it will take approximately 3 * (70+10) = 240 watts. If you weigh 70 kg, that would mean you would need to produce 240/70 = ~ 3.4 watts/kg of body mass. To put that into context, 3.4 watts/kg for 10 minutes is not a bad amount of power output for a casual recreational cyclist; on a normal walk on flat level ground people average around 1 watt/kg, while a pro cyclist might be able to average in excess of 5 watts/kg for an hour. It has been estimated that Lance Armstrong produced slightly over 6 watts/kg for 40 minutes in climbing Alpe d'Huez during the Tour de France.

A rule of thumb to use to convert speed to power on steep hills is this: On a steep hill, multiply the hill's gradient by your speed in km/h, then by ~ 3. If you measure your speed in mph, multiply by 5 rather than 3. That will give you a ballpark estimate of the watts/kg you need to produce. For example, if you are climbing a 10% hill in a 39/23 gear ratio at 50 rpm on a standard sized bike, you're traveling at ~ 11 km/h (or around 6.5 mph). So 10% * 11 km/h = 1.1, and 1.1 * 3 = 3.3 watts/kg. Alternatively, if you measure speed in mph, 10% * 6.5 mph = .65, and .65 * 5 = 3.25 watts/kg. Basically, all you have to remember for this rule of thumb is the number 3 if you measure speed in km/h, or 5 if you measure speed in mph.

How did I convert your cadence in a particular gear to speed? On a standard normal-sized bike, the "700c" rear wheel has a circumference ~ 2100mm (= ~ 2.1 meters). If you were pedaling at 50 rpm through a 39/23 gear, then (50 rpm) * (39/23) * (60 minutes/hour) * (2.1 meters) = ~ 10700 meters/hour, or 10.7 km/h, or 6.6 mph.

And now, the fuller explanation. The equation to convert speed to power is well-understood. Total power demanded has four parts:

Total power = power needed to overcome rolling resistance + 
              power needed to overcome aerodynamic resistance + 
              power needed to overcome changes in speed (kinetic energy) + 
              power needed to overcome changes in elevation (potential energy)

Of these, the simplest piece is the power needed to overcome changes in elevation which, fortunately in this case, is what you were asking. On a steep hill, your speed is low and the aerodynamic and other resistance forces tend to be small relative to the climbing part. The power needed to account for the change in potential energy is straightforward:

watts(PE) = slope * speed in meters/sec * total mass * 9.8 m/sec^2

or

watts/kg = slope * speed in meters/sec * 9.8 m/sec^2

So, all we need is to get speed in m/s. If you have a cyclecomputer that reads in km/h, you need to divide km/h by 3.6 to get m/s and multiply by 9.8. If your cyclecomputer reads in mph, divide mph by 2.25 and multiply by 9.8. If you do this, you'll see that the resulting constants are approximately 3 (for km/h) and 5 (for mph), as stated in the rule of thumb above.

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I don't see how your first formula has any relevance to anything. On the flat your power output is most strongly dependent on relative wind speed, but you don't even hint at what speed you're talking about. –  Daniel R Hicks Jan 24 '12 at 1:40
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I'm not sure which "first formula" you're referring to. Rick Ant's question specified power on a 10% hill, not on the flat, and we know his speed had to have been the speed implied by 50 rpm in a 39/23 -- thus, if the bike used a standard 700c wheel with a circumference of 2.1 meters, we know what his speed would be. It is as I showed: 50 rpm * 60 minutes per hour * 39/23 * 2.1 meters = 10700 meters/hr = 10.7 km/h, or 6.6 mph. –  R. Chung Jan 24 '12 at 2:46
    
So you were answering the specific question of a 10% hill, not his general question of how to calculate power?? You should have stated that. –  Daniel R Hicks Jan 24 '12 at 4:08
    
thanks for the formula, easy to understand... –  Rick Ant Jan 25 '12 at 1:39
    
so if I use 39x21 53 rpm I will get 3.5m/s or 12.6K/h and 12.6K*0.1*3 is 3.78 Watts/kg total power I need =3.78 * 77kg = 291.06 Watts –  Rick Ant Jan 25 '12 at 1:58

Theoretically, you could only measure power with a specialized instrument, usually an electronic (and expensive) torque meter embedded in a custom crankset or rear hub.

For the state-of-the-art about this, take a look at http://www8.garmin.com/train-with-garmin/power-meter.html. It will point you a lot of other links on the subject.

If you want, like your question suggests, to estimate power based on total ascent (disregarding energy spent to overcome wind and rolling resistence), you could use power (W) = energy (J) / time (s) formula, where energy is the variation of potential energy, calculated with energy (J) = mass (kg) * gravity (9,8 m/s²) * height (m), being height the total ascent, and mass the combined mass of your body and bicycle.

The second formula will give the minimum energy spent (since not only ascent, but also drag will consume energy), so you could convert to food calories if you want. Also, if you brake, the kinetic energy is lost, and you spend more energy to accelerate back to speed.

Power means the rate of energy transfer, or how much energy you can spend for each time unit - in plain language, how strong you are.

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You can use the calculator at http://bikecalculator.com, which will give you a reasonable estimate if you know the average grade of the hill, the day's temperature, and the wind speed/direction (probably not so relevant on a hill). A similar calculator is here so you can compare two methods.

The website http://www.cyclingpowermodels.com has a host of information about power models, including the following excerpt. I couldn't find a power calculator on there though (only the opposite).

Model Validation

Two key questions in the application of any model to the analysis of cycling must be "is it accurate?" and "what are the assumptions?"

Models of the relationship between cycling power and speed have been around for a long time and rely on the physical principles in Newtons Laws of Motion. The principal model of cycling power and speed used throughout this site is an implementation of that proposed in Validation of a Mathematical Model for Road Cycling Power which appeared in the Journal of Applied Biomechanics in 1998. This publication demonstrated the completeness and validity of the model by comparison of model predicted and observed power values. The model calculates the power a cyclist would have to produce in order to achieve a certain speed on a certain course, taking account of key physical and environmental parameters. In places this model is used to compute speed, time or the value of another parameter given a specified power.

The performance of any model is only as good as the accuracy of it's inputs which is why we often go into great detail measuring or estimating major variables such as air density, wind and aerodynamic drag. Any asumptions or modelling approaches will generally be outlined. To some extent the use of consistent power models in the field derivation of aerodynamic drag measurements (i.e. field testing of CdA) can improve the reliability of the models when used with that input.

In practise we have found theoretical values of a cyclists ride time, given a specified power and good parameters inputs, to fall consistently within +/-5% of actual ride time and frequently within +/-2%. In the context of the stated accuracy of most cycling power meters at +/-2% we have great belief in the application of physical models to the analysis of cycling events and, more importantly, the analytical power provded to the rider or coach. The more we use these models the more confidence we have in them - if you have used them feel free to let us know your findings.

Note that gear ratio and cadence is not necessary for the calculation (you can produce the same power output with lower gears and faster cadence or vice versa).

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Do the models account for the known fact that it takes twice as much energy late in the day to achieve the same speed you easily achieved six hours earlier? –  Daniel R Hicks Jan 20 '12 at 12:39
    
:-) Well actually there is some evidence (can't find the reference right now) that the reverse is true: performance follows the circadian rhythm, and peak performance is actually around 6pm for most people. Although obviously not if you've already done a session in the morning! –  tdc Jan 20 '12 at 14:53
    
@tdc There is a complicating factor on the same-power-with-more-than-one-cadence principle, because speed of muscle contraction influences its efficiency. Then, if you ride, say, 20mph at 90rpm, your body spend much less energy than if you rode the same speed with 40rpm or 150rpm. The mechanical work measured from outside might be the same, but the physiological effort would not, theoretically. –  heltonbiker Jan 20 '12 at 15:31
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@heltonbiker I don't think that's right: I think the energy required is the same, it's just that the body has different systems that work at the two different cadences (slow/fast twitch muscles), which use different energy systems in the body. The result is a difference in psychological effort, not physiological effort. –  tdc Jan 20 '12 at 15:41
    
@tdc I'm pretty much inclined to disagree, but I think more scientifical evidence would be needed to support any of the two points of view. If I find some paper or the like, I'll come back and post (even if it contradicts me). Besides, since this is not a forum, I'm afraid it's already offtopic. –  heltonbiker Jan 20 '12 at 16:35

There are good answers already. For practicality I just use this website: http://cycle2max.com to estimate power. You can't use it for training but it's quite useful for comparing different climbs. IIRC, they also tune their algorithm with power meter data.

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