How to convert calories to watts on Strava rides?

Question

I'm trying to find out how I can convert calories into Watt hours on Strava rides. I know the formula that 1 kJ = 1000 Ws but calorie unit is different. When I looked up to the formula to convert calories to joules, I found this:

1 cal(th) = 4.184 J

However, when I calculate with this formula it doesn't give me any meaningful result.

Can you check out below ride and tell me how to find out average power in Watts?

https://www.strava.com/activities/6797539669

I wonder how these calcuations would compare to the empirical readings from a power meter for the same ride. — Criggie, Mar 10 '22 at 10:20
Strava estimated power calculations are much lower comparing to power meter readings: https://youtu.be/7vG8Z906rPo — Ender, Mar 10 '22 at 10:40
"Watts per hour" makes no sense, as watts are a unit of power, i.e. a rate of energy. You can convert between energy in calories and in watt-hours - is that what you meant? — Toby Speight, Mar 10 '22 at 16:45
The correct formula is 1J = 1Ws, where J, W, and s are the units Joule, Watt, and second, respectively. Please correct the formula in the first paragraph. Also note that Americans have the annoying habit of calling the kilo-calorie kcal, or big calorie Cal, simply calorie. Which, of course, is plain wrong. And which easily, and unnecessarily confuses such calculations by a factor of 1000: 1cal = 4.184J and 1kcal = 1Cal = 4.184kJ. — cmaster - reinstate monica, Mar 10 '22 at 17:17
@Criggie: I had assumed that she was riding with a power meter but chose not to publish the data? — Michael, Mar 10 '22 at 17:48
I've made an edit to correct the wrong units in your question. I hope I have preserved your meaning. Maybe use of correct unit already solves your question? — cmaster - reinstate monica, Mar 10 '22 at 19:07
I ride without a power meter but Strava usually plots its estimate and average of that. The fact I can't see it on that ride indicates it's actually recorded but hidden (as @Michael suggests). Another tip is that this is a professional (double lines in the subscriber indicator) — Lamar Latrell, Mar 10 '22 at 23:31
@TobySpeight Watts per hour might be a bit redundant in the cycling use case, but W/s would be the rate (slope) of ramp up or ramp down of power in a ERG mode workout on Zwift or similar. Definitely a thing, at least IMO. 'Acceleration of energy', if you will, although that's a bit confusing — Lamar Latrell, Mar 10 '22 at 23:36
@LamarLatrell I understand Strava calulates that estimated wattage after the fact, becuause it uses the Big Data of other riders and terrain info to work out the gradient at any given point on the route. Your phone/headunit doesn't have access to that data and it would be computationally expensive. — Criggie, Mar 11 '22 at 00:45
@Lamar, that's absolutely right, and I should have chosen my words more carefully to put them into context (i.e. makes no sense when comparing with calories). The good thing is that the question has been edited to make it self-consistent. :) — Toby Speight, Mar 11 '22 at 08:01

score 13 · Accepted Answer · answered Mar 10 '22 at 09:56

13

Turning my comment into an answer:

Watts in a bicycling context are usually measured (or estimated) as mechanical power at the wheel (or crank). Calories burned are the (estimated) total input energy. 2720kcal are 11.3MWs. Over 4h3m that’s an average of 765W total (heat+mechanical) power output. Assuming 22% muscle efficiency that would be 168W average mechanical output power.

Usually calories burned are estimated based on measured mechanical power. The big uncertainty is really the muscle (in)efficiency. I dimly recalled 30% but a quick Google search shows up 18 – 24%. It very much varies between individuals and also depends on intensity.

answered Mar 10 '22 at 09:56

Michael

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I think the exact numbers should be: 2.720 kcal are 11.38MWs. (1 cal = 4,184 J) Over 4h3m that’s an average of 780,55W total (heat+mechanical) power output. Assuming 22% muscle efficiency that would be 171W average mechanical output power. – Ender Mar 10 '22 at 10:09
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A minor nitpick: 22% muscle efficiency is an assumption, and moreover it's something like an average value. Individuals will vary somewhat in what their actual muscle efficiency is. Unfortunately, there's no practical way for you to actually measure yours (lab testing would be it). Some more info here, article is (I believe) open access https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3761728/ – Weiwen Ng Mar 10 '22 at 15:06
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@WeiwenNg: Yes, but a power meter (or Strava) is also making a simple assumption to produce the “calories burned” number. I think van Vleuten in OP’s example simply chose not to publish her power numbers. But maybe her “calories” are still based on real power meter values and can be converted back to actual power if one knew the conversion factor Garmin (or Strava) uses. – Michael Mar 10 '22 at 17:51
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@Michael: Yes, it appears from my rides that I can hide power and the Calories don't change, so it appears they're based on real power data. Also, when I back-calculate what Gross metabolic efficiency must be to match what Strava shows, I get 0.215 (to within the rounding that Strava uses). – R. Chung Mar 12 '22 at 16:01

Weiwen Ng · Answer 2 · 2022-06-16T14:16:01.663

Minor detail for the general audience: the ride in question lacks actual power meter measurements. It's pretty likely that the cyclist in question wanted to keep her power meter data off Strava, as she is a prominent professional road racer.

If there's no power meter data, Strava estimates power using an algorithm. It has speed from the cycling computer (albeit if speed was measured by GPS, there will be some imprecision). It also knows the elevation profile of the route. It should also know the rider's weight - or at least, the weight the rider entered into Strava, which may or may not be current. It will make assumptions about rolling resistance and several things related to aerodynamic drag. Notably, air density changes with temperature, and I believe Strava assumes something like a 15C temperature for all rides. It also assumes the rider's coefficient of drag area - and this will vary depending on your position and to a lesser extent, your bicycle.

Thus, the algorithm-estimated power data are a best guess. For many athletes, they may be good enough. There may be some edge cases where they are a poor guess - e.g. if your position is particularly aerodynamic or un-aerodynamic, if the air pressure markedly departs from the assumption, etc. So, this doesn't affect the arithmetic of getting from reported calories to average power over the ride. It affects more the validity of the estimated power.

Another step in the arithmetic, as others have mentioned, is the assumption of gross efficiency. That is, normally a power meter measures the power you put into the drivetrain. However, our bodies aren't perfectly efficient at converting energy from food into motion about the rider's gross efficiency. So, work done field on Strava, which is in kJ, is the amount of work you did to your drivetrain, measured by the powermeter or the Strava power algorithm. The calories burnt field, in calories, is an estimate of how much energy your body burned to do the reported amount of work on the drivetrain.

One study I found (and cited in a comment) estimated that of its sample of experienced female cyclists, the average gross efficiency was 23.2%, with a standard deviation of 3.5 percentage points at around their functional threshold power. That is, if the sample is representative, 95% of female cyclists should have gross efficiency within about 2 standard deviations of the mean - implying a range of 16.2% to 30.2%. That's quite a big range. Men may have a slightly lower gross efficiency.

This is just one study. There may be other studies on gross efficiency, which I didn't bother to search for. If you are searching on Pubmed, "gross efficiency" may be a standard keyword, which would make it easier to search. Note that not all studies may use the same standard keyword, however. Unlike functional threshold power, I don't believe there's a practical way to measure your own gross efficiency outside of a lab test in an exercise physiology lab.

As a worked example from my workouts, on a recent 78.4 mile ride, Strava reported:

1,916 kJ total work (i.e. measured at the drivetrain)
1,963 calories burnt (i.e. estimated work done by my body)

The original post is correct that calories and kJ actually measure total energy. Using the conversion factor of 4.184 kilojoules (i.e. thousand joules, which the OP missed) to 1 calorie, we do get 8,213.192 estimated kJ burnt by the body.

RChung's comment states that Strava assumes a gross metabolic efficiency around 0.215, i.e. Strava assumes that every calorie burnt by the body will put 0.215 calories of work into the drivetrain. The linked study reports that the average GE for trained male cyclists at 60% of max aerobic power was 0.217 (with a standard deviation of 0.016). I believe that 60% of MAP is around a tempo speed, maybe around 60-70% of functional threshold power. Dividing 1,916 (total work done to drivetrain) by 8,213 (energy expended by body), I get 0.233.

Taking the 1,916 kJ of work done to the drivetrain and dividing by 0.217 (average efficiency of a trained male cyclist), I get 8,829 kJ of energy burned by the body. This should be 2,110 calories.

Now to illustrate the effect of individual variation. About 68% of people are within 1 standard deviation of the mean of a normal distribution, i.e. we'd expect that 68% of trained male cyclists have GEs between 0.201 and 0.233. Those translate to calorie (i.e. expended by body) estimates of 2,278 and 1,965. It's a wider range if you're looking at people within 2 standard deviations of the average.

Basically, if you are on a strict diet, then do remember that even with a power meter, you should refine your intake depending on other measurements, like weighing yourself regularly. I am not currently aware of a practical method to estimate your own gross efficiency.

Another possible cause of errors in estimated power is that Strava doesn't know if the rider is riding solo vs. in a bunch. That makes a huge difference which Strava can not compensate for. — Roel Schroeven, Mar 10 '22 at 23:22
@RoelSchroeven and when in a bunch, where in the bunch? Even for us normal people solo vs taking turns in a pair makes a big difference — Chris H, Mar 11 '22 at 12:34
@FreeMan fair enough. There are a few sources of weather data for Strava, so none of us can be expected to keep up with all the changes — Chris H, Mar 11 '22 at 13:24
@Ender Noted. I saw the name of the cyclist in question and immediately got cross-wired with Strade Bianche. — Weiwen Ng, Mar 11 '22 at 16:03
One can "hide" power in Strava even if using a power meter, and it does not appear to affect the displayed value for Calories; I just checked that on a couple of different rides. So if AvV were collecting power data (likely) but chose to hide them, it appears Strava does not re-estimate the power and base Calories on that. I also checked my own raw data, and it appears Strava makes an assumption of gross metabolic efficiency near 0.215. — R. Chung, Mar 12 '22 at 14:32

score 5 · Answer 3 · edited Mar 12 '22 at 00:28

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Every rider will have a different efficiency which will change things slightly, but a commonly used formula that provides a 'good enough' estimate is:

Caloric intake needed (kcal) = Watts * Hours * 4

We can re-arrange this to estimate Watts from Calories (kcal):

Watts = Calories / (Hours * 4)

I would point out that for the ride you linked average power is not a particularly useful metric - there is a lot of climbing/descending. It is likely the climbing was done at a much higher power and the descending much lower.

Edit: Including the full maths for the commenters and downvoters.

100 Watts = 100 J/s

100 J/s * 3600 (seconds per hour) = 360000 J/hr

360000 / 4.186 = 86001calories = 86kcal (Calories) per hour

Note the difference between calorie with a small c and the dietary unit with a big C

Cyclists have an efficiency of between 18-25% depending on the individual - the number in the middle of that range is 21.5%

Kcal required to generate 100W at the pedals for 1 hour = 86 / 0.215 = 400kcal

edited Mar 12 '22 at 00:28

MaplePanda

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answered Mar 10 '22 at 09:42

Andy P

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No, no, no. Something's missing in your "good enough" estimate. An efficiency should be added, because one kcal is approximately 1.163Wh and surely not 4Wh. I know we're on bicycles.stackexchange, but still : the question already contains completely wrong units, the answers shouldn't also include wrong physics equations. – Eric Duminil Mar 10 '22 at 18:27
@EricDuminil I have edited the post to include the maths (and SI units) since it seems you were struggling. I hope you reconsider the downvote – Andy P Mar 11 '22 at 08:55
@cmaster-reinstatemonica i'm not sure if the comment was directed at me or Eric, but note that I have updated the answer with the full maths – Andy P Mar 11 '22 at 08:56
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Yes, my comment was directed at you, I would have included "at"Eric otherwise. And thanks for the clarification. And sorry for not doing the math myself, as it's not exactly complicated... – cmaster - reinstate monica Mar 11 '22 at 09:12
I'm not struggling with the units, thank you very much. The first equation in your answer is still wrong by a factor of 4. You need to include an efficiency somewhere, explicitly in this equation. Right now, it looks like pi = 1. – Eric Duminil Mar 11 '22 at 09:30
@EricDuminil kindly explain why it is wrong by a factor of 4 AFTER i've shown you the maths to prove it is correct. Calories(kcal) = Watts * Hours * 4 -> 400 = 100 * 1 * 4 -> 400 = 400 – Andy P Mar 11 '22 at 09:34
@EricDuminil remember you're able to add a separate answer, with your view on how the maths should be. Estimates can be notoriously hard to agree on. – Criggie Mar 11 '22 at 09:59
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That looks like a decent match to Strava, to me (reading a few hours after the edit adding the maths). Strava claims I averaged 133W for just under 10 riding hours, and burnt 4291 kcal. Working backwards Strava's assumption comes out to about 26%. I'm pretty certain I've found them to be 25.0% before – Chris H Mar 11 '22 at 12:47
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@EricDuminil the rule of thumb ("good enough estimate") effectively takes advantage of a lot of factors being close to 4 (x10^3): the number of J in one kcal, the number of seconds in an hour, and the inverse of human efficiency. Multiply 2 of those together, divide by the 3rd and you've got 4 (to within an error far smaller than the error in Strava's estimation of power). While you might prefer a different presentation, the approximate formula just rolls up all the constants into one – Chris H Mar 11 '22 at 12:51
1 cal is 4.1868J, by definition. So 1kcal is 4186.8J, or 1.163Wh. The problem with your first equation is that it contains symbols which look like units, but actually represent physical quantities expressed in the corresponding units. So when you write "kcal", you actually mean "food intake (expressed in kcal)" or "food_intake / kcal", and "Watts" is actually "mechanical power (expressed in W)" or "mechanical_power / W". – Eric Duminil Mar 11 '22 at 19:36
@EricDuminil it should be perfectly fair to equate the nutritionists kcal with work. After all, the nutritional value is just the chemical enthalpy determined by complete oxidation in a calorimeter. Andy already mentioned that there is an efficiency factor. That can be unit less though since we have a defined way to get from enthalpy to mechanical work (cf statistical physics and thermodynamics). – gschenk Mar 11 '22 at 23:43
@gschenk: I give up. I'm apparently the only one bothered by the way the first equation is written. "kcal = 4Wh" is simply wrong, and cannot be saved by post-scriptums, comments, bicycles, statistical physics or the grand unified theory. – Eric Duminil Mar 11 '22 at 23:51
@ChrisH: Yes, your explanation makes sense, and the way the first equation is presented is indeed the problem. – Eric Duminil Mar 11 '22 at 23:53
@EricDuminil Fixed? – MaplePanda Mar 12 '22 at 00:28
@MaplePanda: It's a small step in the right direction. – Eric Duminil Mar 12 '22 at 11:40
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If you're interested, I asked a question on physics stackexchange (https://physics.stackexchange.com/q/699915/148854) on how to write the above rule of thumb. https://physics.stackexchange.com/a/699927/148854 has good tips. – Eric Duminil Mar 21 '22 at 13:36

score 0 · Answer 4 · answered Jun 16 '22 at 18:30

I think there's a basic misunderstanding in the question that none of the answer highlights. The existing answers seem to highlight only that there are two energy measures, energy input and energy output, but seem to miss that who asked this question obviously doesn't know what's the difference between energy and power.

Calories and joules are a measure of energy. Energy is something that for example food contains. More food, more energy.

Watts are a measure of power, the rate of energy transfer. One watt is one joule per second.

Also you should note that calorie can refer to either real calorie or kilocalorie. To get kilocalories from calories, divide by 1000; to get calories from kilocalories, multiply by 1000. If someone says kilocalories you can trust it. If someone says calories it may either refer to real calories or kilocalories. You will need to try both and see which unit gives reasonable values.

So if in a ride of 15 minutes you burn 120 kilocalories, that's 4.184 * 120 * 1000 = 502080 joules. 15 minutes is 900 seconds. So the rate of using additional food energy is 502080/900 = 557.87 watts (or joules per second). Of course in this ride the cyclist is also using energy needed for basic metabolism, the 557.87 watts is the rate at which additional food energy, over that of basic metabolism, is used.

You may leave it there if you are interested in energy input. However, if you want energy output, you need to know how efficiently a human converts food energy into mechanical energy. I assume 25% is a reasonable conversion efficiency so that would give 0.25 * 557.87 W = 139.47 W. From that, we can observe that the values I invented weren't for a racing cyclist but rather a casual Sunday rider, as a racing cyclist would produce more power.

Note that both 557.87 W and 139.47 W are correct. The 557.87 W is the rate at which the cyclist converts extra food energy into both mechanical energy and heat. 139.47 W is that mechanical energy. The heat would be then 557.87 W - 139.47 W = 418.40 W. The produced heat is why cyclists sweat.

So to calculate watts from calories, you also need the duration of the ride, just calories isn't enough. If you don't know that, you can't say at how many watts the cyclist is producing power. It may be the ride is a short-duration intense ride and power production rate is extreme, something only racing cyclists can achieve. It may also be the ride is a long-duration casual ride and power production rate is practically nothing, easily achievable by a Sunday cyclist. Both of those two rides can have the same amount of calories burnt.

How to convert calories to watts on Strava rides?

4 Answers4

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