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The bikes at my gym show your current power output in Watts and I'm trying to figure out what this means in some tangible form.

  • I know 1 Watt = 1 Joule / second.
  • I know 1 Joule = 0.239 calories.

So for example, if I average 200 Watts over an hour ride, what does this mean of these?

  • I pushed 200 Watts for the whole 60 minutes burning a total of
    (200 * 0.239) = 47.8 calories (seems very low!), or

  • I was pushing 200 Watts every second giving me:
    (200 Watts * 0.239 calories per Watt) * 60 seconds * 60 Minutes = 172080 calories per hour (seem ridiculous), or

  • It was 200 Watts per minute
    (200 Watts * 0.239) * 60 Minutes = 2868 calories per hour which still seems really high.

8

Generally, given the mechanical efficiency compared to human efficiency, you can multiply your watts x 4 and get your calories burned. It's going to vary some, as some people are more efficient than others, but that is a good way to get pretty close. It's based on both human and mechanical efficiency of the bike being around 20-25%.

The math:

200 watts at the powermeter x 3600 (seconds in an hour) = 720,000 joules = 720 kJ (kiloJoules)

1 kCal (A food Calorie is actually 1000 calories, small c, or 1 kCal) = 4.18 kJ

So 720 / 4.18 = 172.24 kCal

Now, take that and divide by the efficiency factor, which is around 19-26% depending on the person. Take the middle, 23%, and you get:

172.24 / .23 = 748.9 kCals for your hour of work, which is right around 4*200. You may burn slightly less or slightly more depending on your own efficiency.

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    Would be interesting to wrap the body in a heat absorber that also measures heat energy transferred per second in watts so you can directly measure both mechanical power and heat power. – user3015682 Feb 20 '18 at 22:55
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    @user3015682 mmm...not sure that would work. Heat dissipation is key in high intensity exercise. If you wrap the body, you will prevent some of that and thus impair power output. – JohnP Feb 21 '18 at 2:50
  • Something that absorbs, measures, and transfers the heat away from your body. I have no idea how one would be constructed, but it would be neat to have such a device. Ya overheating is definitely an issue with intense workouts. – user3015682 Feb 22 '18 at 3:16
2

I calculated a bit and your second calculation should be right. It only seems ridiculous, because you've got 172080 calories, which is 172.08 kcal. It still is really low for an hour of biking, though.

(Some notes on my calculation: 200 Watts for 1h is 0.2kwH = 0,72 MJ = 172 kcal)

You might want to use this calculator for future calculations, as it's quite easy to use and just much faster.

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    It's not low, it's just the mechanical output. Total energy consumed it considerably higher, just that the vast majority (75%+) of it is waste heat, so multiply by 4-5 to get the total calories/kilojoules used. – Compro01 Mar 14 '14 at 14:38
  • @Compro01: I suspected something like that, thanks for clarifying. – user8119 Mar 16 '14 at 15:57
1

JohnP gave a good explanation of the calculations involved.

Note that wattages and calories on exercise equipment are generally inflated, and often inflated significantly. 200 watts/hour is a pretty hard effort; I ride about 3000 miles/year, and my best hour effort for last year was about 215 watts.

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1

There is a difference between Joules (Watts*time) produced and Joules (Calories) burned. The difference is the bodies efficiency and is approx 25% or a factor of 1/4.

Then there is the constant factor between Joules and Calories which is 4.18.

So by mere coincidence Power [Watts] * Time [Seconds] ~= Calories burned.

So Output 200W * 3600 sec = 720'000 J = 720kJ ~= 720 kCal burned.

In other words, if you have a Powermeter that displays Joules or Ws this is by all necessary accuracy the same Number as the Calories you have burned.

What various fitness devices in the Gym display has more often a close relation to Marketing than Physics.

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