You know total kJ is just ((average wattage * 60 * (minutes ridden)) / 1000), right?

I know this is pretty basic, but after getting into a discussion with people at my tri club, lots didn’t realize this, or maybe they just never thought about it.

I averaged 241 watts for my 30 minute cool down ride last night, and Garmin says I used 434kJ of energy (effectively 434 cal burned).

241 watts * 60 seconds in a minute * 30 minutes ridden = 433,800 joules, which is 433.8kJ as there are 1,000 joules in a kilo joule.

I think it’s useful if you have a goal of, say, using 800kJ on your hour ride. That means you only need to average 222 watts on your hour ride to burn 800 calories. Easy!

Sorry if this is too low level for everyone, but the more you know!

and TSS is just

(sec x NP® x IF®)/(FTP x 3600) x 100
.

Yes, but what’s NP and IF?

What is the equation for IF?

What is the equation for IF?

NP/FTP

If you can compute NP in your head on the fly, then you should probably switch to a career in mathematics.

But most head units can display it

Didn’t realize this. So guy going 400 watts for an hour literally burned double the calories of the guy going 200 calories for an hour?

Yep.

New weight loss plan: double FTP
.

Another math application of this realization is the ability to get wattage out of gym exercise equipment to help with pacing.

So if you burn 1000 calories in an hour (in my gym, the machines all report cal/hour), that’s about 1000kJ/hour, and thus about a 277W average.

Are you saying there is 100% efficiency between calories burned and energy produced at the power meter?

I used 434kJ of energy (effectively 434 cal burned).

Are you sure?

http://en.wikipedia.org/wiki/Food_energy

In the International System of Units, energy is measured in joules (J) or its multiples; the kilojoule (kJ) is most often used for food-related quantities. An older metric system unit of energy, still widely used in food-related contexts, is the calorie; more precisely, the “food calorie”, “large calorie” or kilocalorie (kcal or Cal), equal to 4.184 kilojoules.

434 kJ / 4.184 = 103.7 Cal

Are you saying there is 100% efficiency between calories burned and energy produced at the power meter?

No, the conversion (or, lack of conversion in this case) from joules to calories just happens to approximate the efficiency of the human body.

It isn’t exact but it is close.

Not at all. There’s a lot of energy lost as heat. Plus base metabolism, etc.

1 food calorie is 4.184kJ, as someone posted above.

The human body is about 20-25% efficient (depends on lots of environmental and personal factors). So your 4.184kJ of food energy gets turned into about 1kJ of actual work done in the environment and about 3kJ of heat.

Thus, just about 1 food calorie will be burnt by 1kJ of work done, so it makes it a good approximation. If you know your kJ spent, you can estimate your calories burnt. If you know your calories burnt, then you can estimate your kJ spent and thus your wattage.

what jackmott said

when you consider we are about 25% efficient at cycling, ie the 400w you produce at ftp(cuz we are on slowtwitch) requires the body to expand ~1600w to produce that 400w to the pedal. more or less

That’s not terrible except for these points

  1. Does an ex equipment company have an incentive to overstate calories burned? Yes they do “OMG Becky I burned 1000 calories on that machine and I hardly sweated at all.”
  2. Do they have any incentive to understate calories burned? None that I can see
  3. Is anyone verifying that their stated calories burned are accurate? No, no one is checking that.

what jackmott said

when you consider we are about 25% efficient at cycling, ie the 400w you produce at ftp(cuz we are on slowtwitch) requires the body to expand ~1600w to produce that 400w to the pedal. more or less
Better wear your stretch lycra.

You’d typically express that in terms of the total energy expended, not the power. If you did 200W for an hour, you’d have put 720kJ into the bike and somewhere between 2160 and 2880 kJ as heat to support the rest of your body, for a total of 2880 to 3600 kJ, and that converts to between 689 to 861 kcal of food energy, depending on where you fall in the 20 to 25% efficiency range.

Part of the reason to break this down somewhat pedantically, is to illustrate that the approximation is only good for figuring out your basic fueling needs. If you’re trying to dial your exercise precisely, as part of a long term weight management program, then you need to know more exactly what is your metabolic efficiency on the bike. The difference from 689 to 861 calories is more than a can of Coke.

True on all points. But I don’t really care, as I’m not really a “must optimize my TSS and IF and NP” guy. It’s an estimate, not a scientific measurement, just like this entire thread.

In my case, I estimate the elliptical in my apartment is about 25% optimistic. If I do 20 minutes as hard as I can on it, it’ll give me an estimated wattage about 25% higher than the 20-minute cycling watts I could do at the same time.

I think kJ may be one of the most under utilized metrics, if not the most underutilized metric that a LC triathlete could monitor.

IMO it’s way more important to know kJ in a IM or half then what TSS your attaining.

For Calorie counting purposes I find that I cannot eat the same amount of calories I burn on the bike (converted from KJ) and hold my weight steady.

It’s normally about half of that for me.

jaretj