A Metabolic Equivalent of Task (METs) is:

1 MET := Consumption of 3.5 mL_O2/kg/min ~= 1 kcal/kg/hr

Most online calculators that use MET to estimate caloric burn use:

Eq. 1) Cal_kcal = Duration_min * (Mets_MET * 3.5 * Mass_kg / 200)

But none of them derive it or provide a proof to get from 3.5 mL_O2/kg/min => 1 kcal/kg/hr! Where did "200" come from? Intuitively, knowing the units of MET kcal/kg/hr, you'd likely simply perform this following conversion instead:

Eq. 2) Cal_kcal = Duration_hr * Mets_MET * Mass_kg

But this differs by 5% from Eq. 1. Please derive Eq. 1 and explain all units/constants so that we may understand why it is used in almost all MET calculator (and not Eq. 2).


These links are the closest I've ever seen to a derivation: Metabolic Equations for Anaerobic Exercise? and https://fitness.stackexchange.com/a/25564/40079. However, there is no source, explanation of why 5/1000 is used (vs 4.X), or what the units all are. Even more confusing, they both conclude by using Eq. 2 not Ex. 1 like all online calculators!

1 Answer 1


You won't get to eqn 1 starting from your definition of MET.

MET is VO2 in milliliters per minute per kilogram body mass divided by 3.5. So using the conventional value of: 1 liter of O2 consumed is roughly equivalent to burning 5 kcal we get:

a) E = MET * 3.5 / 1000 ml/l * 5 kcal * duration_min * mass_kg

which is

b) E = MET * 3.5 / 200 ml/l*kcal * duration_min * mass_kg

which is eqn 1 in your question. That is why you get a deviation of 5% since its not 1 kcal/kg/h per MET as given by your definition but 1.05 kcal/kg/h per MET.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.