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Physical Data


Based on: http://www.braydenwm.com/calburn.htm

Calories and Power as a Function of Heart Rate

enter gender m/f
enter weight pounds
enter age years
enter avg heart rate bpm
enter VO2max enter 0 if unknown
enter duration minutes
Calories burned per hour
Total Calories burned
Watts average power absolute ( ~24% efficiency)
Watts/kg average power relative


Reference: Prediction of energy expenditure from heart rate monitoring during submaximal exercise.

Discussion
According to the reference, energy expenditure is a linear function of heart rate over the range of approximately 90 to 150 bpm. In other words, it is a straight line. As you would expect, everyone has their "own" straight line, characterized by a slope and intercept. Early heart rate monitors with a Calorie (Kcal) option used an experimentally obtained "average" line and applied that "average" line to everyone. Today, individual physiological parameters, such as weight, age, resting heart rate, max heart rate, VO2max, etc. are used to minimize the total error between a "measurement class" and a "test class" in deriving an optimum set of coefficients for predicting an individual's slope and intercept. The equations below are taken from the reference paper and modified to reflect calorie output. Companies like Polar, use equations of this form with different coefficients obtained from their statistical analysis of experimental data. Details can be found in their patent applications.

Using VO2max:

Men: C/min = (-59.3954 + (-36.3781 + 0.271 x age + 0.394 x weight + 0.404 x VO2max + 0.634 x HR))/4.184
Women: C/min = (-59.3954 + (0.274 x age + 0.103 x weight + 0.380 x VO2max + 0.450 x HR)) / 4.184

Without VO2max:

Men: C/min = (-55.0969 + 0.6309 x HR + 0.1988 x weight + 0.2017 x age) / 4.184
Women: C/min = (-20.4022 + 0.4472 x HR - 0.1263 x weight + 0.074 x age) / 4.184

where weight is in kilograms and C stands for Kcals.

In order to assess the accuracy of the predictive equations, I have entered my individual physiological parameters (male, age = 53 , weight = 158) and plotted the predicted values (red line) against measured values (blue line) for my heart rate versus power. Using a bike trainer with a watt meter, I measured my power output for different heart rates, holding each data point for at least 10 minutes. For me, there was good agreement between the measured and predicted curves. This may not be the case for others. In fact , there was terrible agreement between the Kcals given by my Sigma Sport heart rate monitor and these equations. Note that the measured curve is approximately linear as predicted by the model equations. If I repeated these measurements I would find variation in my power output versus heart rate, depending on how I felt that day.

One last item. If you believe the Calorie (Kcal) function of your heart rate monitor is accurate, you can convert Kcals/ hour to Watts and thus use it as a power meter. However, I would take any results with a grain of salt, as most heart rate monitors are not very accurate when it comes to estimating calories burned as a function of heart rate. You can increase the accuracy of estimating your power by holding your heart rate steady during the measuring process. Assuming 24% effieciency,

Watts=(Kcals/hour)*1000/3600.

I have provided a calculator script below which does the calculations for 2 different efficiency values. The efficiency of your body in transforming chemical energy to mechanical energy for cycling is anywhere between 20 to 25%, with most people falling in the middle of that range. Yes, that includes pros too. Yes, they produce gobs of power, but they require gobs of calories to do it. 24% is a commonly used efficiency because it makes the numbers come out nice. All that being said, if you run the numbers and find you are putting out more power than a member of the pro peleton, don't believe it. Assume the Kcal/hour output of your heart rate monitor is in error. :-)

red=predicted, blue=measured

Convert Calories Burned per Hour to Watts

enter Kcal/Hour
standard assumption of ~24% efficiency Watts
assuming ~22% efficiency Watts