## Kcal Estimates from Activity Counts using the Potential Energy Method

• Published on 01/01/1998

A problem exists in applying the kinetic energy equation KE mv2 to estimates of energy expenditure based on activity counts over long epochs. The instantaneous velocity should be obtained over short time periods (on the order of 0.1 second) and squared to obtain the instantaneous kinetic energy. The instantaneous KE is computed from I/2 In V2 but the square of the average of v is not the same as the average of V2. Since we can only estimate the average velocity from the activity counts integrated over long epochs we should not use the KE equation. An alternative method is to calculate by estimating the “energy potential hill” which is climbed with each step. The Model 7164 summed magnitude algorithm rectiñes the acceleration signal and sums each sample. When the summation value (Epoch Activity Counts) is divided by the number of samples (typically 600 for one minute epochs), one has the average acceleration over the epoch. If the average acceleration is integrated twice over time, one gets distance, In the case for a waist worn Actigraph, the distance measure is in the vertical plane. We can use the potential energy equation, PE = mgh, to calculate the work done in raising and lowering the body against gravity.

The example that follows here, shows how one can use Activity Counts to estimate caloric expenditure in kcals.

A Model 7164 worn tightly on the hip while walking 4.8 km/hr (3 mi/hr) yields an average of 3000 +/-500 activity counts per minute. The average acceleration over one minute is >>>>>>

3000 counts/mín X 0.01664 g/oount 600 samples/min = 0.0832 g/sample

Substituting m/sec2 the average acceleration is 0.82 m/sec2

The vertical distance moved over one second >>>>>>>>

h = 1/2 acceleration t2 = 0.41 meters Note: This is equivalent to a vertical displacement of 0.117 meters

(4.6 inches) with each step and a step rate of 1.75 steps/sec. (1.75 steps X 2 X 0.117m = 0.41m)

From the computed vertical distance moved against gravity over each second, an estimate of energy

expended from Work performed in raising and lowering the body against gravity can be calculated from:

PE = mgh

Energy expenditure (in Joules/sec) for a body mass of 90 kg >>>>>>

90 kg X 9.8 m/s2 X 0.41 m = 361 Joule/sec or 361 Watts

Energy expenditure (in kilo calories per mín) is >>>>>>

361 Joules/sec / 4,184 Joules/kcal X 60 sec/min ~= 5 kcals/min

From this analysis a simple equation can be written to convert Activity Counts to

Kcaìs/min = 0.0000191 * counts/min * body mass in kg

While this conversion factor seems quite simple, testing in activities such as walking and running reveals

the PAEE kcal estimates to be very close to published values for those activities.

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