One of my earliest posts was about how efficient walking is and how little energy it takes to walk around. I illustrated this by the observation that it takes only the energy contained in two heaped teaspoons of sugar to allow an average adult to walk for a kilometre at a comfortable pace. After mentioning this as part of a tutorial I gave last month at the Gait and Clinical Movement Analysis in Portland several of us sat on chatting about whether this is a small amount of energy or not.
There is a problem in trying to think about how efficient walking is in that efficiency is generally defined as the energy output by a system divided by the energy that is input to a system. The problem in relation to walking over level surfaces is that it doesn’t necessarily take any energy to move an object from one point to another if it is at the same height and moving at the same speed at the end of the movement as it was at the start. Think of a perfect wheel, once we use a relatively small amount of energy to get it rolling, it will continue to roll along a level surface without requiring any energy input. If the energy output by the system is zero then it makes the calculation of efficiency a rather pointless exercise. Nothing divided by two teaspoons of sugar is nothing but so is nothing divided by one teaspoon of sugar or nothing divided by a hundred teaspoons. How can we get a handle on whether two teaspoons is a lot of energy or not?
One way might be to calculate the gradient of a slope we would have to be walking down in order for the loss in height to be provide the energy for walking rather than our bodies burning up food. Ralston’s classic paper of 1958 calculated the nutritional energy cost of walking (the amount of food energy that needs to be consumed) to be about 3.3 Joules/kg/m (assuming 1 cal = 4.186J) and more recent work that I’ve published agrees. If that energy all came from a loss of potential energy (mass x height x acceleration due to gravity) then it is quite easy to calculate that this would require a loss of 0.33m for every metre walked (=3.3/9.81). The gradient we would have to walk down would be 1 in 3 which sounds very steep to me.
But things are not quite as simple as this. The efficiency with which food energy is converted to mechanical energy is estimated to be about 20% so the mechanical work that 3.3 J/kg/m represents is about 20% of the nutritional energy cost. This energy is thus only really equivalent to walking down an 1 in 15 gradient. It is also important to remember that about half of the gross energy cost of walking comes from the basal metabolism that is required to keep your body functioning regardless of whether you are walking or not. On this basis the effective gradient should perhaps be reduced even further to 1 in 30.
So does this help? Well the 1 in 3 slope that we arrived at when just thinking about the nutritional cost is quite steep and perhaps serves as a reminder that the energy density of foods such as sugar is very high. We shouldn’t assume that just because we’ve got a a small amount of sugar that we have a small amount of energy. On balance, however, I think the 1 in 30 slope that arises when we take account of the basal metabolism and the efficiency of conversion of food to mechanical energy is a fairer reflection of how efficient walking is. This slope looks quite gentle and I think the overall conclusion that the walking is reasonably efficient is justified. The gradient isn’t however so small that it can just be ignored. The mechanical cost of walking appears to be the equivalent of raising the body mass by about 4cm more than is necessary for every stride (assuming a stride is about 1.3m long).