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SPEED: THE MOST EFFICIENT HIKING SPEED By Ol' Creosote I recently acquired a most informative book: "EXERCISE PHYSIOLOGY" by McArdle & Kazch. Since backpacking means load packing and I yearn so spare my bad back the pain of hauling a heavy pack full of grub, I read the book eagerly. My goal is to learn to walk as efficiently as is possible for me so I can get by on less weight of food. I wanted to find if there is experimental evidence that there is an optimal speed at which a hiker should walk so consume the least energy per kilometer, or per mile, and if so, what is that speed? On page 149 of said book there appears a graph showing the experimental relation between walking speed and energy expended per minute. The authors interpreted the graph to mean that efficiency declines with speed, especially at the higher speeds. Extrapolating that rule backwards leads to the absurd conclusion that zero walking speed gives the most efficiency. A more careful analysis seems in order. Look at the graph. With experimental values so neatly clustered along a smooth curve, it seemed to me a more mathematical treatment might reveal a more consistent interpretation. If you are not into analytic geometry and calculus, skip the following equations and digest the conclusions. EQUATIONS: The curve seems to be of the form: Y = C + mx2 When X = o, Y = C, Also, m = (Y-c)/x2, Substituting from the two known points on the curve yielded a close fitting equation: Kcal/Min= 1.7 +0.11 (Km/hr2) To check the fit of this formula to the experimental data, I substituted all the Km/hr values along the X-axis and computed corresponding values for the Y-axis. I plotted the formula values on the experimental curve (the "+" marks), and as you can see, the benefit is about as perfect |
as you might hope for. (Actually I
used m=0.1085,not .11) CONCLUSION (I): In other words, the walkers were using 1.7 Kcal/min and getting nowhere waiting for the start. Then for each walking speed they consumed extra energy proportional to the square of the speed. When the walking speed counts double against efficiency, it can easily be overemphasized. FOR EXAMPLE The equation indicates a walker going four Km/Hr would use 3.44 Kcal/Min. Going twice that speed, he would burn 8.64 Kcal/Min, considerably more than twice as much energy/Min. With the formula, I could compute the Kcal/Min for any hiking speed, but that is not what I needed to know. I need so know what speed to hike to use the least calories per mile, not per minute. The least energy per minute is at zero speed, a 100% waste, if you want to get somewhere. Obviously, if I begin to walk, I immediately start making distance for my slight additional expenditure of energy, that is a big percentage difference gain (from zero distance to some distance). But if I keep going faster and faster, a point of diminishing returns will soon be reached and I will be a big loser. Some where in between wasteful zero speed and wasteful excess speed, there is an optimum. I could approximate it by dividing all of the Y values by their X values and taking the smallest. That would be a lot of crank turning for an approximate answer. Fortunately, there is a mathematical way of computing the minimum point exactly. I have to divide the energy by the speed algebraically, then solve the ratio for the rate of change. EQUATIONS: Dividing: Y/X (or Kcal/Min//Km/Hr) gives energy/dist. E = Y/X = (1.7 + 0.11X2)/X Differentiating and equating dE to zero: gives: 0 = 0.11- 1.7/X X = square root of 1.7/0.11 = 3.96 Km/Hr or 2.5 Mi/Hr |