If I do a certain hike or run, it's going to cost a certain amount of effort, which could be measured by how many calories I burn. How does this effort depend on the amount of climbing? Is there a conversion factor, like a certain number of calories per 100 meters of elevation gain?
A related question, posed in terms of a specific hike, was "How many calories does hiking burn? "
The previous question motivated me to study this question more deeply, and to write some open-source software to do the relevant calculations, along with a scientific paper testing the results against a large volume of real-world data from people running races. Along the way, I learned that a lot of what people believe about this subject seems to be wrong.
People have traditionally tried to quantify this kind of thing using two numbers that can be pretty easily estimated from a paper topo map: the horizontal distance and the total elevation gain, i.e., the sum of all the elevation increases, not counting any of the elevation decreases. So for example, if we start from sea level, climb to the summit of a mountain that is 3000 meters tall, and then come back down to our starting point, the total gain is not zero (because we don't count the descent), and is greater than or equal to 3000 meters. It can be greater because you may do some up-down-up-down stuff rather than just steadily climbing to the top and then steadily descending back down.
Using the horizontal distance makes sense, because laboratory studies of people running and walking on treadmills show that distance is an extremely important factor. It simply takes energy to put one foot in front of the other.
But the total gain turns out to be a very poor measure of energy expenditure. The energetic cost of running or walking does depend on the slope i, but for the values of i usually encountered in the real world, this slope dependence is not very big. Even a trail that people perceive as extremely steep will typically have a slope of only about 0.03, i.e., 3 meters of elevation gain for every 100 meters of horizontal travel. Furthermore, most hiking and running routes are loops or out-and-backs, so that you end up at the same elevation where you started. Except on extremely steep downhill grades, going downhill is more efficient than walking on flat ground. The result is that the effects of any climbing and descending tend to cancel out unless the terrain is fairly steep. In mathematical terms, the energy cost per meter of horizontal travel is a function E(i), where i is the slope, and although this function has some curvature, the curvature is not very strong, so in most cases the average of E(i) and E(-i) is pretty close to E(0), the cost of flat hiking.
When I studied race data, I found that the treadmill data usually provided a much better predictor of people's times than a traditional-style rule in which elevation gain is the only factor. However, the efficiency of downhill running in real-world conditions, as measured by race times, seemed not to be anywhere near as efficient as you would think based on the energy measurements in treadmill experiments. This may be because of factors like safety and trail etiquette. Because of this I came up with a model that seems to fit the data better. This model is incorporated as the default in the code.
Here are three illustrative examples of the output of the model, for a 66 kg person:
A. Run 20 km on flat terrain. Cost: 1130 calories.
B. Run 10 km up a steady grade, gaining 500 meters of elevation, and then run back down, for a total distance of 20 km. Cost: 1224 calories.
C. Run 1 km up a hill 500 meters tall, then down the back of the hill, another 1 km. (The climb and the descent are both steady.). Run another 18 km on the flats, for a total distance of 20 km and 500 meters of total gain, the same as in example B. Cost: 1478 calories.
D. Run 20 km up a steady grade, gaining 500 meters of elevation. Ride home in a car. Cost: 1286 calories.
If we describe these runs in the traditional way, then A is 20 km with no gain, while B, C, and D are each 20 km with 500 meters of gain. However, the energy costs of B, C, and D are all different. I've tried to find a simple statistic that would help people to more accurately characterize how hard a certain run or hike would be. What I came up with was something I call the "climb factor," or CF. It's defined as the fraction of your energy that was spent on climbing. For example, if you compare runs A and C above, the CF for run C is (1478-1130)/1478, or about 24%. Studying the stats for one of my favorite trail runs near my house, which I consider fairly hilly, I was demoralized to learn that its CF was only 3%.
References
Baumel, Bob, "Hill effect to second order," Measurement News, January 1989, #33, p. 36, http://www.runscore.com/coursemeasurement/MeasurementNews/033_89a.pdf
Baumel, Bob and Jones, Alan, "Uphills, downhills, and the Boston marathon," Measurement News, March 1990, #40, p. 15, http://www.runscore.com/coursemeasurement/MeasurementNews/040_90a.pdf
Crowell, "From treadmill to trails: predicting performance of runners," https://www.biorxiv.org/content/10.1101/2021.04.03.438339v1 , doi:10.1101/2021.04.03.438339
Looney et al., "Estimating Energy Expenditure during Level, Uphill, and Downhill Walking, Medicine & Science in Sports & Exercise, 2019, doi:10.1249/mss.0000000000002002
Minetti et al. "Energy cost of walking and running at extreme uphill and downhill slopes," J. Applied Physiology 93 (2002) 1039, http://jap.physiology.org/content/93/3/1039.full
Short answer: Each 100m of elevation gain is as much effort as 0.8km of running (in terms of calories). Depending on speed, terrain, inclination, climbing efficacy etc. it may go down to an equivalency running/hiking 0.4km.
Long answer:
Each 100m of elevation gain equals adding 0.4km to your hike, according to several hiking calculators: Example:
In other words: the additional 250m of elevation gain equal the calories of extra 1km.
With running, adding elevation seems to add more effort. Highly trained mountain-running athletes claim that on an average mountain half marathon, each 1000m of elevation will add 30 to 50 minutes of time (let´s say 40 min on average). This "penalty" calculates to an equivalent of adding 0,8km per 100hm elevation gain to your run as follows:
Activitiy with endurance-VO2max Dist./ Elev. / Time / Speed
==> Claim is that the athlete can keep runnint at his best half-marathon speed of 13km/h for 90minutes but also for 130min, as long as it is flat (and burn the same amount of calories per km as long as it is flat).
Example: 10km + 250m elevation gain: 10km + 2,5 (100m) * 0.8km = 12km In other words: the additional 250m of elevation gain equal the calories of extra 2km.
Further reading:
Let's take a different approach and look at the physics number of moving the mass of a 66 kg person 1000 m up.
The equation for energy expenditure is force x distance or, in this case, mass x acceleration x distance. Plugging in the numbers at 66 x 9.8 x 1000, you get
Wow! That's a lot. However, that is not calories which are the energy costs of heating 1 g of water 1 degree. And in fact, a food calorie is a kilocalorie
For historical reasons, two main definitions of calorie are in wide use. The small calorie or gram calorie (usually denoted cal) is the amount of heat energy needed to raise the temperature of one gram of water by one degree Celsius (or one kelvin).1 The large calorie, food calorie, or kilocalorie (Cal, calorie or kcal), most widely used in nutrition,4 is the amount of heat needed to cause the same increase in one kilogram of water.5 Thus, 1 kilocalorie (kcal) = 1000 calories (cal). By convention in food science, the large calorie is commonly called Calorie (with a capital C by some authors to distinguish from the smaller unit).6 In most countries, labels of industrialized food products are required to indicate the nutritional energy value in (kilo or large) calories per serving or per weight.
Calorie relates directly to the metric system, and therefore to the SI system. It is regarded as obsolete within the scientific community, since the adoption of the SI system, but is still in some use.4 The SI unit of energy is the joule. One calorie is defined as exactly 4.184 J, and one Calorie (kilocalorie) is 4184 J.
So, 646800J / 4184 => 154 calories, assuming 100% thermodynamic efficiency, are necessary to raise 66 kg of mass up 1000 m.
That's regardless of the horizontal distance so its relevance to walking vs running vs bouldering up El Capitan is unclear. But it does give a baseline on the order of magnitude effort involved.
What about that 100% efficiency? Well, turns out human muscle efficiency is mostly about 20%.
Or...
Efficiency The efficiency of human muscle has been measured (in the context of rowing and cycling) at 18% to 26%. The efficiency is defined as the ratio of mechanical work output to the total metabolic cost, as can be calculated from oxygen consumption. This low efficiency is the result of about 40% efficiency of generating ATP from food energy, losses in converting energy from ATP into mechanical work inside the muscle, and mechanical losses inside the body. The latter two losses are dependent on the type of exercise and the type of muscle fibers being used (fast-twitch or slow-twitch). For an overall efficiency of 20 percent, one watt of mechanical power is equivalent to 4.3 kcal per hour. For example, one manufacturer of rowing equipment calibrates its rowing ergometer to count burned calories as equal to four times the actual mechanical work, plus 300 kcal per hour,[18] this amounts to about 20 percent efficiency at 250 watts of mechanical output. The mechanical energy output of a cyclic contraction can depend upon many factors, including activation timing, muscle strain trajectory, and rates of force rise & decay. These can be synthesized experimentally using work loop analysis.
So, you could roughly estimate that 150 (K)cal of useful work requires 750 K (cal) of energy input.
Or, going back to the original question, if you weighed 66kg you might be expected to burn on the order of 75 extra calories per 100 m elevation change going up. With, and I admit this is speculation, a relatively limited effect from going back down rather than stopping at the top.
Compare this now to the A vs D answer by Ben. 1130 for 20k no elevation, 1876 for 20k with a 1000m elevation gain. 1876 - 1130 => 746 k cal.
Considering how proportionally significant the sheer physics component of moving mass up in these calculations are, I rather doubt any findings reporting near 0 effect in calorie expenditure from 0 net elevation gain circuits.
That would require using exclusively stored energy to go back down and that makes intuitively very little sense in the context of muscle-driven effort. Unlike say pumping water up into a dam like is sometimes considered for renewables energy storage.
