# What is the flat ground distance equivalent of 500 metres running uphill with an incline of 10%?

To explain the question: I ran 500 meters uphill. This costs energy, which can be measured in Joules etc.

My question is: how far should I run on flat ground to expend the same amount of energy?

• Since you are working against gravity here, this will depend on your weight obviously. The running makes it interesting because this won't be linear anymore, but I wonder if it's such a bad estimate to make this into the trivial physics problem you know from middle school: You elevated your body by 50 m, so you expend 50m * g * [your weight in kg] of joules to lift the object. Now add some factor to account for friction and your body not being perfect. Since the m * g * h is literally something you learn as a 13 year old, you should include it into your question as part A of solving this issue – Raditz_35 Aug 23 '19 at 16:33
• @Raditz_35, applying m * g * h would also result in the insight that running on flat ground doesn't cost energy at all, and that running up a 50m elevation costs as much energy on a 2% incline as it does on a 100% incline. In other words, it's not very helpful. You could use it for determining a lower boundary... for a 70kg person, this would result in an additional energy expenditure of at least 8 kcal. – UnbescholtenerBuerger Aug 23 '19 at 17:00
• my weight is 55kg. and I never had physics in school as a subject. – GwenKillerby Aug 23 '19 at 17:26
• Please note that what I wrote was not an answer but a comment to improve your question. How much energy you use while running is easily googled. There will be a large error bar, but well, you can assume it's the same error bar for running uphill, they should cancel each other out mostly. Btw, that's a good observation for someone who never had physics: You are correct, that formula says that there is no difference in potential energy when running on flat ground. mgh will give you the difference, not the total. You can then translate that difference to running time/distance you have to add – Raditz_35 Aug 23 '19 at 19:03

Approximately 750 m.

Any physical activity has a MET value associated to it. The definition is not important; what matters for us is the following:

(...) MET can be thought of as an index of the intensity of activities: for example, an activity with a MET value of 2, such as walking at a slow pace (e.g., 3 km/h) would require twice the energy that an average person consumes at rest (e.g., sitting quietly).

A fortunate coincidence is that running on flat ground gives MET values that roughly coincide with the speed in km/h. For example, running at 12 km/h has a MET value of approximately 12 (it's actually 12.5). To compare that with running uphill, we need a MET table with values for different inclines. One is given by American College of Sports Medicine: Guidelines for Graded Exercise Testing and Exercise Prescription, 2nd Edition, Lea & Febiger, Philadelphia (1980) and reprinted in this article (the grades for outdoors are wrong, they're all halved). I found a formula that reproduces all those values with less than 0.7% error:

``````MET = 1 + (0.957*v - 0.064)*(1 + 0.045*g)
``````

where v is the speed in km/h and g is the grade in percentage (e.g., for 7% incline, g = 7).

We really don't need to be too precise because there are so many other factors involved, so there's no problem simplifying the formula above to, say:

``````MET = v*(1 + 0.05*g)
``````

That is, add 5% to the calories burned when running for every 1% of incline. In your case, it's 10% incline, so +50% calories. That's the same as running +50% of the distance: 500 + (50% of 500) = 750.

You could also try an online calculator (but some of them are way off). I tried this one with some typical values and the OP's weight:

• 10 km/h, 10%, 55 kg, 2.5 min (which gives 500 m): net energy of 33 kcal
• 10 km/h, 0%, 55 kg, 3.75 min (which gives 750 m): net energy of 34 kcal