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I How to calculate the calories burnt during walking

  1. Jan 24, 2017 #1
    Based on Laws of Physics, how to calculate the calories burnt by a person, weighing 60 kg during walking at a speed of 4 kmh–1 for 1 hour.
    Surprisingly different answers were given by online calculators. Though I am not a Physicist, please let me know the formula or principle behind the calculation and I know dietary Calorie = 1000 calories.
    is this logic
    =1/2 × 60 kg × (4000/3600 ms–1)2 ~ 37 J
     
    Last edited: Jan 24, 2017
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  3. Jan 24, 2017 #2

    PeroK

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    It's not really a theroretical physics question because it depends mostly on physiology and muscular inefficiencies. Two related questions:

    How many calories do you burn if you sit in a chair for an hour?

    How many calories do you burn if you cycle at 4km/h for one hour?
     
  4. Jan 24, 2017 #3

    BvU

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    There isn't much simple physics in this calculation: the physical work to actually move something horizontally for an hour has a lower limit of zero.
    The physics would then be in maintaining body temperature and that's indeed rather broad in terms of Joules, as you can understand (wind, clothing).

    Is there no common range of values at all ? Examples ?

    I personally don't trust the makers of exercise machines: they never seem to be able to convert calories/hour into Watt (Joule/second).
     
  5. Jan 24, 2017 #4
    Thank you BvU. Can we use simple Kinetic energy equation energy require to move mass of 60 kg at a speed of 4km/h. Though it is not accurate, is it logic way. Please forgive me I am not Physicist.
     
  6. Jan 24, 2017 #5
    Thank you BvU. Can we use simple Kinetic energy equation energy require to move mass of 60 kg at a speed of 4km/h. Though it is not accurate, is it logic way. Please forgive me I am not Physicist.

    =1/2 × 60 kg × (4000/3600 ms–1)2 ~ 37 J
    is this logic
     
    Last edited: Jan 24, 2017
  7. Jan 24, 2017 #6

    PeroK

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    The energy required for that is simply zero! It takes no energy to maintain constant motion. Essentially.
     
  8. Jan 24, 2017 #7
    =1/2 × 60 kg × (4000/3600 ms–1)2 ~ 37 J
    is this logic
     
  9. Jan 24, 2017 #8
    =1/2 × 60 kg × (4000/3600 ms–1)2 ~ 37 J
    is this logic
     
  10. Jan 24, 2017 #9

    PeroK

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    Which is independent of the time. That's the initial energy to accelerate to ##4km/h##. The Kinetic energy to remain at that speed for 1 second, 1 minute, 1 hour or whatever is zero. That's your problem. All the energy required to maintain 4km/h is wasted through one inefficiency of motion or another.
     
  11. Jan 24, 2017 #10

    russ_watters

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    I always figured one was input and the other output, but I've never bothered to look at the actual comparison/conversion.
    [Edit] Pulling some numbers from memory and converting, I guess my bike is telling me I'm about 20% efficient.
     
    Last edited: Jan 24, 2017
  12. Jan 24, 2017 #11
    =1/2 × 60 kg × (4000/3600 ms–1)2 ~ 37 J
    is this logic
     
  13. Jan 24, 2017 #12
    Assume if speed is constant for 1 hour, then how can we proceed? By simple assumptions is it possible to get some values. I wonder is it that much difficult... sorry
     
  14. Jan 24, 2017 #13

    PeroK

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    You're not listening. No, it's not possible for the reasons given. Constant motion requires essentially no energy. All the energy is used up by muscular inefficiencies. The only way to get a figure is to study muscles (or get some empirical data).
     
  15. Jan 24, 2017 #14
    Yeah I understand once triggered it moves. So, by assuming if there is no factors influencing, whether this should be considered as energy required.
    Correct me if I am wrong. Sorry for annoying.
     
  16. Jan 24, 2017 #15

    PeroK

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    Obviously it can't be. You have a very small amount of energy required to get to 4km/h, and then an unknown energy to maintain this.

    You can compare with a bicycle where a constant motion of 4km/h would require almost no effort. On a well-oiled bicycle you could coast perhaps 50m before you stopped. If you are walking, there is no coasting mechanism. You have to keep moving your muscles all the time. That's where the energy goes.

    PS this is why the bicycle was such a great invention. It has no engine and you are doing all the work, but to move at, say, 10km/h is easy on a bike, but hard work if you are running.
     
  17. Jan 24, 2017 #16
    You wan't to know the 'calories burnt' ..... the most accurate way to measure this must be to measure the CO2 output (and O2 consumed) when on a treadmill ... this can be compared with what it takes to burn carbohydrates , a simple chemical equation ...

    I seem to remember the usable output from an athlete (to drive a man powered airplane) is 300w .... a horsepower (output from a horse) is 750w ... I think both are around 10% efficient.
     
  18. Jan 24, 2017 #17

    Andy Resnick

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    As this thread shows, your question is not easy to answer. Your calculation equates kinetic energy and 'expended energy' using the rationale that when you stop trying to walk, you more or less immediately stop moving. An alternative approach is to use expended power P = Fv, where v is the velocity and F is the frictional force. Neither of these are 'well-respected' approaches to answering your question because gait is a very complex motion:

    http://www.footeducation.com/foot-a...-foot-and-ankle/biomechanics-of-walking-gait/

    https://books.google.com/books?id=1...onepage&q=biomechanics walking energy&f=false

    Figure 5.6 in the second link has some information about energy expenditure. Practically speaking, measurements of expended energy are performed by measuring O2 consumption. It's common for a wide range of values to be reported, since many physiological factors (age, weight, overall fitness) impact energy expenditure.
     
  19. Jan 24, 2017 #18
    Thank you for your explanation and your time.
     
  20. Jan 24, 2017 #19
    Thank you for your detailed explanation and your time.
     
  21. Jan 24, 2017 #20
    Thank you Perok and Resnik,
    I think the question is recoined to get a straight uncomplicated answer
    "how to calculate the energy for an object weighing 60 kg to move in a frictionless surface at a speed of 4 kmh–1 for 1 hour and by assuming without the influence of any external factors" In that case may I use
    =1/2 × 60 kg × (4000/3600 ms–1)2 ~ 37 J
     
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