How much energy do I use when swinging a sledge hammer?

  1. I love to split firewood, mainly for exercise, but I wanted to know how much energy I'm burning with each swing. I'm trying to find out if the method I'm using to figure this out is correct. I designed an electronic device with infrared sensors that are one foot apart. As I swing the sledge hammer to split the wood, it goes passed the first sensor which starts a timer that measures in ten-thousanths of a second. When it passes the second sensor it stops the timer. The timer now contains the total time that the sledge hammer took to travel one foot. If I take the reciprocal of this time I have the velocity in feet per second that the sledge hammer was traveling. The sledge hammer weighs 16 pounds. So it has a mass of m=W/32=16/32=0.5 slugs. Now if I apply the formula for Kinetic Energy which is K.E.=(mv^2)/2. This now gives me the total energy that the sledge hammer had at impact. Since the sledge hammer started at rest, then I gave it 100% of this energy. If I convert this energy in foot-pounds into calories by multiplying it by the conversion factor 0.3239, this should give me the total calories I burnt during this one swing. Is this method correct or am I doing something wrong and misapplying physics?
  2. jcsd
  3. NascentOxygen

    Staff: Mentor

    It's true that you imparted a good deal of energy to the hammer head as it swung downwards, but gravity assisted you. However, gravity is merely returning the potential energy that you imparted to the hammer on its upswing. So if you determine the kinetic energy just before impact, and consider that that's the energy you gave it for the whole blow, the result should be close enough, I think.
  4. interesting question..I have wondered similarly....

    I suspect you are burning more calories than is apparent at the hammer head.
    For one thing you are of course burning maybe 1500 or 2,000 calories daily just sitting around doing nothing....just staying alive. In other words, I suspect your conversion factor of 0.3239 is too low. You are also imparting energy to both arms and burning calories in muscle tensing and relaxation associated with your work.

    I don't know how the biological production of useful energy output works but a quick check here

    suggests there is a lot to consider....seems too complicated...

    It would be interesting to have some rule of thumb for the efficiency of human effort:

    I vaguely recall reading that for extended periods people cannot produce more than 1/10 or 1/16 hp for extended periods ??,ok, , found something in wikipedia:

    [although I'm not sure just what "produce" means.... I assume "useful output"??

    Suggest you test your result against this as a rough check. [convert hp-hr to calories,
    using maybe 0.2 HP depending on how "trained" you consider yourself]
  5. Thank you NascentOxygen and Naty1 for your replies. I am no athlete. I workout only once or twice a week. But in the summer I can spend many hours splitting firewood. That is when I get my best workouts. I got the 0.3239 conversion factor from my HP 48G calculator. I had it convert 1 ft/lb of energy into calories and that is the number it gave as the result. I'm not sure if the calories used in physics are the same as dietary calories which are sometimes called kcals. The main purpose of this thread is to find out if the method I'm using to calculate the calories I'm burning is reasonably accurate. What I have done now is to take the test apparatus that I used to calculate the time it took for the sledge hammer to travel one foot and turn this apparatus into an exercise machine. Unlike actually splitting real firewood which is done mainly during the warmer weather, I can use this machine to exercise all year round. That is the reason I need it to be reasonably accurate. The machine will keep a running total of the calories burned from all the swings. If it is not accurate then the total calories burned will be a bogus number and unusable. I am now ready to program the microcontroller to calculate total calories. I just want to make sure that I'm doing it right so I don't waste my time writing a program that will be useless.
  6. Kenny: why not wait a few days here and see if anyone replies more specifically.

    If not, try posting here:

    BIOLOGY, in these forums, and ask about biological caloric consumption producing useful
    work output.
    Somebody calculates biological "calories burned" I'm pretty sure because
    that is an output on some treadmills with electronic monitoring....
  7. NascentOxygen

    Staff: Mentor

    As they say, that goes without saying. :smile:

    Just balancing the sledge hammer (I call it 'block buster', it's like a heavy, very blunt axe?) in midair burns up calories, but adds nothing to the potential energy of the hammer. So if Kenny_M wished to increase his kcal burnt with each blow, he could just pause for 10 secs with the hammer held in midair at the top of its sweep, or he could raise it with exaggerated slowness. Neither of these efforts add to the K.E. of the falling hammer, but will burn more calories. Just how many more is anyone's guess.

    If I were a fearless gambler, I'd say the human effort (in kcals) would be at least double that recorded in Kenny-M's falling hammer apparatus, but then I don't gamble. The odds always seem stacked against me! :smile:

    When I lived in the countryside, one of the few physical exertions from which I derived enjoyment was chopping up blocks of firewood with the blockbuster, too! The other being bike riding along rough forest tracks.
  8. sophiecentaur

    sophiecentaur 13,286
    Science Advisor
    Gold Member

    We have had two similar threads on this topic recently and there is no simple answer. All you can say is that the hammer ended up with a certain amount of Kinetic Energy (which could be measured). What energy you actually 'put into' the event is much less certain and depends upon your technique and training.
  9. I thank all of you again for your replies. I had no idea this would be so complex. I tried a different approach to getting an answer that I think all of you will find interesting. I went to my local university and tried to speak to both a physics and biology professor. Unfortunately the biology professor had left and would be back after the holidays. But I did get to speak to the physics professor. He explained the difference between kcals in biology and calories in physics. He said that kcals are the energy available to the body from the oxidation of food. It has nothing to do with the energy our body uses during work. Then he explained the physics version of energy which I found most interesting. He said in physics, no matter how much energy you use to do work, you are only credited with the work done in the direction of motion. My calculation is based on the sledge hammer traveling vertically downward. He said the sledge hammer is actually traveling in rotational motion and therefor has rotational acceleration. He also said that if the hammer hits when it is perpendicular (when the handle is horizontal), then 100% of the hammer's energy will be used in splitting the wood. But if the hammer is not perpendicular, then there will be a horizontal and vertical component of force and only the vertical component will be used to calculate the energy because it is in the direction of motion. Then he said something that brightened up my day. He said that even though physics only accounts for the energy in the direction of motion, my body has to burn enough calories to give the sledge hammer all its energy. So my calculation should be fairly accurate. It will actually burn more calories than my calculation because the human body is not 100% efficient. He also said some other things but I don't remember what they were.
    Last edited: Dec 23, 2011
  10. I do not understand what that means...I'd call it simply wrong.
    An explanation which matches my understanding is in the first section here:

    So this sure seems the appropriate measure of input energy.

    His explanation here has little to do with your question since when chopping wood we all attempt to hit the wood vertically....Your conclusion is unwarranted and is explained in my initial post. Your use more calories that is measured via the ax head...the issue is "how much more?"
  11. sophiecentaur

    sophiecentaur 13,286
    Science Advisor
    Gold Member

    Your calculation will be pretty INaccurate, actually because the Work Done On an object can easily be zero (i.e holding a sledgehammer, stationary, out in front of you) but you will be sweating away if you want to do this for more than a few seconds and using a lot of energy supply up. Efficiency, in strictly mechanical terms here is Zero.
    You can do better than that, of course, particularly when you match the output of your muscles to a suitable task. Riding a bicycle uphill in a suitable gear is a good example of this.
    It's not really an satisfactorily answerable question.
  12. You still need to accelerate the hammer, so you can't take the velocity at one point of the swing and multiply it by the length of the swing.
  13. sophiecentaur

    sophiecentaur 13,286
    Science Advisor
    Gold Member

    A given final Kinetic Energy of the head can be achieved with many combinations (integrals) of different force times distance at different parts of the travel. Some paths may be more efficient for a particular person's physiology than others. Humans are very non-ideal / non-linear machines.
  14. Not particularly scientific but this might be of interest to you. A lot of digital exercise machines have METs calculations built-in. (Metabolic Equivalent)

    I would place your MET somewhere between 8 and 10 and this can then be used to approximate energy burn in joules.

    I post this on New Year's Eve so think of all those people discovering this for the first time as part of their new years resolution to lose weight. They'll be in the gym on Jan 3rd with shiny new shoes, towels and water bottles, staring in a confused manner at the machines, and most will be gone by February!
  15. I understand everyone's point. This is much more complicated than I expected it to be. There is no way that I will be able to get an exact amount of calories burned per swing. At this time I will be using the formula I showed previously. I spent the last week and a half programming an 8051 microcontroller to perform the required math for that formula. I programmed it in assembly language so it was a formidable task for me. It is all debugged now and works fine. I will use this until a better equation comes along that I am capable of understanding and programming into a microcontroller. It may not be accurate but at least it gives me some reference point as to how many calories i'm burning. I thank everyone for your comments and suggestions. And if anyone should come across a more accurate formula, then please let me know. And I hope it doesn't contain any calculus. I'm not smart enough to program a microcontroller to perform calculus.
  16. NascentOxygen

    Staff: Mentor

    If you were to paint a white dot on the axe head (even better, use luminous paint), then you may be able to photograph the axe swing in dim lighting using a long exposure and a nearby flashing light (stroboscope) to get a multiple exposure photo that would show the increasing speed of the axe as it falls. The region of the swing where the speed is undergoing the most rapid change is where you are doing most work. I reckon you'll find that most of your energy is expended while the axe swings through about 50 degrees of an almost perfect arc.
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