Accurate Formula To Determine Calories Burned Jogging

All,

Here is a simple formula to figure out ACCURATEY how to determine how many calories your burning while Jogging. I did my research after I became concerned that the user input calories burned was not accurate.

Take 0.75 x (your weight in pounds). Multiply this total times the number of miles ran, and it accurately indicates total loss. For example, a jogger who weighs 150 pounds and runs four miles would burn roughly 450 calories. We get this number by taking 150 x 0.75, which equals 112.5, then multiplying by four, since the jogger ran four miles.
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Replies

  • Awesome, thanks for the information.
  • vickiele1
    vickiele1 Posts: 394 Member
    Thanks for sharing this information. I wonder if there are other equations/formulas for other kinds of exercise. For example my nordic track recumbent bike indicates the calories indicated on the screen are based on a 185 lb male. Well, I don't weigh 185 lbs nor am I a male - it doesn't let me input my own data. So, I try to use the built in HRM to estimate my average heart rate and then use a calculator to estimate calorie burn. I usually try to use a lesser amount than is indicated because I don't trust the numbers and would rather be estimating too low rather than too high.

    Vickie
  • clioandboy
    clioandboy Posts: 963 Member
    thanks for this,

    I wonder is there anything to determine different speeds?

    do I use the same or similar cals if I sedately run 3 miles in say 45 mins or if I do the same 3 miles in 30mins ? cheers
  • Heliconia
    Heliconia Posts: 166 Member
    bump
  • TrainingWithTonya
    TrainingWithTonya Posts: 1,741 Member
    We use the following formulas in my exercise physiology program at university.

    ACSM Walk: 3.5 + [speed (meters per minute) x 0.1] + [speed (meters per minute) x % grade x 1.8] = VO2 in ml/kg/min
    Meters per minute is miles per hour x 26.8.

    ACSM Run: 3.5 + [speed (meters per minute0 x 0.2] + [speed (meters per minute) x % grade x 0.9] = VO2 in ml/kg/min

    Bike: (1.8 x work rate) / [weight (kilograms) + 7] = VO2 in ml/kg/min
    Work rate is resistance in kilograms x flywheel distance x revolutions per minute

    Arm Bike: (3 x work rate) / [weight (kilograms) + 3] = VO2 in ml/kg/min

    Stepper: (0.2 x frequency of steps) + (1.33 x1.8 x Height of step x frequency of steps) + 3.5 = VO2 in ml/kg/min

    Once you have the VO2, you can convert it to METs by dividing it by 3.5 (because one MET is 3.5ml/kg/min). You can then multiply the METs by your weight in kilograms to get calories per hour.
  • FreeRiderInNOVA
    FreeRiderInNOVA Posts: 33 Member
    Yes. Notice when you create your own personal jogging exercise entry, you are prompted to enter it on the right, and there you can change the minutes exercised based on your personalized entry you just created, and it will change based on that.
  • vickiele1
    vickiele1 Posts: 394 Member
    We use the following formulas in my exercise physiology program at university.

    ACSM Walk: 3.5 + [speed (meters per minute) x 0.1] + [speed (meters per minute) x % grade x 1.8] = VO2 in ml/kg/min
    Meters per minute is miles per hour x 26.8.

    ACSM Run: 3.5 + [speed (meters per minute0 x 0.2] + [speed (meters per minute) x % grade x 0.9] = VO2 in ml/kg/min

    Bike: (1.8 x work rate) / [weight (kilograms) + 7] = VO2 in ml/kg/min
    Work rate is resistance in kilograms x flywheel distance x revolutions per minute

    Arm Bike: (3 x work rate) / [weight (kilograms) + 3] = VO2 in ml/kg/min

    Stepper: (0.2 x frequency of steps) + (1.33 x1.8 x Height of step x frequency of steps) + 3.5 = VO2 in ml/kg/min

    Once you have the VO2, you can convert it to METs by dividing it by 3.5 (because one MET is 3.5ml/kg/min). You can then multiply the METs by your weight in kilograms to get calories per hour.

    Okay, I'm a pretty intelligent person, but this is a foreign language to me. Can you provide an example. Say, I intend to run 5 miles per hour at an incline of 2 on the treadmill. I weigh 148.8 lbs - help me do this equation - because I don't get metric and this explanation is anything but clear.

    Thanks

    Vickie
  • reneelee
    reneelee Posts: 877 Member
    bump
  • jekl
    jekl Posts: 2
    can this apply to the elliptical trainer as well?
  • jekl
    jekl Posts: 2
    For example if I did 60 minutes/5.5 miles- average HR 157 intensity 9-10 (50% of that machine I think) 195 lbs 5'7'. Machine says I burned 600- this program says over 800- I kind of doubt all of those...seems high
  • FreeRiderInNOVA
    FreeRiderInNOVA Posts: 33 Member
    Yeah, thats high. I dont know how to find an accurate formula for that other than google it. Probably your best bet.
  • For example if I did 60 minutes/5.5 miles- average HR 157 intensity 9-10 (50% of that machine I think) 195 lbs 5'7'. Machine says I burned 600- this program says over 800- I kind of doubt all of those...seems high

    According to Calburn you would have burned 582.24.

    http://www.braydenwm.com/calburn.htm

    I use Calburn when I'm figuring things out because it takes my HRM into account. Those numbers don't seem high at all.
  • TrainingWithTonya
    TrainingWithTonya Posts: 1,741 Member
    We use the following formulas in my exercise physiology program at university.

    ACSM Walk: 3.5 + [speed (meters per minute) x 0.1] + [speed (meters per minute) x % grade x 1.8] = VO2 in ml/kg/min
    Meters per minute is miles per hour x 26.8.

    ACSM Run: 3.5 + [speed (meters per minute0 x 0.2] + [speed (meters per minute) x % grade x 0.9] = VO2 in ml/kg/min

    Bike: (1.8 x work rate) / [weight (kilograms) + 7] = VO2 in ml/kg/min
    Work rate is resistance in kilograms x flywheel distance x revolutions per minute

    Arm Bike: (3 x work rate) / [weight (kilograms) + 3] = VO2 in ml/kg/min

    Stepper: (0.2 x frequency of steps) + (1.33 x1.8 x Height of step x frequency of steps) + 3.5 = VO2 in ml/kg/min

    Once you have the VO2, you can convert it to METs by dividing it by 3.5 (because one MET is 3.5ml/kg/min). You can then multiply the METs by your weight in kilograms to get calories per hour.

    Okay, I'm a pretty intelligent person, but this is a foreign language to me. Can you provide an example. Say, I intend to run 5 miles per hour at an incline of 2 on the treadmill. I weigh 148.8 lbs - help me do this equation - because I don't get metric and this explanation is anything but clear.

    Thanks

    Vickie

    Okay, here is your example. Since you are running, then you are going to use the following formula:

    ACSM Run: 3.5 + [speed (meters per minute0 x 0.2] + [speed (meters per minute) x % grade x 0.9] = VO2 in ml/kg/min

    First we'll determine the various things you need. Speed in meters per minute is mph x 26.8, so 5 x 26.8 = 134. Percent grade is the 2% incline from the treadmill, which is 0.02 in decimal form.

    3.5 + [134 x 0.2] + [134 x 0.02 x 0.9] = 3.5 + 26.8 + 2.412 = 32.712 ml O2/kg/min

    32.712 / 3.5 = 9.35 METs

    148.8 lbs / 2.2046 = 67.5 kilograms

    9.35 METs x 67.5 kg = 631 calories per hour

    Now, if you only ran for half an hour, you'd divide that 631 calories in half to get a 30 minute workout at that level.
  • LuckyLeprechaun
    LuckyLeprechaun Posts: 6,296 Member
    whoa. That seems HARD. I'm not knocking you, I'm actually way impressed that you 1. know all that and 2. can understand it

    What about just using a HRM?
  • vickiele1
    vickiele1 Posts: 394 Member
    We use the following formulas in my exercise physiology program at university.

    ACSM Walk: 3.5 + [speed (meters per minute) x 0.1] + [speed (meters per minute) x % grade x 1.8] = VO2 in ml/kg/min
    Meters per minute is miles per hour x 26.8.

    ACSM Run: 3.5 + [speed (meters per minute0 x 0.2] + [speed (meters per minute) x % grade x 0.9] = VO2 in ml/kg/min

    Bike: (1.8 x work rate) / [weight (kilograms) + 7] = VO2 in ml/kg/min
    Work rate is resistance in kilograms x flywheel distance x revolutions per minute

    Arm Bike: (3 x work rate) / [weight (kilograms) + 3] = VO2 in ml/kg/min

    Stepper: (0.2 x frequency of steps) + (1.33 x1.8 x Height of step x frequency of steps) + 3.5 = VO2 in ml/kg/min

    Once you have the VO2, you can convert it to METs by dividing it by 3.5 (because one MET is 3.5ml/kg/min). You can then multiply the METs by your weight in kilograms to get calories per hour.

    Okay, I'm a pretty intelligent person, but this is a foreign language to me. Can you provide an example. Say, I intend to run 5 miles per hour at an incline of 2 on the treadmill. I weigh 148.8 lbs - help me do this equation - because I don't get metric and this explanation is anything but clear.

    Thanks

    Vickie

    Okay, here is your example. Since you are running, then you are going to use the following formula:

    ACSM Run: 3.5 + [speed (meters per minute0 x 0.2] + [speed (meters per minute) x % grade x 0.9] = VO2 in ml/kg/min

    First we'll determine the various things you need. Speed in meters per minute is mph x 26.8, so 5 x 26.8 = 134. Percent grade is the 2% incline from the treadmill, which is 0.02 in decimal form.

    3.5 + [134 x 0.2] + [134 x 0.02 x 0.9] = 3.5 + 26.8 + 2.412 = 32.712 ml O2/kg/min

    32.712 / 3.5 = 9.35 METs

    148.8 lbs / 2.2046 = 67.5 kilograms

    9.35 METs x 67.5 kg = 631 calories per hour

    Now, if you only ran for half an hour, you'd divide that 631 calories in half to get a 30 minute workout at that level.

    Thanks so much for the example - now I get it - I almost had it, but missed a step - so I get it now. Thanks so much for explaining it to me.

    Vickie
  • TrainingWithTonya
    TrainingWithTonya Posts: 1,741 Member
    whoa. That seems HARD. I'm not knocking you, I'm actually way impressed that you 1. know all that and 2. can understand it

    What about just using a HRM?

    Yeah, I didn't realize I had to be a mathematician to be an exercise physiologist, either, until getting in depth in the program. Thankfully, my Daddy insisted on me getting an accounting degree before I went to follow my sports dreams so that I can now follow the math we gym rats need. LOL

    As for a heart rate monitor, they are great. But they have their issues, too. We use them in our program so we don't have to take a manual pulse while doing exercise testing, but not for determining calorie burns. The reasons for that are, A) each manufacturer uses their own proprietary formula for determining calories burned which may or may not have been scientifically proven accurate (Generally my prof's trust the Polar though more then other brands) and B) medications can effect heart rate which will effect the calorie burns from a HRM. Since ours is a clinical program, we focus on working with a lot of various illnesses and injuries which means working with a population of people on meds, so that is probably the number one reason why we don't use HRM's for calorie burns. And just FYI, caffeine is one of the things that our patients have to be off of for at least 12 hours before testing due to the fact that we are doing EKG at the same time as our treadmill test because EKG is reading heart rhythms and caffeine increases heart rate without exertion. So, someone who is relying on a HRM for calorie burns but has caffeine the day they exercise (morning coffee anyone?) will get a higher calorie burn reading then what they actually burned because of the increase in heart rate even at rest because of the caffeine. If you don't use any caffeine (even chocolate or tea), take any meds (cold meds are bad about effecting heart rate), and have a high quality HRM that is properly programmed for you (one wrong number in the programming can throw it off too), then you should be okay with a HRM for estimating calorie burns. That's all any of them are really, though, is estimates. In the real world we can't be 100% accurate with calorie burns. The only way to get accurate calorie burns is in a lab.
  • rnroadrunner
    rnroadrunner Posts: 402 Member
    bump. ????
  • butron
    butron Posts: 1 Member
    I tried to simplify and translate the formula for running into pagan units ;-). Even then, instead of dealing with miles/hour, I deal with total distance in miles (just like the original simple formula at the beginning of this forum). This is a simple change but, in my view, a sensible one. Using miles, pounds and percentage of incline (where an incline of 2% is simply a 2) the formula is:

    Calculate MET = 1 + (1.53 + 0.07 x Incline%) x total distance
    Total calories consumed: MET x Weight / 2.2046

    It does not matter if the speed changes along the way or the time it takes to run the distance. It does matter that the incline be kept constant, otherwise one should use a time weighted average for the incline. The latter statement is for mathematicians, for non mathematicians just eyeball your average incline and you will be pretty close.
  • Mr_Knight
    Mr_Knight Posts: 9,532 Member
    Calculate MET = 1 + (1.53 + 0.07 x Incline%) x total distance
    Total calories consumed: MET x Weight / 2.2046

    I would change the first line to:

    Calculate MET = 1 + (1.53 + 0.07 x (Incline%-2)) x total distance

    Reason is that 1-2% incline is generally accepted as what is needed to more closely replicate non-treadmill running, so the first 2% of incline don't really count.
  • RoyBeck
    RoyBeck Posts: 947 Member
    I'm a runner not a jogger :)

    But seriously how accurate is this? I'm not so sure.