Stationary bike watts per hour

bufger
bufger Posts: 763 Member
in Fitness and Exercise
I'm trying to get a more accurate handle on my excercise calories now I'm in a routine and have found my current limits.

213lbs, 6'1, 34 year old male.
140 watts for 30 minutes straight (distance was just short of 10 miles). The stationary bike said I had burned 270kcals and MFP probably says I've burnt a thousand..

What calculation would you use to incorporate all of my stats so I can work it out in the future too?

I appreciate any time spent helping. I realise I can just eat back half of my MFP calories but I'm measuring food down to a gram and I'd like to measure energy expenditure more accurately too :)

Replies

  • aokoye
    aokoye Posts: 3,495 Member
    I'm just gonna put this here - http://mccraw.co.uk/powertap-meter-convert-watts-calories-burned/ . There have also been a number of threads about this over the past handful of months but that link sums them all up assuming the stationary bike's power meter is accurate (which in a lot of cases means calibrated on a regular basis).
  • NorthCascades
    NorthCascades Posts: 10,968 Member
    Assuming your power meter (the thing that says you put out 140 watts) is accurate, you have indeed burned about 250 kcal. You can eat them all back safely.

    Bikes are one of the most efficient forms of transportation ever invented. Which is another way of saying they don't burn gobs of calories. But they're a good cardiovascular workout with almost no impact.
  • ttippie2000
    ttippie2000 Posts: 412 Member
    Ohhhhhh, bicycling has impact. Big time. You just save it up and have it all at once. My turn came when an elderly lady failed to notice she was driving in the bike lane. Got me a new bike though...
  • bufger
    bufger Posts: 763 Member
    aokoye wrote: »
    I'm just gonna put this here - http://mccraw.co.uk/powertap-meter-convert-watts-calories-burned/ . There have also been a number of threads about this over the past handful of months but that link sums them all up assuming the stationary bike's power meter is accurate (which in a lot of cases means calibrated on a regular basis).

    That's perfect, thanks
  • robertw486
    robertw486 Posts: 2,399 Member
    sijomial wrote: »
    As a stationary bike doesn't actually move (!!!) you can't use MFP speed related cycling estimates (they are also very high for road cycling too).

    I ride a fat tire and heavy mountain bike, and have the aero profile of a brick. And the MFP numbers would likely still have me gaining weight if I ate back the estimates. The speed ranges they use in their estimate alone would just toss the numbers all over the place.


    As for the original question, hard to beat a wattage number unless it's not a reliable meter source.
  • Jthanmyfitnesspal
    Jthanmyfitnesspal Posts: 3,522 Member
    MFP will gives me 943kcals/hr for "Bicycling, 16-20 mph, very fast (cycling, biking, bike riding)." (I don't know if it is using my weight in that calculation. Yes, that's a high number.

    The thing is that the resistance of an indoor bike is totally arbitrary, so you may not be simulating the effort of outdoor riding. Also, there is a huge difference between the effort needed to go 16mph and 20mph. I don't even think it's linear with speed due to the non-linearity of wind resistance. Plus, outdoors, you will be climbing hills and stuff.

    Most exercise bikes (at least at a gym) will give you a calorie burn number. If it's a reasonable good model, I'd go with whatever it estimates rather than MFP or some other formula.
  • NorthCascades
    NorthCascades Posts: 10,968 Member
    ^ It takes 8x the power to go 2x the speed because air resistance increases with the cube of speed.
  • tbright1965
    tbright1965 Posts: 852 Member
    edited July 2018
    NorthCascades wrote: »
    ^ It takes 8x the power to go 2x the speed because air resistance increases with the cube of speed.

    Hmmm, I thought it increased based on v squared, not cubed... But it has been 30 odd years since I took college physics.

    Edited to add: tfd8a3jxe0v8.png

  • Jthanmyfitnesspal
    Jthanmyfitnesspal Posts: 3,522 Member
    NorthCascades wrote: »
    ^ It takes 8x the power to go 2x the speed because air resistance increases with the cube of speed.

    The real answer is that it's some non-linear function of speed, usually approximately proportional to speed-squared:

    http://bikecalculator.com/what.html

    Or, you can believe NASA:

    https://www.grc.nasa.gov/WWW/K-12/airplane/drageq.html
  • NorthCascades
    NorthCascades Posts: 10,968 Member
    Oops! Thanks for the correction, both of you.
  • jjpptt2
    jjpptt2 Posts: 5,650 Member
    And I thought I over thought things...
  • mjbnj0001
    mjbnj0001 Posts: 1,266 Member
    jjpptt2 wrote: »
    And I thought I over thought things...

    "Physics is Fun." LOL.
  • BabyJens
    BabyJens Posts: 1 Member
    NorthCascades had it right the first time. It is a cubic relation between aerodynamic power and speed. The well known square relation is between aerodynamic FORCE and speed. However, P=F*V so the relation becomes cubic for power. Technically it is a dot product, but for our purposes we can treat it as pure multiplication (you only care about drag force so that is defined to exactly oppose your direction of travel and thus the dot product reduces to a standard product). To get total power loss you need to also include rolling and friction resistances.

    Also, watts per hour has units of [E/T^2], which is a change in power output with time. Calories per hour and watts are both units of power but watts per hour is not.
  • lorrpb
    lorrpb Posts: 11,463 Member
    BabyJens wrote: »
    NorthCascades had it right the first time. It is a cubic relation between aerodynamic power and speed. The well known square relation is between aerodynamic FORCE and speed. However, P=F*V so the relation becomes cubic for power. Technically it is a dot product, but for our purposes we can treat it as pure multiplication (you only care about drag force so that is defined to exactly oppose your direction of travel and thus the dot product reduces to a standard product). To get total power loss you need to also include rolling and friction resistances.

    Also, watts per hour has units of [E/T^2], which is a change in power output with time. Calories per hour and watts are both units of power but watts per hour is not.

    Hopefully that explains why cycling uphill is so hard rather than that I’m weak. 🤔🤣
  • sijomial
    sijomial Posts: 19,809 Member
    lorrpb wrote: »
    BabyJens wrote: »
    NorthCascades had it right the first time. It is a cubic relation between aerodynamic power and speed. The well known square relation is between aerodynamic FORCE and speed. However, P=F*V so the relation becomes cubic for power. Technically it is a dot product, but for our purposes we can treat it as pure multiplication (you only care about drag force so that is defined to exactly oppose your direction of travel and thus the dot product reduces to a standard product). To get total power loss you need to also include rolling and friction resistances.

    Also, watts per hour has units of [E/T^2], which is a change in power output with time. Calories per hour and watts are both units of power but watts per hour is not.

    Hopefully that explains why cycling uphill is so hard rather than that I’m weak. 🤔🤣
    @lorrpb
    It actually explains why wearing a baggy jacket and having an upright riding position doesn't matter much at low speed but matters a hell of a lot the faster you go (or the stronger the headwind).

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