Stationary bike watts per hour

bufger

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

aokoye

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

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

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

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

sijomial

Just some points....
Your body weight is insignificant for a stationary bike non weight bearing exercise. If you start doing part of your stationary bike workout standing cycling that changes (also means the power to calories estimate in under-stated).

Your weight does become significant when cycling outdoors, especially on hills and stop/start rides. Apps like Strava take your weight into account and try to back calculate your power.

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).

The simple formula is average watts x 3.6 for an hour = net calories. (140w x 3.6 = 504cals for an hour, 252 for 30mins.)

Eating back half MFP exercise calories is a mathematical abomination! People would be better served using more appropriate estimating methods for their own particular exercise routines.

robertw486

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

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

^ It takes 8x the power to go 2x the speed because air resistance increases with the cube of speed.

tbright1965

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:

Jthanmyfitnesspal

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

Oops! Thanks for the correction, both of you.

jjpptt2

And I thought I over thought things...

mjbnj0001

jjpptt2 wrote: »

And I thought I over thought things...

"Physics is Fun." LOL.

BabyJens

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

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

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).