Question about calorie defecits and timing.

24

Comments

  • Pokerface
    Pokerface Posts: 7,960
    I don't use a Garmin to 'estimate' calories used. I use a Powertap to much more precisely define how many calories I have used.
  • Chiggy
    Chiggy Posts: 261
    Watts and kCals/min are directly related through J/s.

    For example, 200 Watts is 2.868 kCals/min.

    Back in the mid nineties, Audax UK used to advise their members to consume about 8 kCals/min if their finishing speed was to be 'mid range'. Mid range on a Rando is 22.5 kmh ( 14 mph ).
    At 14mph on a typical Audax bike, this is about 90 Watts average for the 125 mile ride.
    90 Watts is 1.3 kCals/min.
    Dressed is shorts and S/S jersey at these speeds, the rider would be burning at less than 20% efficiency, so 6 times 1.3 is 7.8kCals/min.

    At 14 mph, its a 4 1/2 minute mile, so 1 mile will be 36 kCals/mile. 4500 for a 200 Rando!

    At 20mph average, the Watts increases to 220 W. That's 3.15 kCals/min, and at 25% efficiency in a 3 minute mile, its 37.8 kCals/mile. Not a million miles away from the 36 kCals/mile of the sedate audax rider.

    At 30 mph average, Watts is up at 550. 25% efficiency through a 2 minute mile gives 63 kCals/mile, which is what I use to calc my cals needs on my SWorks working like a man possesed. :lol:
  • P_Tucker
    P_Tucker Posts: 1,878
    So you can translate from watts to speed just like that? 20mph on the flat is the same for Backstedt and Pantani, is it? Astonishing news.
  • Chiggy
    Chiggy Posts: 261
    Pokerface wrote:
    I don't use a Garmin to 'estimate' calories used. I use a Powertap to much more precisely define how many calories I have used.

    What a PowerTap doesn't tell you is your efficiency. Carbon Hydrogen Oxygen molecule consumption vs resultant power at the tyre.

    Any calorific estimations without CO2 carbon balance is pissing in the wind.
    This is the reason I 'overcook' my calorific needs to the 'worst case scenario', so I am confident of getting round the 300 km Audax route.
  • Anonymous
    Anonymous Posts: 79,667
    But different people need different power levels to attain different speeds. And terrain, wind, traffic conditions further affects that, so you are still guessing.

    You seem to post a bit about absolutes, e.g. 200 watts is steady for everyone.

    It isn't. It's like that daft 'healthy man' graph you've referred to. If you weigh 50kg then 200w is a good threshold. If you weigh 100kg it's rubbish.
  • P_Tucker
    P_Tucker Posts: 1,878
    Chiggy wrote:
    Pokerface wrote:
    I don't use a Garmin to 'estimate' calories used. I use a Powertap to much more precisely define how many calories I have used.

    What a PowerTap doesn't tell you is your efficiency. Carbon Hydrogen Oxygen molecule consumption vs resultant power at the tyre.

    Any calorific estimations without CO2 carbon balance is pissing in the wind.
    This is the reason I 'overcook' my calorific needs to the 'worst case scenario', so I am confident of getting round the 300 km Audax route.

    I'm pretty sure your Garmin doesn't know sh!t about your "Carbon Hydrogen Oxygen molecule consumption".
  • Chiggy
    Chiggy Posts: 261
    P_Tucker wrote:
    So you can translate from watts to speed just like that? 20mph on the flat is the same for Backstedt and Pantani, is it? Astonishing news.

    Of course it isn't.

    Marco's coffin is a lot less aerodynamic than Magnus' bike.

    The only data I have around me I can quote is related to myself on my bike. I don't have an extensive library of every cyclist in the world's Cd and roadload data.

    Any cyclist who wants to calc their kCals requirements NEEDS to evaluate their own roadload data.

    Use your brain, and think of another pro cyclist who is still alive.
  • Anonymous
    Anonymous Posts: 79,667
    Chiggy wrote:
    P_Tucker wrote:
    So you can translate from watts to speed just like that? 20mph on the flat is the same for Backstedt and Pantani, is it? Astonishing news.

    Of course it isn't.

    Marco's coffin is a lot less aerodynamic than Magnus' bike.

    The only data I have around me I can quote is related to myself on my bike. I don't have an extensive library of every cyclist in the world's Cd and roadload data.

    Any cyclist who wants to calc their kCals requirements NEEDS to evaluate their own roadload data.

    Use your brain, and think of another pro cyclist who is still alive.

    I have yet to see a post from your good self where you aren't spouting tosh... Trying to patronise with fancy terms when simple ones will do (and do better).
  • Chiggy wrote:
    If you're quoting 'mechanical work', I presume you are talking about the results of a PowerTap recording??
    This, I beg to suggest is the 'work' you did to move the bike forward and up the hills.

    Did you get an estimation of your heat losses? How many calories did you use keeping your skin at the temperature your brain wanted it to be?
    As we've already ascertained, the human can be 33% efficient at the best climatic conditions, so multiply your MECHANICAL result by 3 to get a realistic figure for calories used.

    How does 13800 grab you?
    It is well established that trained cyclists have a gross metabolic efficiency (energy metabolised / mechanical work done) ranging from ~ 19-24%. Professional cyclists are no more or less efficient in general than your average club rider.

    Power meters such as Powertaps or SRMs (which is what I use) accurately record the mechanical work done in joules. You then convert that to an estimate of energy metabolised by dividing by the 19-24% range.

    That expresses the metabolic requirement in joules, however most refer to calories when talking about metabolism, so you'll then need to convert from joules to calories, the conversion of which I believe was posted by someone else but it's ~ 4.18kJ per Cal (kcal).

    Hence when you convert mechanical work performed in kJ to how much energy has been metabolised, you simply multiply the kJ of mechanical work as reported by an accurate power meter by 1.10 +/- 0.05 to get an indicative range.

    Some cyclists are fortunate to know their GME more accurately than others, as it can be determined. This is typically done with a VO2 max test by examining the gas exchange versus power output.

    Of course there will be acute variations in GME, as things such as heat, hydration, cadence for instance have an impact, but over the longer range, a cyclist will have a typical GME.

    Anyone relying on a Garmin device, or a heart rate monitor to estimate energy utilisation is living is a fantasy land. Data provided by such devices are, at best, light comic relief.
  • Chiggy wrote:
    My understanding is the 50/50 fat/CHO point is at about 70% MHR.
    Any HR below 70% is deemed 'fat burning' because when HR has stabilised, more fat than CHO is used for fuel.
    Oh dear.
  • Pokerface wrote:
    Your weight doesn't necessarily mean you'll burn more calories than a lighter person.
    A heavier person will typically have a higher base metabolic rate.
    Pokerface wrote:
    It means it's harder to go uphill for you but depending on how you ride up that hill - you will burn the same as a lighter person. Essentially it's all about how much power you use (which is determined by how fast/hard you ride - and the terrain). The lighter guy might ride uphill twice as fast and use more energy than you, even though he is lighter.
    The energy required to get up a hill is mostly that required to increase potential energy, plus some extra for overcoming wind, rolling and other very minor resistance forces.

    The energy required to move a mass up a given vertical ascent is not related to the rate at which the ascent happens. Any extra energy required to go fast over going slow is simply due to the forces that scale with speed (air & rolling resistance) and on a steep climb (e.g. 8+%) these represent <10% of the overall energy requirement.

    Hence a heavier person will need to use more energy to climb the same hill than a lighter person, no matter how fast each of them do it. One could fly up like a lightweight Pro, and another twiddle their way up in the granny gear. The big guy will have used more energy.

    Think about it. Energy = power x time. Increase the power (and hence speed), then you reduce the time taken.
    Pokerface wrote:
    Don't know that there is a way to determine fat burning percentages. Isn't the 'fat burning zone' a myth?
    The zone is a myth but it is possible to calculate the relative contribution metabolised by gycogen and FFA. It does require one to use a metabolic cart, and do the gas exchange analysis.
  • P_Tucker
    P_Tucker Posts: 1,878
    Chiggy wrote:
    My understanding is the 50/50 fat/CHO point is at about 70% MHR.
    Any HR below 70% is deemed 'fat burning' because when HR has stabilised, more fat than CHO is used for fuel.
    Oh dear.

    :lol:
  • Pokerface
    Pokerface Posts: 7,960
    Chiggy wrote:
    Pokerface wrote:
    I don't use a Garmin to 'estimate' calories used. I use a Powertap to much more precisely define how many calories I have used.

    What a PowerTap doesn't tell you is your efficiency. Carbon Hydrogen Oxygen molecule consumption vs resultant power at the tyre.

    Any calorific estimations without CO2 carbon balance is pissing in the wind.
    This is the reason I 'overcook' my calorific needs to the 'worst case scenario', so I am confident of getting round the 300 km Audax route.


    As Alex explained, the efficiency number will be +- 5%

    So, even if I'm consistently 5% out - that's only 50 cals for every 1000 it 'says' I burn.

    That's hardly p!ssing in the wind.
  • G-Wiz wrote:
    Anyway, thanks to Pokerface the penny has dropped. I don't need to replace all the calories because some of them come from fat, which is the whole point. As long as I top up the glycogen stores I should be OK. That also explains why I only seem to need to eat the same after an intense sufferfest hour as after an easy 3 hour ride, because they both do about the same to glycogen stocks.

    So, the next logical question is:

    Is there a way to estimate the % of calories coming from fat at a given intensity, and how do I know where the food eaten on the ride is going? Presumably even when you're completely drained, some of it goes back to fat?
    Don't get sucked into thinking about selectively burning and replacing fats and glycogen. it doesn't work like that.

    What matters for weight changes is the overall calorie balance (irrespective of what fuels substrate was metabolised, and what balance of CHO/fat you eat).

    Eat a healthy balanced diet with all the appropriate food types with the necessary macro and micro nutrients, and maintain a sustainable calorie deficit for the period you seek to lose body mass.

    If you are not losing weight, you are simply eating too much for the amount of exercise you do.
  • Pokerface
    Pokerface Posts: 7,960
    Hence when you convert mechanical work performed in kJ to how much energy has been metabolised, you simply multiply the kJ of mechanical work as reported by an accurate power meter by 1.10 +/- 0.05 to get an indicative range.


    Alex - I normally just look at the work figure (kJ) as the amount of calories used for the ride.

    Are you saying that cals with be slightly higher (1.05-1.15 range)?

    If so, a nice conservative estimate is just to stick with the kJ number?


    Thanks
  • P_Tucker
    P_Tucker Posts: 1,878
    Pokerface wrote:
    Hence when you convert mechanical work performed in kJ to how much energy has been metabolised, you simply multiply the kJ of mechanical work as reported by an accurate power meter by 1.10 +/- 0.05 to get an indicative range.


    Alex - I normally just look at the work figure (kJ) as the amount of calories used for the ride.

    Are you saying that cals with be slightly higher (1.05-1.15 range)?

    If so, a nice conservative estimate is just to stick with the kJ number?


    Thanks

    As he stated, efficiency is 19-24%, 1 joule = 0.239 calories so dividing one by t'other gives a range of 0.995 to 1.257.
  • Pokerface
    Pokerface Posts: 7,960
    P_Tucker wrote:
    Pokerface wrote:
    Hence when you convert mechanical work performed in kJ to how much energy has been metabolised, you simply multiply the kJ of mechanical work as reported by an accurate power meter by 1.10 +/- 0.05 to get an indicative range.


    Alex - I normally just look at the work figure (kJ) as the amount of calories used for the ride.

    Are you saying that cals with be slightly higher (1.05-1.15 range)?

    If so, a nice conservative estimate is just to stick with the kJ number?


    Thanks

    As he stated, efficiency is 19-24%, 1 joule = 0.239 calories so dividing one by t'other gives a range of 0.995 to 1.257.

    SO I was right in saying - I can stick with the kJ number as a good guess for calories burned - and am only going to be 5% out either way (at the extreme)?
  • Anonymous
    Anonymous Posts: 79,667
    edited December 2010
    Yes. Just use a direct 1 to 1 comparison, it's all gravy anyhoo.
  • P_Tucker
    P_Tucker Posts: 1,878
    Pokerface wrote:
    P_Tucker wrote:
    Pokerface wrote:
    Hence when you convert mechanical work performed in kJ to how much energy has been metabolised, you simply multiply the kJ of mechanical work as reported by an accurate power meter by 1.10 +/- 0.05 to get an indicative range.


    Alex - I normally just look at the work figure (kJ) as the amount of calories used for the ride.

    Are you saying that cals with be slightly higher (1.05-1.15 range)?

    If so, a nice conservative estimate is just to stick with the kJ number?


    Thanks

    As he stated, efficiency is 19-24%, 1 joule = 0.239 calories so dividing one by t'other gives a range of 0.995 to 1.257.

    SO I was right in saying - I can stick with the kJ number as a good guess for calories burned - and am only going to be 5% out either way (at the extreme)?

    Well no, you could be 25.7% out if you choose your ratio as 1. If you choose slap bang in the middle, it'll be +/- 13%
  • Pokerface
    Pokerface Posts: 7,960
    P_Tucker wrote:
    Pokerface wrote:
    P_Tucker wrote:
    Pokerface wrote:
    Hence when you convert mechanical work performed in kJ to how much energy has been metabolised, you simply multiply the kJ of mechanical work as reported by an accurate power meter by 1.10 +/- 0.05 to get an indicative range.


    Alex - I normally just look at the work figure (kJ) as the amount of calories used for the ride.

    Are you saying that cals with be slightly higher (1.05-1.15 range)?

    If so, a nice conservative estimate is just to stick with the kJ number?


    Thanks

    As he stated, efficiency is 19-24%, 1 joule = 0.239 calories so dividing one by t'other gives a range of 0.995 to 1.257.

    SO I was right in saying - I can stick with the kJ number as a good guess for calories burned - and am only going to be 5% out either way (at the extreme)?

    Well no, you could be 25.7% out if you choose your ratio as 1. If you choose slap bang in the middle, it'll be +/- 13%


    Your math doesn't equal Alex's math :?
  • P_Tucker
    P_Tucker Posts: 1,878
    Pokerface wrote:
    P_Tucker wrote:
    Pokerface wrote:
    P_Tucker wrote:
    Pokerface wrote:
    Hence when you convert mechanical work performed in kJ to how much energy has been metabolised, you simply multiply the kJ of mechanical work as reported by an accurate power meter by 1.10 +/- 0.05 to get an indicative range.


    Alex - I normally just look at the work figure (kJ) as the amount of calories used for the ride.

    Are you saying that cals with be slightly higher (1.05-1.15 range)?

    If so, a nice conservative estimate is just to stick with the kJ number?


    Thanks

    As he stated, efficiency is 19-24%, 1 joule = 0.239 calories so dividing one by t'other gives a range of 0.995 to 1.257.

    SO I was right in saying - I can stick with the kJ number as a good guess for calories burned - and am only going to be 5% out either way (at the extreme)?

    Well no, you could be 25.7% out if you choose your ratio as 1. If you choose slap bang in the middle, it'll be +/- 13%


    Your math doesn't equal Alex's math :?

    Yes it does, the confusion is because you have to divide by the 0.19/0.24 which widens the gap between them.
  • Pokerface
    Pokerface Posts: 7,960
    P_Tucker wrote:
    Yes it does, the confusion is because you have to divide by the 0.19/0.24 which widens the gap between them.

    I'll start again.


    Alex said:

    Hence when you convert mechanical work performed in kJ to how much energy has been metabolised, you simply multiply the kJ of mechanical work as reported by an accurate power meter by 1.10 +/- 0.05 to get an indicative range.


    Therefore, if I take my work done (let's say 1000kJ) and convert it to cals - I get somewhere between 1050 and 1150. Makes sense to me as that covers the -+5% range in typical efficiency (the 19-24% figure stated elsewhere).

    So - using my 1000kJ figure - show me, using your calculations, what it works out to.



    EDIT - I used the other set of calculations he listed - and came up with a range of 996-1260 cals (roughly).

    Thus my confusion.
  • I was just using a middle-of-the-range range as a way to provide a reasonable estimate and a number that was easy to remember (1.1).

    You could use kJ mechanical work = Cal metabolised (1:1) to make it easier if you like, although you are more likely to slightly under state, than overstate calories metabolised. Still, if you apply your BMR to the time spent training, then it'll be thereabouts.

    End of the day most people don't properly estimate their calorie intake and that's where the biggest impact is made.
  • Pokerface
    Pokerface Posts: 7,960
    I was gladly (always) underestimate calories burned rather than overestimate! Especially as I aim to lose weight :)
  • Chiggy
    Chiggy Posts: 261
    Pokerface wrote:
    Chiggy wrote:
    Pokerface wrote:
    I don't use a Garmin to 'estimate' calories used. I use a Powertap to much more precisely define how many calories I have used.

    What a PowerTap doesn't tell you is your efficiency. Carbon Hydrogen Oxygen molecule consumption vs resultant power at the tyre.

    Any calorific estimations without CO2 carbon balance is pissing in the wind.
    This is the reason I 'overcook' my calorific needs to the 'worst case scenario', so I am confident of getting round the 300 km Audax route.


    As Alex explained, the efficiency number will be +- 5%

    So, even if I'm consistently 5% out - that's only 50 cals for every 1000 it 'says' I burn.

    That's hardly p!ssing in the wind.

    What Alex doesn't know is what clothes you was wearing, how much you was sweating and how cold the frigging wind was. How can he quote figures for efficiency?

    I wouldn't trust Alex to specify me a wetsuit for diving in the North sea.
  • Anonymous
    Anonymous Posts: 79,667
    And what is the scope for deviation that that provides?
  • P_Tucker
    P_Tucker Posts: 1,878
    NapoleonD wrote:
    And what is the scope for deviation that that provides?

    Wait a sec, he's probably checking his Garmin.
  • Pokerface
    Pokerface Posts: 7,960
    Chiggy wrote:
    What Alex doesn't know is what clothes you was wearing, how much you was sweating and how cold the frigging wind was. How can he quote figures for efficiency?

    I wouldn't trust Alex to specify me a wetsuit for diving in the North sea.


    Let's just say I'd trust ANYTHING that Alex says over anything you say. Any day.



    Muppet.
  • Chiggy wrote:
    What Alex doesn't know is what clothes you was wearing, how much you was sweating and how cold the frigging wind was. How can he quote figures for efficiency?

    I wouldn't trust Alex to specify me a wetsuit for diving in the North sea.
    Neither would I, since I know nothing about diving in the North Sea.

    I do however know a bit about cycling.

    I quote a typical range* of efficiency for cyclists as per the science on this, findings which have been reliably repeated many times. It is well established public information for those that care to investigate.

    * clearly this has escaped you, as has my previous comment about acute variations in efficiency.
  • As an example of differences in efficiency due to changes in conditions, this example showed a 0.9% reduction in average GE (from 20.5% to 19.6%) with a rise in ambient temperature from ~ 15C to ~ 35C.
    http://www.ncbi.nlm.nih.gov/pubmed/17661069

    But still within the typical range.