Wattage and calories?

I ride on my trainer at easy pace 15mph - 20 miles - 80mins.
According to Kurt's website wattage is governed by:
P = (5.244820) * S + (0.01968) * S^3
P - power in watts
C - speed in mph

P = 146 W
So letsee... 1W = 1 J/s
80 mins at 146W means 146*60*80 = 701280 J
1 J = 0.24 cal (small calorie)
so it seems that I burn 168307cal or 168kC

But when I plug in same numbers into SportTracks it shows 1650
calories burnt. that is factor of 10 difference... I am missing
something
very obvious here?

On 2008-12-02, Woland99 wrote:
I ride on my trainer at easy pace 15mph - 20 miles - 80mins.
According to Kurt's website wattage is governed by:
P = (5.244820) * S + (0.01968) * S^3
P - power in watts
C - speed in mph

P = 146 W
So letsee... 1W = 1 J/s
80 mins at 146W means 146*60*80 = 701280 J
1 J = 0.24 cal (small calorie)
so it seems that I burn 168307cal or 168kC

But when I plug in same numbers into SportTracks it shows 1650
calories burnt. that is factor of 10 difference... I am missing
something
very obvious here?

You need to multiply by about 4-- 146W is your output power and the
human body is only about 25% efficient.

But that's only a factor of 4 not of 10. Your calculations look correct
to me.

On Dec 2, 9:11 am, Ben C wrote:
On 2008-12-02, Woland99 wrote:



I ride on my trainer at easy pace 15mph - 20 miles - 80mins.
According to Kurt's website wattage is governed by:
P = (5.244820) * S + (0.01968) * S^3
P - power in watts
C - speed in mph

P = 146 W
So letsee... 1W = 1 J/s
80 mins at 146W means 146*60*80 = 701280 J
1 J = 0.24 cal (small calorie)
so it seems that I burn 168307cal or 168kC

But when I plug in same numbers into SportTracks it shows 1650
calories burnt. that is factor of 10 difference... I am missing
something
very obvious here?

You need to multiply by about 4-- 146W is your output power and the
human body is only about 25% efficient.

But that's only a factor of 4 not of 10. Your calculations look correct
to me.

OK - I checked Wikipedia "The efficiency of human muscle has been
measured (in the context of rowing and cycling) at 14% to 27%. The
efficiency is defined as the ratio of mechanical work output to the
total metabolic cost"

So letsay it is 14% and lets throw in 0.9 efficient gears.
That means 1650*0.14*0.9 = 230kCal. That is closer to 168kCal
but still off.

Woland99 wrote:

But when I plug in same numbers into SportTracks it shows 1650
calories burnt. that is factor of 10 difference... I am missing
something very obvious here?

I think the very obvious thing you're missing is that SportTracks'
calculation is unreliable.

"Woland99" wrote in message
...
On Dec 2, 9:11 am, Ben C wrote:
On 2008-12-02, Woland99 wrote:



I ride on my trainer at easy pace 15mph - 20 miles - 80mins.
According to Kurt's website wattage is governed by:
P = (5.244820) * S + (0.01968) * S^3
P - power in watts
C - speed in mph

P = 146 W
So letsee... 1W = 1 J/s
80 mins at 146W means 146*60*80 = 701280 J
1 J = 0.24 cal (small calorie)
so it seems that I burn 168307cal or 168kC

But when I plug in same numbers into SportTracks it shows 1650
calories burnt. that is factor of 10 difference... I am missing
something
very obvious here?

You need to multiply by about 4-- 146W is your output power and the
human body is only about 25% efficient.

But that's only a factor of 4 not of 10. Your calculations look correct
to me.

OK - I checked Wikipedia "The efficiency of human muscle has been
measured (in the context of rowing and cycling) at 14% to 27%. The
efficiency is defined as the ratio of mechanical work output to the
total metabolic cost"

So letsay it is 14% and lets throw in 0.9 efficient gears.
That means 1650*0.14*0.9 = 230kCal. That is closer to 168kCal
but still off.

I think you need to check the assumptions in SportTracks. Do you have a
choice of bike type. You might be able to get those sort of numbers if it
assumes you are riding a mountain bike off road but not on a decent road
bike riding on the hoods. My sums give me a figure of around 700 kCal for
that trip in terms of the energy requirement to power the bike plus you
could add around 95 kCal required to power you. Giving just less than 800 in
total.

I agree with your other replies that having corrected for body efficiency (I
happen to use 21%) then you are in the right ball park. If you need any
futher assurance then look up the energy consumption of cycling on the web
at 15 mph on a road bike. Or for that matter any excercise regarded as "easy
pace" and you will find that 600 kCal per hour is about par. To put into
some form of perspective that's equivalent to jogging at around 6 mph or 10
minute miling if you prefer which I think every body would agree is easy
pace.

Graham

On Dec 2, 2:01 pm, "Robert Chung"
wrote:
I think the very obvious thing you're missing is that SportTracks'
calculation is unreliable.

That is actually Garmin Edge 305 - not ST....

Looking at my last workout - 20 miles at 15mph.
P = (5.244820) * S + (0.01968) * S^3
P - power in watts
C - speed in mph

Wattage disspated by trainer P = 146 W
1W = 1 J/s
so 80 mins at 146W means 146*60*80 = 701280 J
1 J = 0.24 cal (small calorie)
So 168kCal.

Assume 0.9 gear efficiency (is it reasonable)
Assume 0.14 body efficiency (low end?)
Energy that body expanded to put 168kCal into moving trainer:
168/(0.9*0.14) = 1333kCal

According to Garmin Edge 305 - calories burnt were 1672kCal

So now we have deficiency of 339kCal.... can this be how
much your body needs for 80mins at about 130bpm HR (75 max HR)?

On Dec 2, 2:42 pm, "graham" wrote:
"Woland99" wrote in message

...



On Dec 2, 9:11 am, Ben C wrote:
On 2008-12-02, Woland99 wrote:

I ride on my trainer at easy pace 15mph - 20 miles - 80mins.
According to Kurt's website wattage is governed by:
P = (5.244820) * S + (0.01968) * S^3
P - power in watts
C - speed in mph

P = 146 W
So letsee... 1W = 1 J/s
80 mins at 146W means 146*60*80 = 701280 J
1 J = 0.24 cal (small calorie)
so it seems that I burn 168307cal or 168kC

But when I plug in same numbers into SportTracks it shows 1650
calories burnt. that is factor of 10 difference... I am missing
something
very obvious here?

You need to multiply by about 4-- 146W is your output power and the
human body is only about 25% efficient.

But that's only a factor of 4 not of 10. Your calculations look correct
to me.

OK - I checked Wikipedia "The efficiency of human muscle has been
measured (in the context of rowing and cycling) at 14% to 27%. The
efficiency is defined as the ratio of mechanical work output to the
total metabolic cost"

So letsay it is 14% and lets throw in 0.9 efficient gears.
That means 1650*0.14*0.9 = 230kCal. That is closer to 168kCal
but still off.

I think you need to check the assumptions in SportTracks. Do you have a
choice of bike type. You might be able to get those sort of numbers if it
assumes you are riding a mountain bike off road but not on a decent road
bike riding on the hoods. My sums give me a figure of around 700 kCal for
that trip in terms of the energy requirement to power the bike plus you
could add around 95 kCal required to power you. Giving just less than 800 in
total.

I agree with your other replies that having corrected for body efficiency (I
happen to use 21%) then you are in the right ball park. If you need any
futher assurance then look up the energy consumption of cycling on the web
at 15 mph on a road bike. Or for that matter any excercise regarded as "easy
pace" and you will find that 600 kCal per hour is about par. To put into
some form of perspective that's equivalent to jogging at around 6 mph or 10
minute miling if you prefer which I think every body would agree is easy
pace.

Graham

Lets do the math again:

Looking at my last workout - 20 miles at 15mph.
Using Kurt Kinetic trainer with following formula:
P = (5.244820) * S + (0.01968) * S^3
P - power in watts
C - speed in mph

Wattage dissipated by trainer P = 146 W
1W = 1 J/s
so 80 mins at 146W means 146*60*80 = 701280 J
1 J = 0.24 cal (small calorie)
So 168kCal.

Assume 0.9 gear efficiency (is it reasonable?)
Assume 0.14 body efficiency (low end?)
Energy that body expanded to put 168kCal
into moving trainer:
168kCal/(0.9*0.14) = 1333kCal

Calories burnt according to to Garmin Edge 305 -
for the very same workout (which apparently takes
into account body weight and HR - max and avg) -
1672kCal

So now we have deficiency of 1672 - 1333 = 339kCal
can this be how much your body needs for 80mins
at about 130bpm HR (75 max HR)?

Lets look at BMR - basal metabolic rate:
For man:
P = 6.2377*M + 12.7084*H - 6.7550*A + 66.473
P BRM in kCal/day
M mass in pounds
H height in inches
A age in years

Taken from:
http://en.wikipedia.org/wiki/Basal_metabolic_rate

so we have P = 2340kCal/day
So in 80mins (at rest it) would have been:
2340 * 1.33/24 = 130kCal

So there are still about 210kCal no accounted for.
Perhaps that the energy that body needs to function for
80mins at 75% maxHR ?

On Dec 2, 8:47*am, Woland99 wrote:
I ride on my trainer at easy pace 15mph - 20 miles - 80mins.
According to Kurt's website wattage is governed by:
P *= (5.244820) * S + (0.01968) * S^3
P - power in watts
C - speed in mph

P = 146 W
So letsee... 1W = 1 J/s
80 mins at 146W means 146*60*80 = 701280 J
1 J = 0.24 cal (small calorie)
so it seems that I burn 168307cal or 168kC

But when I plug in same numbers into SportTracks it shows 1650
calories burnt. that is factor of 10 difference... I am missing
something
very obvious here?

heart rate determining effort

Woland99 wrote:
On Dec 2, 2:01 pm, "Robert Chung"
wrote:
I think the very obvious thing you're missing is that SportTracks'
calculation is unreliable.

That is actually Garmin Edge 305 - not ST....

Assume 0.9 gear efficiency (is it reasonable)
Assume 0.14 body efficiency (low end?)

In that case, I think the very obvious thing you're missing is that the
Garmin 305's calculation is unreliable. Do you enter your weight and an
estimate of your CdA into the 305? What gross efficiency does it assume?

0.9 drivetrain efficiency is probably low unless you were using an old
Sturmey-Archer 3-speed, or were in an extreme gear. For a modern derailleur
system not in an extreme gear a 5% loss is a better ballpark.

0.14 is probably low as an estimate of gross metabolic efficiency. A better
ballpark is from .19 to maybe .24.

Both of those make the difference larger.

(BTW, there's a minor error on the Kurt Kinetic page: the coeff for mph^3 is
0.19168)

On Dec 2, 5:45 pm, "Robert Chung"
wrote:
(BTW, there's a minor error on the Kurt Kinetic page: the coeff for mph^3 is
0.19168)

Thanks. You meant 0.019168.
Or else I am God of Watts - doing 740W for an hour ;-)