Calories Burned

Don Kirkman said:

Relative to a discus^H^H^H argument in another news group,
I'm looking for clues to published research findings
supporting the rule of thumb that one mile = 100 calories.
I'd have thought the Cooper Institute might have done
something like this, but couldn't see it on their site.

Does anybody have URLs or titles I can track? TIA

100 calories per mile will only work if you're a certain
weight, namely
137.8 lb. If you weight more, yippie, you burn more calories
per mile; if you weigh less, bummer. The rough formula
is Kcal = Km x Kg, all metric, so you have to convert to
pounds and miles. There are more complicated formulas
that take grade (up/down hill) into account, but this
one gives the rule of thumb you requested. URLs should
be easily locatable using your friend and mine, Google.

--
ø¤º°`°º¤ø,,,,ø¤º°`°º¤ø,,,,ø¤º°`°º¤ø,,,,ø¤º°`°º¤ø¤º°`°º¤ø,,,-
,ø¤º eNo "If you can't go fast, go long." ø¤º°`°º¤ø,,,,ø¤º°-
`°º¤ø,,,,ø¤º°`°º¤ø,,,,ø¤º°`°º¤ø¤º°`°º¤ø,,,,ø¤º

From the American College of Sports Medicine:

VO2 (ml/kg/min) = 0.2*V + 0.9*V*G +3.5

v=velocity in meters/min (1mph = 26.8 m/min) G = grade (as a
decimal so a 1% grade is 0.01)

Take the value from above and multiply by your mass in
kilograms (1kg=2.2 lbs)

Take that value and divide by 1000 (to convert ml to liters)

Take that and multiply by 5.0 kcal/liter (a rough
approximation of submaximal energy consumption)

Or:

Kcals = ((0.2*V + 0.9*V*G +3.5)* M)/1000)* 5.0

where M = body mass in kg.

So a 154 pound person running at 7 mph on flat ground would
expend 14.3 kcal/min. Since running at that speed would
cover a mile in 8:30, the energy expenditure would be 122
kcals. Of course that assumes that running economy is the
same when in fact running economy can vary greatly and that
as you become more proficient, your economy improves and you
actually expend slightly less energy running a mile even at
the same pace.

"eNo" <[email hidden]> wrote in message news:NhFec.15$UF5.14@dfw-
service2.ext.ray.com...

Quoted message said:

Don Kirkman said:

Relative to a discus^H^H^H argument in another news
group, I'm looking for clues to published research
findings supporting the rule of thumb that one mile =
100 calories. I'd have thought the Cooper Institute
might have done something like this, but couldn't see it
on their site.

Does anybody have URLs or titles I can track? TIA

100 calories per mile will only work if you're a certain
weight, namely
137.8 lb. If you weight more, yippie, you burn more
calories per mile; if you weigh less, bummer. The
rough formula is Kcal = Km x Kg, all metric, so you
have to convert to pounds and miles. There are more
complicated formulas that take grade (up/down hill)
into account, but this one gives the rule of thumb you
requested. URLs should be easily locatable using your
friend and mine, Google.

--
ø¤º°`°º¤ø,,,,ø¤º°`°º¤ø,,,,ø¤º°`°º¤ø,,,,ø¤º°`°º¤ø¤º°`°º¤ø,-
,,,ø¤º eNo "If you can't go fast, go long." ø¤º°`°º¤ø,,,,-
ø¤º°`°º¤ø,,,,ø¤º°`°º¤ø,,,,ø¤º°`°º¤ø¤º°`°º¤ø,,,,ø¤º

Sam said:

From the American College of Sports Medicine:

VO2 (ml/kg/min) = 0.2*V + 0.9*V*G +3.5

v=velocity in meters/min (1mph = 26.8 m/min) G = grade (as
a decimal so a 1% grade is 0.01)

Take the value from above and multiply by your mass in
kilograms (1kg=2.2 lbs)

Take that value and divide by 1000 (to convert ml to
liters)

Take that and multiply by 5.0 kcal/liter (a rough
approximation of submaximal energy consumption)

Or:

Kcals = ((0.2*V + 0.9*V*G +3.5)* M)/1000)* 5.0

where M = body mass in kg.

Hmm. Neat. Let's try it on our 137.8 pounder on a flat
surface and running at 10 minute/mile (6.0 miles/hr =
160.9 m/min).

Kcals = ((0.2*160.9m/min + 3.5) * 62.5kg) / 1000) * 5 =
11.1 Kcals???

Looks like this formula is off by about a factor of 10...

--
ø¤º°`°º¤ø,,,,ø¤º°`°º¤ø,,,,ø¤º°`°º¤ø,,,,ø¤º°`°º¤ø¤º°`°º¤ø,,,-
,ø¤º eNo "If you can't go fast, go long." ø¤º°`°º¤ø,,,,ø¤º°-
`°º¤ø,,,,ø¤º°`°º¤ø,,,,ø¤º°`°º¤ø¤º°`°º¤ø,,,,ø¤º

It seems to me I heard somewhere that Sam wrote in article
<[email hidden]>:

Quoted message said:

From the American College of Sports Medicine:

Quoted message said:

VO2 (ml/kg/min) = 0.2*V + 0.9*V*G +3.5

Quoted message said:

v=velocity in meters/min (1mph = 26.8 m/min) G = grade (as
a decimal so a 1% grade is 0.01)

Quoted message said:

Take the value from above and multiply by your mass in
kilograms (1kg=2.2 lbs)

Quoted message said:

Take that value and divide by 1000 (to convert ml to
liters)

Quoted message said:

Take that and multiply by 5.0 kcal/liter (a rough
approximation of submaximal energy consumption)

Quoted message said:

Or:

Quoted message said:

Kcals = ((0.2*V + 0.9*V*G +3.5)* M)/1000)* 5.0

Quoted message said:

where M = body mass in kg.

Quoted message said:

"eNo" <[email hidden]> wrote in message news:NhFec.15$UF5.14@dfw-
service2.ext.ray.com...

Quoted message said:

Don Kirkman wrote:

Quoted message said:

Quoted message said:

Quoted message said:

Relative to a discus^H^H^H argument in another news
group, I'm looking for clues to published research
findings supporting the rule of thumb that one mile =
100 calories. I'd have thought the Cooper Institute
might have done something like this, but couldn't see
it on their site.

Quoted message said:

Quoted message said:

Quoted message said:

Does anybody have URLs or titles I can track? TIA

Quoted message said:

Quoted message said:

100 calories per mile will only work if you're a certain
weight, namely
137.8 lb. If you weight more, yippie, you burn more
calories per mile; if you weigh less, bummer. The
rough formula is Kcal = Km x Kg, all metric, so you
have to convert to pounds and miles. There are more
complicated formulas that take grade (up/down hill)
into account, but this one gives the rule of thumb
you requested. URLs should be easily locatable using
your friend and mine, Google.

I have the formulas--in fact, I've used them in the
discussion in question-- and have Googled extensively
without locating what I'm asking for: published peer
reviewed research articles that *support* or *underlie* the
formulas. All I come up with is variations on the handful of
basic formulas, without any citation of the clinical basis
for any of them.
--
Don [email hidden]

"eNo" <[email hidden]> wrote in message
"]news:[email hidden]...

Quoted message said:

Sam said:

From the American College of Sports Medicine:

VO2 (ml/kg/min) = 0.2*V + 0.9*V*G +3.5

v=velocity in meters/min (1mph = 26.8 m/min) G = grade
(as a decimal so a 1% grade is 0.01)

Take the value from above and multiply by your mass in
kilograms

(1kg=2.2

Quoted message said:

Quoted message said:

lbs)

Take that value and divide by 1000 (to convert ml to
liters)

Take that and multiply by 5.0 kcal/liter (a rough
approximation of submaximal energy consumption)

Or:

Kcals = ((0.2*V + 0.9*V*G +3.5)* M)/1000)* 5.0

where M = body mass in kg.

Hmm. Neat. Let's try it on our 137.8 pounder on a flat
surface and running at 10 minute/mile (6.0 miles/hr =
160.9 m/min).

Kcals = ((0.2*160.9m/min + 3.5) * 62.5kg) / 1000) * 5 =
11.1 Kcals???

Looks like this formula is off by about a factor of 10...

That's 11.1 Kcal per minute or 111 Kcal per mile at your 6.0
mile/hr pace.

Matthew

"eNo" <[email hidden]> wrote in message
"]news:[email hidden]...

Quoted message said:

Sam said:

From the American College of Sports Medicine:

VO2 (ml/kg/min) = 0.2*V + 0.9*V*G +3.5

v=velocity in meters/min (1mph = 26.8 m/min) G = grade
(as a decimal so a 1% grade is 0.01)

Take the value from above and multiply by your mass in
kilograms

(1kg=2.2

Quoted message said:

Quoted message said:

lbs)

Take that value and divide by 1000 (to convert ml to
liters)

Take that and multiply by 5.0 kcal/liter (a rough
approximation of submaximal energy consumption)

Or:

Kcals = ((0.2*V + 0.9*V*G +3.5)* M)/1000)* 5.0

where M = body mass in kg.

Hmm. Neat. Let's try it on our 137.8 pounder on a flat
surface and running at 10 minute/mile (6.0 miles/hr =
160.9 m/min).

Kcals = ((0.2*160.9m/min + 3.5) * 62.5kg) / 1000) * 5 =
11.1 Kcals???

Looks like this formula is off by about a factor of 10...

the units are kcal per MINUTE. So if you figure it
takes 10 min to cover a mile at 6mph, that is 111
kcals. Always good to the units and I should have
included them in the final equation.

Quoted message said:

--
ø¤º°`°º¤ø,,,,ø¤º°`°º¤ø,,,,ø¤º°`°º¤ø,,,,ø¤º°`°º¤ø¤º°`°º¤ø,-
,,,ø¤º eNo "If you can't go fast, go long." ø¤º°`°º¤ø,,,,-
ø¤º°`°º¤ø,,,,ø¤º°`°º¤ø,,,,ø¤º°`°º¤ø¤º°`°º¤ø,,,,ø¤º

"Don Kirkman" <[email hidden]> wrote in message
"]news:[email hidden]...

Quoted message said:

It seems to me I heard somewhere that Sam wrote in article
<[email hidden]>:

Quoted message said:

From the American College of Sports Medicine:

Quoted message said:

VO2 (ml/kg/min) = 0.2*V + 0.9*V*G +3.5

Quoted message said:

v=velocity in meters/min (1mph = 26.8 m/min) G = grade
(as a decimal so a 1% grade is 0.01)

Quoted message said:

Take the value from above and multiply by your mass in
kilograms (1kg=2.2 lbs)

Quoted message said:

Take that value and divide by 1000 (to convert ml to
liters)

Quoted message said:

Take that and multiply by 5.0 kcal/liter (a rough
approximation of submaximal energy consumption)

Quoted message said:

Or:

Quoted message said:

Kcals = ((0.2*V + 0.9*V*G +3.5)* M)/1000)* 5.0

Quoted message said:

where M = body mass in kg.

Quoted message said:

"eNo" <[email hidden]> wrote in message news:NhFec.15$UF5.14@dfw-
service2.ext.ray.com...

Quoted message said:

Don Kirkman wrote:

Quoted message said:

Quoted message said:

> Relative to a discus^H^H^H argument in another news
> group, I'm

looking

Quoted message said:

Quoted message said:

Quoted message said:

> for clues to published research findings supporting
> the rule of thumb that one mile = 100 calories. I'd
> have thought the Cooper Institute might have done
> something like this, but couldn't see it on their

site.

Quoted message said:

Quoted message said:

Quoted message said:

> Does anybody have URLs or titles I can track? TIA

Quoted message said:

Quoted message said:

100 calories per mile will only work if you're a
certain weight, namely
137.8 lb. If you weight more, yippie, you burn more
calories per mile; if you weigh less, bummer. The
rough formula is Kcal = Km x Kg, all metric, so you
have to convert to pounds and miles. There are more
complicated formulas that take grade (up/down hill)
into account, but this one gives the rule of thumb
you requested. URLs should be easily locatable
using your friend and mine, Google.

I have the formulas--in fact, I've used them in the
discussion in question-- and have Googled extensively
without locating what I'm asking for: published peer
reviewed research articles that *support* or *underlie*
the formulas. All I come up with is variations on the
handful of basic formulas, without any citation of the
clinical basis for any of them.
--
Don [email hidden]

Do a PubMed search for Balke or better Margaria (from 1963)
in the J of Applied Physiology. The formula still stands the
test of time pretty well.

Quoted message said:

From the American College of Sports Medicine:

VO2 (ml/kg/min) = 0.2*V + 0.9*V*G +3.5

Actual VO2 consumption or CO2 production is MEASURED in a
special exercise room or airtank apparatus and extrapolated
to various running conditions. The primary ingredient is
weight and distance traveled. Other factors total to less
than 20% variation.

El Paisano said:

"eNo" <[email hidden]> wrote in message news:q6Vec.5$od.4@dfw-
service2.ext.ray.com...

Quoted message said:

Sam said:

From the American College of Sports Medicine:

VO2 (ml/kg/min) = 0.2*V + 0.9*V*G +3.5

v=velocity in meters/min (1mph = 26.8 m/min) G = grade
(as a decimal so a 1% grade is 0.01)

Take the value from above and multiply by your mass in
kilograms

(1kg=2.2

Quoted message said:

Quoted message said:

lbs)

Take that value and divide by 1000 (to convert ml to
liters)

Take that and multiply by 5.0 kcal/liter (a rough
approximation of submaximal energy consumption)

Or:

Kcals = ((0.2*V + 0.9*V*G +3.5)* M)/1000)* 5.0

where M = body mass in kg.

Hmm. Neat. Let's try it on our 137.8 pounder on a flat
surface and running at 10 minute/mile (6.0 miles/hr =
160.9 m/min).

Kcals = ((0.2*160.9m/min + 3.5) * 62.5kg) / 1000) * 5 =
11.1 Kcals???

Looks like this formula is off by about a factor of 10...

That's 11.1 Kcal per minute or 111 Kcal per mile at your
6.0 mile/hr pace.

Matthew

Ah, how could I miss that if the formula is based on
velocity (not distance), it is on a per minute basis, so
therefore, yet another conversion is necessary to arrive at
what the original poster requested, namely, *calories per
mile*. Anyway, the 111 Kcal result isn't all that far from
the 100 Kcal estimate the pure work (Distance * Mass)
formula yields.

--
ø¤º°`°º¤ø,,,,ø¤º°`°º¤ø,,,,ø¤º°`°º¤ø,,,,ø¤º°`°º¤ø¤º°`°º¤ø,,,-
,ø¤º eNo "If you can't go fast, go long." ø¤º°`°º¤ø,,,,ø¤º°-
`°º¤ø,,,,ø¤º°`°º¤ø,,,,ø¤º°`°º¤ø¤º°`°º¤ø,,,,ø¤º

It seems to me I heard somewhere that Sam wrote in article
<[email hidden]>:

Quoted message said:

"Don Kirkman" <[email hidden]> wrote in message
"]news:[email hidden]...

Quoted message said:

It seems to me I heard somewhere that Sam wrote in
article
<[email hidden]>:

Quoted message said:

Quoted message said:

Quoted message said:

From the American College of Sports Medicine:

Quoted message said:

Quoted message said:

Quoted message said:

VO2 (ml/kg/min) = 0.2*V + 0.9*V*G +3.5

Quoted message said:

Quoted message said:

Quoted message said:

v=velocity in meters/min (1mph = 26.8 m/min) G = grade
(as a decimal so a 1% grade is 0.01)

[...]

Quoted message said:

Quoted message said:

Quoted message said:

"eNo" <[email hidden]> wrote in message news:NhFec.15$UF5.14@dfw-
service2.ext.ray.com...
> Don Kirkman wrote:

Quoted message said:

Quoted message said:

Quoted message said:

> > Relative to a discus^H^H^H argument in another news
> > group, I'm

looking

Quoted message said:

Quoted message said:

> > for clues to published research findings supporting
> > the rule of thumb that one mile = 100 calories. I'd
> > have thought the Cooper Institute might have done
> > something like this, but couldn't see it on their

site.

Quoted message said:

Quoted message said:

Quoted message said:

> > Does anybody have URLs or titles I can track? TIA

[...]

Quoted message said:

Quoted message said:

I have the formulas--in fact, I've used them in the
discussion in question-- and have Googled extensively
without locating what I'm asking for: published peer
reviewed research articles that *support* or *underlie*
the formulas. All I come up with is variations on the
handful of basic formulas, without any citation of the
clinical basis for any of them.

Quoted message said:

Do a PubMed search for Balke or better Margaria (from 1963)
in the J of Applied Physiology. The formula still stands
the test of time pretty well.

Thanks. Just read this and it's late in the day, but I'll
follow up hoping for success. :-)
--
Don [email hidden]

In article <[email hidden]>,

rick++ said:

Quoted message said:

From the American College of Sports Medicine:

VO2 (ml/kg/min) = 0.2*V + 0.9*V*G +3.5

Actual VO2 consumption or CO2 production is MEASURED in a
special exercise room or airtank apparatus and extrapolated
to various running conditions. The primary ingredient is
weight and distance traveled. Other factors total to less
than 20% variation.

Do treadmills and other aerobic machines measure calories
based upon some such formula, or do they directly measure
the energy absorbed from the line? In other words, do they
estimate the work done by the user or measure the work done
by the motor?

--
***********************************************************-
*************
Terry R. McConnell Mathematics/215 Carnegie/Syracuse, N.Y.
13244-1150 [email hidden] 229B Physics Bldg
barnyard.syr.edu~tmcOpen ↗
***********************************************************-
*************

As best as I can tell they use some sort of formula. I have
never been able to find a treadmill manufacturer that
provides the information on the formula used.

My experience has been that the estimations are not
very good.

"Terry R. McConnell" <[email hidden]> wrote in message
"]news:[email hidden]...

Quoted message said:

In article
<[email hidden]>, rick++

Quoted message said:

Quoted message said:

From the American College of Sports Medicine:

VO2 (ml/kg/min) = 0.2*V + 0.9*V*G +3.5

Actual VO2 consumption or CO2 production is MEASURED in a
special exercise room or airtank apparatus and
extrapolated to various running conditions. The primary
ingredient is weight and distance traveled. Other factors
total to less than 20% variation.

Do treadmills and other aerobic machines measure calories
based upon some such formula, or do they directly measure
the energy absorbed from the

line?

Quoted message said:

In other words, do they estimate the work done by the user
or measure the work done by the motor?

--
*********************************************************-
***************
Terry R. McConnell Mathematics/215 Carnegie/Syracuse, N.Y.
13244-1150 [email hidden] 229B Physics Bldg
barnyard.syr.edu~tmcOpen ↗
*********************************************************-
***************

"eNo" <[email hidden]> wrote in message
"]news:[email hidden]...

Quoted message said:

El Paisano said:

"eNo" <[email hidden]> wrote in message news:q6Vec.5$od.4@dfw-
service2.ext.ray.com...

Quoted message said:

Sam wrote:

>From the American College of Sports Medicine:
>
>VO2 (ml/kg/min) = 0.2*V + 0.9*V*G +3.5
>
>v=velocity in meters/min (1mph = 26.8 m/min) G = grade
>(as a decimal so a 1% grade is 0.01)
>
>Take the value from above and multiply by your mass in
>kilograms

(1kg=2.2

Quoted message said:

>lbs)
>
>Take that value and divide by 1000 (to convert ml to
>liters)
>
>Take that and multiply by 5.0 kcal/liter (a rough
>approximation of submaximal energy consumption)
>
>Or:
>
>Kcals = ((0.2*V + 0.9*V*G +3.5)* M)/1000)* 5.0
>
>where M = body mass in kg.

Hmm. Neat. Let's try it on our 137.8 pounder on a flat
surface and running at 10 minute/mile (6.0 miles/hr =
160.9 m/min).

Kcals = ((0.2*160.9m/min + 3.5) * 62.5kg) / 1000) * 5 =
11.1 Kcals???

Looks like this formula is off by about a factor
of 10...

That's 11.1 Kcal per minute or 111 Kcal per mile at your
6.0 mile/hr

pace.

Quoted message said:

Quoted message said:

Matthew

Ah, how could I miss that if the formula is based on
velocity (not distance), it is on a per minute basis, so
therefore, yet another conversion is necessary to arrive
at what the original poster requested, namely, *calories
per mile*. Anyway, the 111 Kcal result isn't all that far
from the 100 Kcal estimate the pure work (Distance * Mass)
formula yields.

No it is not and if you consider the great variation in
running economy, it could be close.

It seems to me I heard somewhere that Don Kirkman wrote in article
<[email hidden]>:

Quoted message said:

Relative to a discus^H^H^H argument in another news group,
I'm looking for clues to published research findings
supporting the rule of thumb that one mile = 100 calories.
I'd have thought the Cooper Institute might have done
something like this, but couldn't see it on their site.

Quoted message said:

Does anybody have URLs or titles I can track? TIA

Just to close out the thread, thanks to all for the helpful
replies. I learned a lot in searching through the URLs, but
unfortunately facts weren't enough to faze the main
disputant (whom you can read about in the thread "Subject:
Watch out when you chose a doctor" started by James216440
[Message-ID:
<[email hidden]>]).
--
Don [email hidden]