On Thu, 26 Jun 2003 15:52:40 +0000 (UTC), deshi_austinpower{at}yahoo.com
(kramer) wrote:

>Dear friends,
>
> This is what I am trying to understand. We know that E=mgh+mv2  . Now
>during stair climbing, the part of the equation that wud be valid is
>E=mgh ( that is due to height displacement. there will be some
>insignificant contribution from mv2 too.. but I think it is very
>insignificant comparing to mgh part.). Now I calculated if I weigh 77
>kg ( ~ 170 lb) and climb 10 storied building of total approximate
>height of 24.384 meter ( 80 ft) I have to use 18462 J ( 4430 cal) from
>E= m (77 kg)*g (9.81 m/s2) * h (24.384 mt) = 18464 J = 4430 cal .( 1
>Joule  = .24 cal) . And if I repeat this process 5 times I shud be
>using 22154 cal which means loss of 6.32 lb of body fat ( 1 lb fat
>contains ~ 3500 cal).
>

> Now I would like to know if there is anything wrong with this
>calculation. it seems to be it is just too easy to lose all those lbs
>so easily. Thanks in advance for your thoughtful comments.
>
>Kramer

Are you remembering that a nutritional Calorie (spelled with an upper
case 'C') is really a kilocalorie?

Try climbing those stairs 5000 times to lose 6 pounds of fat!

This is a problem I assign to my Animal Physiology students.  The
moral is that you don't exercise to lose fat.  You exercise to build
cardiovascular fitness.  You eat less to lose fat.  (Yes, exercise is
an important adjunct to weight loss, but not because of the above
calculation!)
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