Re: formula for energy expenditure

[rides2far@juno.com (Angela C. McGhee)]

Could you please translate that into coffee cans of Omalene?

Angie

On Thu, 05 Mar 1998 15:10:05 -0800 Susan Evans Garlinghouse
<suendavid@worldnet.att.net> writes:
>Hiya,
>
>A bunch of people have been asking me lately about replacing the
>calories burned by the horse during a ride and I thought everyone 
>might
>be interested in seeing a nifty formula that was published in the
>journals from Pagan and Hintz (both widely respected) that pretty
>accurately calculates the calories expended by a horse during 
>exercise.
>This is the formula that is much more accurate than the formulae
>currently used in the NRC books to calculate energy required for work
>above maintenance levels.
>
>It's interesting, if nothing else, to see the calories required for an
>average horse to cover 50 miles---a 900 lb horse carrying a 165 lb 
>rider
>and covering 50 miles in 6 hours actual riding time burns up 18.3
>Mcals.  The same horse carrying a 210 lb rider and covering the same
>fifty miles in four hours would burn 26 Mcals.  A 1000-lb thoroughbred
>carrying 112 pounds and running a mile in two minutes (sorry, Tom, I
>don't know average times for a mile) would burn 4.6 Mcals in just two
>minutes, which is pretty damn impressive considering he's providing
>almost all of that energy in the form of glucose (endurance horses get
>at least a large portion of their energy requirements from fats, which
>are in greater supply.)  And since feed efficiency is only around 60%
>efficient for a typical ration (that means that only 60% of the 
>calories
>ingested actually get utilized in energy production, the rest is 
>burned
>in metabolism or lost heat, etc), that means that you would have to 
>feed
>30.5, 7.7 and 43.4 Mcals, respectively, to make up the energy burned
>during that ride.  Incredible, considering that the same horse would
>only use 13.4 Mcals a day for daily maintenance.  Another good 
>argument
>for making sure your horse is in good body condition before you start 
>a
>heavy ride season---he's going to need those extra stores of fat!
>
>(And, by the way, please don't take this as a recommendation to try to
>feed back those calories right after the ride, or we'll all going to
>have a whole lot of colicky, foundering horses).  
>
>Anyway, here's the formula.  The engineers and math whizzes on the 
>list
>can take it and run with it.  For the rest of us, below are directions
>on how to actually get useable numbers.  Again, this is just for fun,
>not an indication that anyone should make major changes to their 
>horse's
>diet...unless you're riding 50 miles every weekend, only providing 10
>lbs of hay a day and scratching your head as to why ol' Flash is 
>looking
>a little peaky. :-D
>
>The formula is: Y = e (superscript)3.20 + .0065x; where x is the speed
>is meters/minute, and Y equals the calories expended per kg per 
>minute.
>
>Here's how to actually get an answer if you're not a math genius---you
>will need a calculators that does logarithms, natural logs, etc, just
>the cheapy kind they give out at the gas station won't do it.
>
>First you need to calculate the average speed ("x")you're traveling 
>at. 
>It's okay if you change speed alot, the formula still works.  Figure 
>out
>how many miles you traveled and how long it took you to get there. 
>Convert the miles to meters by multiplying miles times 1609.35 (the
>number of meters in a mile).  Fifty miles is 80,467.5 meters.  Divide
>this number by how many minutes you took getting there.  This will 
>give
>you your average speed in meters per minute.  For example, if you took
>six hours to travel fifty miles than x= 80,467.35 divided by 360 =
>223.54 meters/min.
>
>Multiply x by .0065 and add 3.20.  For example, 223.54 times .0065, 
>plus
>3.2 equals 4.65.  If you have a memory function on the calculator, put
>this number into the memory, or at least write it down.
>
>Find on the calculator the inverse function for the natural log.  The
>primary function on the caculator button will say "LN", and the
>inverse/2nd function will have an "e" with an "x" superscript.  Take 
>the
>number you just calculated above (in the example, the 4.65), push the
>2nd function key, and then the LN key to get the inverse function.  In
>the above example, the result would be 104.89.  This number is the
>number of calories your horse is burning PER minute, PER kilogram of
>weight getting moved down the trail.  This is "Y".  Put this number 
>into
>memory or write it down.
>
>To calculate the total expenditure of energy, you need to know the
>kilograms.  Take your horse's weight in pounds and add the number of
>pounds he's carrying in rider and tack weight.  For example, a 900 
>pound
>horse carrying 165 pounds of rider and tack totals1065 pounds.  Divide
>this number by 2.2 to convert to kilograms. 1065 pounds equals 484
>kilograms.
>
>Take the number of kilograms and multiply it by the Y number you
>calculated above, and then calculate THAT number by the minutes you 
>were
>riding.  In the on-going example, if you took six hours to do a 50 
>mile
>ride, you rode for 360 minutes.  So the total expenditure of calories
>was 18,650,000 calories.
>
>To convert this number into units you're more familiar with, divide 
>this
>number by 1,000,000 (one million) to give you Mcals, or by 1000 to 
>give
>Kcals, the unit most people are thinking of when they think calories, 
>as
>in "that piece of cake is 500 calories".
>
>Divide the number of Mcals by .60.  For example, 18.65 Mcal divided by
>.60 equals 31.08 Mcals.  This is the number of extra caolries you 
>would
>have to actually feed just to replace the calories burned during this
>ride, IF you wanted to maintain weight---remember, the horse has 
>already
>supplied the energy he needed from fat stores, as well as a small 
>amount
>from what he ate during the ride.  Hopefully, if you're competing, 
>then
>you are already feeding more than just a flake of hay.  Just to give a
>general idea, you would have to feed 20 pounds of corn to supply the
>18.65 Mcals, or about 15 cups of vegetable oil, or 23 pounds of oats. 
>So if you're competing alot, you might use this formula to give you 
>just
>a general idea of how many extra calories you need to be providing in
>the daily ration (obviously, you would spread the extra calories over 
>a
>number of days, and make sure the horse keeps exercising, so you don't
>run into azoturia problems).  If you don't provide enough calories, 
>the
>horse isn't going to drop dead or his ears fall off, he's just going 
>to
>lose weight throughout the season, which IS eventually going to affect
>his performance.
>
>This formula seemed to be pretty accurate in the study, but probably 
>is
>only a rough estimate for endurance horses that will be expending more
>or less energy depending on factors like terrain, temperature, the 
>skill
>of the rider and individual factors like temperament, keeping 
>qualities,
>type of feed provided and feed efficiency factors.  But for those 
>riders
>that like knowing the details of what's going on, this formula might 
>be
>something interesting for you to keep around just for grins.
>
>Seeya,
>
>Susan Garlinghouse
>
>

_____________________________________________________________________
You don't need to buy Internet access to use free Internet e-mail.
Get completely free e-mail from Juno at http://www.juno.com
Or call Juno at (800) 654-JUNO [654-5866]