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Re: RC: Cannon bones/Tendon size
You talk like some of the engineers that I work with.
I showed your post to a couple of them, explained what you where talking about and they looked at your numbers and agreeded. Think I will remeasure tonight.
guess I should have got my degree instead of stopping 12 hrs short.
In a message dated Wed, 3 Nov 1999 11:24:53 AM Eastern Standard Time, Truman Prevatt <firstname.lastname@example.org> writes:
> I got thinking about this some time back. If you take a solid round column,
> its abiltiy to withstand a load is directly proportional to its cross sectional
> area and not its circumference. Of course it is very difficult to directly
> measure cross sectional area of a horses cannon and is easier to measure
> circumference so it is logical that circumference is the measure of choice.
> The cross sectional area is directly proportional to the square of the
> circumference and not to the circumference. Hence to double the cross
> sectional area you only need the square root of two as large a circumference or
> to double the cross sectional area you would need to multiply the circumference
> by 1.414.
> So if 8 inches is correct for a 1000 pound horse then for a 1150 pound horse
> you would need about 8.6 instead of the 9.2 which would be the estimate if
> you took the proportianl change in the circumference in instead of the cross
> sectional area.
> And of course in any analysis we are assuming all horses have the same type of
> gait (which they don't) and all bones are the same density - which they aren't.
> So I would think the best first order approximation is 8*Sqrt(weight
> (lbs)/1000) in inches for the nominal size of the cannon bone as a function of
> the horses weight.
> For a 750 pound horses this would be 6.9, for 900 , 7.6, for 1200, 8.8, etc.
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