Gunner's Postulate of T*R*E*A*T*S

Gunner did so woofeth....

... I have a theory, which I humbly submit as Gunner's First
Postulate of T*R*E*A*T*S. The time it takes to time to eat a T*R*E*A*T
t(E), is:

t(E) = V * T * (D + 1) * g
---------------------- * C * G
W * t(LT)


Where V is the volume of the T*R*E*A*T (in cubic centimeters)
T is the Toughness of the T*R*E*A*T, on a scale where air is
a zero and a beef bone is a one.
(Reference values: piggy earz = 0.82 to 0.86
Beggin' Strips = 0.28
Nylabone = 1066
Cheese = 0.001)

D is the number of distractions, weighted by their interest
level. E.g., if Dad leaves the room, D = 1; if he opens the
refrigerator, D = 4, etc.
g is the gravitational constant, 980 cm/sec^2
W is the weight of the doggy eating the T*R*E*A*T, in kilograms
t(LT) is the time, in seconds, since the last T*R*E*A*T that
doggy enjoyed
C is the Chomp Factor, a measure of the doggy's chewing power,
ranked from zero (swallows everything whole) to 1 (toothless).
and G is Gunner's constant, 1.80x10^5 kg/cm^3.

(I'm a *German* shorthair, so all the units are metric.)

So, s'pose a 34 kg doggy is given a 20 cm^3 piggy ear (a fresh one,
T = 0.82). There are no distractions, and this poor doggy hasn't had
a piggy ear for 24 hours. From previous tests, we know his Chomp
Factor is 0.2. Then you can calculate t(E) = 197 seconds,
just over 3 minutes. But if a ccccccat comes into the room and sits
on top of the nonsmellybox (D = 2), then it takes t = 590 seconds,
nearly ten minutes.

If that same doggy is given a block of cheese 2 cm on a side (8 cm^3)
and the ccccat is taken out and shot, the doggy will eat the cheese in
0.096 seconds. A 100cm^3 nylabone, on the OTher paw, would take 14.8
days of solid chewing to disappear.


To use this formula, you need to get a test T*R*E*A*T, like a piggy
ear, and measure how long it takes to eat it without any
distractions. (It may take years to get a distraction-free test,
depending how serious you are about your T*R*E*A*T*S.) Values of T for
various treats can be found in the Canine Rover Company Handbook. Use
this information to calculate your personal Chomp Factor. From then
on, you can scientifically predict how fast you would destroy any
T*R*E*A*T.


For Gretta, let's guess she weighs about ten kilos (such a petite
little girl!) We'll assume an average T of 0.84, and that it had
been twenty-four hours since her last T*R*E*A*T. Being in the car
counts as 1 distraction.

Then I calculate her Chomp Factor as 0.26 - pretty respectable for a
buppy!


But to refine this theory I need to do lots of experiments. Please,
send me all your T*R*E*A*T*S - it's for the future of science!

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