KK,
Your question is not a sign of lunacy; it's just an attempt to explore the bounds of the equation.
To illustrate my point that it is only good for rate 1 turns, I have run the shortcut equation and a more accurate one through a spreadsheet. The results were as follows:
Comparing the results for Rate 1 turns at various speeds
TAS TAN AOB = TAS x ROT AOB = (TAS/10)+7 . .(Kts) g . . Error . .180 Kts, 26 actual AOB, 25 prediected, -1 error. .200 Kts, 28.4 actual AOB, 27 predicted, -1.4 error. .250 Kts, 34.1 actiual AOB,32 predicted, -2.1 error. .300 Kts, 39 actual AOB, 37 predicted, -2 error . .350 Kts, 43.4 actual AOB, 42 predicted, -1.4 error. .400 Kts, 47.2 actual AOB, 47 [redicted, -0.2 error. .450 Kts, 50.6 actual AOB, 52 predicted, +1.4 error. .500 Kts, 53.5 actual AOB, 57 predicted, +3.5 error. .550 Kts, 56.1 actual AOB, 62 predicted, 5.9 error. .600 Kts, 58.4 actual AOB, 67 predicted, 8.6 error
This shows that the shortcut equation is reasonably accurate throughout a fairly wide speed range.
. .Comparing results for various rates of turn at 250 Kts TAS.
Turn rate TANAOB = TAS x ROT AOB=((N xTAS)/10)+7 (N) g . . . .Rate 0.5, 18.8 actual,19.5 predicted, 0.7 error. .Rate 1.0, 34.1 actual, 32 predicted, -2.1 error. .Rate 1.5, 44.6 actual, 44.5 predicted, -0.1 error. .Rate 2.0, 53.7 actual, 57 predicted, 3.3 error. .Rate 2.5, 59.6 actual, 69.5 predicted, 9.9 error. .Rate 3.0, 63.9 actual, 82 predicted, 18.1 error. .Rate 3.5, 67.2 actual, 94.5 predicted, 27.3 error. .Rate 4.0, 69.8 actual, 107 predicted, 37.2 error
The shortcut equation is reasonably accurate only within a very narrow band (up to rate 2 turn), and then quickly diverges such that its results are nonsensical (more than 90 degrees AOB) at rate 3.5 and above.
. .CAT DRIVER,
Your problem is a bit more complicated.
First, we need to know the rate at which heat is put into the system. This is unlikely to be a linear function, because it will be proportional to the hump rate. So let's use the derivative dh/dHR.
However, hump rate is not likely to be constant either. For very young dogs, it will increase rapidly, whereas for very old dogs it will quickly subside. To account for this we require another derivative, dHR/D ageofdedog.
Next we need to consider the rate at which the system cools due to the escape of heat. Once again, it will not be linear so we need to use the derivative dc/dt.
Therefore, we now have an equation something like:
Temperature = dh/DHR . dHR/D agededog . dc/dt
Then of course we need to consider the quantity of water in the system. The problem here is that however much we might try to maintain constant experimental conditions, it is inevitable that someone will come along and take out some of the water.
This phenomenon is admirably illustrated by JT's conundrum about the sunken submarine. For some desperate people extracting water is the only way to get it up!!
Is this perhaps the problem you really wanted to share with us?
[ 16 February 2002: Message edited by: Keith Williams. ]</p>