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On Sun, 06 Feb 2005 04:56:37 GMT, Steven O. <null@null.com> wrote:
[snip]
could be ( - a Ts ), so that
ln R = a T - a Ts = a (T - Ts)
exp[ ln R ] = exp[ a (T - Ts) ]
exp[ a (T - Ts) ] = 1 + a (T - Ts) approx., so that
R = 1 + a (T - Ts)
If the Prof will buy into that, then
R1/R2 = [ a (T1 - Ts) ] / [ a (T2 - Ts) ]
R1 = { [ a (T1 - Ts) ] / [ a (T2 - Ts) ] } R2
As I said, just a guess on my part, FWIW.
[snip]
I'll make a guess here that your "constant of integration"But here is how the math works out for me:
a = (1/R)(dR/dT)
dR/R = a dT Take indefinite integral of both sides....
ln R = a T + Ts, where Ts is said reference temperature
could be ( - a Ts ), so that
ln R = a T - a Ts = a (T - Ts)
exp[ ln R ] = exp[ a (T - Ts) ]
exp[ a (T - Ts) ] = 1 + a (T - Ts) approx., so that
R = 1 + a (T - Ts)
If the Prof will buy into that, then
R1 = a (T1 - Ts) and R2 = a (T2 - Ts)Assume R1 corresponds to T1, and R2 to T2, then....
ln R1 = a T1 + Ts, and ln R2 = a T2 + Ts
R1/R2 = [ a (T1 - Ts) ] / [ a (T2 - Ts) ]
R1 = { [ a (T1 - Ts) ] / [ a (T2 - Ts) ] } R2
As I said, just a guess on my part, FWIW.