- #1
Phrak
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The topic of this post is about the applicability of mathematical formalism in applied physics, and especially how physical units as well as physical variables might have proper attention.
I've come upon an engineering problem requiring integration of 1/R. R is in centimeters. It has associated units.
To simplify, I have a radially symmetric problem in thermal resistance, that can be stated as
[tex]R_{Th} = K \int^{R_b}_{R_a}dR \frac{1}{R}[/tex]
R is the radial dimension.
So I went to a table of indefinite integrals and found this
[tex]\int \frac{dx}{x} = ln(x)[/tex]
Applying this template function in the usual way I get
[tex]R_{Th} = K [ln(R_{b}) - ln(R_a)][/tex]
Now, I'm stuck. I can find ln(R), but I don't know how many centimeters are in one log centimeter.
How many cm are in one ln(cm)?
I've rewritten my log tables to say
[tex]\int \frac{dx}{x} = ln(x/c)\ , c>0[/tex]
Anything wrong with this? (Please don't nit-pick over x vs. |x|.)
I've come upon an engineering problem requiring integration of 1/R. R is in centimeters. It has associated units.
To simplify, I have a radially symmetric problem in thermal resistance, that can be stated as
[tex]R_{Th} = K \int^{R_b}_{R_a}dR \frac{1}{R}[/tex]
R is the radial dimension.
So I went to a table of indefinite integrals and found this
[tex]\int \frac{dx}{x} = ln(x)[/tex]
Applying this template function in the usual way I get
[tex]R_{Th} = K [ln(R_{b}) - ln(R_a)][/tex]
Now, I'm stuck. I can find ln(R), but I don't know how many centimeters are in one log centimeter.
How many cm are in one ln(cm)?
I've rewritten my log tables to say
[tex]\int \frac{dx}{x} = ln(x/c)\ , c>0[/tex]
Anything wrong with this? (Please don't nit-pick over x vs. |x|.)
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