MHB How to Convert 1/(ft*°F) to 1/(m*°C)?

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To convert 1/(ft*°F) to 1/(m*°C), first convert feet to meters using the factor 0.3048. The temperature conversion requires recognizing that an increase of 1°F corresponds to an increase of 5/9°C. This means that the conversion involves both linear distance and temperature change. Understanding these relationships allows for accurate unit conversion in the equation.
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Hi everyone,

so I have equation, let's say X and I want to convert it to SI units. Now X is in 1/(ft*°F) and I need 1/(m*°C). Ok, so I simply convert ft to m= 0.3048, but what to do about °F? 1°F is -17.2°C which will not work.
Any suggestions?

Thanks,
Miroslav
 
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mk747pe said:
Hi everyone,

so I have equation, let's say X and I want to convert it to SI units. Now X is in 1/(ft*°F) and I need 1/(m*°C). Ok, so I simply convert ft to m= 0.3048, but what to do about °F? 1°F is -17.2°C which will not work.
Any suggestions?

Thanks,
Miroslav

Hi Miroslav,

You must understand x as one unit per foot and per increase of one °F. (Absolute temperatures would not make sense : what if the temperature is 0°F ?).

You can then do as usual : one ft is 0.3048 m, and an increase of one °F is an increase of 5/9 °C (or °K).
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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