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Thread: Conversion of formula (or equation) given in metric form to US customary unit form

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    Conversion of formula (or equation) given in metric form to US customary unit form

    Is there a procedure (or process) that can be followed to transform a formula or an equation that uses metric units to an equivalent one that uses US customary units? While I know the various literal variable letters would be the same, the tricky part I believe would be constants in the form of numerical coefficients and such.

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    Quote Originally Posted by rww88 View Post
    Is there a procedure (or process) that can be followed to transform a formula or an equation that uses metric units to an equivalent one that uses US customary units? While I know the various literal variable letters would be the same, the tricky part I believe would be constants in the form of numerical coefficients and such.
    It may be helpful if you can give a specific example of the sort of formula you are asking about, and the units you want to change it to, so we can be sure we are answering the right question.

    Many formulas (especially in physics) are designed around a specific set of units (or, really, the units are designed for the formulas) so that there are no visible coefficients. When you want to use different units, you can replace the variable with an expression that transforms the units you want to use into those required for the formula as provided. For example, if the formula takes r in centimeters and you want it in inches, you would replace r in the formula with (2.54R), where R is the new variable measured in inches, because this will represent the number of centimeters. (You can eventually go back to using r for the variable, but it is important to remember that this would be a different variable than the original. It's dangerous to use the same letter to mean two different things.)

    It can also be useful to note that the coefficient has its own dimensions; my 2.54 is really 2.54 cm/in.

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    Thank you for your rapid response

    Thank you; your statements make complete sense. Here is an example equation: The maximum gas velocity V (in m/s) in a pipe is given by V =175 (1/density)**0.43 where the units of gas density are kg/m**3. What would this equation transform to if the gas density was in lb/ft**3 and the desired velocity units were ft/s ?

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    Quote Originally Posted by rww88 View Post
    Is there a procedure (or process) that can be followed to transform a formula or an equation that uses metric units to an equivalent one that uses US customary units? While I know the various literal variable letters would be the same, the tricky part I believe would be constants in the form of numerical coefficients and such.
    As Dr.Peterson indicated, using units can sometimes make life easier when transforming from one system of units to another. For example, to convert a distance formula from metric [m=meters] to US [yd=yards] we would have something like this for a (usually unnecessarily) detailed process

    D (yd) = d (m) * 1.0936 (yd/m) = 1.0936 d (m) (yd/m) = 1.0936 d (yd) (m/m) = 1.0936 d (yd)

    -Ishuda

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    Quote Originally Posted by rww88 View Post
    Thank you; your statements make complete sense. Here is an example equation: The maximum gas velocity V (in m/s) in a pipe is given by V =175 (1/density)**0.43 where the units of gas density are kg/m**3. What would this equation transform to if the gas density was in lb/ft**3 and the desired velocity units were ft/s ?
    I assume you are using "**" to mean an exponent, so the problem is this:

    The maximum gas velocity V (in m/s) in a pipe is given by [tex]V =175 \left(\dfrac{1}{\rho}\right)^{0.43}[/tex] where the units of gas density are kg/m3. What would this equation transform to if the gas density was in lb/ft3 and the desired velocity units were ft/s ?

    Nice example!

    If we define new variables D = density in lb/ft3, and U = velocity in ft/s, we would replace [tex]\rho[/tex] with

    [tex]\rho = D \dfrac{lb}{ft^3} \times \dfrac{0.4536 kg}{1 lb} \times \left(\dfrac{3.28 ft}{1 m}\right)^3 = 16 D kg/m^3[/tex]

    and V with

    [tex]V = U \dfrac{ft}{s} \times \dfrac{1 m}{3.28 ft}= \dfrac{U}{3.28}m/s[/tex]

    making the new equation

    [tex]\dfrac{U}{3.28} =175 \left(\dfrac{1}{16D}\right)^{0.43}[/tex]

    which, solving for U to make it a formula, becomes

    [tex]U =574 \left(\dfrac{1}{16D}\right)^{0.43}[/tex]

    Or at least I think so. Check it out with some specific numbers.

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    It checks!

    Air at 15 degrees C and sea level has a density of 1.225 kg/m3. So, V is 175(1/1.225)**0.43 = 160.376 m/s. Converting the density results in an equivalent density of 0.07647 lb/ft3. So in this case V = 574[1/(16)(0.07647)]**0.43 = 526 ft/s. When 160.376 m/s is converted to ft/s the result is 526. It checks!

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    Quote Originally Posted by rww88 View Post
    Air at 15 degrees C and sea level has a density of 1.225 kg/m3. So, V is 175(1/1.225)**0.43 = 160.376 m/s. Converting the density results in an equivalent density of 0.07647 lb/ft3. So in this case V = 574[1/(16)(0.07647)]**0.43 = 526 ft/s. When 160.376 m/s is converted to ft/s the result is 526. It checks!
    Good. Of course, the important thing (beyond my having gotten the numbers right) is that you have the idea for other conversions.

    This was a really good example, involving conversions at both input and output, which I might have forgotten if I'd made up my own.

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