# 3.2: Calculation of Mass from Measured Weight

This practically important process is straightforward for the traditional aeronautical and structural unit systems, but we can be confused when working in the SI system if the particular sensor used measures weight in the kilogram force (kgf) unit. Let us consider an example: suppose we have a laboratory scale with two different modes of weight measurement, pound force (lb) and kilogram force (kgf). We place an object on the scale with the display set to the lb mode, and the reading is $$W$$ = 15.50 lb. We next change the display to the kgf mode, and the reading changes to $$W$$ = 7.031 kgf. Let us calculate the mass in the three different unit systems using $$m$$ = $$W/g$$.

$\text { lb-inch-sec system: } m=\frac{15.50 \mathrm{lb}}{386.1 \mathrm{inch} / \mathrm{sec}^{2}}=0.04015 \frac{\mathrm{lb}-\mathrm{sec}^{2}}{\mathrm{inch}} \equiv 4.015 \mathrm{e}-2 \frac{\mathrm{lb}-\mathrm{sec}^{2}}{\text { inch }}$

$\text { lb-ft-sec system: } m=\frac{15.50 \mathrm{lb}}{32.17 \mathrm{ft} / \mathrm{sec}^{2}}=0.4818 \mathrm{slug} \equiv 4.818 \mathrm{e}-1\, \mathrm{slug} \equiv 4.818 \,\mathrm{deci} \text { -slug }$

For the SI calculation, we need to recognize that the kgf is not a consistent SI force unit, even though many weight scales sold and used in the United States have been calibrated in this unit since the 19th century. The only consistent SI force unit is the newton (N). So let us ignore the $$W$$ = 7.031 kgf reading for now and use, instead, the $$W$$ = 15.50 lb reading in association with the well-known conversion, 1 lb = 4.448 N:

$\text { SI system: } m=\frac{15.501 \mathrm{b} \times 4.448 \frac{\mathrm{N}}{1 \mathrm{b}}}{9.807 \mathrm{m} / \mathrm{sec}^{2}}=\frac{68.95 \mathrm{N}}{9.807 \mathrm{m} / \mathrm{sec}^{2}}=7.031 \mathrm{kg}$

So, in the SI system, the mass in kg has the same numerical value as the weight in kgf, which, in the future, will relieve us from having to calculate a conversion. This equality is precisely true only at elevations on Earth where $$g$$ is exactly 9.807 m/s2, but it is close enough for most engineering calculations using weights measured anywhere on Earth’s surface.