Basics of Unit Conversion – Fusion of Engineering, Control, Coding, Machine Learning, and Science

In this video, we explain the basics of unit conversion. The video accompanying this post is given below.

One of the first things that future engineers should learn is how to express physical variables in different units. The universally accepted system of units is the International System of Units (abbreviated as SI ).

The SI unit for length is the meter. The SI symbol for meter is $[m]$ . In engineering and scientific reports, the standard notation is to surround units by square brackets. For example, we usually write $1[m]$ for 1 meter. Next, we learn to convert miles to meters.

Problem 1: Express $124 [mi]$ ( $[mi]$ is the symbol for mile) in meters.
Solution: First, we need to know how many meters one mile has. According to Mr. Google, $1[mi] = 1609.34 [m]$ . Consequently, we have

(1) $\begin{align*}& 124 [mi] = 124 \cdot 1 [mi] = 124 \cdot 1609.34 [m] = 199 558.16 [m] \\& =199.55816 \cdot 10^{3} [m]\end{align*}$

The SI unit for mass is the kilogram, denoted by the symbol $[kg]$ . Notice that kilogram is not the unit for weight. The unit for weight is Newton ( $[N]$ is the SI symbol). In the US, we often express mass in pounds (the symbol is $[lb]$ ). Let us learn to convert pounds to kilograms.

Problem 2: A person’s mass is $88 [kg]$ . Express the mass in pounds.
Solution: First, we need to find how many pounds we have in one kilogram. Again, according to Mr. Google, $1[kg]=2.20462 [lb]$ . Consequently, we have

(2) $\begin{align*}88 [kg] = 88 \cdot 1[kg] = 88 \cdot 2.20462 [lb] =194.00656 [lb] \end{align*}$

Okay, let us now learn to solve a bit more complex unit conversion problems.

The velocity of an object is expressed in meter per second. The symbol is $[m/s]$ or in the fraction form $[\frac{m}{s}]$ . In Europe, speed limits are expressed in kilometer per hour ( $[km/h]$ or $[\frac{km}{h}]$ ). On New York highways, the speed limit is usually $55 [mi/h]$ (55 miles per hour).

Problem 3: Convert $55 [mi/h]$ to
a) [km/h]
b) [m/s]

Solution: a) According to Mr. Google, $1 [mi] = 1609.34 [m] = 1.60934 [km]$ . Consequently

(3) $\begin{align*}55 \frac{[mi]}{[h]}= 55 \frac{1.60934 [km]}{[h]}= 88.5137 \frac{[km]}{[h]}\end{align*}$

b) $1$ hour has $60 [min]$ (the symbol for minutes is $[min]$ ), and $1 [min]$ has $60 [s]$ (the symbol for seconds is $[s]$ ). Consequently, $1 [h] =3600 [s]$ , and we can write

(4) $\begin{align*}55 \frac{[mi]}{[h]}= 55 \frac{1609.34 [m]}{3600 [s]} =24.58713\bar{88} \frac{[m]}{[s]}\end{align*}$

in the last equation $\bar{88}$ stands for $8888888\ldots$ .

Finally, let us solve the following problem.

Problem 4: Express $5.6 [cm]$ (centimeter) in

a) $[mm]$ (milimeter)
b) $[\mu m]$ (micrometer)
c) $[km]$

Solution: a) $1 [cm]$ has $10 [mm]$ , consequently $1 [mm] = 1/10 [cm]= 10^{-1} [cm]$ . From here, we have $5.6 [cm] = 56 [mm]$ . b) $1 [m] = 10^{6} [\mu m]= 10^{2} [cm]$ . From this equation, we have:

(5) $\begin{align*}10^{6} [\mu m]=10^{2} [cm]\end{align*}$

(6) $\begin{align*}1 [cm]=\frac{10^{6}}{10^{2}} [\mu m]=10^{4} [\mu m]\end{align*}$

Consequently, $5.6 [cm] = 5.6 \cdot 10^{4} [\mu m]$ . c) $1 [km]= 10^{3} [m] = 10^{3} \cdot 10^{2} [cm]=10^{5} [cm]$ . Consequently, $1 [cm] =10^{-5} [km]$ , and $5.6 [cm] = 5.6 \cdot 10^{-5} [km]$ .

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