Unit conversion factors
- These factors are equivalent to one
- They can have units but be dimensionless
- They are not numerically equal to one in most cases.
- Units can be crossed out
Creating a Conversion Factor
Start with a quantity equality
Now use algebra rules to create a fraction equal to one
{{2.54 cm}\over{1 inch}} = {{1 inch}\over{1 inch}}
{{2.54 cm}\over{1 inch}} = 1
We can also divide both sides by 2.54 cm instead.
{{1 inch}\over{2.54 cm}} = {{2.54 cm}\over{2.54 cm}}
{{1 inch}\over{2.54 cm}} = 1
Multiplying a length by these factors doesn’t change the length since the factor is equal to one.
We can also thing of multiplying this factor as dividing by a length to get a number and then multiplying by an equal length with different units.
Different Dimensions
- You cannot convert a quantity with one dimension to another dimension with a unit conversion factor
- There may be a linear relationship between those two quantities that looks like a “conversion” but isn’t
Unit conversion factors
These two quantities are equal.
3\;\textrm{feet} = 1\;\textrm{yard}
If we divide both sides by 3 feet, we get
1 = \frac{1\;\textrm{yard}}{3\;\textrm{feet}}
Note that this quantity has units of yard per feet, but has no dimensions since it is a length divided by a length. If you multiply a quantity of feet by this, you won’t change the quantity but you will change the units.
Alternate Technique
Some folks find the unit conversion fractions to be hard to read. An alternate unit conversion method is starting with the equation and then multiplying both sides.
To convert 10 meters to feet:
1 m = 3.28 ft 10 \cdot 1 m = 10 \cdot 3.28\; ft 10 m = 32.8 ft
To convert 30 miles per hour to meters per second:
1\; mile = 1609\; meters 1\; hour = 3600\; seconds
We can divide these two equations to get a combined unit conversion equality or equation.
1\; mile \div 1\; hour = 1609\; meters \div 3600\; seconds 1\; \textrm{mile per hour} = 0.447\; \textrm{meters per second}
Now we can use this for conversions.
30 \cdot 1\; \textrm{mile per hour} = 30 \cdot 0.447\; \textrm{meters per second} 30\; \textrm{miles per hour} = 13.4\; \textrm{meters per second}
Mass Example
Convert the quantity of 150 pounds to kilograms.
150\;{\color{blue}\cancel{\textrm{pounds}}} \cdot \frac{1\;\textrm{kilogram}} {2.2\;\color{blue}\cancel{\textrm{pound}}} = 68\;\textrm{kilograms}
Speed Example
Convert the quantity of 100 miles per hour to meters per second. Note that we make two unit conversions. One from miles to meters and another from hours to seconds.
Once we have this set up, we perform the computation on a calculator or other device.
\frac{100\;{\color{blue}\cancel{\textrm{miles}}}} {\color{green}\cancel{\textrm{hour}}} \cdot \frac{1600\;\textrm{meters}} {\color{blue}\cancel{\textrm{mile}}} \cdot \frac{\color{green}\cancel{\textrm{hour}}} {60\;\color{purple}\cancel{\textrm{min}}} \cdot \frac{\color{purple}\cancel{\textrm{min}}} {60\;\textrm{sec}} = 44\;\frac{\textrm{meters}} {\textrm{sec}}
Some students prefer to perform this computation in steps.
\frac{100\;{\color{blue}\cancel{\textrm{miles}}}} {\textrm{hour}} \cdot \frac{1600\;\textrm{meters}} {\color{blue}\cancel{\textrm{mile}}} = \frac{160,000\;\textrm{meters}}{\textrm{hour}} \frac{160,000\;\textrm{meters}}{\color{green}\cancel{\textrm{hour}}} \cdot \frac{\color{green}\cancel{\textrm{hour}}} {60\;\textrm{min}} = \frac{2667\;\textrm{meters}}{\textrm{min}} \frac{2667\;\textrm{meters}}{\color{purple}\cancel{\textrm{min}}} \cdot \frac{\color{purple}\cancel{\textrm{min}}} {60\;\textrm{sec}} = 44\;\frac{\textrm{meters}} {\textrm{sec}}
“Per Per” Example
In August 2021, Sonoma County is recording 20 new cases per day per 100,000 people. If we have a population of 5,000 people, how many cases per week do we expect?
\frac{20\;\textrm{cases}}{{\color{green}\cancel{\textrm{day}}}\cdot 100,000\;\color{blue}\cancel{\textrm{people}}} \cdot 5,000\;{\color{blue}\cancel{\textrm{people}}} \cdot \frac{7\;\color{green}\cancel{\textrm{day}}}{\textrm{week}} = 7\;\textrm{cases per week}
It is recommended that students spend in total 3 hours per week per unit. Let’s assume all of our classes are 4-units, we can then convert this to minutes per day per class.
\frac{3\;\color{blue}\cancel{\textrm{hours}}} {\color{purple}\cancel{\textrm{week}}\cdot\color{green}\cancel{\textrm{unit}}} \cdot \frac{60\;\textrm{min}} {\color{blue}\cancel{\textrm{hour}}} \cdot \frac{4\;\color{green}\cancel{\textrm{units}}} {\textrm{class}} \cdot \frac{\color{purple}\cancel{\textrm{week}}} {7\;\textrm{days}} = 103\; \frac{\textrm{minutes}} {\textrm{day}\cdot\textrm{class}}
Two units in the denominator
This is often confusing. A few thought experiments can help.
- Is energy per temperature per mass the same as energy per mass per temperature?
- Is J/g/C the same as J/C/g?
- Is 1/2/3 the same as 1/3/2?
- Is \frac{J}{gC} the same as \frac{J}{Cg}
Food Example
Estimation: How many calories worth of food does the caf need to buy each week?
Converting Body Units
Let’s say you measure that your stride is 5 feet and a distance is 6 strides. You can then convert strides to feet and meters.
6 strides x 2.5 ft/stride = 15 feet
6 strides x 2.5 ft/stride x 1 meter/3.28 ft = 4.6 meters
Multiple Units
In the example above, you may see the quantity written in several different ways:
- 3 hours per week per unit
- 3 hours/week/unit
- 3 hours/week \cdot unit
- this one has an implicit parenthesis around the week and unit
Another example is that humans need approximately 2000 kcalories per person per day.
Dimensional analysis example
We can use the dimensions of the relevant quantities to deduce the form of an equation
Energy has dimensions of mass times length squared divided by time squared.
E = \frac{\textrm{[mass]}\textrm{[length]}^2}{[\textrm{time}]^2}
If we know that the mass and velocity are related to energy, we can find how by:
- m mass of the object
- v velocity of the object
That is we think the energy formula must be of the form:
E \propto m^a v^b
We can substitute these units
m = [\textrm{mass}] v = \frac{\textrm{[length]}}{\textrm{[time]}}
Set equal to the units for energy
\textrm{[mass]}^a \frac{\textrm{[length]}^b}{[\textrm{time}]^b} = \frac{\textrm{[mass]}\textrm{[length]}^2}{[\textrm{time}]^2}
and deduce that
E \propto m v^2
These units match the formula for kinetic energy and for mass-energy equivalence.