**Units and dimension analysis
**

In many situations, to help see how to combine numbers, you can associate a measurement *unit,* also called a *dimension, * with each number.

**Example 1 **

Start with an example that you could probably do without dimension analysis. A car travels at 50 miles in 3 hours. What's its average speed?

**Solution:** Since the problem is stated using miles and hours, you'll measure speed in miles per hour. *Per* means *for each.* It is represented mathematically by a fraction. For *miles per hour* you write the fraction miles/hour. Writing the dimension as a fraction tells you that you'll divide the number of miles by the number of hours. That is, you have 50 miles/3 hours, or 50/3 miles/hour.

**Example 2**

This is only slightly more difficult: You run for 3 hours at 5 miles/hour. How far do you run?

**Solution:** Yes, you multiply the 3 by 5. The dimension of hours cancels to give miles. That is, you travel (3)(5) miles, or 15 miles.

**Example 3**

You run for 3 miles at ^{5 miles/hour. How long do you run?
}

**Solution:** You'll measure
"how long" in hours. To get from miles and miles/hour to hours, you divide
.

So to find how long you run, you divide 3 miles by 5 mi/hour, to get 3/5 hours.

**Example 4**

The 3/5 hours of the previous problem is how many seconds?

**Solution:** You know there are 60 seconds in each minute. That is, there are 60 seconds per minute, or 60 sec/min. Similarly, there are 60 min/hour. To find how many seconds there are in 3/5 hours, you multiply, and the parts of the fractions cancel:

Note how numerators and denominators of the units cancel when multiplied. Arithmetic then gives you 2160 seconds.

**Example 5**

The sun is about 93 million miles from the earth. Light travels at a speed of 186,000 miles per second. How many minutes does a ray of light take to travel from the sun to the earth?

**Solution: ** If you divide distance by speed to get time, as you did in Example 3, you have
.

The result will be in seconds. But what is the result? With all of these zeros, you should probably use scientific notation. You have:

But the question asks for minutes, not seconds. To translate into minutes, do you multiply or divide? Again, dimensions help: you need to take

**Example 6**

The volume of a building is 103 cubic meters. The area of each floor is 73 square meters. What's the building's height?

**Solution:** The volume of a "box" is the area of its base times its height. So the height of the building is its volume divided by the area of its base.

A standard notation for cubic units and square units is helpful. Cubic meters represented by m^{3}, square meters by m^{2}. If you divide cubic meters by square meters, you get m^{3}/m^{2}= m^{1} = m. In this case, you can find the height of the building to be 103 m^{3}/73m^{2} = 1.37m.

**Example 7**

How many centimeters is the height in the previous example?

**Solution: ** There are 1000 centimeters per meter. That is, there are 10^{3} cm/m. To get the number of cm given the number of m, you just
multiply cm/m by m; the m's will cancel to give cm. In the previous example, you find the height of the building to be
(103 m) (1000 cm/m) = 103000 cm.

**Example 8**

How many feet tall is the building of the previous example if one foot equals about 0,3048 meters?

**Solution:** There are 0.3048 m/ft. You know the number of meters in the height of the building, and you want to find the number of feet, so you divide:

**Example 9**

A discharge of *R* megawatts into a streamflow of *F* cubic feet per second raises the water temperature
*T* = 15.2 *R*/*F* degrees Fahrenheit.

What are the dimensions, the measurement units, involved?

**Solution: **You know the units of *R* (megawatts), *F* (cubic feet per second), and *T* (degrees F), so the real question concerns the dimensions of the constant 15.2. You can write cubic feet per second as ft^{3}/sec, or cfs. Then the units on the fraction R/F are megawatts/cfs. You multiply the constant 15.2 by this fraction to get degrees, so for cancellation to take place the constant must have units degrees-cfs/megawatt. To check, write it out to see how the units cancel:

**Example 10**

Each person in a community of 10,000 people uses 170 gallons of water each day. If 6.54 gallons make up one cubic foot of water, then how many cubic feet of water does the entire community use in a year?

**Solution: ** The first fact can be expressed as 170 gallons per person per day. When there's repeated division, as indicated by per, the units continue to go in the denominator of the fraction: 170 gallons/person-day. If you multiply by 10,000 persons (people), the person units will cancel, and you'll get

(170)(10,000) gallons/day.Since there are 365 days/year, multiplying will cancel the dimensions of days to give you

(170)(10,000)(365) gallons/year.Since each cubic foot of water is 6.54 gallons, you have 6.54 gallons/ft

Now you can do the calculation to find the number of cubic feet per year used by the entire community.

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