__Dimensional Analysis__

Dimensional analysis is a powerful tool for solving problems and making conversions. It can allow you to do calculations without memorizing specific formulas. Dimensional analysis will be used repeatedly, especially in the first four chapters.

__Example__

Provided with the following conversion factors, 5028 ft = 1 mi, 2.54 cm = 1 inch, determine the number of meters in 10.2 miles?

10.2 miles
´ [5028 ft/1 mi] ´ [12 inches/1 ft] ´ [2.54 cm/1 in] ´ [1 m/100 cm] = 15631
m = 1.56 ´ 10^{4} m

To use dimensional analysis to solve problems we need to have at hand the following pieces of information:

**(a) What data and units
are given to start the problem?**

**(b) What quantity is
needed as an answer?**

**(c) What conversion
factors are available to go from the given data to the answer?**

If you use appropriate conversion factors and cancel units you can check to see if you have done the problem correctly, by determining the units of your answer.

In the previous example we converted from one unit of length to another unit of length. Dimensional analysis can also be used when the starting point and ending point are different types of units, provided we use quantities with two types of units as conversions.

For Example

- Density ® Relates mass and volume
- Velocity ® Relates length and time

__Example__

If water
has a density of 1.00 g/cm^{3}, what would be the mass of
water (in kg) inside a swimming pool 10 ft wide, 10 ft long and 8
ft deep?

Q) What quantity are we starting with?

A) The
volume of a swimming pool 10 ft ´ 10 ft ´ 8 ft = 800 ft^{3}

Q) What quantity do we want to determine?

A) The mass of water inside the pool.

[8 ´ 10^{2} ft^{3}]
´ [12 in/1 ft]^{3}
´ [2.54 cm/1 ft]^{3}
´ [1.00 g/cm^{3}]
´ [1 kg/1000 g] =
22,653 kg = 2
´ 10^{4}
kg

__Example__

(68 from
BL&B). The distance from the earth to the moon is 240,000
miles (2.40 ´ 10^{5} mi)
and the Concorde flies at 2400 km/hr (2.4 ´ 10^{3}
km/hr). How many hours would it take to fly to the moon on the
Concorde?

What quantity are we starting with?

2.40 ´ 10^{5}
miles to the moon.

What quantity do we want to determine?

How many hours will it take to get to the moon.

2.40 ´ 10^{5} mi ´ [5280 ft/1 mi] ´ [12 in/1 ft] ´ [2.54 cm/1 in] ´ [1 m/100 cm] ´ [1 km/1000 m] ´ [2.4 ´ 10^{3}
km/1 hr]

= 160.9
hours = 1.6 ´ 10^{2} hrs
= 6.7 days