### 1.2 You won’t believe what you can do with math!

You’ve been told many times that meteorology is a math-intensive field. It is. But for this course, you already know much of the math, and what you haven’t seen, you will see in vector calculus. To get ready for the meteorology and atmospheric science in this course, you will need to refresh your ability to solve simple math problems, including solving simple problems in differential and integral calculus. At the same time, we will remind you about the importance of correctly specifying significant figures and units in your answers to the problems. The goal of this first lesson is to boost your confidence in the math you already know.

### How many figures should be in my answer?

Suppose you are asked to solve the following word problem:

**In the radar loop, a squall line is oriented in the north-south direction and is heading northeast at 57 km hr ^{-1}. In the last frame of the loop, the line is 17 km west of the Penn State campus. You are out running and know that you can make it back to your apartment in 25 minutes. Will you get back to your apartment before you get soaked?**

You reason that the line is moving northeast, and thus, at an angle of 45^{o} relative to the east. Therefore, the eastward motion of the squall line is just the velocity times the cosine of 45^{o}. That gives you the eastward speed. You decide to divide the distance by the eastward speed to get the amount of time before the line hits campus. You plug the numbers into your calculator and get the following result:

$\text{time}=\frac{17\text{km}}{\text{(}57\text{km/h)}\cdot \mathrm{cos}\left(\mathrm{45\xba}\right)}=0.42178\text{hours}=25.3070\text{minutes}$

According to your calculation, you will make it back with 0.3 minutes (18 seconds) to spare. But can you really be sure that the squall line will strike in 25.3070 minutes? Maybe you should figure out how many significant figures your answer really has. To do that, you need to remember the rules:

#### Significant Figures Rules

- Non-zero numbers (1,2,3,4,5,6,7,8,9) are ALWAYS significant.
- Zeroes are ALWAYS significant:
- between non-zero numbers
- SIMULTANEOUSLY to the right of the decimal point AND at the end of the number
- to the left of a written decimal point and part of a number $\ge $ 10

- In a calculation involving multiplication or division, multiply numbers as you see them. Then the answer should have the same number of significant figures as the number with the fewest significant figures.
- In a calculation involving addition and subtraction, the number of significant figures in the answer depends on the number of significant figures to the right of the decimal point when all the added or subtracted numbers are put in terms of the same power-of-ten. Add or subtract all the numbers. The answer has the same number of significant figures to the right of the decimal point as the number with the least number of significant figures to the right of the decimal point.
- The number of significant figures is unchanged by trigonometric functions. For other special functions:,
- Logarithms: When taking log of a number with X significant figures, the resulting value should have X number of digits to the right of the decimal point. We can generally apply the same rule for natural logarithms (ln).
- Anti-logarithms: When taking the anti-logarithm (eg. 10
^{value}) of a number with X values to the right of the decimal point, the resulting value should have X significant figures. We can generally apply the same rule for exponentials (e).

- Exact numbers never limit the number of significant figures in the result of a calculation and therefore can be considered to have an infinite number of significant figures. Common examples of exact numbers are whole numbers and conversion factors. For example, there are exactly 4 sides to a square and exactly 1000 m in a km.
- For multi-step calculations, any intermediate results should keep at least one extra significant figure to prevent round-off error. Calculators and spreadsheets will typically keep these extra significant figures automatically.
- When rounding, numbers ending with the last digit > 5 are rounded up; numbers ending with the last digit < 5 are rounded down; numbers ending in 5 are rounded up if the preceding digit is odd and down if it is even.

Number(s) | Answer | Number of Significant Figures | Reason |
---|---|---|---|

$25+.3$ | 25 | 2 | 25 has only 2 significant figures |

$25\xb70.3$ | 8 | 1 | $25\xb70.3=7.5$ , round to 8 because 0.3 has only 1 significant figure |

$1.5\left({10}^{3}\right)+3.24\left({10}^{2}\right)$ | $1.8\left({10}^{3}\right)$ | 2 | $1.5\left({10}^{3}\right)+0.324\left({10}^{3}\right)=1.824\left({10}^{3}\right)$ , then drop 2 to get $1.8\left({10}^{3}\right)$ |

$1.5\left({10}^{3}\right)+3.86\left({10}^{2}\right)$ | $1.9\left({10}^{3}\right)$ | 2 | $1.5\left({10}^{3}\right)+3.86\left({10}^{2}\right)=1.886\left({10}^{3}\right)$ , round up then drop 2 to get $1.9\left({10}^{3}\right)$ |

$\frac{\left(57.3+6.41\right)}{15.6}$ | 4.08 | 3 | $\frac{63.71}{15.6}=4.0840$ , trim to 3 significant figures to get 4.08 |

$200\left(3.142\right)$ | 600 | 1 | 200. has 3 significant figures; 200 (no decimal point) has 1 but is ambiguous |

$152\left({e}^{-.52}\right)$ | 90 | 2 | number in exponent has only 2 significant figures |

Check out this video (11:23): Unit Conversions & Significant Figures for a brief (1 minute) explanation of those rules! Start watching at 9:14 for the most relevant information. Note a minor error starting at 9:50 in which "60" should actually have a decimal point following the zero.

### What are the typical types of variables?

There are two types of variables – scalars and vectors. Scalars are amount only; vectors also have direction.

#### Dimensions and units are your friends.

Most variables have dimensions. The ones used in meteorology are:

**L**, length**T**, time**Θ**, temperature**M**, mass**I**, electric current

Some constants such as π have no units, but most do.

The numbers associated with most variables have units. The system of units we will use is the International System (SI, from the French Système International), also known as the MKS (meter-kilogram-second) system, even though English units are used in some parts of meteorology.

We will use the following temperature conversions:

$K{=}^{o}C+273.15$

$\left(\frac{5}{9}\right)\left({}^{o}F-32\right){=}^{o}C$

We will use the following variables frequently. Note the dimensions of the variables and the MKS units that go with their numbers.

Type | Variable | Dimensions | MKS Units | Common Unit Name |
---|---|---|---|---|

Scalar | length (x or ...) | L | m | |

area (A) | L^{2} |
m^{2} |
||

volume (V) | L^{3} |
m^{3} |
||

speed (u, v, w) | L/T | m/s | ||

energy (E) | ML^{2}/T^{2} |
kg m^{2}/s^{2} |
J = Joule | |

power (P) | ML^{2}/T^{3} |
kg m^{2}/s^{3} |
W = Watt | |

density (ρ) | M/L^{3} |
kg/m^{3} |
||

pressure (p) | M/LT^{2} |
kg/ms^{2} |
Pa = Pascal | |

electrical potential | ML^{2}/T^{3}A |
kg m^{2}/s^{3}A |
V = Volt | |

temperature (T) | Θ | K | ||

Vectors | velocity (v) |
L/T | m/s | |

momentum (mv) |
ML/T | kg m/s | ||

acceleration (a) |
L/T^{2} |
m/s^{2} |
||

force (F) |
ML/T^{2} |
kg m/s^{2} |
N = Newton |

#### Pressure is used for many applications.

**p = (normal force)/area = (mass x acceleration)/area = ML/T ^{2}L^{2} = M/LT^{2}**

**1 Pa = 1 kg m ^{–1} s^{–2}; 1 hPa = 100 Pa = 1 mb = 10^{–3} bar (hPa = hecto-Pascal)**

**1013.25 hPa = 1.01325 x 10 ^{5} Pa = 1 standard atmospheric pressure = 1 atm**

#### Wind speed is another frequently used variable.

The knot (kt) is equal to one nautical mile (approximately one minute of latitude) per hour or exactly 1.852 km/hr. The mile is nominally equal to 5280 ft and has been standardized to be exactly 1,609.344 m.

Thus, **1 m/s = 3.6 km/hr ≈** **1.944 kt **and** 1 kt ≈ 1.151 mph**.

**surface winds are typically 10 kts ~ 5 m/s**

**500 mb winds are ~50 kts ~ 25 m/s**

**250 mb winds are ~100 kts ~ 50 m/s**

#### Temperature is a third frequently used variable.

**Kelvin (K) must be used** in all physical and dynamical meteorology calculations. **Surface temperature is typically reported in ^{o}F (or ^{o}C for METARS) and in ^{o}C for upper air soundings**.

#### Water vapor mixing ratio is another frequently used variable.

$$\text{w=}\frac{{\text{massH}}_{2}\text{O}}{\text{massdryair}}$$

Usually the units for water vapor mixing ratio are g kg^{-1}. **In the summer w can be 10 g kg ^{-1}; in the winter, it can be 1-2 g kg^{-1}**.

Dimensions truly are your friend. Let me give you an example. Suppose you have an equation **ax + b = cT**, and you know the dimension of **b**, **x** (a distance), and **T** (a temperature), but not **a** and **c**. You also know that each term in the equation – the two on the left-hand side and the one on the right-hand side – must all have the same units. Therefore, if you know **b**, you know that the dimensions of **a** must be the same as the dimensions of **b** divided by **L** (length) and the dimensions of **c** must be the same as the dimensions of **b** divided by **Θ**.

Also, if you invert a messy equation and you're not sure that you didn’t make a mistake, you can check the dimensions of the individual terms and if they don’t match up, it’s time to look for your mistake. Or, if you have variables multiplied or divided in an exponential or a logarithm, the resulting product must have no units.

So, how do you find the dimenstions of units of an integral equation? One way to look at this problem is the figure out the dimension or the units of the integral' s solution. Or, if you do not want to solve the integral, then you just assign the dimensions or units of each term in the integral expression, even the differentials such as dx, dt. The integrals symbols themselves have no dimensions or units. For example, take the integral of force over a distance, which is the integral of F dx. Pay no attention to the integral symbol. The dimensions of F are MLT^{-2} and the dimensions of dx are L, so when you multiply them together, you get MLT^{-2} *L = ML^{2} T^{-2}, which has the same units as energy!

To find the dimensions or units of a derivative, the same thought process should be applied: either figure out the dimensions or units that the actual derivative would have, or if you do not want to solve the derivative, assign dimensions or units to every term in the equation that can have dimensions and units. For example, in dx/dt, dx is just a length and dt is just a time, so assign them those dimensions or units, which in this example is L/T or m/s. On the other hand, d/dt by itself has just the dimension of 1/T because the expression "d" has no units by itself. You now see that finding the dimensions or units for integral or derivative expressions is just as easy as for algebraic expressions.

Always write units down and always check dimensions if you aren’t sure. That way, you won’t crash your spacecraft on the back side of Mars. View the following video (2:42).

#### Quiz 1-1: Significant figures, dimensions, and units.

Now it's time to to take a quiz. I *highly* recommend that you begin by taking the Practice Quiz before completing the graded Quiz. Practice Quizzes are *not* graded and *do not affect your grade* in any way–except to make you more competent and confident to take the graded Quizzes : ).

- In Canvas, find
**Practice Quiz 1-1**. You may complete this practice quiz as many times as you want. It is not graded, but it allows you to check your level of preparedness before taking the graded quiz. - When you feel you are ready, take
**Quiz 1-1**. You will be allowed to take this quiz only**once**. This quiz is timed, so after you start, you will have a limited amount of time to complete it and submit it. Good luck!