METEO 300
Fundamentals of Atmospheric Science

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

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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 45o relative to the east. Therefore, the eastward motion of the squall line is just the velocity times the cosine of 45o. 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:

time= 17 km  ( 57 km/h) cos( 45º ) =0.42178 hours =25.3070 minutes MathType@MTEF@5@5@+=faaagCart1ev2aqaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaabshacaqGPbGaaeyBaiaabwgacqGH9aqpdaWcaaqaaiaaigdacaaI3aaabaGaaGynaiaaiEdacqGHflY1ciGGJbGaai4BaiaacohadaqadaqaaiaaisdacaaI1aaacaGLOaGaayzkaaaaaiabg2da9iaaikdacaaI1aGaaiOlaiaaiodacaaIWaGaaG4naiaaicdacaqGGaGaaeyBaiaabMgacaqGUbGaaeyDaiaabshacaqGLbGaae4Caaaa@501D@

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

  1. Non-zero numbers (1,2,3,4,5,6,7,8,9) are ALWAYS significant.
  2. Zeroes are ALWAYS significant:
    1. between non-zero numbers
    2. SIMULTANEOUSLY to the right of the decimal point AND at the end of the number
    3. to the left of a written decimal point and part of a number MathType@MTEF@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbqedmvETj2BSbqefm0B1jxALjharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8FesqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9Gqpi0dc9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyyzImlaaa@2FDF@ 10
  3. 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.
  4. 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.
  5. The number of significant figures is unchanged by trigonometric functions. For other special functions:,
    1. 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).
    2. Anti-logarithms: When taking the anti-logarithm (eg. 10value) 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).  
  6. 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.
  7. 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.
  8. 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.
Examples
Number(s) Answer Number of Significant Figures Reason
25+.3 25 2 25 has only 2 significant figures
25·0.3 8 1 25·0.3=7.5 , round to 8 because 0.3 has only 1 significant figure
1.5 10 3 +3.24 10 2 1.8 10 3 2 1.5 10 3 +0.324 10 3 =1.824 10 3 , then drop 2 to get 1.8 10 3
1.5 10 3 +3.86 10 2 1.9 10 3 2 1.5 10 3 +3.86 10 2 =1.886 10 3 , round up then drop 2 to get 1.9 10 3
57.3+6.41 15.6 4.08 3 63.71 15.6 =4.0840 , trim to 3 significant figures to get 4.08
200 3.142 600 1 200. has 3 significant figures; 200 (no decimal point) has 1 but is ambiguous
152 e .52 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.

Unit Conversions and Significant Figures
Click here for transcript of the Significant Figures video.

Now to the magic of figuring out how many sig figs your answer should have. There are two simple rules for this. If it's addition or subtraction it's only the number of figures after the decimal point that matters. The number with the fewest figures after the decimal point decides how many figures you can have after the decimal in your answer. So 1,495.2+1.9903 you do the math. First you get 1,497.1903 and then you round to the first decimal, because that first number only had one figure after the decimal. So you get 1,497.2. And for multiplication, just make sure the answer has the same sig figs as your least precise measurement. So 60 x 5.0839 = 305.034, but we only know two sig figs so everything after those first two numbers is zeroes: 300. Of course then we'd have to point out to everyone that the second zero but not the third is significant so we'd write it out with scientific notation: 3.0 * 10^2. Because science! Now I know it feels counterintuitive not to show all of the numbers that you have at your fingertips, but you've got to realize: all of those numbers beyond the number of sig figs you have? They're lies. They're big lying numbers. You don't know those numbers. And if you write them down people will assume that you do know those numbers. And you will have lied to them. And do you know what we do with liars in chemistry? We kill them! Thank you for watching this episode of Crash Course Chemistry. Today you learned some keys to understanding the mathematics of chemistry, and you want to remember this episode in case you get caught up later down the road: How to convert between units is a skill that you'll use even when you're not doing chemistry. Scientific notation will always make you look like you know what you're talking about. Being able to chastise people for using the wrong number of significant digits is basically math's equivalent of being a grammar Nazi. So enjoy these new powers I have bestowed upon you, and we'll see you next time. Crash Course Chemistry was filmed, edited, and directed by Nick Jenkins. This episode was written by me, Michael Aranda is our sound designer, and our graphics team is Thought Bubble. If you have any questions, comments or ideas for us, we are always down in the comments. Thank you for watching Crash Course Chemistry.

Credit: Crash Course "Unit Conversion & Significant Figures: Crash Course Chemistry #2." YouTube. February 19, 2013.

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

5 9 o F32 = o C

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

Variables With Associated Dimensions and MKS Units
Type Variable Dimensions MKS Units Common Unit Name
Scalar length (x or ...) L m
area (A) L2 m2
volume (V) L3 m3
speed (u, v, w) L/T m/s
energy (E) ML2/T2 kg m2/s2 J = Joule
power (P) ML2/T3 kg m2/s3 W = Watt
density (ρ) M/L3 kg/m3
pressure (p) M/LT2 kg/ms2 Pa = Pascal
electrical potential ML2/T3A kg m2/s3A V = Volt
temperature (T) Θ K
Vectors velocity (v) L/T m/s
momentum (mv) ML/T kg m/s
acceleration (a) L/T2 m/s2
force (F) ML/T2 kg m/s2 N = Newton

Pressure is used for many applications.

p = (normal force)/area = (mass x acceleration)/area = ML/T2L2 = M/LT2

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 105 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 oF (or oC for METARS) and in oC for upper air soundings.

Water vapor mixing ratio is another frequently used variable.

w =  mass H 2 O mass dry air MathType@MTEF@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaabEhacaqGGaGaaeypaiaabccadaWcaaqaceaadbGaaeyBaiaabggacaqGZbGaae4CaiaabccacaqGibWaaSbaaSqaaiaaikdaaeqaaOGaae4taaqaaiaab2gacaqGHbGaae4CaiaabohacaqGGaGaaeizaiaabkhacaqG5bGaaeiiaiaabggacaqGPbGaaeOCaaaaaaa@47B5@

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 = ML2 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).

When NASA Lost a Spacecraft Because it Didn't Use Metric
Click here for transcript of the NASA video.

Remember when NASA lost a spacecraft because it's simultaneously used Imperial and metric measurements on the same mission? The Mars Climate Orbiter disappeared 15 years ago this month and here's a very brief recap of exactly what went wrong. The Mars Climate Orbiter launched on December 11, 1998 on a mission to orbit Mars. This first interplanetary weather satellite was designed to gather data on Mars' climate and also serve as a relay station for the Mars Polar Lander, a mission that launched a few weeks later. But you can't just launched a spacecraft towards Mars and trust that it's going to get where it's going. You to have to monitor its progress. Many spacecraft have reaction wheels to keep them oriented properly and navigation teams behind interplanetary spacecraft that constantly monitor the angular momentum and adjust trajectory to make sure it gets exactly where it needs to go. In the case of the Mars Climate Orbiter, monitoring its trajectory and angular momentum involved a few steps. First, data from the spacecraft was transferred to the ground by telemetry. There it was processed by a software program and stored in an angular momentum desaturation file that process data was what scientists used to adjust the trajectory. Adjustments that were made by firing the spacecraft's thrusters. Every time the thrusters were fired, the resulting change in velocity was measured twice: once by software program on the spacecraft and once by software program off the ground. And here's where the problem comes in. It turned out that the two systems the processing software on the spacecraft and the software on the ground we're using two different units of measurements. The software on the spacecraft measured impulse, or the changes by thrusters in newton seconds a commonly accepted metric unit of measurement, while the processing software on the ground use the Imperial pound seconds. And it was unfortunately the ground computer's data that scientists used to update the spacecraft trajectory and because one pound of force is equal to 4.45 Newton's every adjustment was off by a factor of 4.45. For a spacecraft traveling tens of millions of miles to destination a number of seemingly small errors really add up. During the Mars Climate Orbiters nine-month cruise to Mars seven errors were introduced into its trajectory that meant that when it reached the red planet it was 105 miles closer to the Martian surface than expected. This turned out to be an unsurvivably low altitude for its Mars encounter when the spacecraft fire its main engine for the orbit insertion burn that was designed to put it into an elliptical orbit nothing happened. NASA lost contact quite abruptly with the spacecraft. So while we know the root cause of just what went wrong we'll never know exactly what happened to the Mars Climate Orbiter. The loss of the Mars Climate Orbiter very sadly happened in space. Leave your spacey questions and comments below, and don't forget to subscribe.

Credit: Scientific American Space Lab. "When NASA Lost a Spacecraft Because It Didn't Use Metric - It Happened in Space #21." YouTube. September 18, 2014.

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 : ).

  1. 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.
  2. 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!