Published on *METEO 300: Fundamentals of Atmospheric Science* (https://www.e-education.psu.edu/meteo300)

In previous lessons, we were able to explain physical and chemical processes using only algebra and differential and integral calculus. Thermodynamics, moist processes, cloud physics, atmospheric composition, and atmospheric radiation and its applications can all be quantified (at this level of detail) with fairly simple mathematics. However, to understand and quantify the dynamics of the atmosphere requires more math skill.

This lesson introduces you to the math and mathematical concepts that will be required to understand and quantify **atmospheric kinematics**, which is the description of atmospheric motion; and **atmospheric dynamics**, which is an accounting of the forces causing the atmospheric motions that lead to weather. Weather is really just the motion of air in the horizontal and the vertical and the consequences of that motion. The motion is caused by wind and wind has both direction and speed, which are best described by vectors.

The Earth is a spinning, slightly squashed sphere. The atmosphere is a tenuous thin layer on this orb, so from a human’s limited view, the Earth appears to be flat. For some applications, a simple Cartesian coordinate system, with three dimensions in the *x*, *y*, and *z* directions, seems like a good way to mathematically describe motion. For processes that occur on the larger scale, where the Earth’s curvature is noticeable, we must resort to using coordinates that are natural for a sphere.

The way wind direction is described sprang out of wind observations, and is now firmly implanted in the psyche of every weather enthusiast: easterly, northerly, westerly, and southerly. This wind convention, however, is quite different than that used in the equations that govern atmospheric motion, which are the basis of weather forecast models. Here we will see that a conversion between the two conventions is straightforward but requires some care.

Finally, we will see that movement of air can either be described by fixed observers on the ground (called the **Eulerian framework**) or by someone riding along with a moving parcel of air (called the **Lagrangian framework**). These two points-of-view are very different, but we will see that they are related to each other by **advection**, which is just the movement of air with different properties (such as temperature, pressure, and relative humidity) from some place upwind of the place where you are.

With this math and these concepts you will be ready to take on atmospheric kinematics and dynamics.

By the end of this lesson, you should be able to:

- calculate partial derivatives
- implement vector notation, the dot product, the cross product, and the del operator
- explain the different coordinate systems and how they are used
- convert between math and meteorological wind directions
- calculate temperature advection at any point on a map of isotherms (lines of constant temperature) and wind vectors

If you have any questions, please post them to the Course Questions discussion forum. I will check that discussion forum daily to respond. While you are there, feel free to post your own responses if you, too, are able to help out a classmate.

Here is another chance to earn **one point of extra credit: Picture of the Week**!

- You take a picture of some atmospheric phenomenon—a cloud, wind-blown dust, precipitation, haze, winds blowing different directions—anything that strikes you as interesting.
- Add a short description of the processes that you think are causing your observation. A Word file is a good format for submission.
- Use your name as the name of the file. Upload it to the
**Picture of the Week Dropbox**in this week's lesson module. To be eligible for the week, your picture must be submitted by 23:59 UT on Sunday of each week. - I will be the sole judge of the weekly winners. A student can win up to three times.
- There will be a Picture of the Week dropbox each week through Lesson 11. Keep submitting entries!

In your first calculus class you learned about derivatives. Suppose we have a function *f* that is a function of *x*, which we can write as *f*(*x*). What is the derivative of *f *with respect to *x*?

$$\frac{df(x)}{dx}$$

What about a new function that depends on two variables, *h*(*x,y*)? This function could, for example, give the height *h* of mountainous terrain for each horizontal point (*x,y*). So what is the derivative of *h* with respect to *x*? One way we determine this derivative is to fix the value of *y *=* y _{1}*, which is the same as assuming that

$${\left(\frac{dh}{dx}\right)}_{y=\text{constant}}\equiv \frac{\partial h}{\partial x}\text{}$$

This is called the **partial derivative** of *h* with respect to *x*. It’s pretty easy to determine because we do not need to worry about how *y* might depend on *x*.

Let $h={\left(x-3\right)}^{2}\mathrm{cos}\left(y\right)$ . What is the partial derivative of *h* with respect to *x*?

$$\frac{\partial h}{\partial x}=\frac{\partial \left({\left(x-3\right)}^{2}\mathrm{cos}(y)\right)}{\partial x}=2\left(x-3\right)\mathrm{cos}(y)$$

We can also find the partial derivative of *h* with respect to *y*. Can you do this?

$$\frac{\partial h}{\partial y}=\frac{\partial \left({\left(x-3\right)}^{2}\mathrm{cos}(y)\right)}{\partial y}=-{\left(x-3\right)}^{2}\mathrm{sin}(y)$$

So you can see that the $\partial h/\partial x$ may be different for each value of *y* and $\partial h/\partial y$ may be different for each value of *x*. Thus, even if you are not entirely familiar with partial derivatives and their notation, you can see that they are no different from ordinary derivatives but you take the derivative for just of one variable at a time.

(7:29)

Remember that a scalar has only a magnitude while a vector has both a magnitude and a direction. The following video (12:33) makes this difference clear.

Typically the vectors used in meteorology and atmospheric science have two or three dimensions. Let’s think of two three-dimensional vectors of some variable (e.g., wind, force, momentum):

$$\begin{array}{l}\overrightarrow{A}=\overrightarrow{i}{A}_{x}+\overrightarrow{j}{A}_{y}+\overrightarrow{k}{A}_{z}\hfill \\ \overrightarrow{B}=\overrightarrow{i}{B}_{x}+\overrightarrow{j}{B}_{y}+\overrightarrow{k}{B}_{z}\hfill \end{array}$$[3]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaerbdfwBIjxAHbqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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@51F4@

Sometimes we designate vectors with **bold lettering**, especially if the word processor does not allow for arrows in the text. When Equations [8.3] are written with vectors in bold, they are:

$$\begin{array}{l}A=i{A}_{x}+j{A}_{y}+k{A}_{z}\hfill \\ B=i{B}_{x}+j{B}_{y}+k{B}_{z}\hfill \end{array}$$[3]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaerbdfwBIjxAHbqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaabauaaaOqaauaadaqaceaaaeaaqqaaaaaaaaGySf2yRbWdbiaahgeacqGH9aqpcaWHPbGaamyqa8aadaWgaaWcbaWdbiaadIhaa8aabeaak8qacqGHRaWkcaWHQbGaamyqa8aadaWgaaWcbaWdbiaadMhaa8aabeaak8qacqGHRaWkcaWHRbGaamyqa8aadaWgaaWcbaWdbiaadQhaa8aabeaaaOqaa8qacaWHcbGaeyypa0JaaCyAaiaadkeapaWaaSbaaSqaa8qacaWG4baapaqabaGcpeGaey4kaSIaaCOAaiaadkeapaWaaSbaaSqaa8qacaWG5baapaqabaGcpeGaey4kaSIaaC4AaiaadkeapaWaaSbaaSqaa8qacaWG6baapaqabaaaaaaa@508C@

**Be comfortable with both notations.**

In the equations for vectors, *A _{x}* and

Sometimes we want to know the magnitude (length) of a vector. For example, we may want to know the wind speed but not the wind direction. The magnitude of $\stackrel{\rightharpoonup}{A}$ , or *A*, is given by:

$\left|\overrightarrow{A}\right|=\sqrt{\left({A}_{x}^{2}+{A}_{y}^{2}+{A}_{z}^{2}\right)}$

We often need to know how two vectors relate to each other in atmospheric kinematics and dynamics. The two most common vector operations that allow us to find relationships between vectors are the **dot product** (also called the scalar product or inner product) and the **cross product** (also called the vector product).

The dot product of two vectors * A* and

$$\begin{array}{c}\overrightarrow{A}\u2022\overrightarrow{B}={A}_{x}{B}_{x}+{A}_{y}{B}_{y}+{A}_{z}{B}_{z}\\ =\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|\mathrm{cos}\beta \\ =\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|\text{if}\overrightarrow{A}\text{isparallelto}\overrightarrow{B}\\ =0\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}\overrightarrow{A}\perp \overrightarrow{B}\end{array}$$[3]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaerbdfwBIjxAHbqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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@80D1@

The dot product is simply the magnitude of one of the vectors, for example ** A**, multiplied by the projection of the other vector,

Also note that the unit vectors (a.k.a., direction vectors) have the following properties:

$$\begin{array}{l}\overrightarrow{i}\xb7\overrightarrow{i}=\overrightarrow{j}\xb7\overrightarrow{j}=\overrightarrow{k}\xb7\overrightarrow{k}=1\\ \overrightarrow{i}\xb7\overrightarrow{j}=\overrightarrow{i}\xb7\overrightarrow{k}=\overrightarrow{j}\xb7\overrightarrow{k}=\overrightarrow{j}\xb7\overrightarrow{i}=\overrightarrow{k}\xb7\overrightarrow{i}=\overrightarrow{k}\xb7\overrightarrow{j}=0\\ \overrightarrow{i}\xb7\overrightarrow{A}={A}_{x}\\ \overrightarrow{B}\xb7\overrightarrow{A}=\overrightarrow{A}\xb7\overrightarrow{B}\end{array}$$

Note that the dot product of the unit vector with a vector simply selects the magnitude of the vector's component in that direction ($\overrightarrow{i}\xb7\overrightarrow{A}={A}_{x}$ ) and that the dot product is commutative $(\overrightarrow{A}\xb7\overrightarrow{B}=\overrightarrow{B}\xb7\overrightarrow{A})$ .

Equation [8.4] can be rearranged to yield an expression for $\mathrm{cos}\beta $[3]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaerbdfwBIjxAHbqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaabauaaaOqaaabbaaaaaaaaIXwyJTgapeGaci4yaiaac+gacaGGZbGaeqOSdigaaa@3ADE@ in terms of the vector components and vector magnitudes:

$$\mathrm{cos}\beta =\frac{{A}_{x}{B}_{x}+{A}_{y}{B}_{y}+{A}_{z}{B}_{z}}{\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|}$$[3]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaerbdfwBIjxAHbqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaabauaaaOqaaabbaaaaaaaaIXwyJTgapeGaci4yaiaac+gacaGGZbGaeqOSdiMaeyypa0ZaaSaaa8aabaWdbiaadgeapaWaaSbaaSqaa8qacaWG4baapaqabaGcpeGaamOqa8aadaWgaaWcbaWdbiaadIhaa8aabeaak8qacqGHRaWkcaWGbbWdamaaBaaaleaapeGaamyEaaWdaeqaaOWdbiaadkeapaWaaSbaaSqaa8qacaWG5baapaqabaGcpeGaey4kaSIaamyqa8aadaWgaaWcbaWdbiaadQhaa8aabeaak8qacaWGcbWdamaaBaaaleaapeGaamOEaaWdaeqaaaGcbaWdbmaaemaapaqaa8qaceWGbbWdayaalaaapeGaay5bSlaawIa7amaaemaapaqaa8qaceWGcbWdayaalaaapeGaay5bSlaawIa7aaaaaaa@539B@

The cross product of two vectors * A* and

$\begin{array}{l}\overrightarrow{A}\times \overrightarrow{B}=\left(\begin{array}{ccc}\overrightarrow{i}& \overrightarrow{j}& \overrightarrow{k}\\ {A}_{x}& {A}_{y}& {A}_{z}\\ {B}_{x}& {B}_{y}& {B}_{z}\end{array}\right)\\ \\ \overrightarrow{A}\times \overrightarrow{B}=\left({A}_{y}{B}_{z}-{A}_{z}{B}_{y}\right)\overrightarrow{i}-\left({A}_{x}{B}_{z}-{A}_{z}{B}_{x}\right)\overrightarrow{j}+\left({A}_{x}{B}_{y}-{A}_{y}{B}_{x}\right)\overrightarrow{k}\end{array}$

The magnitude of the cross product is given by:

$$\begin{array}{l}\left|\overrightarrow{A}\times \overrightarrow{B}\right|=\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|\mathrm{sin}\beta \hfill \\ \begin{array}{l}\text{\hspace{1em}}\text{\hspace{1em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=0\text{if}\overrightarrow{A}\text{isparallelto}\overrightarrow{B}\\ \text{}=\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|\text{if}\overrightarrow{A}\perp \overrightarrow{B}\end{array}\hfill \end{array}$$[3]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaerbdfwBIjxAHbqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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@836A@

where $\beta $ is the angle between * A* and

Note that the cross product is a vector. The direction of the cross product is at right angles to * A* and

The following video (2:06) reminds you about the right-hand rule for cross products.

It follows that the cross products of the unit vectors are given by:

$$\begin{array}{l}\overrightarrow{i}\times \overrightarrow{j}=\overrightarrow{k}\text{\hspace{1em}}\text{\hspace{1em}}\overrightarrow{j}\times \overrightarrow{k}=\overrightarrow{i}\text{\hspace{1em}}\text{\hspace{1em}}\overrightarrow{k}\times \overrightarrow{i}=\overrightarrow{j}\\ \\ \overrightarrow{i}\times \overrightarrow{j}=-\overrightarrow{j}\times \overrightarrow{i}\end{array}$$

Note finally that $\overrightarrow{A}\times \overrightarrow{B}=-\overrightarrow{B}\times \overrightarrow{A}$.

We sometimes need to take derivatives of vectors in all directions. For that we can use a special vector derivative called the Del operator, $\overrightarrow{\nabla}$.

Del is a vector differential operator that tells us the change in a variable in all three directions. Suppose that we set out temperature sensors on a mountain so that we get the temperature, *T*, as a function of *x*, *y*, and *z*. Then $\overrightarrow{\nabla}$*T* would give us the change of *T* in the *x*, *y*, and *z* directions.

$\overrightarrow{\nabla}=\overrightarrow{i}\frac{\partial}{\partial x}+\overrightarrow{j}\frac{\partial}{\partial y}+\overrightarrow{k}\frac{\partial}{\partial z}$

The Del operator can be used like a vector in dot products and cross products but not in sums and differences. It does not commute with vectors and must be the partial derivative of some variable, either a scalar or a vector. For example, we can have the following with del and a vector * A*:

$$\begin{array}{l}\stackrel{}{}\end{array}$$

$$\begin{array}{l}\overrightarrow{\nabla}\xb7\overrightarrow{A}=\frac{\partial {A}_{x}}{\partial x}+\frac{\partial {A}_{y}}{\partial y}+\frac{\partial {A}_{z}}{\partial z}\text{,whichisascalar}\\ \overrightarrow{\nabla}T=\overrightarrow{i}\frac{\partial T}{\partial x}+\overrightarrow{j}\frac{\partial T}{\partial y}+\overrightarrow{k}\frac{\partial T}{\partial z}\text{,whichisavectoreventhough}T\text{isascalar}\\ \overrightarrow{A}\xb7\overrightarrow{\nabla}T={A}_{x}\frac{\partial T}{\partial x}+{A}_{y}\frac{\partial T}{\partial y}+{A}_{z}\frac{\partial T}{\partial z}\text{,whichisascalar}\end{array}$$

- Find
**Practice Quiz 8-1**in Canvas. 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 8-1**. You will be allowed to take this quiz only**once**. Good luck!

In meteorology and other atmospheric sciences, we mostly use the standard *x*, *y*, and *z* coordinate system, called the **Cartesian coordinate system**, and the **spherical coordinate system**. Let’s review some of the main points of these two systems.

The Cartesian coordinate system applies to three dimensions (as seen in the figure below). The convention is simple:

- The zero point,
*x*=*y*=*z*= 0 or (0,0,0), is arbitrary. *x*increases to the east;*x*decreases to the west.*y*increases to the north;*y*decreases to the south.*z*increases going up;*z*decreases going down.- A distance vector extending from the origin to
*(x,y,z)*as**L**=**i**x +**j**y +**k**z.

Unit vectors (length 1 along standard coordinates) are ** i** (east);

Often we will consider motion in two dimensions as being separate from motions in the vertical. We usually denote the horizontal with a subscript *H*; for example, * L_{H} = i x + j y*, where

We like this coordinate system because it works well over relatively small scales on Earth, perhaps the size of an individual state, where Earth's curvature is not important. However, it does not work so well for large-scale motion on Earth, which is spherical.

Credit: W. Brune

Credit: W. Brune

When we place the Cartesian coordinate system on a sphere, note that *x* always points toward the east, *y* always points toward the North Pole, and *z* always points up along the direction of Earth’s radius (as seen in the figure above).

Life would be much easier if the Earth were flat. We could then use the Cartesian coordinate system with no worries. But the Earth is a sphere, which implies that to accurately describe motion, we must take the Earth’s spherical shape into account.

Credit: Artima Developer [5]

We use the following terms:

*r*= distance from the center of the Earth**$\Phi $**= latitude (–90^{o}to +90^{o}, or –$\pi $ /2 to +$\pi $/2)**$\lambda $**= longitude (–180^{o}to 180^{o}, or –$\pi $ to +$\pi $)- State College PA is at $\Phi \text{}=\text{}{40.8}^{o}and\text{}\lambda =-{77.9}^{o}$ .

Note that 1^{o} of latitude is always 111 km or 60 nautical miles, but 1^{o} of longitude is 111 km only at the equator. It is smaller in general and equal to 111 km x cos($\Phi $). Note that 1 nm = 1.15 miles.

To find the horizontal distance between any two points on Earth's surface, we first need to find the angle of the arc between them and then we can multiply this angle by Earth's radius to get the distance. To find the angle of the arc, $\Delta \sigma $ , we can use the **Spherical Law of Cosines**:

$$\Delta \sigma =\mathrm{arccos}\left(\mathrm{sin}{\varphi}_{1}\cdot \mathrm{sin}{\varphi}_{2}+\mathrm{cos}{\varphi}_{1}\cdot \mathrm{cos}{\varphi}_{2}\cdot \mathrm{cos}\Delta \lambda \right)$$

where the latitude and longitude of the two points are ${\Phi}_{1},\text{}{\lambda}_{1}and\text{}{\Phi}_{2},\text{}{\lambda}_{2}$ respectively and $\Delta \lambda =|{\lambda}_{1}-{\lambda}_{2}|$ is the absolute difference between the longitudes of the two points. Note that the angle of the arc must be in radians, where $2\pi \text{}radians\text{}=\text{}{360}^{o}$. To find the distance, simply multiply this arc angle by the radius of the Earth, 6371 km.

Show that 1^{o} of latitude = 111 km distance.

Distance = 6371 km * (1/360)*2π = 111.2 km

In summary, we will use a Cartesian coordinate system *when our scales of interest are not too large (synoptic scale or smaller)*, but will need to use spherical coordinates *when the scale of interest is larger than synoptic scale*.

For another explanation of these two systems, visit this Coordinate Systems website [6].

Three different vertical coordinates are used in meteorology and atmospheric science: height, pressure, and potential temperature.

We have already introduced the vertical coordinate ** z,** which is a height, usually in m or km, above the Earth's surface in the Cartesian coordinate system;

Another useful vertical coordinate is pressure, which decreases with height. Pressure is often a useful vertical coordinate for calculating dynamics. To a good approximation, pressure falls off exponentially with height, *p = p _{o}exp(-z/H)*, so that ln

altitude (km) | altitude (kft) | pressure level (hPa or mb) |
---|---|---|

0 | 0 | 1000 |

1.5 | 4.4 | 850 |

3.0 | 9.9 | 700 |

5.5 | 18.3 | 500 |

A third important vertical coordinate is potential temperature, $\theta $ (Equation 2.58). This quantity is the temperature that an air parcel would have if it were brought to a pressure of 1000 hPa without any exchange of heat with its surroundings. This vertical coordinate has a nice property: air parcels tend to move on surfaces of constant potential temperature because moving on such a surface requires no energy. This coordinate is particularly useful in the stratosphere, where the rapid increase with altitude tends to keep air motion stratified.

Credit: Justin Otto [7] via flickr

Meteorologists talk of northeasterlies and southerlies when they describe winds. These terms designate directions that the winds *come from*. But when we think about the dynamic processes that cause the wind, we use the conventions for direction that are common in mathematics and in coordinate systems like the Cartesian coordinate system. The conversion between the two conventions—math and meteorology—is not simple. However, we will show you a simple way to do the conversion (see the second figure below).

The wind vector is given by **U** = **i **u + **j **v + **k **w**. **The wind vector points to the direction the wind *is going*.

The subscript “H” will be used to denote horizontal vectors, such as the horizontal velocity, *U**_{H} = i u + j v * (though note that sometimes the symbols

Credit: NOAA National Weather Service

The meteorology wind convention is often used in meteorology, including station weather plots [8]. The wind vector points to the direction the wind *is coming from*. The angle is denoted by delta, $\delta $, which has the following directions:

direction wind is coming from | angle $\delta $ |
---|---|

north (northerlies or southward) | 0^{o} |

east (easterlies or westward) | 90^{o} |

south (southerlies or northward) | 180^{o} |

west (westerlies or eastward) | 270^{o} |

Meteorology angles, designated by $\delta $, increase clockwise from the north (*y*) axis. Math angles, designated by $\alpha $ , increase counterclockwise from the east (*x*) axis.

Credit: W. Brune

In the diagram on the left, the wind is southwesterly, the meteorology angle (measured clockwise from the north or y-axis) $\delta \text{}=\text{}{225}^{o}$[3]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaerbdfwBIjxAHbqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaabauaaaOqaaabbaaaaaaaaIXwyJTgapeGaeqiTdqMaaeiOaiabg2da9iaabckacaaIYaGaaGOmaiaaiwdapaWaaWbaaSqabeaapeGaam4Baaaaaaa@3ED2@ , and the math angle (measured counterclockwise from the east or x-axis) $\alpha \text{}=\text{}{45}^{o}$. If the wind is northerly (southward), the wind vane points to the north, the wind blows to the south, $\delta \text{}=\text{}{0}^{o},\text{}and\text{}\alpha \text{}=\text{}{270}^{o}$. If the wind is westerly (eastward), $\delta \text{}=\text{}{270}^{o},\text{}and\text{}\alpha \text{}=\text{}{0}^{o}$.

Note that in all cases, we can describe the relationship between the math and the meteorology angles as:

$$math\text{}angle\text{}=\text{}27{0}^{o}-\text{}meteorology\text{}angle$$

When the meteorology angle is greater than 270^{o}, the math angle will be negative but correct. However, to make the math angle positive, simply add 360^{o}.

Drawing a figure like those shown in the figure above often helps when you are trying to do the conversion. The following video (2:17) explains the conversion between meteorology and math wind angles using the figure above.

- Find
**Practice Quiz 8-2**in Canvas. 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 8-2**. You will be allowed to take this quiz only**once**. Good luck!

The **gradient** of a variable is just the *change in that variable as a function of distance*. For instance, the temperature gradient is just the temperature change divided by the distance over which it is changing: $\Delta T/\Delta distance$. The gradient is a vector and thus has a direction as well as magnitude.

Credit: Unisys

Consider the surface temperature contour plot from NOAA for 8 September 2012 in the figure above. A strong temperature variation is draped across the eastern and southern US from New York down to Texas. How do we quantify this temperature variation? First we have to specify where we want to measure the temperature gradient. Then we simply need to choose isotherms on either side of the point, take the difference between the isotherms, figure out how far apart they are in horizontal distance, and divide the temperature change between the isotherms by the distance between the isotherms. The direction for the gradient is on the normal (perpendicular to the isotherms) from the lower temperatures to the higher temperatures. It’s pretty easy to figure out where the gradient vector points just by quick examination, but it is a little harder to figure out what the gradient magnitude and actual direction are.

Now watch this video (2:12) on finding distances:

Mathematically, if we know the algebraic expression for the temperature change, such that *T = T*(*x,y*), we can find the gradient by using the **del operator**, which is also called the **gradient operator**.

Recall the del operator:

$$\overrightarrow{\nabla}=\overrightarrow{i}\frac{\partial}{\partial x}+\overrightarrow{j}\frac{\partial}{\partial y}+\overrightarrow{k}\frac{\partial}{\partial z}$$

If we are looking only at changes in *x* and *y*, then we can define a horizontal del operator:

$$ \vec{\nabla}_{H}=\vec{i} \frac{\partial}{\partial x}+\vec{j} \frac{\partial}{\partial y} $$

At any point, we can determine the gradient of the temperature:

$$ \vec{\nabla}_{H} T=\vec{i} \frac{\partial T}{\partial x}+\vec{j} \frac{\partial T}{\partial y} $$

Note that this quantity has dimensions of $\theta /L$ and a magnitude and a direction. The gradient direction is always normal to the isolines and pointing in the direction of an increase. We can define the normal vector, which is just the unit vector in the direction of the increasing temperature. We will call this normal vector * n*.

Credit: H.N. Shirer

We can calculate a gradient for every point on the map, but to do this we need to know the change in the temperature over a distance that is centered on our chosen point. One approach is to calculate the gradients in the *x* and *y* directions independently and then determine the magnitude by:

$\left|{\overrightarrow{\nabla}}_{H}T\right|=\sqrt{{\left(\frac{\partial T}{\partial x}\right)}^{2}+{\left(\frac{\partial T}{\partial y}\right)}^{2}}=\left|\frac{\partial T}{\partial n}\right|$

and the direction by:

$\mu =\text{}{\mathrm{tan}}^{-1}\left(\frac{\raisebox{1ex}{$\partial T$}\!\left/ \!\raisebox{-1ex}{$\partial y$}\right.}{\raisebox{1ex}{$\partial T$}\!\left/ \!\raisebox{-1ex}{$\partial x$}\right.}\right)$

We can program a computer to do these calculations.

However, often we just want to *estimate* the gradient. The gradient can be determined by looking at the contours on either side of the point and computing the change in temperature over the distance. These partial derivatives can be approximated by small finite changes in temperatures and distances, so that $\partial $ is replaced by $\Delta $ in all places in these equations. We can calculate gradients by using “centered differences” as shown in the figures below.

Credit: H.N. Shirer

We then calculate the magnitude with Equation [8.12] and the direction with Equation [8.13], where we replace the partial derivatives with the small finite differences in all places in these equations.

Credit: H.N. Shirer

The magnitude and direction are:

$$\begin{array}{l}\left|{\nabla}_{H}T\right|=\sqrt{{\left(\frac{\Delta T}{\Delta x}\right)}^{2}+{\left(\frac{\Delta T}{\Delta y}\right)}^{2}}=\sqrt{{\left(\frac{4{}_{}{}^{\text{o}}\text{F}}{45\text{nm}}\right)}^{2}+{\left(\frac{-4{}_{}{}^{\text{o}}\text{F}}{84\text{nm}}\right)}^{2}}=0.1{}_{}{}^{\text{o}}\text{F}\text{/nm}\hfill \\ \mu ={\mathrm{tan}}^{-1}\left(\frac{\raisebox{1ex}{$\Delta T$}\!\left/ \!\raisebox{-1ex}{$\Delta y$}\right.}{\raisebox{1ex}{$\Delta T$}\!\left/ \!\raisebox{-1ex}{$\Delta x$}\right.}\right)={\mathrm{tan}}^{-1}\left(\frac{\raisebox{1ex}{$-4$}\!\left/ \!\raisebox{-1ex}{$84$}\right.}{\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$45$}\right.}\right)=-{28}^{\text{o}}\text{,whichpointstothesoutheast}\hfill \end{array}$$[3]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaerbdfwBIjxAHbqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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@B384@

When you calculate the arctangent, keep in mind that the tangent function has the same values every 180^{o}, or every π in radians. If you get an answer for the arctangent that is 45^{o}, how do you know whether the angle is really 45^{o} or 45^{o} + 180^{o} = 225^{o}? The gradient vector always points toward the higher temperature air, so always choose the angle so that the gradient vector points toward the warmer air.

Recap of the process for calculating the temperature gradient:

- Determine the distance scale by any means that you can. Sometimes it is given to you; sometimes you can scale off a ruler; sometimes you just estimate it using the size of known boundaries.
- Determine the spacing between the isotherms.
- Find the temperature change in the
*x*and*y*directions using the centered difference method. These two numbers, $\Delta T/\Delta x\text{}and\text{}\Delta T/\Delta y$, are needed to calculate both the gradient magnitude and the gradient direction. Note that $\Delta T/\Delta x\text{}and\text{}\Delta T/\Delta y$ can be either positive or negative. - Calculate the magnitude by finding the square root of the squares of the gradients in the
*x*and*y*directions (i.e., $\Delta T/\Delta x\text{}and\text{}\Delta T/\Delta y$). - Calculate the direction of the gradient vector by finding the arctangent of the
*y*-gradient divided by the*x*-gradient. Pay attention to the direction—make sure that it points toward the warmer air.

Now watch this video (3:52) on finding gradients:

A word about finding gradients in the real world. Sometimes the centered differences method is difficult to apply because the gradient is too much east–west or north–south. For instance, in the temperature map at the beginning of this section, the *x*-gradient is hard to determine by the centered difference method in the Oklahoma panhandle and the *y*-gradient is hard to determine in central Pennsylvania because in both cases, the temperature hardly changes. In these cases, you could say that the gradient in that direction equals 0, but then your computer program might have a hard time finding the arctangent. One way around this problem is to put in a very small number for the gradient in that direction, say 1 millionth of your typical gradient numbers, to do the calculation.

A second word about finding gradients in the real world. When you are finding temperature gradients from a temperature map, it is sometimes hard to determine the temperature gradient at some locations because the isotherms are not evenly spaced and can be curvy. Don't despair! Use your best judgment as to what the gradients are. Check your answers for the magnitude and direction of the temperature gradient vector by estimating the magnitude and direction by eyeballing the normal to the isotherms at that location and pointing the gradient vector to the warmer air. If your calculated direction is 160 degrees when your eyeball check says about 220 degrees, check your math again.

- Find
**Practice Quiz 8-3**. 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 8-3**. You will be allowed to take this quiz only**once**. Good luck!

Credit: Nicholas A.Tonelli [9] via flickr

Suppose you were driving underneath an eastward-moving thunderstorm at the same speed and in the same direction as the storm, so that you stayed under it for a few hours. From your point-of-view, it was raining the entire time you were driving. But from the point-of-view of people in the houses you passed, the thunderstorm approached, it rained really hard for twenty minutes, and then stopped raining.

The people in houses formed a network of observers, and if they talked to each other, they would find that the thunderstorm moved rapidly from west to east in a path that rained on some houses and missed others. This point-of-view is called the **Eulerian** description because it follows the thunderstorm through a fixed set of locations. You, on the other hand, followed along with the thunderstorm; any changes you saw were due to changes in the thunderstorm’s intensity only. Your point-of-view was a **Lagrangian** description because you stayed with the storm.

We can now generalize these ideas to any parcel of air, not just a thunderstorm. An air parcel is a blob of air that hangs more-or-less together as it moves through the atmosphere, meaning that its mass and composition are conserved (i.e., not changing) as it moves. It has a fairly uniform composition, temperature, and pressure, and has a defined velocity.

Air parcels do not live forever. They are just air moving in air, so they change shape as they bump into other air parcels and they mix until they disappear. Larger air parcels tend to live longer than smaller air parcels. Even though air parcels do not live forever, they are very useful concepts in explaining the differences between the Eulerian and Lagrangian frameworks.

**Eulerian Framework**:

- Observes atmospheric properties and their changes at
*fixed points*in space - Tracks motion by reporting those observations from time-to-time
- Is the way most weather observations are taken and the way the most weather prediction models do computations

**Lagrangian Framework**:

- Observes atmospheric properties and their changes in a
*moving*air parcel - Tracks changes within the air parcel as it moves
- Is a conceptual or numerical approach and is difficult to realize with observations

Here's an interesting fact. Radiosonde measurements are considered Eulerian because they measure properties like temperature and humidity at a fixed location. However, the horizontal velocity is measured by a radiosonde’s lateral movement with the horizontal wind and so is actually a Lagrangian measurement. A cup anemometer or sonic anemometer is an example of a Eulerian method of measuring wind velocity.

**(3 discussion points)**

In the lesson, I presented an example of a driver in a car that happened to be traveling at the same speed as a rainstorm so that to the driver it was raining the entire time, while to the observers in the houses that the driver passed, the rain showers were brief. We could have given a second case in which there was widespread rain so that both the car driver and the house occupants observed constant rainfall for several hours. In the first case, the advection exactly matched the local rainfall change wherever the driver was, so that the driver was in rain constantly, but the house occupants saw rain only for a short while. In the second case, there was no gradient in the rainfall rate, so even if the storm was moving, neither the driver nor the house occupants observed a change in rainfall.

**Think of one event or phenomenon from the Eulerian and Lagrangian points-of-view, one in which the event or phenomenon looks very different from the two points-of-view. These events do not have to be weather related. Be creative.**

- You can access the
**Eulerian and Lagrangian Points-of-View Discussion**in Canvas. - Post a response that answers the question above in a thoughtful manner that draws upon course material and outside sources.
- Keep the conversation going!
**Comment on at least one other person's post**. Your comment should include follow-up questions and/or analysis that might offer further evidence or reveal flaws.

This discussion will be worth 3 discussion points. I will use the following rubric to grade your participation:

Evaluation | Explanation | Available Points |
---|---|---|

Not Completed | Student did not complete the assignment by the due date. | 0 |

Student completed the activity with adequate thoroughness. | Posting answers the discussion question in a thoughtful manner, including some integration of course material. | 1 |

Student completed the activity with additional attention to defending his/her position. | Posting thoroughly answers the discussion question and is backed up by references to course content as well as outside sources. | 2 |

Student completed a well-defended presentation of his/her position, and provided thoughtful analysis of at least one other student’s post. | In addition to a well-crafted and defended post, the student has also engaged in thoughtful analysis/commentary on at least one other student’s post as well. | 3 |

Changes over time in properties, such as temperature and precipitation, can be expressed in Lagrangian and Eulerian frameworks, and often the changes are different in the two frameworks (as in the precipitation change for the thunderstorm example). We can express these changes mathematically with time derivatives. Suppose we have a scalar, *R*, which could be anything, but let’s make it the rainfall rate. It is a function of space and time:

$$R=R(x,y,z,t)$$

To find the change in the rainfall rate *R* in an air parcel over space and time, we can take its differential, which is an infinitesimally small change in *R*:

$$dR=\frac{\partial R}{\partial t}dt+\frac{\partial R}{\partial x}dx+\frac{\partial R}{\partial y}dy+\frac{\partial R}{\partial z}dz$$

where *dt* is an infinitesimally small change in time and *dx*, *dy*, and *dz* are infinitesimally small changes in *x*, *y*, and *z* coordinates, respectively, of the parcel.

If we divide Equation [8.15] by *dt*, this equation becomes:

$$\frac{dR}{dt}=\frac{\partial R}{\partial t}+\frac{\partial R}{\partial x}\frac{dx}{dt}+\frac{\partial R}{\partial y}\frac{dy}{dt}+\frac{\partial R}{\partial z}\frac{dz}{dt}$$

where *dx/dt*, *dy/dt*, and *dz/dt* describe the velocity of the air parcel in the *x*, *y*, and z directions, respectively.

Let’s consider two possibilities:

**Case 1:** The air parcel is *not* moving. Then the change in *x*, *y* and *z* are all zero and:

$$\frac{dR}{dt}=\frac{\partial R}{\partial t}$$

So, the change in the rainfall rate depends only on time. $\frac{\partial R}{\partial t}$ is called the Eulerian or **local time derivative**. It is the time derivative that each of our weather observing stations record.

**Case 2: **The air parcel is moving. Then the changes in its position occur over time, and it moves with a velocity, $\overrightarrow{U}=\overrightarrow{i}u+\overrightarrow{j}v+\overrightarrow{k}w$ , where:

$$\begin{array}{l}\frac{dx}{dt}=u\text{\hspace{1em}}\frac{dy}{dt}=v\text{\hspace{1em}}\frac{dz}{dt}=w\\ \frac{dR}{dt}=\frac{\partial R}{\partial t}+u\frac{\partial R}{\partial x}+v\frac{\partial R}{\partial y}+w\frac{\partial R}{\partial z}\text{\hspace{1em}}\end{array}$$

A special symbol is given for the derivative when you follow the air parcel around. It is called the **substantial, or total, derivative** and is denoted by:

$$\frac{DR}{Dt}=\frac{\partial R}{\partial t}+u\frac{\partial R}{\partial x}+v\frac{\partial R}{\partial y}+w\frac{\partial R}{\partial z}$$

Mathematically, we can express this equation in a more general way by thinking about the dot product of a vector with the gradient of a scalar as we did in an example of the del operator:

$$\frac{DR}{Dt}=\frac{\partial R}{\partial t}+\overrightarrow{U}\xb7\overrightarrow{\nabla}R$$

where the second term on the right hand side is called the advective derivative, which describes changes in rainfall that are solely due to the motion of the air parcel through a spatially variable rainfall distribution.You should be able to show that equation [8.19] is the same as equation [8.18].

We can rearrange this equation to put the local derivative on the left.

$$\frac{\partial R}{\partial t}=\frac{DR}{Dt}-\overrightarrow{U}\xb7\overrightarrow{\nabla}R$$

The term on the left is the local time derivative, which is the change in the variable *R* at a fixed observing station. The first term on the right is the total derivative, which is the change that is occurring in the air parcel as it moves. The last term on the right, $-\overrightarrow{U}\xb7\overrightarrow{\nabla}R$, is called the *advectio*n of *R*. Note that advection is simply the negative of the advective derivative.

To go back to the analogy of the thunderstorm, the change in rainfall that you observed driving in your car was the total time derivative and it depended only on the change in the intensity of the rain in the thunderstorm. However, for each observer in a house, the change in rainfall rate depended not only on the intensity of the rainfall as the thunderstorm was over the house but also on the movement of the thunderstorm across the landscape.

*R* can be any scalar. Rainfall rate is one example, but the most commonly used are pressure and temperature.

Equation [8.20] is called **Euler’s relation** and it relates the Eulerian framework to the Lagrangian framework. The two are related by this new concept called advection.

Let’s look at advection in more detail, focusing on temperature.

We generally think of advection being in the horizontal. So often we only consider the changes in the *x* and *y* directions and ignore the changes in the *z* direction:

$$\text{horizontaltemperature}\text{\hspace{0.17em}}\text{advection}=-{\overrightarrow{U}}_{H}\xb7{\overrightarrow{\nabla}}_{H}T=-\left(u\frac{\partial T}{\partial x}+v\frac{\partial T}{\partial y}\right)$$[3]@5@5@+=faaahmart1ev3aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaebbnrfifHhDYfgasaacH8srps0lbbf9q8WrFfeuY=ribbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@6C52@

So what’s with the minus sign? Let’s see what makes physical sense. Suppose *T* increases only in the *x*-direction so that:

$$\frac{\partial T}{\partial y}=0\text{and}\frac{\partial T}{\partial x}0$$

If *u > 0* (westerlies, blowing eastward), then both *u* and $\frac{\partial T}{\partial x}$ are positive so that temperature advection is negative. What does this mean? It means that colder air blowing from the west is replacing the warmer air, and the temperature at our location is decreasing from this advected air. Thus $\frac{\partial T}{\partial t}$ should be negative since time is increasing and temperature is decreasing due to advection.

If the temperature advection is *negative*, then it is called *cold-air advection*, or simply cold advection. If the temperature advection is *positive*, then it is called *warm-air advection*, or simply warm advection.

Some examples of simple cases of advection show these concepts (see figure below). When the wind blows along the isotherms, the temperature advection is zero (Case A). When the wind blows from the direction of a lower temperature to a higher temperature (Case B), we have cold-air advection. When the wind blows as some non-normal direction to the isotherms, then we need to multiply the magnitude of the wind and the temperature gradient by the cosine of the angle between them. We can estimate the temperature advection by doing what we did for the gradient, that is, replace all derivatives and partial derivatives with finite $\Delta s$ .

Credit: H.N. Shirer

When the isotherms with the same temperature difference are further apart on the map (see figure below), then the horizontal temperature advection will be less than when the isotherms are closer together, if the wind velocity is the same in the two cases.

Credit: H.N. Shirer

In summary, to calculate the temperature advection, first determine the magnitude and the direction of the temperature gradient. Second, determine the magnitude and direction of the wind. The advection is simply the negative of the dot product of the velocity and the temperature gradient.

Watch this video (2:20) on calculating advection:

- Find
**Practice Quiz 8-4**in Canvas. 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 8-4**. You will be allowed to take this quiz only**once**. Good luck!

You now have all of the math tools that you will need to understand the lessons on kinematics and dynamics. For some of you, the concept of partial derivatives is new, but you see how easy it is to apply. We have reviewed vectors and a little vector calculus that you will need in the next few lessons. Dot products and cross products will appear frequently in the next few lessons, so make sure that you are comfortable with them and their applications. We introduced a new operator, the “del” or “gradient” operator, which is essential for describing weather observations and how atmospheric properties vary in space. There are two major coordinate systems that are used to describe atmospheric motion: Cartesian and spherical. And there are three different coordinates that are used to define the vertical distance starting at Earth’s surface and rising radially: height *z* (m); pressure or logarithm of pressure *p* (hPa); and potential temperature $\theta $ (K). You will become skilled in converting between the meteorological wind directions shown on weather maps and in station weather plots and the wind direction needed to do calculations of wind motion and its effects. We introduced the concepts of the Eulerian and Lagrangian frameworks and showed that they are related by advection of a scalar, which is simply the dot product of the wind velocity and the gradient of the scalar.

Once you successfully complete the activities in this lesson, you will be ready to learn about atmospheric kinematics (the description of air movement) and atmospheric dynamics (the study of why air moves the way that it does).

You have reached the end of Lesson 8! Double-check that you have completed all of the activities before you begin Lesson 9.

**Links**

[1] https://www.youtube.com/channel/UCUDlvPp1MlnegYXOXzj7DEQ

[2] https://www.youtube.com/channel/UCEik-U3T6u6JA0XiHLbNbOw

[3] mailto:MathType@MTEF

[4] https://www.youtube.com/channel/UCQbibyTf5Awx_y8SABfwcLA

[5] https://www.artima.com/forums/flat.jsp?forum=123&thread=157443

[6] http://hyperphysics.phy-astr.gsu.edu/hbase/coord.html

[7] https://www.flickr.com/photos/8604504@N03/5704952607/in/photolist-9G8nDR-f11Hzr-qfCw1A-eHheNa-8AqNas-bGEHZi-5ozWUm-8ys9gD-nStkfA-9YWWVQ-8nJ8he-c9ivJh-dXW2BR-4rWzw9-bxpivx-oC4xgr-iuByub-bGEDYP-dpY8pn-bmEEaF-5izUTL-6cGjEk-bmEDRg-btKQBS-9fRjZF-ambmr4-6oQC31-eFiZaz-bMknDR-pcwe3S-bx3TvN-aDiNMp-7ujXFV-6TeevC-4zKUA8-dMKrMy-4oJTtx-c9ivRm-pmWZ7U-6bF9UU-9jfA5o-fsr5fJ-fdw6a1-bEPJh6-4j48du-9Fzphk-2w81rU-6sFyEJ-6p5MjE-6inrCa

[8] http://www.srh.weather.gov/srh/jetstream/synoptic/wxmaps.htm#ww

[9] https://www.flickr.com/photos/nicholas_t/944963485/in/photolist-2rvbHT-9Ki1Ra-aF4PjB-h6A8C-51ZwTp-524JGE-6JkDSw-6JkFi1-6Jgz9c-51Zy2V-ahwt3K-9TMeKV-6Jgv2a-5ts6mT-354nNy-6JgxT6-6JkyYQ-6Jky9y-6Jguhp-6JgAxz-6JkCpN-6JgrX8-2jicnb-5dxezb-a2A331-2UGJcp-a2A3vb-9VM5xj-a2xasi-a2A3p3-a2A31f-cp5oyW-hHRdBL-2mLnq-2mEYwT-oNW8V-bykcXb-6SLBmU-8hVaBW-8vpvGG-bDUh14-6Jgw28-qKXnv-cw2K5N-cpipd5-iawch-5twtzJ-cw2JMJ-713rG9-6ZYsz2