Wednesday, Jan. 29, 2014

A couple of songs from Dessa before class: "551" and  "Alibi" (3:05), but not enough time for "Skeleton Key" .

The Practice Quiz is next week and a Practice Quiz Study Guide is now available online.  Because I'm guessing a little bit about how much material we will get through between now and the end of class next Monday, there may be some small changes at the end of the study guide.  Reviews are also planned for Monday and Tuesday afternoon next week but the precise times and locations aren't yet known.

A 1 inch x 1 inch x 52 inch long steel bar was passed around in class. 



You were supposed to estimate it's weight.  The fact that it was 1" by 1" is significant.  More about the bar later in today's notes.

And then a bit of a detour in response to a student question at the end of class on Monday.  We had been talking about particulate matter in the atmosphere and how it can affect (reduce) visibility.  The student asked whether particles in the air can have an effect on sunsets.



The answer is basically yes.  And actually you don't need particles.  Air can have quite an effect on the color of the setting sun by itself.  Air molecules selectively scatter the shorter wavelengths of light (i.e. scatter much more of the violet blue and green light than the warmer colors like yellow, orange, and red).  As the sun sets it takes a longer and longer path through the atmosphere and more and more of the violet, blue and green light is removed.  The transmitted light becomes yellow sometimes orange. (photo is from free-photos.co)

The sketch above at left shows how the color of the sunlight changes as it passes through the atmosphere.  The light eventually turns orange as the violet blue and green light is removed by scattering.  Clouds in the sky will reflect this orange light after the sun has set as seen in the picture above at right (source of that photo).  It is safe to look at this reflected light, you shouldn't ever look directly at the sun).



In this picture (source of this image), sunlight is shining through smoke from a forest fire.  The smoke particles are scattering and perhaps absorbing sunlight.  The sun is much less intense than it would have been if it were just shining through air.



This picture was Earth Science Picture of the Day for May 15, 2011.  Air and particles are working to change the color and reduce the intensity of the sunlight.  The article that accompanied the photograph mentions that scattering of violet, blue, and green light by air molecules is not enough to produce an image of the sun as red as this.  Particulates are slso needed.







Strong volcanic eruptions will sometimes send dust and sulfur dioxide (which then forms small drops of sulfuric acid) into the stratosphere where it can remain for a year or more.  The dust can spread around the globe and can create colorful and longer lasting sunsets.  The purple color seems to be characteristic of volcanic sunsets.  My understanding is that the purple color comes from a mixture of the reddish light from an ordinary sunset mixed with blue and violet light coming from scattering of sunlight by the volcanic dust higher up in the atmosphere.

The photo was taken by Jason Reilley and shows the effects of dust from the 2011 Puyehue-Cordon Caulle volcano in Chile on sunsets in Tasmania, Australia (source of the image).  Sunsets like this were seen in Tucson following the 1982 El Chichon eruption in Mexico.



If you look closely at this image, taken a split second before the sun set, you'll notice that the sun is green.  This is a fairly rare phenomenon known as the green flash and produced by refraction of light, not scattering. (source of the photo at right).  You'll be able to read more and write a 1S1P report on rainbows, mirages, and the green flash later in the semester.

Keep your eyes open.  The thin, high altitude cirrus clouds in the sky for the past several days have produced a halo (photo above at left, source), cloud iridescence (right photo, source) and probably much more.  Neither of these figures was shown in class.



I brought back the blocks of brick, wood, and lead to quickly remind you about mass, weight, and density.  They have different masses which is shown in the figure by the number of dots filling each of the objects.  Different masses meant they also had different weights.  That was something you could feel by picking them up.  The three blocks all had about the same volume (not exactly but almost).  Different masses squeezed into about the same volume meant different densities.

You could figure out which object was which by picking them up and seeing how much they each weighed.

We returned briefly to an alternate definition of mass
 

It would be harder to start a lead car moving than a car made out of wood.



You could use this to distinguish between the lead, wood, and brick in outer space where the blocks would be weightless.  You could give them each a push.  It would take more effort to get the lead brick moving than the piece of wood.  Once moving it would also take more work to slow it down to a stop.



Let's add a 1 inch x 1 inch column of air to our picture.  Other than being invisible it's really no different from the other objects.

Now we're ready to define (and hopefully understand) pressure.  It's a pretty important concept.  A lot of what happens in the atmosphere is caused by pressure differences.  Pressure differences cause wind.  Large pressure difference (such as you might find in a tornado or a hurricane) create powerful and destructive storms. 



The air that surrounds the earth has mass.  Gravity pulls downward on the atmosphere giving it weight.  Galileo conducted (in the 1600s) a simple experiment to prove that air has weight.  The experiment wasn't mentioned in class.

Atmospheric pressure depends on, is determined by, the weight of the air overhead.  This is one way, a sort of large scale representation, of understanding air pressure.

Pressure is defined as force divided by area.  In the case of atmospheric pressure the weight of a column of air divided by the area at the bottom of the column (as illustrated above). 

Under normal conditions a 1 inch by 1 inch column of air stretching from sea level to the top of the atmosphere will weigh 14.7 pounds.  Normal atmospheric pressure at sea level is 14.7 pounds per square inch (psi, the units you use when you fill up your car or bike tires with air).

Now back to the iron bar.  It was a little hard to tell but I think that a lot of people felt it weighed more than 20 pounds.  The bar actually weighs 14.7 pounds.  When you stand the bar on end, the pressure at the bottom would be 14.7 psi.




So the weight of a 1" x 1" steel bar 52 inches long is the same as a 1" x 1" column of air that extends from sea level to the top of the atmosphere 100 or 200 miles (or more) high.  The pressure at the bottom of both would be 14.7 psi.

Psi are perfectly good pressure units, but they aren't the ones that most meteorologists use.


Typical sea level pressure is 14.7 psi or about 1000 millibars (the units used by meteorologists and the units that we will use in this class most of the time) or about 30 inches of mercury.    Milli means 1/1000 th.  So 1000 millibars is the same as 1 bar.  You sometimes see typical sea level pressure written as 1 atmosphere.

Inches of mercury refers to the reading on a mercury barometer.  Mercury (13.6 grams/cm3)  is denser than steel ( about 7.9 grams/cm3 ) so it would only take about a 30 inch tall column of mercury to equal the weight or pressure of the atmosphere.




Each of these columns would weigh 14.7 pounds.  The pressure at the base of each would be the same. 

A mercury barometer used to measure pressure is, we'll find, just a balance.  You balance the weight of a very tall column of air with the weight of a much shorter column of mercury.

Here's the story I threatened to include in the online notes.
You never know whether something you learn in ATMO 170A1 will turn up.  I lived and worked for a short time in France (I had a really great time and go back occasionally now to try to ride my bike up some of the famous Tour de France mountain stages).  Here's a picture of a car I owned when I was there (this one is in mint condition, mine was in far worse shape)



It's a Peugeot 404.  After buying it I took it to the service station to fill it with gas and to check the air pressure in the tires.  I was a little confused by the air compressor though, the scale only ran from 0 to 3.  I'm used to putting 30 psi or so in my car tires (about 90 psi in my bike tires).  After staring at the scale for a while I finally realized the numbers were pressures in "bars" not "psi".  Since 14.7 psi is equivalent to 1 bar, 30 psi would be about 2 bars.  So I filled up all the tires and carefully drove off (one thing I quickly learned was you have to watch out for in France is the "Priority to the right" rule). 

You can learn a lot from bricks. 

For example the photo below (taken in my messy office) shows two of the bricks from class.  One is sitting flat, the other is sitting on its end.  Each brick weighs about 5 pounds.  Would the pressure at the base of each brick be the same or different in this kind of situation? 





Pressure is determined by (depends on) weight so you might think the pressures would be equal.  But pressure is weight divided by area.  In this case the weights are the same but the areas are different.  In the situation at left the 5 pounds must be divided by an area of about 4 inches by 8 inches = 32 inches.  That works out to be about 0.15 psi.  In the other case the 5 pounds should be divided by a smaller area, 4 inches by 2 inches = 8 inches.  That's a pressure of 0.6 psi, 4 times higher.  Notice also these pressures are much less the 14.7 psi sea level atmospheric pressure.

14.7 psi might not sound like much.  A single brick laid on its big side produces a pressure of 0.15 psi.You would need a stack of 100 bricks (500 pounds of bricks) to equal atmospheric pressure.


The main reason I brought the bricks was so that you could understand what happens to pressure with increasing altitude.  Here's a drawing of the 5 bricks stacked on top of each other.

At the bottom of the pile you would measure a weight of 25 pounds (if you wanted to find the pressure you'd divide 25 lbs by the 32 square inch area on the bottom of the brick).  If you moved up a brick you would measure a weight of 20 pounds, the weight of the four bricks that are still above.  The pressure would be less.  Weight and pressure will decrease as you move up the pile.

The atmosphere is really no different.  Pressure at any level is determined by the weight of the air still overhead.  Pressure decreases with increasing altitude because there is less and less air remaining overhead. 



At sea level altitude, at Point 1, the pressure is normally about 1000 mb.  That is determined by the weight of all (100%) of the air in the atmosphere.

Some parts of Tucson, at Point 2, are 3000 feet above sea level (most of central Tucson is a little lower than that around 2500 feet).  At 3000 ft. about 10% of the air is below, 90% is still overhead.  It is the weight of the 90% that is still above that determines the atmospheric pressure in Tucson.  If 100% of the atmosphere produces a pressure of 1000 mb, then 90% will produce a pressure of 900 mb. 

Pressure is typically about 700 mb at the summit of Mt. Lemmon (9000 ft. altitude at Point 3) because 70% of the atmosphere is overhead..

Pressure decreases rapidly with increasing altitude.  We will find that pressure changes more slowly if you move horizontally.  Pressure changes about 1 mb for every 10 meters of elevation change.  Pressure changes much more slowly normally if you move horizontally: about 1 mb in 100 km.  Still the small horizontal changes are what cause the wind to blow and what cause storms to form.

Point 4 shows a submarine at a depth of about 30 ft. or so.  The pressure there is determined by the weight of the air and the weight of the water overhead.  Water is much denser and much heavier than air.  At 30 ft., the pressure is already twice what it would be at the surface of the ocean (2000 mb instead of 1000 mb).

I was just reading about the fairly new sport of free diving and found a story about someone that died after trying to set a new record.  Here's a link to the article and to the graphic shown in class.

We didn't cover any of the remaining material in class today.

I did want to explain one more page from the Class Notes.  But we were running short on time and didn't want to rush through it.  I've stuck it in below just in case you want to read ahead.

The figure (p. 26 in the ClassNotes) attempt to explain why the rate of pressure change as you move or down in the atmosphere depends on air density.  In particular why does air pressure decrease more quickly when you move upward through high density air than if you move upward through low density air.



There's a lot going on in this picture, we'll examine it step by step.

1.The sea level pressure is the same, 1000 mb, in both pictures.  Since pressure is determined by the weight of the air overhead, the weight of the air overhead in the left picture is the same as in the right picture.  The amount (mass) of air above sea level in both pictures is the same.

2.  There is a 100 mb drop in pressure in both air layers.  Pressure has decreased because air that was overhead (the air between the ground the level of the dotted line) is now underneath.  Because the pressure change is the same in both pictures the weight of the air layers are the same.  The thin layer at left has the same weight as the thicker layer at right.  Both layers contain the same amount (mass) of air.

3.  Both layers contain the same amount (mass) of air.  The air in the layer at left is thinner.  The air is squeezed into a smaller volume.  The air in the layer at left is denser than the air in the layer at right.

4.  To determine the rate of pressure decrease you divide the pressure change (100 mb for both layers) by the distance over which that change occurs.  The 100 mb change takes place in a shorter distance in the layer at left than in the layer at right.  The left layer has the highest rate of pressure decrease with increasing altitude.

So both the most rapid rate of pressure decrease with altitude and the densest air are found in Layer A.

The fact that the rate of pressure decrease with increasing altitude depends on air density is a fairly subtle but important concept.  This concept will come up 2 or 3 more times later in the semester.  For example, we will need this concept to explain why hurricanes can intensify and get as strong as they do.