On the test, be sure to look at your numbers to see if they make sense!   I have given you several examples in class, here, in the book, etc.   Note that they are all somewhat similar - there are no 10,00,000 m high mountains, no temperatures of 200 degrees, etc...   So if at some step you seem to find that the air will cool by 600 degrees, or have a VP of 33,000mb, or rise to 99,000 meters - something does not add up!   It sounds simple, but students often forget this important step...   The answers you'll need to give will be all within range of those we've dealt with already - mountains of about 2-5,000 m, air between about -20 to +30 degrees C, vapor pressures between 1mb and 15mb, and relative humidities between 5 and 100%.

All units will be metric; I'll  note this, and remember to put all units on answers (mb, %RH, degrees C, meters, etc.).

Remember what you need to calculate, and what you can get from the VP diagram!

The diagram given will probably start at sea level (is there a fish?) on the winward side - but don't assume that; look for the elevation to be sure!

It's not really so complicated!

Think of what you have to do - divide the problem into two halves - the windward side and the leeward side.

Let's use this example: air starts out on the windward side at sea level (0m), 10 degrees C, 9mb of vapor pressure.   Remember that I might (to make you all more thoughtful and miserable) give you the relative humidity instead on the actual VP, so that would require finding the actual VP from the RH.   You'd look on the chart (which you'll be provided with on the test), find the saturation VP at the starting point (here 10 degrees c), then take whatever percentage of that (the RH).

The first half is the windward side, and you are interested in finding the temperature at the top of the mountain.   But to find this, you  need to determine where the adiabatic cooling rate will turn from dry to wet.   To determine this, though, you need to know the dewpoint of the air that starts out at sea level - because when this air has risen enough to reach dewpoint, it will thereafter cool at the MAR for the rest of the way up the mountain.

The air at the bottom of the mountain has a VP of  9mb; looking at the diagram, you see that this air has a dewpoint of about 5 degrees (based, as you recall, on the VP of the air), and the temperature of the air at sea level is 10 degrees.   The difference between the temperatures at sea level and where the air will reach dew point is 5 degrees.   Since you know that the DAR is 10 deg./1000m, you know that to cool 5 degrees. the air needs to rise 500m.

When it reaches 500m, the RH will be 100%, and the air will be at dewpoint.   From here, the air will of course continue to rise to get over the mountain, but will be cooling at the MAR - 6 deg./km.   If the mountain top is, say, 3000m, it means that the air must rise another 2500m, which means that it will cool another 15 degrees ([6 deg/1000 m] x 2500 m = 15).   So that means that the air at top will be -10 degrees (5 - 15 = -10).

Since the air will reach dewpoint far below this elevation as it rises, it will of course still be at dewpoint (100% RH) at the top.    At the top the air is at 100% RH and -10 degrees; this means that the air has a VP of about 3mb (you can see this by looking at the chart).

So now the second half: the leeward side of the mountain.   Going down the leeward side, the air warms at the DAR, since immediately upon warming just above the temperature at the top, it goes above dewpoint temperature again (below 100% humidity), so no condensing takes place.   Remember, the only time you use the MAR is on the windward side, between the condensation level and the top.   You use the DAR from the base level to the condensation level and on the whole leeward side.

If you want the RH and temperature at 2 km on the leeward side, you need to figure how much the air will warm; this is of course directly related to how much it will lower and warm at the DAR.   In this case it falls 1000m, so it warms by 10 degrees.   So the temperature at 2 km is zero.

The RH is the percentage of VP in the air compared to what it can hold at a given temperature.   So you need to know the VP at 2 km, which is the same as at the top (see note above concerning the reason for DAR on the leeward side).   At the top the air was at 100% RH and -10 degrees; this means that the air had a VP of about 3mb.   At zero degrees (the temp at 2 km), the saturation VP (aka maximum VP) is about 6mb.   So the RH at 2 km is 50% - has 3mb, can hold at that temperature 6mb.

Here's a worked through example, with answers.   Suppose we have air at the bottom of a mountain that is 25 degrees C, with a vapor pressure of 10mb.   The mountain is 3000m high.

1. determine the DP of the air at the bottom (the starting VP will give you this by looking at the graph).   Air with 10mb of VP has a DP of about 7 degrees C; the temperature it starts at is irrelevant.

2. Find the elevation at which that the air, rising and cooling at the DAR, will reach this dewpoint temperature.   If it's got to reach 7 degrees, it'll have to cool 18 degrees (25-7=18).   So, cooling at the DAR, it'll rise 1800 meters (18 x [10 degrees/1000m] = 1800).   So at 1800 m the air will reach saturation, at a temperature of 7 degrees.   It'll stay at 100% humidity as long as the air keeps cooling.

3. From this elevation, (the condensation level), find out how much more the air will rise before getting to the top of the mountain.   Then determine how much this will cause the air to cool at the MAR.   This will get you the temperature at the top of the mountain.  In our example, it will rise another 1200m to get from 1800 to 3000 m; cooling at 6 degrees/1000m, this air will cool by 7.2 degrees (1200 x [6 degrees/1000m). n  So it cools from 7 degrees to -.2 (7-7.2 = -.2).

4. Determine the VP of the air at the top of the mountain... Remember it's still at 100% RH when it reaches the top of the mountain, so you need only look at the chart to determine the maximum VP.   Since it's -.2 at the top, the VP is about 6mb. (looking at the chart, you see that -.2 on the bottom side matches about 6mb on the left side)
 

5. Going down the other side, the air warms at the DAR.   If I ask for some info for any point on the leeward side, you'll be able to tell me that too.   Say I ask the RH and temperature at 600m.   You can tell the temperature by taking the difference in elevation (3000 m - 600 m = 2400); the air will descend 2400m, and warming at the DAR will warm by 24 degrees from the top (2400 x [10 degrees/1000m] = 24).   So at 600m the air temperature will be 23.8 (--.2 + 24 = 23.8).

The RH at this point can be determined easily since you remember that the VP at the top was 6mb, and didn't decrease as the air descended and warmed, since the VP will only decrease if it is cooling below DP...   So the VP here is the same as the top: 6mb.   The air at 23.8, however, has a maximum VP (aka saturation VP) of about 29.5 mb.   So the RH is 6 / 29.5 = .20 (20%).
 

Back to the Geography 1111L page
 

          Nam risu inepto res ineptior nulla est.
          [Untimely grinning is the silliest sin]

                                        -Cattalus