Dew Point and Temperature vs altitude

In a recent post we discussed a very simple (clearly over simplified) method of coming up with a way to estimate cloud ceilings in special circumstances. A point of the discussion was that the dew point lapse rate, as used in this model, was less than the environmental air temp lapse rate. We came up with something like -1º /1000 ft for the dew point drop with altitude, compared to an average value of the environmental air temp lapse rate, which we took to equal the Standard Atmosphere value of -3.6º/1000 ft.

So now we have to face the truth to see if this model can be justified at all by looking at real soundings and then real cloud height measurements. Soundings are the measured values of T, DP, and pressure (among other parameters) as a function of altitude. Soundings are nicely presented at the University of Wyoming. As you surf around the world looking at these, the  first thing you notice is the the temperature and dew point changes with altitude are all over the place. First reaction would be there is no way at all to predict this behavior.

But we are looking at very special cases, namely we must  have low clouds there in the first place, which means the air temp must drop to the dew point at some altitude less than 7,000 feet (2134 m). In other words, we were not predicting clouds at a certain altitude based on T and DP, but instead saying that if we do have low clouds, then we might estimate the ceiling or cloud base from the T and TP on the surface.

So step one in the data search is to find those cases were we do see the T and DP coming together at some height less than 2000 m. Below is a pic explaining the diagrams (details here), followed by some pictures taken at random. After that we analyze then in the light of our past discussions.

The dew point is always on the left. Background lines are theoretical values of the dry rate (green, 9.8C ~ 5F) and the moist rate (blue, 6C~3F). Notice that the dry rate slope is about 10º shallower, and this 10º corresponds to a lapse rate difference of about 2º F. Notice, too, there are no low clouds in this case; it is even an inversion. For a while the temp is increasing with altitude.

Here are some examples taken at random.  Again, only criteria was that T and DP met below 2000m. In each of the pictures, we marked the temp with a green line, the DP with a red one, then we duplicated a segment of the green line and moved it to the base of the DP so you can see the difference in slope more clearly.

We see first that indeed in these cases the DP lapse rate is lower than the temperature. To get a better feeling we can analyze the slopes of the lines we marked in the figures (1 top, 8 bottom), and from these compute the lapse rates.

Sample	slope		lapse rate °C/km		lapse rate °F/1000ft		Delta
	DP	T	DP	T	DP	T	T-DP
1	349	333	-3.8	-9.9	-2.1	-5.4	-3.3
2	350	341	-3.4	-6.7	-1.9	-3.7	-1.8
3	351	341	-3.1	-6.7	-1.7	-3.7	-2.0
4	346	339	-4.8	-7.4	-2.6	-4.1	-1.4
5	347	337	-4.5	-8.2	-2.5	-4.5	-2.1
6	352	336	-2.7	-8.6	-1.5	-4.7	-3.2
7	344	338	-5.6	-7.8	-3.0	-4.3	-1.2
8	349	340	-3.8	-7.0	-2.1	-3.9	-1.8
Averages=			-3.9	-7.8	-2.2	-4.3	-2.1

This brief analysis seems to imply that a 2º DP lapse rate is better than the 1º we came up with, but the closing rate used to estimate cloud ceilings is reasonably close. This shows an average of 2.1 and we had 2.6.  But we have to admit this is all very crude analysis. It can only show there is some ballpark value and that the standard values we see in pilot's license training materials and other books (usually 2.5) might not be unreasonable.

There is still another way we can look into the usefulness of this approach and that is to check actual metar reports from airports. They report T, DP, and cloud height, which is presumably measured with some form of a ceilometer–a laser device for measuring cloud height.  We have started this list, but we are already detecting the limits of this analysis, namely they ceilometer data are not being reported to a very high precision.  In fact, it is lower than some standards say it should be.

station	T (ºC)	DP (ºC)	Cloud height (m)	k (ºF/1000 ft)
KCHS	18.3	15	300	1.8
MIAMI	26.7	23.3	600	1.9
YBRK	19.6	18.5	600	0.6
KMFL	30.6	23.9	600	3.7
8557	14	7.8	600	3.4
FZAB	31	20.4	600	5.8
KLAX	21.7	15.6	600	3.3
KLGA	12.8	4.4	1000	4.6
DULUTH	4.4	-5	1500	5.2
KDLH	7.2	-2.2	1500	5.2
			Average =	3.5

We need to get a lot more of this data before any conclusions. We have limited the cloud height to below 2000 m, but what we find in the data is the next step up is 2500 m, and the other 6 stations found with cloud heights all reported 2500m, which can't be right. Reminds me of the old days when all ship reports had wind only from the cardinal and intercardial directions!

So far we did a crude analysis and predicted k = 2.6, but this came from an estimated DP lapse rate of 1ºF/1000 ft, which was then subtracted from an estimated average T lapse rage of 3.6. This is not consistent with the soundings which showed averages of 2.2 and 4.3 for an average k = 2.1 ±1

The 10 metar data so far gives k= 3.5 ±2.  We do not have a lot of data, but the data included all that matched our criteria: in soundings, we took all we found that had T and DP meeting before 2000m and the metar data we have taken all we found that gave measured cloud heights below 2000m. So the statistics are low, but should form some realistic sample. (PS it could be I am misinterpreting the metar data. i will check that.)

Therefore I must conclude that  I cannot see that there is not a simple way to predict useful cloud heights based on surface values of T and DP.  So we have to change our textbook from saying this is a formula for estimating cloud height, to something like this is the formula some books say can be used to estimate cloud height, but it must have large uncertainties.

This is the end of this discussion for now.  I have to wait to see if someone who knows about these matters might shed some light on this topic... i am a bit in the dark here.

Relative Humidity and Dew Point as a Function of Altitude -- A Way to Estimate Cloud Ceilings

Air temperature drops as we rise in altitude above the surface. At some point the air temp drops to the dew point of the air at which point the water vapor in the air condenses into liquid water, and this water we see condensed onto specs of dust in the air makes up the clouds. All air contains some amount of water vapor, varying from just a fraction of a percent (by weight) for cold dry desert air, on up to some 3% for hot steaming jungles.... and, sure enough, all air contains some dust!

If we consider the Standard Atmosphere (a set of average values invented by aircraft design engineers to facilitate their work) as a starting point, we can find tables of properties showing how air temperature decreases with altitude in this model atmosphere. It does so at a rate (sometimes called the average lapse rate) of -3.6° F per 1,000 ft, starting at 59° F at the surface, height = 0. Thus at 2,000 ft, the temperature has dropped to 52° F.

There is no Standard Atmosphere dew point or relative humidity; it assumes the air is dry. So we have to add the water ourselves if we want to think about that. To do this, we need a careful look at the definitions and some properties of water. We tend to think of relative humidity as a property of the air and end up with such phrases as the "air can hold so much water vapor at this temperature," etc, but this is very misleading. These two gases (air and water vapor) are essentially independent, and it is actually the properties of the water that matter here,  not the properties of the air.

If you had a jar of dry air, and poured some water into the bottom of it, water molecules on the surface will evaporate from the surface and periodically condense back into the surface until it reaches some equilibrium value where evaporation and condensation is equal. At that point the air in the jar will contain a certain amount of water vapor. That amount depends only on the temperature of the water, which we are assuming is the same as the temperature of the air at this equilibrium state. At 59° F this amount of water vapor is 12.8 grams per cubic meter.  Raise the water temperature to 86° F and this goes up to 30.4 grams per cubic meter. Drop the temperature to 41° F and this drops to 6.8 grams per cubic meter.

You could ask why this behaves this way, and indeed we could spend some more time looking at the basics to better understand the trends, but the actual numbers are an inherent property of water, like its freezing point and boiling point.  Whatever they are, they do not depend on the air at all!

If  instead of a jar of air, we had an evacuated jar with no air at all in it, and then introduced water into that system, the same thing would take place: evaporation and condensation until an equilibrium amount of pure water vapor gas was above the surface of the water. The grams per cubic meter would be the same as above for the temperatures listed, but in a pumped out chamber like this we could measure it directly with a pressure gauge.  We would go from no pressure on the gauge before adding the water to 17 mb of pressure on the gauge when the water was at 59 F.  Raise the water temp to 86 F and you will see the pressure rise to 42.4 mb. Drop the water temp to 41 F and the pressure will drop to 8.7 mb. (See Table 3.5 in our book Modern Marine Weather for a list of these values.)

In other words, the amount of water vapor present does not depend on the air at all; it depends only on the temperature of the water.

This unit for water vapor content is called its partial pressure, and in this terminology the equilibrium value (called the saturation value) of water at 59 F is 17 mb, which is equivalent to 12.8 grams of water vapor per cubic meter. This is a nice unit because we do not have to consider volume size in using it.

Relative humidity is defined as the ratio of how much water vapor is present to the maximum that can be present. At 59° F the maximum  is 17 mb, so if there is only 8.5 mb of water vapor, the relative humidity (RH) is 50%.  Relative humidity depends only on the temperature and actual water vapor present. It has nothing to do with the air pressure itself.

If I am at sea level with a pressure of 1013 mb, and the air temp is 59° F, then 8.5 mb of water vapor content means the RH = 50% (8.5/17). If  I am on a mountain peak and the pressure is 942 mb and it also contains 8.5 mb of water vapor and the temp is still 59 F, then that air also has a RH=50% (8.5/17).

But this is not the case we are talking about. We are talking about a case where the surface temperature is 59 and the temperature at 2000 ft is 52. What happens then to the RH?

In a sense, we know the answer without much analysis, because if there are no clouds on the surface and there are low clouds overhead we are looking at the answer! That is, if you have an air mass that has a uniform distribution of water vapor throughout, then the relative humidity has to go up as you go up in the atmosphere.

With no clouds at the surface (ie no fog) then the RH was less than 100% and the temp of the air was higher than the dew point. As we go up in altitude, the temp drops and when it reaches the dew point the RH is 100% and clouds are formed. So the RH must go up as we go up in the atmosphere. But this is only half of our answer.

If we want to predict cloud bases (or at least have some rough guideline for doing so), for example, we need to know when the temp drops to the dew point. We know how the air temp is dropping with altitude, but what about the dew point? Is it staying the same, rising or dropping?

We have several guides to how the air temperature drops with altitude. As noted above, in a standard atmosphere it drops at -3.6° F per 1,000 ft. Meteorologists remind us that this is some sort of average, and that theoretically air that is not yet saturated would show a temperature drop of -5.5° F per 1000 ft, called the dry lapse rate, and that air that was already saturated will display a slower temperature drop with altitude called the moist lapse rate of 3.0° F per 1000 ft. The difference can be attributed to how the moisture content changes the air temperature as the water vapor condenses on the way up adding heat to system. The actual temp drop in a real atmosphere is called the environmental lapse rate, and it could be quite different from any of these or even reversed as in a temperature inversion.

We are not getting into these details for now, we will just use the standard atmosphere. For those who want to see real air temp profiles, refer to the excellent live data compilation from the University of Wyoming.

At this stage I have to say I really do not know the best way to think about what I assume is called the dew point lapse rate. There is a thorough set of related equations online from Vaisala.

The following is the logic used in Modern Marine Weather, which is based on numerical values of the relative humidity, which can be evaluated from Table 3.5 or from  a dew point calculator such as the nice one Vaisala offers online.

Consider a standard atmosphere with 50% RH.  At sea level we have P = 1013 mb, T=59°, saturation = 17 mb, water vapor = 8.5 mb, RH =50%, with a DP = 40.4° F.

At 2,000 ft, P = 942 mb, T=52° F.

Then we assume the relative water vapor content is the same at 2,000 ft as it is at the surface, so the absolute water vapor content of the thinner air at 2,000 ft would be (942/1013) x 8.5 = 7.9 mb.

Then we look up saturation vapor pressure at 52º F, which is 13.3 mb, and conclude that RH=7.9/13.3 = 59%. The relative humidity went up as expected, but we are clearly not at the cloud base yet.

Then we figure the DP for T= 52, RH=59 and that is 38.5º F. You can follow though these results by interpolating our table or the Vaisala calculator.

Thus at 2,000 ft above the surface, the RH rose from 50 to 59 and the DP dropped from 40.4 to 38.5.

Then from that type of analysis, I have concluded that the dew point drops at a rate of about 1º F per 1,000 ft, i.e. (40.4-38.5)/2 — compared to the air temp drop of 3.6º F per 1,000 ft.

So in a standard atmosphere, the temp is approaching to the DP at a rate of 2.6º (3.6 − 1.0)  per 1,000 ft, and we can make an equation for the cloud ceiling based on the surface T and DP (in ºF) as:

Ceiling height /1000 ft  = (T - DP) / 2.6

That, I am sorry to say,  is a very long answer to the questions we have received about where does that formula come from!  Please post your comments and questions.