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.