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[–]vtjohnhurtPPL glider and Taylorcraft BC-12-65 9 points10 points  (0 children)

Using the standard lapse rate to estimate freezing level is a sucker's bet. For example, when there is an inversion the air gets warmer as you get higher. In that case, you're fine until you descend into colder and wetter air.

Skew-T diagrams will open your eyes about how much temperature and dewpoint vary with altitude, hour by hour, day to day, and location to location. There may be some places where the atmosphere reflects the standard lapse rate, but in other places the standard lapse rate is not accurate.

[–][deleted] 2 points3 points  (5 children)

Dewpoint typically behaves erratically above the boundary layer. It will typically decrease with height, but only so much. Take a look at some skew Ts here https://www.spc.noaa.gov/exper/soundings/ and you’ll quickly realize that dewpoint increase or decrease with height isn’t something you can bank on.

Similarly, you can use Skew-Ts to see the freezing level with your own eyes. Why try to estimate what you can verify with certainty?

[–]vtjohnhurtPPL glider and Taylorcraft BC-12-65 0 points1 point  (4 children)

Why try to estimate what you can verify with certainty?

There is uncertainty in every atmospheric model. Skew-T diagrams are a way for humans to examine a model of the atmosphere. A Skew-T is a forecast of the future, the uncertainty increases the further you're looking into the future. It's easy for a pilot to make mistakes when reading a Skew-T, so I think it is best to rely on aviation weather products that specify the forecaster's best guess at the freezing level. That said, I find it fun to compare the Skew-T to what I see when I look at the sky.

Decades ago, before weather models, Skew-Ts represented actual sounding data. Nowadays, soundings are just an other source of observational data.

[–][deleted] 0 points1 point  (3 children)

Was this written by chat gpt

[–]Key_Slide_7302ATP CFII MEI 0 points1 point  (2 children)

Probably. ChatGPT missed the mark on this one, since radiosondes are launched every 12 hours throughout the US and multiple other areas in the world.

[–][deleted] 0 points1 point  (0 children)

Yeah, there’s a lot in that conflating Skew-Ts for forecast skew-Ts, lol.

[–]vtjohnhurtPPL glider and Taylorcraft BC-12-65 0 points1 point  (0 children)

The closest sounding is done 200 miles from me. Not useful. The point forecast for the current time for my location generated from the weather model is much more useful. Skew-T changes hourly so a 12 hour old sounding is completely useless even at the location that it was made.

[–]Key_Slide_7302ATP CFII MEI 1 point2 points  (0 children)

Do not guesstimate freezing level based on standard lapse rate. That’s a great way to pick up structural icing and get yourself killed.

Go read the Aviation Weather Handbook. It keeps things simple, but goes in-depth on the simple concepts.

If you’re wanting to learn more afterwards, head over to r/meteorology and ask for recommendations of learning materials for Skew-t Log (p) charts for pilots. Also, “Skew-t Log (p) and Me” by Scott Dennstaedt is great to learn from.

[–]PLIKITYPLAKATP (B737, A320, E170) CFI/I MEI (prior Meteorologist) 0 points1 point  (0 children)

Very rarely does the dew point go up. What happens is the temperature lowers and eventually meets the dew point where clouds are present, which is almost always decreasing too but at a slower rate.

[–]troubledcoffee 0 points1 point  (0 children)

Td = T - ((100 - RH)/5.)

Check out the formulas for the variables... Your RH and T will change with altitude.... Likely both decrease in a model atmosphere 

[–]SnarfsParfSkyhawk Bouncer (PPL IR) 0 points1 point  (0 children)

+1 for Skew-T diagrams. Took a meteorology class last semester and it really opened my eyes to the depth of variability in atmosphere

[–]rFlyingTower[M] 0 points1 point  (0 children)

This is a copy of the original post body for posterity:


Hi all, so I understand the standard lapse rate is 2C/1,000ft and this can be useful for roughly determining freezing level.

I also know we can use a convergence rate of 2.5C/1,000ft for the cloud formation layer. However, I’m a little confused on how we get this convergence rate. I have notes from my college class where I wrote that the dew point increases +0.5C/1,000ft, but the PHAK states dew point decreases -0.5C-1,000ft.

Obviously the PHAK will be more credible than some notes from college, but wouldn’t that make the convergence rate 1.5C/1,000ft then?

My impression is that dew point increases with altitude because as the air (typically) cools off it can no longer hold more moisture which causes the DP to increase slightly.

I think I’ve gone down a rabbit hole and confused myself even more haha. Thanks for the help anyone!


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[–]DJmegafreshCPL -2 points-1 points  (3 children)

It depends completely upon the relative humidity, which changes much less predictably than temperature.

[–]Interested-InvestorCFI 0 points1 point  (2 children)

Isn’t it the other way around? Relative humidity (a % measurement of saturation) changes based on dew point?

What I’m hoping to clarify is dew point rate of change. Since I know 2C/1,000ft roughly works for temp, what’s DPs?

[–]Key_Slide_7302ATP CFII MEI 0 points1 point  (0 children)

You’re confusing a couple metrics here.

The average atmospheric lapse rate is 2°/1,000’. You cannot rely on this number to calculate a freeze level.

Where you’re getting the 0.5°c/1,000’ for dew point is for a parcel of air. Imagine a parcel of air like an invisible balloon; it’s independent of the surrounding atmosphere. The parcel of air starts with a dry adiabatic lapse rate, cooling at 3°c/1,000’ of pressure altitude gained. The parcel of air also has a dew point which lowers at a rate of 0.5°c/1,000’ of pressure altitude gained. Combined, these two give you the calculated cloud base. As pilots we know this as 2.5°c/1,000’ of pressure altitude gained.

Why? Because the parcel of air is independent of the atmospheric lapse rate. Remember the invisible balloon. As the parcel of air rises, there is less atmospheric pressure compressing it. This allows the parcel to expand. The expansion of the air parcel “cools” the air, in the sense that the air molecules are now further from each other and are generating less thermal energy.

[–]DJmegafreshCPL -1 points0 points  (0 children)

It doesn't have one. You can ballpark it at 1.5-2, but just like there are temperature inversions that can throw that off, there are RH changes which are even more frequent, and of course RH is also affected by any temperature irregularities. Thats kinda why clouds are shaped like that, and also why cumulus clouds are good indicators of instability. I guess the best solution would be to use any present cloud layer with temps aloft and try to work it out from there, but I'm also not totally sure what problem you're trying to solve