how accurate is this ?

SeanD

SeanD

Well-Known Member
Is this ratio accurate on predicting cloud levels?

KLAS weather today
69degrees - 9% dewpoint = 60 x 1000 / 4.4 = 13,636 ft
 
I don't remember the math exactly, but that seems right. I cheat and use my electronic E6B instead :nana2:
 
Yes, that is correct, although, you can skip a step and just divide 60/4.4 and will give you 13.63 which is essentially the same thing.
 
SeanD said:
Is this ratio accurate on predicting cloud levels?

KLAS weather today
69degrees - 9% dewpoint = 60 x 1000 / 4.4 = 13,636 ft
Click to expand...
Almost. It would be correct if it were

69degrees - 9 degrees dewpoint = 60 x 1000 / 4.4 = 13,636 ft

and remember that this is the formula for fahrenheit temperatures.
 
Cool thanks guys, we just went over weather the other day in ground school class. Weather is amazing, never really understood how complex it was and all the challenges it brings when you fly. I think its good to be in this ground school class before accually taking the controls of the airplane.
 
My instructor recommended me the book "Weather Flying " by Rob Buck. It's an exciting read. It's true they say the more you read into weather theory, the more you don't know.
 
That's a really good book. Although I would be hard pressed to call it exciting. It took me about three months to get through it.
 
A lot of whats in it I had to use my imagination a lot but it's one of those books that you'll end up referring to it over and over again.
 
Just went over this in the instrument ground school the other day. The rule of thumb is 2.5 C/1000'.

For the cooling rates, the dry adiabatic lapse rate (DALR) is 2 degrees C/1000', and the saturated adiabatic lapse rate (SALR) is 3 degrees C/1000'. Any air parcel that rises after it becomes "supersaturated" cools at the SALR. Thats around the time it becomes a cloud.

-ColM
 
ColMustard said:
Just went over this in the instrument ground school the other day. The rule of thumb is 2.5 C/1000'.
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Right. Because a degree Celsius is about 5/9 the size of a degree fahrenheit (remember the old conversion formula from grade school?)

5/9 of the 4.4 approximation in the fahrenheit conversion is 2.4444.
 
Windchill said:
Is there a rhyme or reason for dividing by 4.4?
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Sure. You can find the details on some weather sites. This is just a quick summary:

In the type of unstable air where you have cumulus clouds, the rising air cools at a rate of 5.4° F/1,000 ft; the dewpoint decreases at 1° F/1,000 ft. 4.4 is the rate at which they are converging. The base of the clouds is, of course, where the temperature and dewpoint meet.
 
Windchill said:
Is there a rhyme or reason for dividing by 4.4?
Click to expand...


Yea it has to do with the dew pt temp compared to the temperature of the rising air. The dew pt. temp decreases about 1degree fahrenheight per 1,000ft gain in altitude and the parcel air lapse rate cools at 5.4deg. F/1000ft. When the air rises it converges with the dew pt. so eventually the dew pt temp and the rising parcel of air temp are equal and that's when the base of the clouds form. That's why this formula only works with towering clouds or those with vertical development. 5.4 - 1 is 4.4. That formula is basically comparing the temp of rising air and the dew point against the change in dew pt. and temp.
 
Not always the case when temp. and dewpoint meet that you will have clouds.

Thats why the 2 degrees C/1000' lapse rate is not completely accurate. When a parcel of air is warmer than its DP, it can hold more water, and cools at approximately 2 degrees/1000'. After it reaches the DP temperature (aka, condensation level), the air parcel will lapse at both the SALR (3 degrees C) and DALR (2 degrees C).

From this you get a rule of thumb (more like a compromise) of estimating the cloud bases using 2.5 degrees C/1000' as the lapse rate to determine the approximate height of the cloud bases.

Not an exact science though, too many variables present in the atmosphere. Course if we are accurate to 50'-100', I'd be satisfied.

-ColM
 
Im pretty sure the 2.5 degrees comes from adding the dry diabatic lapse rate to the dewpoint lapse rate. Im dang near 100% positive on that just thinking off hand here, too lazy to go look in the book.

To estimate the cloud layer altitude we need to find out where the temp and dewpoint converge. The dry diabatic lapse rate is 2*C/1000 feet, and the dewpoint lapse rate goes up by .5*C/1000 feet. So we add those together and you get your 2.5*/1000 feet. The air temp is going down by 2, and dewpoint is going up by .5, so they converge at 2.5 degress C/1000 feet.

Since most weather information is given to you in *C, I'll convert your figures as an example.

20*(69 *F)-neg15*(9*F)= 35 35/2.5= 14 X 1000= 14,000, close enough.
 
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