Weather reprots: AGL or MSL confusion
Alex Ly
Studying for the PPL written test, I am coming across a question:
It seems that METARS, PIREPS, TAF's, and Area Forecasts (FA's) disclose the
ceiling in ft MSL.
At my home airport, when I listen to the ATIS or the AWOS, it appears
ceilings are given in ft AGL.
Could you guys confirm?
Thanks for any help.
Alex Ly
Jonathan Friedman
Not quite correct.
Area Forecasts and PIREPS give cloud heights in feet MSL. The reason is
that they cover areas with a range of terrain elevations. AGL reports would
be meaningless. Occasionally, there will be an "AGL" or "ceiling"
designation used in an FA, and in that case the clould height(s) are AGL.
METARS, TAF's, ATIS, AWOS, ASOS, etc. all give cloud heights in feet AGL.
Again, because those reports are for specific locations with known terrain
elevations.
Good luck with your written.
Jon Friedman CFI CFII
Alex Ly <koo...@hotmail.com> wrote in message
news:9mp590$ni5$1...@news.tamu.edu...
Alex Ly
Thanks for your help. I have another question <grin>
Today, the altimeter at my airport was 29.84, 0.08 lower than the standard
29.92. Also, density altitude was given by the AWOS as 1,900 ft vs. an
airport elevation of 321 ft.
I believe the altimeter setting of 29.84 corrected for a day with low
pressure around the airport. This low pressure would mean that the pressure
altitude would be 0.08 x 1000 = 80 ft. Add that value to airport elevation
of 321 ft to get a pressure altitude of 401 ft. Am I right so far?
Then, because today's temperature was higher than standard temperature (I am
in Texas), density altitude further corrected the pressure altitude for
temperature, and thus the AWOS announced a density altitude of 1,900 ft.
This means that the trusty Cessna would be performing at standard conditions
at an altitude of 1,900 ft, instead of airport elevation of 321 ft.
Thanks for any insight, again.
Alex Ly
"Jonathan Friedman" <j...@ix.netcom.com> wrote in message
news:9mp8oi$csr$1...@slb6.atl.mindspring.net...
Jonathan Friedman
>I believe the altimeter setting of 29.84 corrected for a day with low
> pressure around the airport.
Correct, but there's also a correction thrown in for temperature variation
from standard and for field elevation.
>This low pressure would mean that the pressure altitude would be 0.08 x
>1000 = 80 ft
Right on.
>Add that value to airport elevation of 321 ft to get a pressure altitude of
>401 ft. Am I right so far?
Not so fast :) You've already (correctly) stated that the pressure altitude
is 80 feet. The pressure altitude you've determined has already taken field
elevation into consideration.
> This means that the trusty Cessna would be performing at standard
>conditions at an altitude of 1,900 ft, instead of airport elevation of 321
ft.
I think you have it right, although I'm not sure what you mean by
"performing at standard conditions at an altitude......". If you mean your
Cessna would perform as if it were flying in standard conditions at 1,900
feet MSL, your are absolutely correct, Sir.
Jon Friedman CFI CFII
Alex Ly <koo...@hotmail.com> wrote in message
news:9mpgse$78t$1...@news.tamu.edu...
Alex Ly
The 80 ft represent the correction factor which has to be applied (added or
subtracted) to field elevation to determine pressure altitude. I don't
believe the 80 ft. is my pressure altitude but rather 80 ft + field
elevation or 80 + 321 or 401 ft. The correction of 80 ft was added since
the altimeter setting today is lower than the standard 29.92.
Does this sound correct?
Alex Ly
"Jonathan Friedman" <j...@ix.netcom.com> wrote in message
news:9mpkiu$tij$1...@slb7.atl.mindspring.net...
Jonathan Friedman
Pressure altitude is determined by setting your altimeter to 29.92 (standard
pressure) and reading the resulting altitude...period.
Keep in mind that, that altitude reading is a function of local atmospheric
pressure as well as your altitude. Your altimeter is effectively indicating
the elevation of the airport. You do not add the airport elevation to the
altimeter reading to compute pressure altitude.
Could it be that you're confusing pressure altitude with true altitude
(altitude above sea level)? The computation that you're suggesting sounds
more like you're attempting to compute the latter.
Jon Friedman CFI CFII
Alex Ly <koo...@hotmail.com> wrote in message
news:9mplff$bun$1...@news.tamu.edu...
Peter Duniho
> Not so fast :) You've already (correctly) stated that the pressure
altitude
> is 80 feet. The pressure altitude you've determined has already taken
field
> elevation into consideration.
No. Pressure altitude is what your altimeter will read if you set it to
29.92.
If you're sitting at 321', and you set the altimeter to 29.92 when the
correct setting is 29.84, your altimeter will read 401'.
Just as Alex said.
Note that the 1" per 1000' rule is just a rule of thumb. Works pretty well,
especially at lower altitudes, but it's not an exact conversion. Definitely
close enough for government work though (to repeat a recently posted cliche
:) ).
Pete
Antonio Aponte
Alex Ly wrote:
>
> This means that the trusty Cessna would be performing at standard conditions
> at an altitude of 1,900 ft, instead of airport elevation of 321 ft.
Minus the term "standard conditions" your statement is true. (Standard
conditions are 59 degrees f. and 29.92 inches of mercury) Let's make it
simple....
You turn the knob on your altimeter until 29.92 is indicated in the
Kollisman window (the little square window on the altimeter) and read
the altimeter. This gives you your pressure altitude (the measurement of
your altitude above or below the 29.92 pressure level that we use as a
reference).
Now that you know your pressure altitude, you can calculate density
altitude which is not really an 'altitude' at all but really a
performance forecast. That is, density altitude tells you how your
airplane *will perform* given the conditions of that particular day.
Some days my sea level airport will act like a sea level airport and
other days--like on hot days--it will act like an airport at a much
higher altitude. Whenever the temperature goes up (or pressre drops) the
air gets thinner and so I cannot expect that my wing, engine, or
propeller will work as well as a cool day. This is what density altitude
is all about and why I say it actually deals with performance rather
than altitude.
Pressure altitude corrected for nonstandard temperature gives you the
density altitude.
Antonio
Peter Duniho
> Pressure altitude is determined by setting your altimeter to 29.92
(standard
> pressure) and reading the resulting altitude...period.
True. As long as you understand that's not the *only* way to determine
pressure altitude. It's certainly the easiest. (Assuming you have a
properly working altimeter handy :) ).
> [...] You do not add the airport elevation to the
> altimeter reading to compute pressure altitude.
Yes you do, if you are computing pressure altitude from a known altimeter
setting and known indicated altitude rather than using an altimeter to
display pressure altitude directly.
> Could it be that you're confusing pressure altitude with true altitude
> (altitude above sea level)? The computation that you're suggesting sounds
> more like you're attempting to compute the latter.
No offense, but it sounds to me like you're the one who's confused. "MSL"
is "(indicated) altitude above sea level", while "true altitude" is
"(indicated) altitude (corrected for non-standard conditions) above sea
level". Hardly any pilot ever uses true altitude, since about all it's
useful for is avoiding mountains, and it's easier to just add a couple
hundred feet to your minimum safe altitude (which is already done for things
like sectional charts anyway).
Pete
Gary L. Drescher
> > pressure around the airport.
>
> Correct, but there's also a correction thrown in for temperature variation
> from standard and for field elevation.
No, Alex had it right--there's no additional correction needed for the
temperature. That's already included in the correction for the pressure.
Temperature is one factor that contributes to pressure; density is another.
You don't have to know how much of each factor there is in order to know
their sum. In order to derive the corrected altimeter setting, you only
need an altimeter; you don't need to know the temperature. But if you want
to know the density altitude given the pressure altitude, you do need to
know the temperature as well, because you want to factor out the component
of the pressure due to nonstandard temperature.
> >This low pressure would mean that the pressure altitude would be 0.08 x
> >1000 = 80 ft
>
> Right on.
>
> >Add that value to airport elevation of 321 ft to get a pressure altitude
of
> >401 ft. Am I right so far?
>
> Not so fast :) You've already (correctly) stated that the pressure
altitude
> is 80 feet. The pressure altitude you've determined has already taken
field
> elevation into consideration.
I think Alex meant to say that the difference between the true altitude and
the pressure altitude is 80', which is correct. The pressure altitude is
indeed 401', as Alex computed. That's what he should see if he sets his
altimeter to 29.84 (assuming he's at the same place and time as the AWOS
source).
Regards,
Gary
Gary L. Drescher
conditions
> > at an altitude of 1,900 ft, instead of airport elevation of 321 ft.
>
>
> Minus the term "standard conditions" your statement is true. (Standard
> conditions are 59 degrees f. and 29.92 inches of mercury) Let's make it
> simple....
I believe the term "standard conditions" refers to pressure and temperature
as a function of true altitude. So 59/29.92 are what you find at sea level
under standard conditions. Alex was correct in referring to standard
conditions at 1900'.
Regards,
Gary
Gary L. Drescher
> > Could it be that you're confusing pressure altitude with true altitude
> > (altitude above sea level)? The computation that you're suggesting
sounds
> > more like you're attempting to compute the latter.
>
> No offense, but it sounds to me like you're the one who's confused. "MSL"
> is "(indicated) altitude above sea level", while "true altitude" is
> "(indicated) altitude (corrected for non-standard conditions) above sea
> level".
That's correct, but it's also correct (as Jonathan stated) that "true
altitude" is simply the actual altitude above sea level (which is why, as
you say, it's useful for avoiding mountains). The correction for
non-standard conditions is to give you the correct altitude above sea level.
Regards,
Gary
Jonathan Friedman
> Yes you do, if you are computing pressure altitude from a known altimeter
> setting and known indicated altitude rather than using an altimeter to
> display pressure altitude directly.
Yes, but he's clearly not attempting to do that. He has pressure alitude
from his altimeter.
> No offense, but it sounds to me like you're the one who's confused.
Not at all.....just trying to brainstorm what it is that Alex is attempting
to do.
Jon Friedman CFI CFII
Peter Duniho <NpOeS...@NnOwSlPiAnMk.com> wrote in message
news:3b907...@news.nwlink.com...
Jonathan Friedman
Man...did I do a booboo!!
That's what I get for attempting higher math at 1am after a 14 hour day.
Sorry for the misinformation, Alex.
Jon Friedman CFI CFII
Peter Duniho <NpOeS...@NnOwSlPiAnMk.com> wrote in message
news:3b907...@news.nwlink.com...
Bill Douglas
altitude from his altimeter. He's trying to get it from the local
altimeter setting. There's a difference.
Bill Douglas
Jonathan Friedman wrote:
>
> Pete
Alex Ly
You say:
"I think Alex meant to say that the difference between the true altitude and
the pressure altitude is 80', which is correct. The pressure altitude is
indeed 401', as Alex computed. That's what he should see if he sets his
altimeter to 29.84 (assuming he's at the same place and time as the AWOS
source)."
I thought I would see 401' if I set my altimeter to 29.92, not 29.84. Just
wondering.
Thanks for all the help.
Alex Ly
"Gary L. Drescher" <GLDre...@hotmail.com> wrote in message
news:gU2k7.87$CR2.2...@typhoon.ne.mediaone.net...
Gary L. Drescher
> Gary,
>
> You say:
>
> "I think Alex meant to say that the difference between the true altitude
and
> the pressure altitude is 80', which is correct. The pressure altitude is
> indeed 401', as Alex computed. That's what he should see if he sets his
> altimeter to 29.84 (assuming he's at the same place and time as the AWOS
> source)."
>
> I thought I would see 401' if I set my altimeter to 29.92, not 29.84.
Just
> wondering.
Oops. Yes, of course.
You're doing a lot better than those of us offering you advice. :)
Regards,
Gary
Antonio Aponte
altimeter setting that you get from a local tower is not actually the
barometric pressure with reference to true (MSL) altitude but an
arbitrary setting that will cause your altimeter to read the local field
elevation.
In the origional question, Alex referred to the local setting ("...I
believe the altimeter setting of 29.84 corrected for a day with low
pressure around the airport....") and I just wanted to remove the
ramifications of that from his question. Therefore, I removed the term
"standard conditions" from his question which, IMO, made his remaining
theoretical calculations correct.
Antonio
Gary L. Drescher
news:3B91EA20...@scattercreek.com...
> This might be true to solve a theoretical problem, but AIUI, the
> altimeter setting that you get from a local tower is not actually the
> barometric pressure with reference to true (MSL) altitude but an
> arbitrary setting that will cause your altimeter to read the local field
> elevation.
Right, the reported altimeter setting isn't the actual barometric pressure
at the field. Rather, it's what the sea-level barometric pressure would be,
assuming a standard lapse rate, given the actual barometric pressure at the
field. And that's the setting that will cause the altimeter to read the
local field elevation, since the altimeter's altitude calibration presumes a
standard lapse rate.
But Alex's question about density altitude didn't invoke a calculation based
on the reported altimeter setting. Rather, he referred to the
directly-reported density altitude of 1900'. So he was right that the plane
performs as though it were at 1900' under standard conditions for 1900'.
And you're right that to calculate the density altitude (which is really a
measure of density, which lift and drag depend on) if it weren't directly
reported, you'd take the pressure altitude (which is really a measure of
pressure, not altitude) and adjust for nonstandard temperature.
Regards,
Gary