Did you ask the AED people about this "feature"?
> So I have been fooling with RDAS data from many
> different sources. Over and over the question of
[quoted text clipped - 29 lines]
> Regards,
> Larry (All opinion. No warranties express or implied) C
Larry Curcio - 31 Mar 2007 14:34 GMT
Hi, John.
Actually, temperature correction is not peculiar
to RDAS. Most altimeters require it. It has always
been an issue with the FAA, and is more so now
that some planes use GPS and others use altimeters.
(I didn't read your email carefully, I've been at work
forever. You gave me a reference. Was this it?)
1)http://www.faa.gov/regulations_policies/rulemaking/historical_documents/1999/med
ia/ac91-xx.pdf
I did not think it news to anyone that temperature
correction is important. I was surprised that, in the
samples I tested, the base temperature appeared to
be 0 degrees C, and not 15 degrees. The most
controversial part of the whole thing is my value
of 128 for the displacement. It seems to work
better than other values I tried, and it is a round
number in binary.
Let me summarize a couple of other points:
1)Now that I think about it, I cannot characterize
all RDAS altimeters in exactly this way. Three
of my samples are from 2003. These were taken
at ~34 degrees C. I have another data set, the
vintage of which I cannot vouch for that is taken
at -15 degrees C. That dataset is not in raw form,
but is in intrpreted form. That is, the altitudes are
already computed. The R^2 statistic between
untransformed inertial altitude and barometric
altitude goes up from .94 to .9996 with the
conversion. The same sort of thing happened
with the other data sets, which were taken at
very different temperatures.
Perhaps later versions of the instrument do things
differently.
2) Check out the formula I suppled in my reply
to Dave Schultz. Note that since pressure is a
ratio, it may be represented in only relative
units. Most units, including RDAS, allow one
to download raw data. One normally adds/subtracts
a displacement from that (in my samples, it
seemed to be 128) and takes a ratio of each
individual reading to a ground reading. (I averaged the
negative time values.) Substitute that into the formula
using your actual base altitude temperature and you get
a calibrated altitude value. You don't need the value that
RDAS gives you, which is likely a (very very good)
polynomial approximation
Do a linear regression on the nominal values and you
can estimate the base temperature used by RDAS. Indeed,
the intercept is an estimate of the reference altitude.
(Problem is the regressor variable, whichever you choose,
has noise, so there is some bias. The intercept tends to be
bigger than it should be.)
Some units claim to be temperature corrected. The data have
seen with RDAS are not consistent with temperature correction,
IMHO - or temperature correction means something other
than the issue detailed here.
See my next (short) post.
Regards,
-Larry
> So I have been fooling with RDAS data from many
> different sources. Over and over the question of
[quoted text clipped - 13 lines]
> standard altimeter equations. These equations involve
> temperature.
In the pressure/altitude conversion equations I have seen the
temperature decreases linearly with altitude. At least at some altitudes.
http://www.grc.nasa.gov/WWW/K-12/airplane/atmos.html
So I am not sure how temperature correction helps. Besides, if I recall
my density altitude stuff correctly the pressure stays the same but the
density decreases. I have yet to find any reasonably authoritative work
that describes how to use a base pressure and temperature other than
that in the standard atmosphere model to calculate altitude.
Using a ratio of apogee pressure to pre-launch pressure in these
equations is an error in my opinion. Instead the two pressures should be
converted to altitude and the difference between the two altitudes used.
While I do not know how the RDAS GUI software performs the conversion,
the rdtools package uses:
log(pressure / 1015.76) / -0.000123
(Note: In this particular case the properties of the logarithm make
using the pressure ratio identical to using the difference in altitudes.
http://www.rocket.flyer.co.uk/rdtools/
Note that this log equation diverges significantly from the standard
atmosphere table at:
http://www.pdas.com/e2.htm
While the equation on the NASA page does not.
(288.14 - 288.08 * pow(P/101.29, 1/5.256)) / 0.00649
> The temperature that corresponds to the best fit with the RDAS
> nominal altitudes is 0 degrees C or 32 degrees F. If this is correct
[quoted text clipped - 11 lines]
> Regards,
> Larry (All opinion. No warranties express or implied) C

Signature
David W. Schultz
http://home.earthlink.net/~david.schultz
------
David Shatzer - the new Douglas Feith?
Larry Curcio - 31 Mar 2007 14:11 GMT
Hi, Dave.
> In the pressure/altitude conversion equations I have seen the
> temperature decreases linearly with altitude. At least at some altitudes.
>
> http://www.grc.nasa.gov/WWW/K-12/airplane/atmos.html
Yup. I have actually measured this on a takeoff
and landing of a jet liner. Standard rate is 6.5 Kelvins
per kilometer. I measured 6.38 over Beijing and 5.78
over Newark. Turns out such lapse rate differences
make for trivial errors in the formula.
FWIW, in the lower stratosphere, the lapse rate is zero.
It eventually reverses higher up, as ozone absorbs light
energy.
> So I am not sure how temperature correction helps.
Altimeter formulas are based on two principles:
the first law of thermodynamics and hydrostatic
equilibrium.
Hydrostatic equilibrium says that at any altitude,
the the pressure on a given horizontal area of air
is balanced by the weight of the column of air
above it. That weight is affected by the temperature
of the air as the temperature lapses. It works out to
Const HC = 0.03418155
Const LapseRate = 0.0065
BaseTemp = BaseTempC + ZeroC
Ratio = PressRead / MeanGroundPressure
Altitude = (BaseTemp / LapseRate) * (1 - (Ratio ^ (LapseRate/HC))
Note that the whole thing is proportional to base temperature.
>Besides, if I recall my density altitude stuff correctly the pressure stays
>the same but the density decreases. I have yet to find any reasonably
[quoted text clipped - 5 lines]
> equations is an error in my opinion. Instead the two pressures should be
> converted to altitude and the difference between the two altitudes used.
Well, this stuff isn't exactly controversial. :-)
I'm a little behind (OK. Some would say I'm a big behind...)
on this thread as I have production issues at work and have
to go in today, Saturday. Will reply to John D. and fill in more there.
The issue is involved.
Good to hear from you.
-Larry Curcio
and summarize to
I have a long treatise on rockets, and within
it is a shorter long treatise on altimeter
formulas, where this stuff is derived. I can
excise this and email it to anyone who wants
it. It is in MSWord. Maybe I can turn it into
a PDF.
It covers this topic in excruciating detail.
It may be a couple of days. Busy busy.