Altimeters and air pressure variation

Sriram Narayan wrote:

That still wouldn't help since the pressure change for a 1000ft
change in
altitude at 18k would be smaller than at sea level. It would have to
have
some non-linear spring compensation as a function of absolute
pressure.

Yes but the non-linearity is very weak. I coded up the formula for the
US standard atmosphere (described below) and plotted it. See
http://www.burningserver.net/rosinsk...atmosphere.jpg
Perhaps the non-linearity is so weak that altimeters neglect it? I
don't know.

The formula for US standard atmosphere can be derived from the ideal
gas approximation (p = rho*R*T) and the hydrostatic approximation
(dp/dz = -rho*g). The final equation is:

z = T0/gamma * (1 - (p/p0)**(R*gamma/g)))

where R=287, T0 = 288K, p0 = 1013.25 mb = 29.92 in, gamma = 6.5 deg/km,
g = 9.8.

The formula assumes that temperature decreases linearly with altitude,
an assumption which becomes invalid above the tropopause. The equation
and its derivation can be found in Wallace & Hobbs, "Atmospheric
Science", pg. 60-61.

Jim Rosinski

Didn't see that, thanks for pointing it out.
At least that explains the note on my altimeter
saying "certified to 20000' " which I hadn't
understood before.

I wonder how altimeters for airliners work, given
the change that happens at 36152' - or indeed the
U2 altimeter. Must be some interesting stuff inside.

John

Peter wrote:
jharper aaatttt cisco dddooottt com wrote:

Dean Wilkinson wrote:


Visit this website and it will answer your questions about the
relationship
between pressure, temperature and altitude... altimeters are
designed to
take the non-linearity into account...

http://www.lerc.nasa.gov/WWW/K-12/airplane/atmosi.html


Nice site, thanks. But presumably there is some standard
atmospheric model that altimeters use? After all nobody actually
cares whether FL300 is really 30000' feet above MSL, as
long as everyone flying there is at the same altitude and,
more importantly, not at somebody else's FL290 or FL310.


Yes, there is a standard model and if you click on the first
link on the cited page you get to:
http://www.lerc.nasa.gov/WWW/K-12/airplane/atmos.html
which gives the equations describing that standard model.


Which implies that there must be some standard mechanical
way of making the translation?


There's a mathematically defined correspondence
between altitude and pressure under the standard
atmosphere assumption. But I doubt if the specific
mechanical means of achieving that correspondence is
specified anywhere. As long as the manufacturer makes
an instrument that is shown to give the right correspondence
to within a specified accuracy why should it matter exactly
how they do it?

I'll ask next time I visit my
avionics shop, but considering what each visit costs I
quite hope this won't be for a while.

"Sriram Narayan" wrote in message news:1105397045.6c0b9af7d0985bd99dd3e30aa7ae44ee@t eranews...



That still wouldn't help since the pressure change for a 1000ft change in
altitude at 18k would be smaller than at sea level. It would have to have
some non-linear spring compensation as a function of absolute pressure.



The USA is just now getting into the RVSM rules which allow vertical separation of 1000 feet. Used to be 2000' minimum
for all of these reasons....

On Mon, 10 Jan 2005 18:09:48 -0800, "jharper aaatttt cisco dddooottt
com" "jharper aaatttt cisco dddooottt com" wrote:

snip

I wonder how altimeters for airliners work, given
the change that happens at 36152' - or indeed the
U2 altimeter. Must be some interesting stuff inside.

http://www.rockwellcollins.com/ecat/...html?smenu=109

http://www.cas.honeywell.com/ats/products/airdata.cfm

Typically the DADC's are corrected for known issues/errors in the
pitot/static system of the specific type aircraft it is installed in,
as well as the "change"s you've noted.

TC

jharper aaatttt cisco dddooottt com opined

At sea level, the change in atmospheric pressure with altitude is
close to 1"Hg/1000'. Logically, this would mean that the air
pressure would drop to zero somewhere not much above 30000'. It
doesn't, because as the density drops the variation with
altitude also changes.

Which brings to mind the question, how does an altimeter deal
with this? As far as I know, it's just a simple aneroid barometer
with a bunch of linkages and gears to turn its expansion into
pointer movement.

My altimeter is marked "accurate to 20000' ". Is this why? Do
altimeters for higher altitudes have some kind of clever
mechanism to deal with the non-linearity of pressure at higher
altitudes.

I asked my acro instructor (10K+ hrs, airforce instructor pilot,
ex U2 pilot so should know a thing or two about high altitudes).
He explained the non-linearity of pressure to me but was
stumped on how this translates to the altimeter mechanism.

A couple of good approximations are

A = 25,000 * ln(30/25000)
and
P = 30 * exp(-A/25000)

For an altimeter, use gears with a varying radius.

-ash
Cthulhu in 2005!
Why wait for nature?

"jim rosinski" wrote in message
oups.com...
Sriram Narayan wrote:

That still wouldn't help since the pressure change for a 1000ft
change in
altitude at 18k would be smaller than at sea level. It would have to
have
some non-linear spring compensation as a function of absolute
pressure.

Yes but the non-linearity is very weak. I coded up the formula for the
US standard atmosphere (described below) and plotted it. See
http://www.burningserver.net/rosinsk...atmosphere.jpg
Perhaps the non-linearity is so weak that altimeters neglect it? I
don't know.

The formula for US standard atmosphere can be derived from the ideal
gas approximation (p = rho*R*T) and the hydrostatic approximation
(dp/dz = -rho*g). The final equation is:

z = T0/gamma * (1 - (p/p0)**(R*gamma/g)))

where R=287, T0 = 288K, p0 = 1013.25 mb = 29.92 in, gamma = 6.5 deg/km,
g = 9.8.

The formula assumes that temperature decreases linearly with altitude,
an assumption which becomes invalid above the tropopause. The equation
and its derivation can be found in Wallace & Hobbs, "Atmospheric
Science", pg. 60-61.

Jim Rosinski


It doesn't look that linear to me. I found a website with a similar graph
and it appears that at sea level and at 10000ft the slope of the curve is at
least 2x different. Your curve is quite a bit more linear (maybe 20%
increase in slope at 10k). There must some sort of mechanical compensation
involved otherwise altimeters would be off quite a bit even at 10k (even
with your curve). Isn't it something like 75ft accuracy requirement for
altimeters?

http://www.atmosphere.mpg.de/enid/16h.html

"jim rosinski" wrote in message
oups.com...
Sriram Narayan wrote:

That still wouldn't help since the pressure change for a 1000ft
change in
altitude at 18k would be smaller than at sea level. It would have to
have
some non-linear spring compensation as a function of absolute
pressure.

Yes but the non-linearity is very weak. I coded up the formula for the
US standard atmosphere (described below) and plotted it. See
http://www.burningserver.net/rosinsk...atmosphere.jpg
Perhaps the non-linearity is so weak that altimeters neglect it? I
don't know.

The formula for US standard atmosphere can be derived from the ideal
gas approximation (p = rho*R*T) and the hydrostatic approximation
(dp/dz = -rho*g). The final equation is:

z = T0/gamma * (1 - (p/p0)**(R*gamma/g)))

where R=287, T0 = 288K, p0 = 1013.25 mb = 29.92 in, gamma = 6.5 deg/km,
g = 9.8.

The formula assumes that temperature decreases linearly with altitude,
an assumption which becomes invalid above the tropopause. The equation
and its derivation can be found in Wallace & Hobbs, "Atmospheric
Science", pg. 60-61.

Jim Rosinski


It doesn't look that linear to me. I found a website with a similar graph
and it appears that at sea level and at 10000ft the slope of the curve is at
least 2x different. Your curve is quite a bit more linear (maybe 20%
increase in slope at 10k). There must some sort of mechanical compensation
involved otherwise altimeters would be off quite a bit even at 10k (even
with your curve). Isn't it something like 75ft accuracy requirement for
altimeters?

http://www.atmosphere.mpg.de/enid/16h.html

"jim rosinski" wrote in message
oups.com...
Sriram Narayan wrote:

That still wouldn't help since the pressure change for a 1000ft
change in
altitude at 18k would be smaller than at sea level. It would have to
have
some non-linear spring compensation as a function of absolute
pressure.

Yes but the non-linearity is very weak. I coded up the formula for the
US standard atmosphere (described below) and plotted it. See
http://www.burningserver.net/rosinsk...atmosphere.jpg
Perhaps the non-linearity is so weak that altimeters neglect it? I
don't know.

The formula for US standard atmosphere can be derived from the ideal
gas approximation (p = rho*R*T) and the hydrostatic approximation
(dp/dz = -rho*g). The final equation is:

z = T0/gamma * (1 - (p/p0)**(R*gamma/g)))

where R=287, T0 = 288K, p0 = 1013.25 mb = 29.92 in, gamma = 6.5 deg/km,
g = 9.8.

The formula assumes that temperature decreases linearly with altitude,
an assumption which becomes invalid above the tropopause. The equation
and its derivation can be found in Wallace & Hobbs, "Atmospheric
Science", pg. 60-61.

Jim Rosinski


It doesn't look that linear to me. I found a website with a similar graph
and it appears that at sea level and at 10000ft the slope of the curve is at
least 2x different. Your curve is quite a bit more linear (maybe 20%
increase in slope at 10k). There must some sort of mechanical compensation
involved otherwise altimeters would be off quite a bit even at 10k (even
with your curve). Isn't it something like 75ft accuracy requirement for
altimeters?

http://www.atmosphere.mpg.de/enid/16h.html

"Ash Wyllie" wrote in message
...
jharper aaatttt cisco dddooottt com opined

At sea level, the change in atmospheric pressure with altitude is
close to 1"Hg/1000'. Logically, this would mean that the air
pressure would drop to zero somewhere not much above 30000'. It
doesn't, because as the density drops the variation with
altitude also changes.

Which brings to mind the question, how does an altimeter deal
with this? As far as I know, it's just a simple aneroid barometer
with a bunch of linkages and gears to turn its expansion into
pointer movement.

My altimeter is marked "accurate to 20000' ". Is this why? Do
altimeters for higher altitudes have some kind of clever
mechanism to deal with the non-linearity of pressure at higher
altitudes.

I asked my acro instructor (10K+ hrs, airforce instructor pilot,
ex U2 pilot so should know a thing or two about high altitudes).
He explained the non-linearity of pressure to me but was
stumped on how this translates to the altimeter mechanism.

A couple of good approximations are

A = 25,000 * ln(30/25000)
and
P = 30 * exp(-A/25000)

For an altimeter, use gears with a varying radius.

Looks good! It is within 2.5% of the actual standard pressure up to 12000
ft. If you change the 25000 to 26000 in the 2nd formula it is within 1% for
the same range. The trick is to build a gear that conforms to this.

Sriram Narayan opined

"Ash Wyllie" wrote in message
...
jharper aaatttt cisco dddooottt com opined

At sea level, the change in atmospheric pressure with altitude is
close to 1"Hg/1000'. Logically, this would mean that the air
pressure would drop to zero somewhere not much above 30000'. It
doesn't, because as the density drops the variation with
altitude also changes.

Which brings to mind the question, how does an altimeter deal
with this? As far as I know, it's just a simple aneroid barometer
with a bunch of linkages and gears to turn its expansion into
pointer movement.

My altimeter is marked "accurate to 20000' ". Is this why? Do
altimeters for higher altitudes have some kind of clever
mechanism to deal with the non-linearity of pressure at higher
altitudes.

I asked my acro instructor (10K+ hrs, airforce instructor pilot,
ex U2 pilot so should know a thing or two about high altitudes).
He explained the non-linearity of pressure to me but was
stumped on how this translates to the altimeter mechanism.

A couple of good approximations are

A = 25,000 * ln(30/25000)
and
P = 30 * exp(-A/25000)

For an altimeter, use gears with a varying radius.

Looks good! It is within 2.5% of the actual standard pressure up to 12000
ft. If you change the 25000 to 26000 in the 2nd formula it is within 1% for
the same range. The trick is to build a gear that conforms to this.

Oops, a typo... A = 25,000 * ln(30/P)


-ash
Cthulhu in 2005!
Why wait for nature?