Altimeters and air pressure variation

"jharper aaatttt cisco dddooottt com" "jharper aaatttt cisco dddooottt
com" wrote in message news:1105391055.635118@sj-nntpcache-3...
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.

Anyone know?



The following is a WAG but that could be the reason that in the Flight
Levels, above 18k feet everyone sets thier altimeter to 29.90.

"Gig Giacona" wrote in message
...

"jharper aaatttt cisco dddooottt com" "jharper aaatttt cisco dddooottt
com" wrote in message news:1105391055.635118@sj-nntpcache-3...

Anyone know?



The following is a WAG but that could be the reason that in the Flight
Levels, above 18k feet everyone sets thier altimeter to 29.90.



Hopefully, 29.92.

"Icebound" wrote in message
...

"Gig Giacona" wrote in message
...

"jharper aaatttt cisco dddooottt com" "jharper aaatttt cisco dddooottt
com" wrote in message news:1105391055.635118@sj-nntpcache-3...

Anyone know?



The following is a WAG but that could be the reason that in the Flight
Levels, above 18k feet everyone sets thier altimeter to 29.90.



Hopefully, 29.92.



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.

It would have to have
some non-linear spring compensation as a function of absolute pressure.

It could also be a non-linear gear compensation, such as using
non-circular gears. Lotsaways it =could= be done, but I don't know
how it =is= done.

Jose
--
Money: What you need when you run out of brains.
for Email, make the obvious change in the address.

I couldn't be done easily with a spring. Springs are very linear except in
very special cases. I think it's done with leaver arms that change
effective length with displacement.

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

"Icebound" wrote in message
...

"Gig Giacona" wrote in message
...

"jharper aaatttt cisco dddooottt com" "jharper aaatttt cisco dddooottt
com" wrote in message news:1105391055.635118@sj-nntpcache-3...

Anyone know?



The following is a WAG but that could be the reason that in the Flight
Levels, above 18k feet everyone sets thier altimeter to 29.90.



Hopefully, 29.92.



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.

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

"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

Sriram Narayan wrote:

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

His graph has more curvature than mine does because his covers a much
greater vertical distance (11000 meters is around 36000 feet). Mine
only goes to 18000 feet. But I agree that a linear assumption will
result in a worst-case error of several hundred feet. Not good enough,
so the nonlinearity must be built into altimeters even if the're only
good to 20000 ft.

Jim Rosinski

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

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

His graph has more curvature than mine does because his covers a much
greater vertical distance (11000 meters is around 36000 feet). Mine
only goes to 18000 feet. But I agree that a linear assumption will
result in a worst-case error of several hundred feet. Not good enough,
so the nonlinearity must be built into altimeters even if the're only
good to 20000 ft.


Oops, thanks for pointing that out. Didn't check the Y-scale carefully.

"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