Quick question about altimeters & pressure
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Quick question about altimeters & pressure
Hi - first post and hope this is in the right place (loads of forums!!)
I was reading a book that says in the ISA, if the atmospheric pressure were to be 300mb/hectopascals, an altimeter would register a height of 30,000 feet.
I've looked into this and I calculate that the altimeter would read 21,397 feet (1013.25hpa - 300hpa = 713.25pa x 30 feet = 21,397)
Can anyone explain - I can't get my head round it.
Thank you,
Jake.
I was reading a book that says in the ISA, if the atmospheric pressure were to be 300mb/hectopascals, an altimeter would register a height of 30,000 feet.
I've looked into this and I calculate that the altimeter would read 21,397 feet (1013.25hpa - 300hpa = 713.25pa x 30 feet = 21,397)
Can anyone explain - I can't get my head round it.
Thank you,
Jake.
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The flaw with your calculation (although not your logic) is that the rate of pressure change with altitude is not uniform; it is around 27hPa at sea level for a few thousand feet and then the rate increases with altitude. 300 hPa and 30,000ft is part of the ISA standard, along with standard pressure at other altitudes (which I'm afraid I don't have to hand). Any pressures within these standards have to be extrapolated.
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Whirls
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Er ... think about it. The 30' is taught as being about right at and close to sea level as an engineering approximation - there's no particular reason to suppose that it's linear all the way up to vacuum, and plenty of reasons to suppose it isn't.
Here's one: if it's 30' per mb all the way up, and it's 1000mb at sea level, then you'll encounter vacuum at 30*1000 = 30,000 feet. Most of us have been for rides in airliners which fly at higher than 30,000 and still have sufficient air for the wings and engines.
(You could try working out the next level of approximation using A level physics, but I suspect you'd fail as you wouldn't be able to work out the temperature gradient without additional information. Personally I'm not going to try.)
Here's one: if it's 30' per mb all the way up, and it's 1000mb at sea level, then you'll encounter vacuum at 30*1000 = 30,000 feet. Most of us have been for rides in airliners which fly at higher than 30,000 and still have sufficient air for the wings and engines.
(You could try working out the next level of approximation using A level physics, but I suspect you'd fail as you wouldn't be able to work out the temperature gradient without additional information. Personally I'm not going to try.)
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Thanks for your help.
I guess the book kind of contradicts itself. It fails to mention the other standard pressure settings at different altitudes and also encourages the reader to assume the average change of pressure with height is 1mb per 30ft upto the tropopause whilst giving an answer that doesn't use the average change.
j-ake
I guess the book kind of contradicts itself. It fails to mention the other standard pressure settings at different altitudes and also encourages the reader to assume the average change of pressure with height is 1mb per 30ft upto the tropopause whilst giving an answer that doesn't use the average change.
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In all honesty, for PPL, that's all you need to know. At CPL/ATPL level, then you need to know the various ISA altitude levels and associated pressure rates of change. Trouble is, after a year, I've forgotten
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You could try working out the next level of approximation using A level physics, but I suspect you'd fail as you wouldn't be able to work out the temperature gradient without additional information.
g is constant at 9.80665 m/s^2 (ISA value)
Density as a function of pressure in the atmosphere can be obtained from [rho = P/R*T] where R = 287.1J/Kg/K (specific gas constant for dry air, ISA value) and T = temperature in Kelvin, obtainable as a function of lapse rate L and height [T = T0 - L * h]
So for example:
1) ISA change of height per mb at MSL:
dh/dP = (-1.225Kg/m^3 * 9.80665m/s^2)^-1
= -0.0832421398m/Pa
= -27.3104133ft/mb
2) ISA change of height per mb at the 300mb pressure level (30065ft=9163.8m[*] geopotential altitude)
T = 288.15K-0.0065K/m*9163.8m = 228.6K
rho = 30000Pa / (287.1 J/Kg/K * 228.6K)
= 0.457100647Kg/m^3
dh/dP = (-0.457100647Kg/m^3 * 9.80665m/s^2)^-1
= -0.22308352m/Pa
= -73.1901312ft/mb
[*] The geopotential altitude in metres for a given tropospheric pressure level in Pa can be obtained by the formula [(1.0 - (P/101325)^0.190263))*(288.15/0.0065)] which is derived by pulling one's hair while rearranging various ISA equations, or from an ISA tabulation.
All very simple stuff really (unless one goes and asks the gurus over at Tech Log)
Last edited by LH2; 8th Jan 2009 at 03:08. Reason: formatting
...along with standard pressure at other altitudes (which I'm afraid I don't have to hand).
400 mb = 24,000' 500 mb = 18,000 - I think.
It was only a guide, anyway, invariably the actual conditions encountered would be different, like headwind out, headwind home - that's Met. for you.
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It doesn't really matter though, as long as everyone's altimeter works in a similar fashion. If your altimeter indicates 30000 and another's indicates 31000 you will still miss each other. The altimeter is just displaying a difference in atmospheric pressure which we, conveniently, interpret as a vertical distance in feet.
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As others say, the pressure does not fall linearly with altitude.
The 30ft/mb is only a very crude approximation which is useful only at low levels, say up to a few thousand feet.
This diagram (ignore the two data lines for now) shows how altitude (shown in metres on this one) varies with millibar pressure.
The 30ft/mb is only a very crude approximation which is useful only at low levels, say up to a few thousand feet.
This diagram (ignore the two data lines for now) shows how altitude (shown in metres on this one) varies with millibar pressure.
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Thanks again - you have to wonder why in a PPL book they suddenly go from talking a few thousand feet to an absurd 30,000 without at least acknowledging there are other principles behind the higher altitude