Corsair2 From United States of America, joined Jan 2001, 248 posts, RR: 0 Posted (8 years 6 months 2 days 8 hours ago) and read 3716 times:
Can someone offer a really good explanation of the difference between pressure altitude and density altitude?
As an example for discussion (using a place not at sea level), suppose I was at Telluride, Colorado (TEX) which has an altimeter setting today of 29.87 and an airport elevation of 9078 ft. I think the pressure altitude is:
29.92-29.87=0.05
0.05*1000= 50 ft
50 ft + 9078 ft = 9128 ft
If this is correct, where does density alt. come into play?
"We have clearance Clarence. Roger, Roger. What's our vector Victor?"
SlamClick From United States of America, joined Nov 2003, 10062 posts, RR: 71 Reply 1, posted (8 years 6 months 2 days 8 hours ago) and read 3705 times:
Your PA calculation looks right on to me.
Density altitude factors in the temperature, and specifically the temperature deviation from standard which is 15o C or 59oF at sea level and the standard adiabatic lapse rate of 2o C per thousand feet of elevation. The standard temperature for TEX should be something like -3o C so if it is warmer than that, the DA is higher than airport elevation, if it is colder than that the DA is lower.
I used to have a formula for this. I found that I never needed it, however, as most larger aircraft have performance charts that find it for themselves. That is, you might enter the chart at your pressure altitude, go across to the temperature and read your solution.
Happiness is not seeing another trite Ste. Maarten photo all week long.
Timz From United States of America, joined Sep 1999, 6468 posts, RR: 8 Reply 2, posted (8 years 6 months 2 days 1 hour ago) and read 3638 times:
Calculating them might be tough, but what they are is simple enough. There's something called the International Standard Atmosphere, a model that specifies what the temperature, pressure and density are at any given altitude on a "Standard Day". Note that they're not independent; if the temperature is 15 C and the pressure is 1013.25 millibars (the Standard sea-level values) then the density has to be Standard too.
So, density altitude is simply the altitude that (on a Standard Day) has the density that you're experiencing at the moment.
If you google International Standard Atmosphere you can find calculators that give the ISA values for any altitude.
Woodreau From United States of America, joined Sep 2001, 890 posts, RR: 7 Reply 3, posted (8 years 6 months 1 day 23 hours ago) and read 3623 times:
Density altitude is simply pressure altitude corrected for non standard temperature.
To figure density altitude, in your scenario, the information you are missing is the temperature at Telluride.
Many pilots that are not accustomed to flying out of high altitude airports make the mistake of thinking that it's only 70 degrees at an airport at 6500ft so that's not too bad and don't bother with figuring out airplane performance. but standard temperature at 6500ft is somewhere around 38F so you are already about 32 degrees above standard temperature and your aircraft performance is much degraded.
Good judgement comes from experience. Experience comes from surviving bad judgement.
Bellerophon From United Kingdom, joined May 2002, 574 posts, RR: 60 Reply 5, posted (8 years 6 months 1 day 21 hours ago) and read 3609 times:
Corsair2
The answers given above by various posters are correct, density altitude (DA) is pressure altitude (PA) corrected for non-standard temperature.
…where does density alt. come into play?...
Well, even with the altimeter setting steady on 29.87” hg, Telluride can still experience a wide variety of outside air temperatures, and these will have an effect on your aircraft’s performance that must be accounted for.
An easy rule of thumb to calculate DA from PA is this:
Add/subtract 1,000ft for each 8°C variation that the actual OAT is above/below ISA.
In order to calculate the density altitude at Telluride, we need to know the OAT, and, as you do not quote an OAT, let us assume that this is -11°C in Winter and +29°C in Summer.
At an elevation of 9,078ft, the ISA temperature at Telluride should be -3°C, and you have already calculated correctly that at an altimeter setting of 29.87” hg, the PA is 9,128ft.
In the Winter case, the temperature deviation is -8°C below ISA, so we should subtract 1,000ft from the PA of 9,128ft, to give a DA of 8,128ft.
In the Summer case, the temperature deviation is +32°C above ISA, so we should add 4,000ft to the PA of 9,128ft, to give a DA of 13,128ft.
A variation in DA of 5,000ft, at the same airfield, at the same altimeter setting, just because the temperature changed! You can see why pilots of all aircraft types need to exercise great care when operating out of hot and high airfields.
Corsair2 From United States of America, joined Jan 2001, 248 posts, RR: 0 Reply 7, posted (8 years 6 months 1 day 8 hours ago) and read 3535 times:
Thanks for all the good posts. From using the density altitude calculator I figured out that a summer day of 68 degrees in Telluride would give a density altitude of over 12000 ft!
"We have clearance Clarence. Roger, Roger. What's our vector Victor?"
SlamClick From United States of America, joined Nov 2003, 10062 posts, RR: 71 Reply 9, posted (8 years 6 months 23 hours ago) and read 3504 times:
I saw that relative humidity factor in the calculator and I've heard other people whom I respect talk about it, but I'd like to know exactly how humidity affects DA. Not "why" as I understand that humid air is less dense, but, numerically - how?
I've written manuals for, flown, instructed on, or consulted in flight ops on A-320, A-330, B-727, B-737, B-757, B-767, BAe-146, DC-3, DC-9 and MD-80 as well as turboprops at half a dozen airlines and I have never once seen a performance chart that had an entry for humidity.
I believe strongly in getting to know the performance charts that are available for your airplane. I once taught an ATP written exam prep course and had two students in this class. One was a fifty-something businessman who owned a C-421 and the other student was a U-2 pilot.
Normally you kind of teach to the middle of the class. In this case there was no middle! The businessman had never once in twenty years with a license so much as looked at the performance charts for one of his airplanes. The U-2 pilot was an instructor pilot in his plane and lived (or died) with those charts. Fortunately he was sharp enough to get what I was teaching him, and still have time and neurons left over to help the other student or we'd have never gotten through the subject.
Still I would bet that the businessman never worked up the nerve to take the test. He probably just continues to stuff his plane full of people and bags and skis and head for the high elevation airports around the west.
Happiness is not seeing another trite Ste. Maarten photo all week long.
Flyf15 From , joined Dec 1969, posts, RR: Reply 10, posted (8 years 6 months 22 hours ago) and read 3496 times:
SlamClick, you're looking at a roughly 500-1000ft increase in density altitude when going from perfectly dry air to saturated air when air temp is around 100 degrees F. When it is around 30 degrees F, the increase is less than 100ft (as there is obviously much less moisture that the air can hold).
So, I'd say that humidity varies density altitude on the same order as pressure variances do. Why don't we compute it when figuring density altitude? Simply...its hard. As you can see from above, density altitude variations not only depend on the humidity, but also the air temperature. Doing it based off of dewpoint would not be as bad, but still more complicated and more intense than makers of density altitude charts care to do.
Timz From United States of America, joined Sep 1999, 6468 posts, RR: 8 Reply 11, posted (8 years 5 months 4 weeks 1 day ago) and read 3453 times:
I'm no expert, but you'd think we could get a pretty good idea of humid air density from a steam table.
At 15 Celsius, water's vapor pressure is 17.1 millibars (says the steam table)-- so at 100% humidity at 15 C, the air pressure is 17.1 millibars less than the total pressure, and at 50% humidity it's 8.55 millibars less-- right? Water vapor is said to be 62.2% as dense as air, so seems like we should be able to work it out.
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