Student needs assistance density altitude calcs

One easy way is to find out the temp at the field you are at. Take the
standard sea level temp of 15C and using the altitude of where you are
at, subtract 2C for every thousand feet. For every 1 degree above
standard, add 120 feet, or 600 feet for every 5C. Add that to the
field altitude. Gives you a great ballpark idea in no time.

Example: 3,000 foot field at 20C
15C - (3 x 2) = 9C (should be standard temp for that altitude)
20C - 9C = 11
11 x 120 = 1320
3000 + 1320 = 4320 density altitude



On Tue, 15 Jun 2004 18:21:34 -0700, The OTHER Kevin in San Diego
skiddz *AT* adelphia *DOT* net wrote:

Anyone got a "formula" for figuring out pressure and density altitudes
quickly and reasonably accurately, specifically if the knowns are

Field Elevation
Barometric Pressure in inches of mercury
Temperature in degress Celsius.

My instructor went over this for quite a while in ground school last
night, but went about it in such a way we all pretty much got lost
halfway through it. I thought perusing my 3 pages of notes would shed
some light on it, but my results aren't "meshing" with the D.A charts
in my P.O H - they're quite a bit off as a matter of fact.

I've called 3 of my fellow classmates and they're just as confused as
I am. A nice easy way to figure it out would be appreciated and maybe
a condensed explanation as well. I've got a pile of textbooks to
peruse for more details..

Thx!

One easy way is to find out the temp at the field you are at. Take the
standard sea level temp of 15C and using the altitude of where you are
at, subtract 2C for every thousand feet to get the standard temp. For
every 1 degree above standard, add 120 feet, or 600 feet for every 5C.
Add that to the field altitude. Gives you a great ballpark idea in no
time.

Example: 3,000 foot field at 20C
15C - (3 x 2) = 9C (should be standard temp for that altitude)
20C (your original temp) - 9C (what standard should be) = 11
(degrees above standard)
11 x 120' = 1320'
3000 + 1320 = 4320 density altitude



On Tue, 15 Jun 2004 18:21:34 -0700, The OTHER Kevin in San Diego
skiddz *AT* adelphia *DOT* net wrote:

Anyone got a "formula" for figuring out pressure and density altitudes
quickly and reasonably accurately, specifically if the knowns are

Field Elevation
Barometric Pressure in inches of mercury
Temperature in degress Celsius.

My instructor went over this for quite a while in ground school last
night, but went about it in such a way we all pretty much got lost
halfway through it. I thought perusing my 3 pages of notes would shed
some light on it, but my results aren't "meshing" with the D.A charts
in my P.O H - they're quite a bit off as a matter of fact.

I've called 3 of my fellow classmates and they're just as confused as
I am. A nice easy way to figure it out would be appreciated and maybe
a condensed explanation as well. I've got a pile of textbooks to
peruse for more details..

Thx!

Shouldn't the temperature correction be applied to the pressure altitude of
the field? e.g. set your altimeter to 29.92 inches and read the press.
altitude and then correct for non std. temp.??
Stu Fields Safari driver
hellothere.adelphia.net wrote in message
...
One easy way is to find out the temp at the field you are at. Take the
standard sea level temp of 15C and using the altitude of where you are
at, subtract 2C for every thousand feet to get the standard temp. For
every 1 degree above standard, add 120 feet, or 600 feet for every 5C.
Add that to the field altitude. Gives you a great ballpark idea in no
time.

Example: 3,000 foot field at 20C
15C - (3 x 2) = 9C (should be standard temp for that altitude)
20C (your original temp) - 9C (what standard should be) = 11
(degrees above standard)
11 x 120' = 1320'
3000 + 1320 = 4320 density altitude



On Tue, 15 Jun 2004 18:21:34 -0700, The OTHER Kevin in San Diego
skiddz *AT* adelphia *DOT* net wrote:

Anyone got a "formula" for figuring out pressure and density altitudes
quickly and reasonably accurately, specifically if the knowns are

Field Elevation
Barometric Pressure in inches of mercury
Temperature in degress Celsius.

My instructor went over this for quite a while in ground school last
night, but went about it in such a way we all pretty much got lost
halfway through it. I thought perusing my 3 pages of notes would shed
some light on it, but my results aren't "meshing" with the D.A charts
in my P.O H - they're quite a bit off as a matter of fact.

I've called 3 of my fellow classmates and they're just as confused as
I am. A nice easy way to figure it out would be appreciated and maybe
a condensed explanation as well. I've got a pile of textbooks to
peruse for more details..

Thx!

Yes, but this is a quick calculation without fiddling with the
altimeter setting. I would bet that against the chart it could be off
by 50-100'. You are only talking a small amount by not using pressure
altitude. This gives you a good idea.

When working high altitude stuff, you should know approximately what
your ship can or can not do at a set altitude (OGE/IGE). This gives
you quick info while flying for that. One tool.


On Tue, 17 Aug 2004 10:11:00 -0700, "Kathryn & Stuart Fields"
wrote:

Shouldn't the temperature correction be applied to the pressure altitude of
the field? e.g. set your altimeter to 29.92 inches and read the press.
altitude and then correct for non std. temp.??
Stu Fields Safari driver
hellothere.adelphia.net wrote in message
.. .
One easy way is to find out the temp at the field you are at. Take the
standard sea level temp of 15C and using the altitude of where you are
at, subtract 2C for every thousand feet to get the standard temp. For
every 1 degree above standard, add 120 feet, or 600 feet for every 5C.
Add that to the field altitude. Gives you a great ballpark idea in no
time.

Example: 3,000 foot field at 20C
15C - (3 x 2) = 9C (should be standard temp for that altitude)
20C (your original temp) - 9C (what standard should be) = 11
(degrees above standard)
11 x 120' = 1320'
3000 + 1320 = 4320 density altitude



On Tue, 15 Jun 2004 18:21:34 -0700, The OTHER Kevin in San Diego
skiddz *AT* adelphia *DOT* net wrote:

Anyone got a "formula" for figuring out pressure and density altitudes
quickly and reasonably accurately, specifically if the knowns are

Field Elevation
Barometric Pressure in inches of mercury
Temperature in degress Celsius.

My instructor went over this for quite a while in ground school last
night, but went about it in such a way we all pretty much got lost
halfway through it. I thought perusing my 3 pages of notes would shed
some light on it, but my results aren't "meshing" with the D.A charts
in my P.O H - they're quite a bit off as a matter of fact.

I've called 3 of my fellow classmates and they're just as confused as
I am. A nice easy way to figure it out would be appreciated and maybe
a condensed explanation as well. I've got a pile of textbooks to
peruse for more details..

Thx!

Assuming hellothere's statements are correct, a simplified version is:

Density Altitude = ((Fe / 500) + Tc + 10) X 120

Where Fe = Feild Elevation and Tc = Field Temp in Celcius

Dennis.



hellothere.adelphia.net wrote:

One easy way is to find out the temp at the field you are at. Take the
standard sea level temp of 15C and using the altitude of where you are
at, subtract 2C for every thousand feet. For every 1 degree above
standard, add 120 feet, or 600 feet for every 5C. Add that to the
field altitude. Gives you a great ballpark idea in no time.

Example: 3,000 foot field at 20C
15C - (3 x 2) = 9C (should be standard temp for that altitude)
20C - 9C = 11
11 x 120 = 1320
3000 + 1320 = 4320 density altitude




Dennis Hawkins
n4mwd AT amsat DOT org (humans know what to do)

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