Atmospheric Model Deviations

[perpetual]

Hiya everyone, just a question maybe someone can help me out.

Sitting here thinking, and brushing up, I've come up with a question.

It's a question about determining atmospheric deviation between the two atmospheric models Jet (JSA) and (ISA) anywhere from FL350 and upward

With givens:

Standard jet atmospheric conditions being = -65C @ FL430
ISA conditions being = -56.5 @ FL430
MSL = +15C for both JSA and ISA (Standard ambient temperature base that we start from etc)
(Isothermic conditions for ISA from FL350 ~ FL650) approx

Given the understanding that a measured temperature decrease is non-existant*[(see note)] above 36,090 Feet for (ISA)

and using lapse rates of:

1.98 / 1000 feet decrease for (ISA)
and
2 / 1000 feet decrease for (JSA)


Would I be correct in showing that:

Temp C for JSA @ FL430 = 43 (-2) = -86 +(+15) = -71C @ FL430
Temp C for ISA @ FL430 = 43 (-1.98) = -85.14 + (+15) = -70.14 + (+56.5) (Isothermal from 36,090 ~ 65,600) = 13.64

And that a true deviation of -71C + (+13.64) = -57.36 + (+56.5) = .86C

Does it sound logical to say that the actual true ambient temperature would be -57.36 and not -56.5?

*Note: appoximately Isothermic from FL350 ~ FL650 with .3C decrease thereafter for ISA

I apologize in advance if wasn't particularly clear.

-Perpetual

[TheFlyingTurkey]

Dude...
Watch TV or something, you've got too much time on your hands.

[Minuteman]

Originally Posted by perpetual

Temp C for JSA @ FL430 = 43 (-2) = -86 +(+15) = -71C @ FL430
Temp C for ISA @ FL430 = 43 (-1.98) = -85.14 + (+15) = -70.14 + (+56.5) (Isothermal from 36,090 ~ 65,600) = 13.64

And that a true deviation of -71C + (+13.64) = -57.36 + (+56.5) = .86C

Does it sound logical to say that the actual true ambient temperature would be -57.36 and not -56.5?

Wow, I've never even heard of the JSA. Do they not believe in the tropopause?

I don't quite understand where "-70.14 + (+56.5) = [-]13.64" comes from (i.e., I don't understand why you calculated a temperature from a lapse rate and then added something to it). Temperature as a function of altitude in the ISA is not a continuous function and you can get away with saying that the ambient temp above 36,090 feet is going to be -56.6°C for as high as jet transports fly.

Are you trying to figure out the difference in temperatures between the JSA and ISA models?

(TV is for morans, let us nerds have our fun. Maybe we'll save you a pound of fuel or something. )

[perpetual]

Hiya Minuteman

I know it may seem frivolous to others and just plain boring, but sure, it’s something that made me think for a second.

The Jet Standard Atmosphere is basically the exact same thing as the ISA with exception to the lapse rate and in addition that the lapse rate continues through the isothermal ISA tropopause layer giving a continued (measured) decrease in temperature upwards. So in other words, while temperature of the given ISA value of -56.5 exists from FL360 upwards to FL650 using ISA, the JSA value is measured from MSL including a continual decrease in temperature even through the tropopause. ( -2C per /1000 foot decrease ).

With that, I simply wondered if I could then determine the actual true deviation between the two models (ISA and JSA) for comparative flight levels.

So by using the JSA established rate of -2C per flight level starting at MSL upwards to FL430 I came up with.

Temp C for JSA @ FL430 = 43 (-2) = -86

Adding the standard temperature of the MSL datum ( positive 15C) to the -86 gave me:

+ (+15) = -71C @ FL430

So with that we established the temperature at FL430 with JSA lapse rates and using the MSL standard temps.

Temp C for JSA @ FL430 = 43 (-2) = -86 + (+15C) = -71C

From there, considering that the established ISA temperature of -56.5 exists at all ISA flight levels from FL360 up to FL650, I wanted to find the JSA deviation between the two.

JSA @ FL430 = -71C
ISA @ FL430 = -56.5
___________________________________
Difference = 14.5C

The mathematical difference of the -71C and the -56.5 was not considered to be in terms of a negative plus a positive, but more so in the sense of simple pure “difference” in place value.

But what would be the true difference if the ISA lapse rate of 1.98 per 1000 Feet decrease was used? (Even throughout the tropopause)

Using:

Temp C for ISA @ FL430 = 43 (-1.98) = -85.14 + (+15) = -70.14C in terms of pure lapse rate (ignoring isothermal conditions)

Now using the ISA standard of (isothermal -56.5C) and (1.98 per 1000 Feet (( -70.14,)) we get a difference of 13.64

From that point the reason (( and I could be completely flawed here so please bare with me )) that I chose to do:

-70.14 + (+56.5) (normal ISA isothermal temp between the tropopause and FL650 etc )

was to find the difference between the max isothermal value of ISA temperatures for the given max ISA flight level (ISA 56.5 @ FL650) in comparison to the calculated decrease by using JSA -2C per 1000 @ ISA/JSA FL430.

Again, the mathematical difference of the -70.14 and the + 56.5 was not considered to be in terms of a negative plus a positive, but more so in the sense of simple pure “difference” between the two.

Put as simply as possible, below is a quick summary of what I have come up with. (( I could be entirely wrong, so if someone can point me in the right direction please feel free ))

43 (-2) = -86.00C
43 (-1.98) = -85.14C (ignoring isothermal values just as JSA)
__________________________
Pure Difference = 0.86C

(-0.86C) + (-56.5C) = -57.36C

So therefore, would the approximate true ambient temperature for ISA @ FL430 be -57.36C and not the said -56.5C (isothermal value) ?


I hope this is a better explanation as to what I am trying to express.

Ps. I love flying turkeys
PPS. And MorAns do watch TV

[MDPilot]

In over 30 years of military and civilian flying, a BS in Aerospace Engineering and a Master's in Aeronautical Science, I've never heard of JSA. Give me a reference to this concept.

[perpetual]

Originally Posted by MDPilot

In over 30 years of military and civilian flying, a BS in Aerospace Engineering and a Master's in Aeronautical Science, I've never heard of JSA. Give me a reference to this concept.

Hiya MDPilot

The metorological information that I am looking at is mostly from JAA publications and is based on a measurement that existed prior to the establishment of the (ISA). From what I can see is that there are quite a few "different" atmospheric model (standards) that are used for specific engineering or testing. In addition these concepts are still being taught for ATPL level theory during ground school.

I have included an example of an ATPL test question below, involving the use of differing (JSA vs ISA) atmospheric modeling along with its answer.

__________________________________________________ ____________
Question:

1.) At FL430 the temperature deviation is given as ISA +3. Express this temperature as a deviation from the Jet Standard Atmosphere.
__________________________________________________ ____________
Given answer:

ISA Temperature (43,000 ft) = - 56.5°C
Deviation from ISA = +3°C
Ambient temperature (43,000 ft) = -53.5°C
JSA temperature (43,000 ft ) = +15°C - (43 x 2°C)
= +15°C - 86°C
= -71°C
JSA temperature deviation = +17.5°C
__________________________________________________ ______________

As you can see this is one of the questions that started me thinking about how to find a particular ambient temp for a related ISA FL in the isothermal region.

Below are some links to either records, publications, or engineering data for its' use.

http://www.naples-air-center.com/pdf...k%20sample.pdf

There is another atmosphere that became popular in the 1960's, that is the Jet Standard Atmosphere. In this atmospheric model, the conditions are the same with the exception that there is no tropopause, the lapse rate being a flat 2°C per 1000ft until the stratosphere.

Here's another that speaks of a non-tropopause containing atmosphere (to an extent )

http://www.ldeo.columbia.edu/~nili/HL1998.pdf

The Charney model has no tropopause and
has nonzero PV gradients in the troposphere.

I know the topic is completely theoretical at best at this point, it was simply the question of needing to find the deviation between each model that sparked my interest further.

Anyway, based on what I wrote the first time, am I at least in the ballpark so to say as far as trying to determine the actual ambient for the given flight level?

And in closing, if this thread gets too rediculous because of it's possible pointlessness, I'll just let my interest for it die

-Perpetual

[TonyC]

I've never heard of JSA before, but a brief session with Google yielded this jewel: Wiljam Flight Training Although I have done planning based on an assumed lapse rate of 2°C per 1000', I have never heard this referred to as the JSA. (Perhaps it was just a useful approximation, or "Rule of Thumb"??)

(OK, I had to Google "Jet Standard Atmosphere" since JSA didn't get me anywhere. )

Originally Posted by perpetual

Hiya everyone, just a question maybe someone can help me out.

Well, I just learned about JSA, but I'm pretty good with math - no, that didn't sound right - - I like to play with math, so I'll give it a whirl.

Originally Posted by perpetual

Sitting here thinking, and brushing up, I've come up with a question.

It's a question about determining atmospheric deviation between the two atmospheric models Jet (JSA) and (ISA) anywhere from FL350 and upward

With givens:

Standard jet atmospheric conditions being = -65C @ FL430

What does this mean, "Standard jet atmospheric conditions"? ISA and JSA are atmospheric models, and have no direct bearing on actual conditions. Models are useful for laboratory study, and for comparing actual conditions to a theoretical model. Sometimes, they're even useful for predicting conditions at one altitude when conditions are known at another. Don't confuse actual conditions with ISA or JSA.

So, on this first point, you have me confused.

Originally Posted by perpetual

ISA conditions being = -56.5 @ FL430

This is a true statement, minus the reference to "conditions". The ISA model temperature at FL430 is -56.5°C.

Originally Posted by perpetual

MSL = +15C for both JSA and ISA (Standard ambient temperature base that we start from etc)

Yes, the ISA model temperature and the JSA model temperature at Mean Sea Level are 15.0°C

Originally Posted by perpetual

(Isothermic conditions for ISA from FL350 ~ FL650) approx

36,090' to 65,617' (or 11 km to 20 km) - again, get rid of that word "conditions"!!

Originally Posted by perpetual

Given the understanding that a measured temperature decrease is non-existant*[(see note)] above 36,090 Feet for (ISA)

...until you get to 65,617 ft, where the temperature rises with height at the rate of 0.3°C per 1,000 ft

Originally Posted by perpetual

and using lapse rates of:

1.98 / 1000 feet decrease for (ISA)
and
2 / 1000 feet decrease for (JSA)


Would I be correct in showing that:

Temp C for JSA @ FL430 = 43 (-2) = -86 +(+15) = -71C @ FL430

Yes...

Originally Posted by perpetual

Temp C for ISA @ FL430 = 43 (-1.98) = -85.14 + (+15) = -70.14 + (+56.5) (Isothermal from 36,090 ~ 65,600) = 13.64

NO! NO! The ISA model states that the temperature between 36,090 and 65,600 remains the same (didn't you just say that?!?!) => -56.5°C

What's the temperature of the ISA model at FL 370? -56.5°C No calculation required.
What's the temperature of the ISA model at FL 380? -56.5°C
What's the temperature of the ISA model at FL 390? -56.5°C
What's the temperature of the ISA model at FL 400? -56.5°C
What's the temperature of the ISA model at FL 450? -56.5°C
What's the temperature of the ISA model at FL 500? -56.5°C
What's the temperature of the ISA model at FL 650? -56.5°C


Now, what's the temperature of the ISA model at 36,090 ft?

ISA Temp = +15.0°C - (1.98 * Altitude (in thousands of feet))
ISA Temp = +15.0°C - (1.98 * 36.09)
ISA Temp = +15.0°C - 71.4582
ISA Temp = -56.4582°C ========> that rounds to -56.5°C !!!! Amazing !!!!



Short version is, no calculation is required to determine the temperature of the ISA model at any altitude between 36,090 ft and 65,600 ft.

Originally Posted by perpetual

And that a true deviation of -71C + (+13.64) = -57.36 + (+56.5) = .86C

No. The -71°C is the correct JSA value, the +13.64 is the number you derived from an incorrect application of the ISA gradient, and +56.5 is the inverse of the correct ISA value.

Originally Posted by perpetual

Does it sound logical to say that the actual true ambient temperature would be -57.36 and not -56.5?

NO, for the reasons stated above.

Originally Posted by perpetual

*Note: appoximately Isothermic from FL350 ~ FL650 with .3C decrease thereafter for ISA

I apologize in advance if wasn't particularly clear.

-Perpetual

Well, I think I muddled through anyhow.

You want to determine the DIFFERENCE between JSA and ISA for a given altitude. Am I correct in that assessment?


If so, you will need 3 different equations, and will apply the correct equation for the altitude in question. Since the lapse rates are different in three bands of the atmosphere in the ISA model, a separate equation will be needed for each band.

Difference implies a subtraction operation. The difference between the JSA value and the ISA value for a given altitude is the JSA value MINUS the ISA value.




DIFFERENCE = JSA Temp - ISA Temp, where:

JSA Temp = +15.0°C - (2.0°C x ALTITUDE) altitude expressed in thousands of feet above MSL



ISA TEMP =

· Sea Level to 36,090 FT: = +15.0°C - (1.98°C x ALTITUDE)
· 36,090FT to 65,600 FT: = -56.5°C
· 65,600 FT and above: = -56.5°C + (0.3°C x (ALTITUDE - 65.6))

DIFFERENCE (Sea Level to 36,090 FT)

= JSA Temp - ISA Temp
= (+15.0°C - (2.0°C x ALT)) - (+15.0°C - (1.98°C x ALT))
= +15.0°C - (2.0°C x ALT) - +15.0°C + (1.98°C x ALT)
= (-2.0°C + 1.98°C) x ALT
= -0.02°C x ALT

DIFFERENCE (36,090 FT to 65,600 FT)

= JSA Temp - ISA Temp
= (+15.0°C - (2.0°C x ALT)) - (-56.5°C)
= +15.0°C - (2.0°C x ALT) + 56.5°C
= +71.5°C - (2.0°C x ALT)

DIFFERENCE (65,600 Ft and above)

= JSA Temp - ISA Temp
= (+15.0°C - (2.0°C x ALT)) - (-56.5°C + (0.3°C x (ALT - 65.6)))
= +15.0°C - (2.0°C x ALT) + 56.5°C - (0.3°C x (ALT - 65.6))
= +15.0°C - (2.0°C x ALT) + 56.5°C - (0.3°C x ALT) + (0.3°C x 65.6)
= +15.0°C - (2.0°C x ALT) + 56.5°C - (0.3°C x ALT) + 19.68°C
= +91.18°C - (2.3°C x ALT)

These three equations yield the temperature difference between the JSA atmospheric model and the ISA atmospheric model for a given altitude. Note that for the two transition altitudes, either equation works. In other words, for 36,090 feet, either the first or the second equation can be used, as they yield the same answer. For 65,600 feet, either the second or the third equation can be used.



Now, let me digest your follow-up posts before I compose a response.

[TonyC]

Originally Posted by perpetual

With that, I simply wondered if I could then determine the actual true deviation between the two models (ISA and JSA) for comparative flight levels.

Yes, you can. See above post.

Originally Posted by perpetual

...

So by using the JSA established rate of -2C per flight level starting at MSL upwards to FL430 I came up with.

Temp C for JSA @ FL430 = 43 (-2) = -86

Adding the standard temperature of the MSL datum ( positive 15C) to the -86 gave me:

+ (+15) = -71C @ FL430

So with that we established the temperature at FL430 with JSA lapse rates and using the MSL standard temps.

Temp C for JSA @ FL430 = 43 (-2) = -86 + (+15C) = -71C

From there, considering that the established ISA temperature of -56.5 exists at all ISA flight levels from FL360 up to FL650, I wanted to find the JSA deviation between the two.

JSA @ FL430 = -71C
ISA @ FL430 = -56.5
___________________________________
Difference = 14.5C

The mathematical difference of the -71C and the -56.5 was not considered to be in terms of a negative plus a positive, but more so in the sense of simple pure “difference” in place value.

Exactly. You have just applied, albeit in a crude fashion, the second equation above.

Originally Posted by perpetual

But what would be the true difference if the ISA lapse rate of 1.98 per 1000 Feet decrease was used? (Even throughout the tropopause)

Using:

Temp C for ISA @ FL430 = 43 (-1.98) = -85.14 + (+15) = -70.14C in terms of pure lapse rate (ignoring isothermal conditions)

Whooooaaaaaaaaa Nelly. Now you're not talking about ISA or JSA, you're talking about PSA, the "Perpetual" Standard Atmosphere! You're establishing a new standard atmosphere from which to make comparisons, base studies, and conduct research. If you can get others to use it also, you might even come up with something useful to do with it. Until then, expect to endure great difficulty when trying to explain to others, that don't know what PSA is, what you're doing and how you're trying to do it.

Originally Posted by perpetual

Now using the ISA standard of (isothermal -56.5C) and (1.98 per 1000 Feet (( -70.14,)) we get a difference of 13.64

From that point the reason (( and I could be completely flawed here so please bare with me )) that I chose to do:

-70.14 + (+56.5) (normal ISA isothermal temp between the tropopause and FL650 etc )

What you're comparing here is ISA and PSA, and then only at a particular altitude, FL430.

Originally Posted by perpetual

was to find the difference between the max isothermal value of ISA temperatures for the given max ISA flight level (ISA 56.5 @ FL650) in comparison to the calculated decrease by using JSA -2C per 1000 @ ISA/JSA FL430.

Now you're really confusing me. I think you're trying to compare your PSA with ISA and JSA at FL430, or you might be trying to find the altitude where the max difference between these models occur. Perhaps you could clarify your goal using the terms ISA, JSA, and PSA.

(To review, ISA is three straight lines with different slopes (-1.98°C, then 0°C, then 0.3°C per thousand feet), JSA is a single straight line (2°C per thousand feet) and PSA is a single straight line (1.98°C per thousand feet))

Originally Posted by perpetual

Again, the mathematical difference of the -70.14 and the + 56.5 was not considered to be in terms of a negative plus a positive, but more so in the sense of simple pure “difference” between the two.

Put as simply as possible, below is a quick summary of what I have come up with. (( I could be entirely wrong, so if someone can point me in the right direction please feel free ))

43 (-2) = -86.00C
43 (-1.98) = -85.14C (ignoring isothermal values just as JSA)
__________________________
Pure Difference = 0.86C

You have correctly computed the difference between JSA and PSA at FL430.

Originally Posted by perpetual

(-0.86C) + (-56.5C) = -57.36C

You have combined the difference of JSA and PSA with the model value of ISA at FL430 and come up with... a rather curious, but meaningless number.

Originally Posted by perpetual

So therefore, would the approximate true ambient temperature for ISA @ FL430 be -57.36C and not the said -56.5C (isothermal value) ?

No. You have MODEL and ACTUAL confused again. The Temperature at FL430 in the ISA model is -56.5°C. The temperature you just derived comes from comparing JSA with your PSA, and then applying a difference to the ISA, which in effect creates yet another model - - the CSA, or CONFUSED Standard Atmosphere!

Originally Posted by perpetual

I hope this is a better explanation as to what I am trying to express.

Ps. I love flying turkeys
PPS. And MorAns do watch TV

It was indeed a different explanation - - I hope this post and my previous post come close to clearing things up for you.




.

[TonyC]

Originally Posted by perpetual

I have included an example of an ATPL test question below, involving the use of differing (JSA vs ISA) atmospheric modeling along with its answer.

__________________________________________________ ____________
Question:

1.) At FL430 the temperature deviation is given as ISA +3. Express this temperature as a deviation from the Jet Standard Atmosphere.
__________________________________________________ ____________
Given answer:

ISA Temperature (43,000 ft) = - 56.5°C
Deviation from ISA = +3°C
Ambient temperature (43,000 ft) = -53.5°C
JSA temperature (43,000 ft ) = +15°C - (43 x 2°C)
= +15°C - 86°C
= -71°C
JSA temperature deviation = +17.5°C
__________________________________________________ ______________

Including the question and “answer” provides interesting insight both into the nature of your quest, and the source of the confusion. The wording of both might contribute to a misunderstanding of the concepts. I’ll assume that the question is posed by a regulatory authority, so I won’t bother to take issue with it. I’ll assume that the “answer” is provided by an agency whose goal, ostensibly, is to teach and prepare pilots to take the test. Therefore, I will approach their wording.

The temperature at a given altitude, in this case, is expressed as a deviation from the ISA atmosphere. In other words, the ACTUAL TEMP is equal to the ISA Temp plus or minus some value. In this case, the deviation is +3°C.

(I said the DEVIATION is +3, not the ISA Deviation. The "answer" gets this correct when it says the "Deviation from ISA is ...." )

ACTUAL TEMP = ISA Temp + Deviation from ISA
ACTUAL TEMP = ISA Temp + 3°C
ACTUAL TEMP = -56.5°C + 3°C
ACTUAL TEMP = -53.5°C

The deviation is the difference between the ACTUAL and the MODEL. The deviation expressed in terms of the JSA model would be found n the same manner.


ACTUAL TEMP = JSA Temp + Deviation from JSA

You determine the JSA Temp using the +15.0°C - (2.0°C x Altitude in thousands of feet) = +15.0°C - (2.0°C x 43) = +15.0°C – 86.0°C = -71.0°C

We now know two values of the equation - - time to “plug and chug”

ACTUAL TEMP = JSA Temp + Deviation from JSA


Or


Deviation from JSA = ACTUAL TEMP – JSA Temp
Deviation from JSA = -53.5°C – (-71.0°C)
Deviation from JSA = -53.5°C + 71.0°C
Deviation from JSA = 17.5°C

(The "Deviation from JSA is 17.5°C. I believe it is misleading to say, as the "Given answer" does, that the JSA Temperature deviation is..." I realize it's nitpicking semantics, but it appears that you may have stumbled over semantics, so it's important.)

So, let’s be careful what we SAY when we have this number, so as not to confuse ourselves. What we have is a Deviation from JSA, so we can say, “The actual Temperature at FL430 is JSA + 17.5°C”.

Originally Posted by perpetual

As you can see this is one of the questions that started me thinking about how to find a particular ambient temp for a related ISA FL in the isothermal region.

The reason the ISA model has an isothermal region is the atmosphere itself has an isothermal region. Your best estimate of the actual ambient temperature will acknowledge this aspect of its structure. Naturally, there are variations that will affect it’s accuracy, such as the seasons and latitudes that affect the height and thickness of the tropopause. But I believe you will not arrive at a better estimate by abandoning the premises of this model. The JSA, as you surely have read, is not praised for it’s superior accuracy in predicting temperatures; it is praised for its simplicity of use.

Originally Posted by perpetual

I know the topic is completely theoretical at best at this point, it was simply the question of needing to find the deviation between each model that sparked my interest further.

Well, I think we’ve shown that it is indeed possible to compare the two models using simple mathematical equations.

Originally Posted by perpetual

Anyway, based on what I wrote the first time, am I at least in the ballpark so to say as far as trying to determine the actual ambient for the given flight level?

And, again, no, comparing the ISA and JSA models will NOT help you to better predict an actual ambient temperature.

Originally Posted by perpetual

And in closing, if this thread gets too rediculous because of it's possible pointlessness, I'll just let my interest for it die

On the contrary, it has been an interesting exercise. And it hasn’t been ridiculous, either. J




.