Calculation of FL
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Calculation of FL
I understand that FL can be derived from knowing ISA sea level T and standard T.
I have a formula (from a set of notes from a defunc but well respected organisation) which must contain a typo. Here it is
Std T = +15(2 x FL)
this formulae is then trnsposed to
FL[(47 + 15)/2]= 310!
When given the values Amb = - 45C
TDev = +2C
StdT = -47
Am I confused or what. Any help gratefully received
I have a formula (from a set of notes from a defunc but well respected organisation) which must contain a typo. Here it is
Std T = +15(2 x FL)
this formulae is then trnsposed to
FL[(47 + 15)/2]= 310!
When given the values Amb = - 45C
TDev = +2C
StdT = -47
Am I confused or what. Any help gratefully received
I think you may be confused. Presumably you're after the temp at altitude given sea level temp and assuming a standard atmosphere ie environmental lapse rate of 2 deg C/1000 ft. Thus if sea level temp is 15 deg C (standard atmosphere), temp at, for example, 7000 ft would be 15-(7*2)=1 dec C.
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Now, see here, young cron:
Flight Level has nowt to do with temperature at all. It is only an expression of vertical position as related to the standard pressure level - 1013.25 Mb. (HPa, if you must). Nothing else.
I think you must have the wrong end, I say the wrong end of stick.
Flight Level has nowt to do with temperature at all. It is only an expression of vertical position as related to the standard pressure level - 1013.25 Mb. (HPa, if you must). Nothing else.
I think you must have the wrong end, I say the wrong end of stick.
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Fred - totally agree but this here set of notes from this here well respected training orgs has got a formula as I've quoted for FL using Temps. They apply it but as I've quoted it, but it does not make sense arithmetically. Thanks for the help
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I too think you are getting something wrong here. Those formulae don't seem to make a hell of a lot of sense either, what is Std T? do you mean SAT/RAT/EAT or ISA?
Like Fred has said, you can calculate temperature at a given FL and give ISA deviation, but you wouldn't normally calculate FL from Temp unless it was some sort of derived exam question relating to Met and lapse rates, as FL is a position relative to pressure.
Like Fred has said, you can calculate temperature at a given FL and give ISA deviation, but you wouldn't normally calculate FL from Temp unless it was some sort of derived exam question relating to Met and lapse rates, as FL is a position relative to pressure.
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I think the formula that you are talking about is to find the deviation from ISA at altitude.
EG. FL 310 + 15 / 2 = 23 (use just the first two figures of the flight level and assume that 23 is a minus figure).
This means that in ISA the temperature at FL 310 should be -23. If the ambient temperature is -20 then what you have is an ambient temperature of ISA -3 (three degrees colder than ISA).
The ISA + or - so many degrees scenario is used by the CAA in many of their exam questions.
EG. FL 310 + 15 / 2 = 23 (use just the first two figures of the flight level and assume that 23 is a minus figure).
This means that in ISA the temperature at FL 310 should be -23. If the ambient temperature is -20 then what you have is an ambient temperature of ISA -3 (three degrees colder than ISA).
The ISA + or - so many degrees scenario is used by the CAA in many of their exam questions.
still learning....
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fb1,
Beg to differ. Take the first 2 digits if the FL, double, then subtract 15. At FL 310 it would be 31 X 2=62. Then subtract 15 to yeild an ISA of -47 at FL 310.
ISA is always SAT, never RAT.
Beg to differ. Take the first 2 digits if the FL, double, then subtract 15. At FL 310 it would be 31 X 2=62. Then subtract 15 to yeild an ISA of -47 at FL 310.
ISA is always SAT, never RAT.
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Flight Level represents the pressure altitude in hundreds of 100 feet.
So FL = Pressure altitude / 100.
So FL 300 = 30000 ft / 100, FL 10 = 1000 ft /100, and FL 0 = 0 ft / 100.
In the ISA the temperature decreases by approximately 2 degrees C per 1000 ft.
This means that for every increase of 10 in FL the temperature decreases by 2 degrees C.
So if we divide the FL by 10 then multiply by 2, we get the amount by which the local temperature is lower than ISA msl.
But at ISA msl the temperature is +15 degrees C, so we can predict the temperature (OAT) at any given FL by taking +15 and subtracting ((FL/10)x2)
This gives the equation OAT (at any given FL in the ISA) = +15 -((FL/10)x 2)
So for example we can predict that at 36000 ft (FL 360) in the ISA
The OAT = +15 -((36 / 100 x 2)
which is OAT = +15 -(36 x 2)
This is OAT at 36000 ft ISA = +15 - 72= -57
THIS CAN ALSO BE USED TO CALCULATE THE FL GIVEN THE OAT.
The equation OAT = +15 -((FL/10) x 2) can be rearranged to give the following:
(OAT -15) / 2 = -FL/10 So FL = 10(15 - OAT)/2) Or FL = 5(15 - OAT)
For example if the OAT is known to be -35 degrees C in the ISA then
FL = 5(15 -(-35))
Which is FL = 5(50) = 250.
So in the ISA if the OAT is -35 then the FL = 250
ACCOUNTING FOR TEMPERATURE DEVIATION
The actual OAT at msl in a real (non ISA) atmosphere is 15 degrees C plus the temperature deviation (Tdev).
Our original equation can therefore be adapted to non ISA conditions by adding Tdev to give the following.
OAT = (+15 + Tdev) - ((FL/10) x 2)
For example if the temperature deviation is known to be plus 2 then:
At msl (FL 0) the OAT = (+15 +2) - ((0/10) x 2) which is 17 degrees C.
And OAT at FL360 = (+15 +2) - ((360/10) x 2) which is - 55 degrees C.
This method can also be applied to give the following equations which take Tdev into account:
At any given OAT, the FL = 5(+15+Tdev - OAT)
and
At any given flight level the OAT = +15+Tdev - ((FL/5)
So FL = Pressure altitude / 100.
So FL 300 = 30000 ft / 100, FL 10 = 1000 ft /100, and FL 0 = 0 ft / 100.
In the ISA the temperature decreases by approximately 2 degrees C per 1000 ft.
This means that for every increase of 10 in FL the temperature decreases by 2 degrees C.
So if we divide the FL by 10 then multiply by 2, we get the amount by which the local temperature is lower than ISA msl.
But at ISA msl the temperature is +15 degrees C, so we can predict the temperature (OAT) at any given FL by taking +15 and subtracting ((FL/10)x2)
This gives the equation OAT (at any given FL in the ISA) = +15 -((FL/10)x 2)
So for example we can predict that at 36000 ft (FL 360) in the ISA
The OAT = +15 -((36 / 100 x 2)
which is OAT = +15 -(36 x 2)
This is OAT at 36000 ft ISA = +15 - 72= -57
THIS CAN ALSO BE USED TO CALCULATE THE FL GIVEN THE OAT.
The equation OAT = +15 -((FL/10) x 2) can be rearranged to give the following:
(OAT -15) / 2 = -FL/10 So FL = 10(15 - OAT)/2) Or FL = 5(15 - OAT)
For example if the OAT is known to be -35 degrees C in the ISA then
FL = 5(15 -(-35))
Which is FL = 5(50) = 250.
So in the ISA if the OAT is -35 then the FL = 250
ACCOUNTING FOR TEMPERATURE DEVIATION
The actual OAT at msl in a real (non ISA) atmosphere is 15 degrees C plus the temperature deviation (Tdev).
Our original equation can therefore be adapted to non ISA conditions by adding Tdev to give the following.
OAT = (+15 + Tdev) - ((FL/10) x 2)
For example if the temperature deviation is known to be plus 2 then:
At msl (FL 0) the OAT = (+15 +2) - ((0/10) x 2) which is 17 degrees C.
And OAT at FL360 = (+15 +2) - ((360/10) x 2) which is - 55 degrees C.
This method can also be applied to give the following equations which take Tdev into account:
At any given OAT, the FL = 5(+15+Tdev - OAT)
and
At any given flight level the OAT = +15+Tdev - ((FL/5)
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Hello there!
I have a quick question which is relavent to this thread. I am aware that the tropopause is reached at around 36,500ft, and that the temperature of the air at this altitude should, according to the ISA formulae used above be -57 degrees-ish, and according to my reference material, the temperature stays roughly the same from there on up. Is this the part of the ISA definition? And if not, for standard calcs and performance graphs etc
should this plateau be taken into account?
I have a quick question which is relavent to this thread. I am aware that the tropopause is reached at around 36,500ft, and that the temperature of the air at this altitude should, according to the ISA formulae used above be -57 degrees-ish, and according to my reference material, the temperature stays roughly the same from there on up. Is this the part of the ISA definition? And if not, for standard calcs and performance graphs etc
should this plateau be taken into account?
still learning....
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cb,
For the aircraft we fly, yes. Actually -56.7 is considered to be the lowest ISA temp we'll experience. For a mission to Mars? Someone else will have to answer.
For the aircraft we fly, yes. Actually -56.7 is considered to be the lowest ISA temp we'll experience. For a mission to Mars? Someone else will have to answer.
Keith, like your formulae, however I think you are inadvertently being a little obscure when you are using OAT in your ISA calculations - you should use ISA TEMP until you introduce the deviation lower down.
Yes, the ISA defintition assumes a sea level temp of 15°C, reducing by 1.98°C up to 36,000 ft, and then remaining constant at -56.5°C.
(The ISA also assumes that the air mass has no moisture, water vapour, dust or other "impurities" and that it confirms perfectly with the gas laws.)
Just as the sea level temperature is usually different from day to day, so is the lapse rate, the humidity and every other assumption about the ISA - including the tropopause height, and the temperature of the tropopause. I have personally seen a temperature indication of -71°C (in winter about 35°S at FL450).
In fact, as you go even higher, the air temperature starts to increase, due to chemical reactions in the mesosphere.
[ 08 November 2001: Message edited by: Checkboard ]
Yes, the ISA defintition assumes a sea level temp of 15°C, reducing by 1.98°C up to 36,000 ft, and then remaining constant at -56.5°C.
(The ISA also assumes that the air mass has no moisture, water vapour, dust or other "impurities" and that it confirms perfectly with the gas laws.)
Just as the sea level temperature is usually different from day to day, so is the lapse rate, the humidity and every other assumption about the ISA - including the tropopause height, and the temperature of the tropopause. I have personally seen a temperature indication of -71°C (in winter about 35°S at FL450).
In fact, as you go even higher, the air temperature starts to increase, due to chemical reactions in the mesosphere.
[ 08 November 2001: Message edited by: Checkboard ]
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Checkerboard,
You are correct in saying that the formulae in my previous post are somewhat obscure. I employed mathematical argument quite simply because the original question was couched in mathematical terms, and required an explanation of the logic of a mathematical formula.
The point of my post was simply to illustrate that it is possible to calculate the (theoretical) FL from outside temperature, and outside temperature from FL. I deliberately left out the complexities of SAT and TAT in order to keep it simple.
The resulting formulae can perhaps better be phrased as follows:
If you know the FL and want to know the outside air temperature, divide the FL by 5, then subtract the result from 15. This will give you what the outside air temperature would be if you were flying in the ISA.
If you then look at your outside air temperature gauge (or whatever it is called in your aircraft) you will probably find that it does not agree with the figure you have just calculated. The difference is caused by a combination of the amount by which temperatures in your atmosphere differ from those in the ISA (Tdev), and (if you are flying too fast) by the kinetic heating effect due to your speed.
If you now want to know the temperature deviation (ignoring kinetic heating) simply subtract the figure you have calculated from the indication on your temperature gauge.
If however you know the outside air temperature and want to calculate the FL , subtract 15, then multiply by 5. Ignore any + or - and you have the FL. Once again any error will be due to a combination of Tdev and kinetic heating.
The original post suggested using the first 2 digits of the FL. This is OK if your FL is three digits but is no good if it is only 2.
For example if we use the first 2 digits of the FL we will get the same result at FL10 as we will at FL100.
You are correct in saying that the formulae in my previous post are somewhat obscure. I employed mathematical argument quite simply because the original question was couched in mathematical terms, and required an explanation of the logic of a mathematical formula.
The point of my post was simply to illustrate that it is possible to calculate the (theoretical) FL from outside temperature, and outside temperature from FL. I deliberately left out the complexities of SAT and TAT in order to keep it simple.
The resulting formulae can perhaps better be phrased as follows:
If you know the FL and want to know the outside air temperature, divide the FL by 5, then subtract the result from 15. This will give you what the outside air temperature would be if you were flying in the ISA.
If you then look at your outside air temperature gauge (or whatever it is called in your aircraft) you will probably find that it does not agree with the figure you have just calculated. The difference is caused by a combination of the amount by which temperatures in your atmosphere differ from those in the ISA (Tdev), and (if you are flying too fast) by the kinetic heating effect due to your speed.
If you now want to know the temperature deviation (ignoring kinetic heating) simply subtract the figure you have calculated from the indication on your temperature gauge.
If however you know the outside air temperature and want to calculate the FL , subtract 15, then multiply by 5. Ignore any + or - and you have the FL. Once again any error will be due to a combination of Tdev and kinetic heating.
The original post suggested using the first 2 digits of the FL. This is OK if your FL is three digits but is no good if it is only 2.
For example if we use the first 2 digits of the FL we will get the same result at FL10 as we will at FL100.
