Im just trying to interpret this old chestnut.

Would it be fair to say that you would use HH for the time of issue and continue utilising this for the next 1.5 hours, then use the values for HH+3 for the next 3 hours etc? Would this be interpolation or am i missing something?

A crude drawing attached!

Thanks

Andrew

Hi Andrew,

Linear interpretation means to spread the difference between the two values evenly over the three hours between the spot times in the TAF. Having said that you of course simply round to the nearest whole value (^oC or hPa)

For example, if you have 3 degrees of temperature difference between the values for HH and HH+3, you can assume the temperature will change by about 1 degree per hour.

If there's only 2 degrees of difference, you can assume that up until half way between HH and HH+3 the HH value will apply and after that the HH+3 value will be applicable.

The regs recently changed with respect to this. It used to be the value given applied to +/- 90 minutes either side of the time in exactly the way you described. Now though the system has gone to the more logical approach of simply spreading the expected change over the whole time period.

Here's a more concrete example from a fictitious TAF for 1820/1908:

TAF XXXXX ....... T 18 18 16 13 Q 1012 1011 1008 1004

You will NOT work out decimal values in the cockpit and you probably won't need to in the exam either. In practice you can eyeball the values quite easily. In my opinion it is easier and more realistic than the previous system.

This table gives you some examples of linear interpolation. Figures in bold are taken directly from the TAF.

___Time(code)___	Temperature	QNH
2000 (HH)	18oC	1012
2145	18oC	1011 (QNH drops by 1hPa over 3 hours and 2145 is more than half way to the next spot value)
2300 (HH+3)	18oC	1011
0030	17oC (Drops 2oC over 3 hours i.e. 1oC per 1.5 hours. So by 0030, the temperature will have dropped 1 degree)	1010 (QNH dropping by 1hPa/hr between 2300 and 0200)
0200 (HH+6)	16oC	1008
0300	15oC (temperature dropping about 1o/hr between 0200 and 0500)	1007 (QNH dropping by 1.25 hPa/hr between 0200 and 0500. Answer rounded 1007)
0415	14oC (temperature dropping about 1o/hr. Forecast temperature will change to 13oC from 0430)	1004 (QNH dropping by 1.25 hPa/hr between 0200 and 0500. QNH will be closer to 1004)
0500 (HH+9)	13oC	1004

Now my head hurts

Cheers,

Rich

Thank Rich,

I was getting little confused on that one, and have my exam tomorrow so thought i better get some clarification.

I believe a question in the Exam Prep modules still uses the old method of 90min either side as you explained.

Thanks again for your help!!!

Andrew

Thank Rich,

I was getting little confused on that one, and have my exam tomorrow so thought i better get some clarification.

I believe a question in the Exam Prep modules still uses the old method of 90min either side as you explained.

Thanks again for your help on both questions!!!!

Andrew