The theory is that you can determine whether you are flying from/to a higher or lower pressure or temperature based on the direction of crosswind. This is Buys Ballot's law. In the Northern Hemisphere, if you stand with your back to the wind, the low pressure system will be on your left. The reverse is true in the Southern Hemisphere. Therefore if you have a crosswind from the left (ie. starboard drift), you are flying towards a lower pressure. Then remember the phrase: "high to low, watch out below!", which will remind you that true altitude decreases as you fly towards a lower temperature or pressure.

Wow ok I get it!! Just clicked. THANK YOU!

need help with a question
Friends, i am stuck with one of a question.can some plz shed some light in solving this question.thanks for the help
Q.Air at T = +16° C and DP = +4° C is forced from sea level over a 10.000 ft mountain range and descends back to sea level on the other side. If the leeward condensation level is observed to be 8.000 ft, what will be the final temperature? A)18° B)22° C)20° D)24° ANS IS B. 
Tried it with:
cloud base = 400x (TTd) SALR= 1.8 degrees/ 1000 ft DALR= 3 degrees/ 1000 ft cools with DALR up to cloud base on windward side, then cools with SALR until the summit, then warms with SALR as it descends on the leeward side until cloud base, then warms with DALR again. But calculating this gives 20 degrees:rolleyes: 
Mass and Balance Question
Hi everyone A mass and balance question I need help with!
At a given mass the CG position is at 15% MAC. If the leading edge of MAC is at a position 625.6 inches aft of the datum and the MAC is given as 134.5 inches determine the position of the CG in relation to the datum. Withn the answer it states  The MAC is 134.5" long. The CG is 15% of this distance back from the leading edge. 15% of 134.5" is 20.17". The leading edge of the MAC is 625.6" aft of the datum, the CG is 625.6 + 20.17 = 645.77" aft. WHERE DO THEY GET 20.17 from!! Or am I just being DUMB! 
Length of MAC = 134.5, so:
1% mac = 134.5/100 =1.345 15% mac = 1.345 * 15 = 20.175 
You need to understand the meanings of the various terms.
The MAC is the Mean Aerodynamic Chord Length. The statement that the MAC is given as 134.5 inches 15% of 134.5 inches is 20.175 inches. So if the CG is at 15% MAC it is 20.175 inches aft of the MAC leading edge. If we now add the MAC leading edge poistion of 625.6 inches aft of datum we get a CG position of 20.175 + 625.6 = 645.775 inches aft of datum. 
Thank you............
..........Mr. Williams for taking the time to so clearly answer the original question.

Pitot questions on recent ATPL exams
Hello all.
I'm looking for recent ATPL questions regarding Pitot blockages. There seems to be some discrepancies between our schools studyguides and real life. As I am a teacher in the subject, it would be interesting to know which stance the current QB has. Our studyguide maintains that a blockage of the ram air opening of a pitot tube will cause the ASI to freeze at current speed whereas in real life, such a blocking would cause the ASI to read 0 because of pressure escaping through the waterdrain. Any thoughts? 
Try it.
Go to an aeroplane with a friend  blow GENTLY into the pitot, then hold your finger firmly over the end. Then have your friend go around and see what is happening in the cockpit. I know what I've got every time I've done that but try it yourself and you'll never forget the answer. G 
As I am a teacher in the subject, Suggest you teach correctly........ 
The altimeter will freeze if a blockage occurs but that would be in the static system. The VSI would read zero. If the pitot gets blocked, the ASI behaves like an altimeter and will increase its readings as you climb. If the static gets blocked, the ASI error will reverse, i.e. it will underread as you climb.
At least that's what the JAA questions expect  real life has nothing to do with it! :) The water drain is usually operated with a spring (every 200 hours on some aircraft), so that implies some sort of seal. It would have to be sealed in normal circumstances (like the alternate static) otherwise the readings would never be correct. 
real life has nothing to do with it! One it did pretty much as the JAA exams expect  froze at the original value (I never went above about 2000ft that trip, so the altimeter thang wasn't really there). The other did something I'd never have predicted  it started reading almost in direct proportion to the RPM gauge. During takeoff ofcourse, that actually seems about right  after five minutes I realised that something odd was happening. Fortunately both totally VFR trips, so all flown on attitude back to an uneventful landing. The "blow & block" test I've learned now to do as part of my preflight whenever I'm doing the first flight of a newbuild aeroplane ! It's also why I know that this cobblers about water drains is just that  if it starts going down, there's a system leak and it needs sorting before flight. It's a problem with "teachers" who have no real knowledge or experience  but aviation suffers from that a lot (as do many schools ofcourse, so they're in good company). G 
Blow and block technique works for me  but you do need two people and be aware when the last flight happened; pitot heat ruins your chances in the pub later. :(

How do you calculate the lowest useable flight level?
a. Lowest QNH and lowest negative temperature below ISA b. Lowest QNH and highest negative temperature below ISA c. Highest QNH and highest temperature above ISA d. Highest QNH and lowest temperature Please Help. 
What temperature and pressure conditions would be safest to ensure that your flight level clears all the obstacles by the greatest margin?
a. Cold temp/low pressure b. Warm temp/high pressure c. Temp less than or equal to ISA and a QNH less than 1013 d. Temp more than or equal to ISA and a QNH greater than 1013 
The worst place to be is in a cold low.

Mass and Balance Questions  Traffic Load
I wonder if there are any brain boxes out there who can shed some light on this question? I have (as far as I am aware) followed the correct proceedure of how to calculate a Traffic Load.
MTOM: 170 000kg ZFM: 112 500kg MLM: 148 500kg DOM: 80 400kg TAXI FUEL: 800KG BLOCK FUEL: 40 000KG TRIP FUEL: 29 000KG. Question asks for the traffic load which can be carried. Here are the choices: A) 32 100kg B) 32 900kg C) 18 900kg D) 40 400kg My calculations: (*1) (I have calculated MTOM fuel to be block fuel (40 000kg)  taxi fuel (800kg) = (32 000kg) ) (*2) (I have calculated MLM fuel to be block fuel (40 000kg)  taxi fuel (800kg)  trip fuel (29 000kg) = 10 200kg MTOM / ZFM / MLM LIMIT 170 000kg / 112 500kg / 148 500kg DOM 80 400kg / 80 400kg / 80 400kg FUEL 39 200kg (*1) /  / 10 200kg (*2) =TRAFFIC LOAD 50 400kg / 31 600kg / 59 900kg THE CORRECT ANSWER SHOULD BE A but I don't get anywher near that!:confused: Can anyone tell me where i have gone wrong? I thought that I had these mass and balance questions sorted :{ I'm currently studying with Bristol, but the question was taken from 'The Daily ATPL' website. I would be so grateful if anyone could help me :) 
The most limiting factor for this flight is:
MZFM  DOM = TL 112500  80400 = 32100 kg Not sure how you got 31 600kg out of that calculation :ok: 
Your problems started with your very first statement:
(I have calculated MTOM fuel to be block fuel (40 000kg)  taxi fuel (800kg) = (32 000kg) ) 
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