Aerocourse ATPL question

chillout · Post by **chillout** » Sun Jul 09, 2006 6:59 pm

Looking for some help with Flight ops question #77/78. How do you get the answer? Thanks in advance.

ahramin · Post by **ahramin** » Mon Jul 10, 2006 12:03 am

What are the questions?

Raven20000 · Post by **Raven20000** » Mon Jul 10, 2006 8:37 am

Question 77 - Which of the following is true with respect to winglets?

a) they reduce profile drag but increase induce drag?
b) they increase form drag but reduce induce drag
c) they increase profile drag and induce drag
d) they increase induce drag but reduce parasite drag

Now b is correct cause the winglets inhibit wing tip vortex development( page 27 ftgu).

Question 78 - An automatic system on large transport category aircraft which utilize inputs from air data computers to provide yaw damping and turn coordination is known as a:

a) Rudder Trim Actuator
b) Yaw Damper
c) Auto Flight Control System
d) Digital Flight Rudder Controller

B being the correct answer again but to explain this one I think just try to remember the question and answer, or if anybody has any usefull tips on this one.

chillout · Post by **chillout** » Mon Jul 10, 2006 2:14 pm

Q#77 An A/C departs an airport that has an elevation of 2000'ASL and levels off at 6000 ASL, which is the MEA. The revised altimeter is 30.12 and OAT is -32C at 6000' Whatwould ground clearance be of A/C when crossing the 4950'ASL ridge S of the airport.
1. At least 1000'
2. Approx 750'
3. Approx 490"
4. Less then 150'
Thanks for your help

ahramin · Post by **ahramin** » Mon Jul 10, 2006 3:04 pm

Without the temperature on the ground you really can't say. If you assume that the temperature on the ground is about the same as in the air then you get

6000 feet - 2000 feet = 4000 feet AGL

Looking at the Altitude Correction Chart in the CAP GEN with 4000 feet and -30°C we get 760 feet.

6000 feet minus 760 feet means you are about 5240 feet MSL.

5240 feet - 4950 feet = 290 feet.

Not close to any of their answers. Typical AeroCourse typo? Or perhaps they think you should use the standard lapse rate to figure out the temperature on the ground (a stupid assumption in these conditions if you ask me)

In which case we get

4000 feet x 2°C/1000 feet = 8°C So the temperature on the ground would be -40°C. In the CAP GEN this gives a correction of 970 feet.

6000 - 970 - 4950 = 80 feet. Still not very close but better than the last one. What does the answer key say?

In any case, what you would actually do is check the temp on the ground (where they are giving you the altimeter setting), do the temperature correction, add 500 feet for mom and the kids, and fly the next highest appropriate altitude.

Raven20000 · Post by **Raven20000** » Mon Jul 10, 2006 5:22 pm

Sorry I was using a older version, but hey at least you got your answers.

birdlegs · Post by **birdlegs** » Mon Jul 10, 2006 10:51 pm

I used my E6B to figure out the true altitude and the best I could do with it was about 550 feet for the clearance, making C the most correct answer. True altitude corrections are useful when the temps are colder than standard, however, the E6B manual cautions that the corrections can be useless if temps are alot below standard. Using the E6B and doing the true altitude check on it doesn't require knowing what the temperature on the ground is. In case you don't have the manual anymore check out the link below and go to page 21.

http://www.asa2fly.com/files/support/E6B_Manual.pdf

airliner · Post by **airliner** » Tue Jul 11, 2006 9:00 am

just memorize the answer and dont question it. the question on the exam is vertually the same.

T.C doesnt give you the option to challange the question.

In short just mack down C and move on to the next. they are not that hard