Altitude Questions
 

Any scientists out there that can figure out the answers to these two questions?

1. Given an indicated altitude of 10,000 feet and an actual OAT of -20C, you set your altimeter to a local station setting of 29.62". If the station elevation is 2500 feet, what is your actual altitude?

A.9250
B.9550
C.9850



2. Given a field elevation of 1500 feet, altimeter 1000 millibars, calculate pressure altitude.

A.1110 feet
B.1890 feet
C.890 feet



I know that the answers to both are (B) but can't for the life of me figure out the methodology used to get there.

1 inch HG = 34 Millibars
1 inch HG = 1000 feet

Appreciate any help. Thanks.:confused:

I finally figured out #2 on my own...

(With a little help from the "search" function)...

1013 - 1000 = 13 x 30 (30 feet/millibar) = 390

1500 + 390 = 1890 feet


Blinded by Science.

The formula you need is

TA = IA + (IA - FE)*ISADEV/°K)

TA=True Altitude
IA=Indicated Altitude = 10,000
FE=Field elevation=2,500
ISADEV=ISA deviation=-15
°K=Temperature in °C at IA +273=253

Inserting your figures in the formula gives

TA=10,000 + (7500*(-15)/253) = 9555

So answer B is correct

Whatever happened to the hectopascal? I though ICAO required that unit instead of milibar.

1 mb = 1 hPa - it is just a label

thanks for the help SFI145.

Or an easier to remember formula for question 1

+/- 4' per degree of ISA Devn x (ht/1000)

Where ht = the height above the temperature datum, which is the airfield.

+4x15(7.5)

So at an indicated altitude of 10000' your altimter will be overeading by 450' giving a true altitude of 9550'.

Q2 is simply at a matter of 'winding' your altimeter up to 1013 to get pressure alt. 13Hp x 30' per Hp = +390' hence B.

Chuckles.

PS but I do like the TA formula with degrees Kelvin, being a naturally enquisitive bloke so I'll write it down and attempt to memorise it:ok: