Terrain Clearance question

[ranklein]

Hi,
I have a question I'm trying to solve..:

An A/C is levelled at FL120 over a mountain range which extends up to 2103 meters AMSL. Using 30ft/mb, what is the approximate clearance over the terrain if reginal QNH is 998mb?

Thanks!

[Old Smokey]

Using actual atmospheric Pressure Lapse Rates (which are logarithmic and not linear), in this case the correction is very close to 27.5 ft / hPa.

That makes the Pressure Altitude at F/L 120 equal to 11581 feet, giving you 4681 feet clearance above your 6900 ft (2103 M) mountain IF the atmosphere is ISA temperature.

At lower than ISA temperatures, the clearance will be less, the opposite for above ISA temperatures.

Regards,

Old Smokey

[ranklein]

Thx old smokey!
So 11581 was the actual altitude (true) after correcting for the 998mb, right? Have you used the 27.5'/mb?How about the transfer from 2103m to 6900'? Do you use 3.28'/1 meter?
Let me give you another one, I hope I'll get the idea then:
A/C flies over mountains 4730m AMSL.
Track is 140(M) ("not sure why they give it") and QNH is 995mb.The required clerance is a minimum of 1500'.What's the minimum FL in cloud?
Here they say assume 1mb=27'

Thanks again!

[aee190]

Hi ranklein,

I'll try to answer your question, however my altimetry is a bit rusty!

Terrain clearance = 1500'

Height of mountain above 995mb level = 4730m = 15514' (using 1m = 3.28')

Difference between 1013mb level and 995 mb level = 18mb * 27' = 486'

Therefore height of aircraft above 1013mb level = 486'+15514'+1500' = 17500' or FL175.

However, according to Instrument Flight Rules, flying on a heading of 140(M) means we have to fly an ODD flight level. Therefore the minimum flight level required is FL190.

P.S. In the UK outside controlled airspace and below FL245 we fly quadrantal flight levels. In this case on a heading of 140(M) we would fly an ODD + 500' flight level and so the minimum flight level would be FL175.

[ranklein]

aee190, It does begining to make sense.
Thx mate

[Old Smokey]

Sorry ranklein, if I seemed rude in not answering a question that was (loosely) addressed to me, aee190 has done a nice job of answering it, and added some good thought of the practical application of Altimetry rules.

With respect to your earlier question, 3.28 ft/M is fine for day to day calculations, I prefer to divide by 0.3048 (which is an exact conversion) rather than multiplying by 3.280839895.... (which is a surd and not an exact conversion), but that's only because I'm pedantic.

The tone of your questions suggest that you might be studying for an exam, beware!, some examiners are pedants (like me ), and in your question....

Obstacle : 4730 M / .3048 = 15518.4 ft,
QNH Corr'n : (1013.25 - 995) = 18.25 X 27 = 492.8 (more if you use 27.5)
Minimum F/L = 15518.4 + 492.8 +1500 = 17511.2 ft.

Now, as aee190 has indicated, F/L 190 would be the correct answer in a U.K. exam, but it's JUST above F/L 175 by 11.2 ft, so if Outside Controlled Airspace, the correct answer would be F/L 195. Beware pedantic exam questions!

Interesting to see a poster from the U.S.A. with an interest in Flight Levels below F/L 180.

Regards,

Old Smokey

[ranklein]

Thanks Old Smokey for the great explanation.
You're right, it's an upcoming exam/s for a non U.S ATP.
Hope to see some of your (and others) replies to other interesting questions I'll submit down the line.

Cheers!!

[aee190]

I agree with Old Smokey that, in my experience, some ATPL exam questions are overly pedantic but usually to avoid ambiguity the conversion factors to be used for a particular question will be stated either in the instructions at the front of the exam paper or in the question itself. My advice, read the questions very carefully. If the worst comes to the worst and you believe the question to ambiguous, you have the option to challenge the examining authority (which I have seen work on more than one occasion).

[ranklein]

New questions..

Thanks, great advice!
Here's another two examples I'd like to make sure I understand:

1. A/C flies at 10000'. What's the range of reception to a VOR sitting at 2000amsl?
The answer used the formula
1.225 * square of altitude
It said 1.225* square10,000 + 1.225* square 2000 = around 177NM.
What's the reason behind adding the two ranges? How do you explain this answer better?

2. Ground speed of 383Kts. DME reads 47NM from a DME station at 1500' AMSL. A/C at 30,000'. What's the EET to the station?
My main concern is the corrected range to the station which is 46.76NM in the answer I have.Is it right?
I thought I should deduct 4.7NM from 47 because at 28,500' above the station that's what the DME reads (28,500/ 6080), so range should be 42.3.

Comments..?

[aee190]

1. VOR's use frequencies in the VHF radio spectrum. As far as I am aware the equation for VHF range is 1.25 x (√transmitter height + √reciever height). Therefore to answer your question the range is approximately 181nm.

2. DME's give slant range to the transmitter, therefore to answer the question the horizontal range to the transmitter needs to be found using good old Pythagoras' theorem. I calculate the horizontal distance to be 46.77nm, now it is just a matter of dividing this distance by the groundspeed of the aircraft to give the EET. I calculate it to be 7 minutes and 20 seconds.

[ranklein]

aee190 thanks.
7:20 is right on.What exact calculation have you done with Pythgorase ?

[aee190]

ranklein, I presume you already know the fundamentals of Pythagoras' theorem.

Hypotenuse = slant range from transmitter = 47nm.
Vertical distance = 30000' - 1500' = 28500' / 6080 = 4.69nm.
Therefore the horizontal distance = √(47² - 4.69²) = 46.77nm.

EET = 46.77 / 383 = 0.1221 hours = 7min 20 sec.