Finding Clearance above a ridge

[skateosiris]

Hi, I am currently working through the Sharper Ege Commercial Pilot exam prep, but I can't seem to figure out this question:

Your Altimeter indicates an altitude of 7000 feet, with an altimeter setting of 29.42 (based on a station at sea level) while flying above a ridge which is 5000 AMSL. The outside air temperature is -20 degrees celsius. Your clearence above the ridge is :

A.)1,950'
B.)1,500'
C.)1,100'
D.)400'

Is this just a matter of finding True Altitude? Any suggestions?

thanks

[Expat]

Of the top of my head, I would say A, based on the cold weather correction factor.

[DHC-1 Jockey]

The formula to find height above a ridge is as follows:
((Indicated Alt - Ground Elevation) / 1000) X Difference between ICAO and actual temp X 4 = amount to be subtracted from indicated altitude.

This gives true altitude. Take your true altitude, subtract it from the height of the ridge and you have your clearance.

[Beefitarian]

Sure but if that ridge isn't going down in relation to the bug on the windscreen start your turn!

"Don't worry I calculated 1873 feet clearance." CRUNCH.

[Expat]

Sure but if that ridge isn't going down in relation to the bug on the windscreen start your turn!

"Don't worry I calculated 1873 feet clearance." CRUNCH.

You do not need to clear the rige, but pass the ...*&^ exam...

[Beefitarian]

Yes the exam is the current issue and I know that well, but hardly anyone gets hurt writing an exam.

[co-joe]

DH 1's formula is not for exam purposes, it's for real life.

TDFS (or temp diff from standard) X 4 (a constant) X altitude (in thousands) = altitude correction, in this case
21 degrees (7x2=14, 15-14=1, -20 is 21 below standard) X 4 X 2 (thousand feet AGL) = 168 ' correction. 2000' AGL - 168 = 1832' AGL

This calculation will always leave you higher than actual altitude but it's just a rule of thumb so for exam you need to use the cold temp correction chart from the CAP but that requires knowing your height AAE which was not given. So I call SQ. (stupid question)

I believe the altimeter setting is just there to throw you off. If it said "you have descended from FL 250 and forgot to reset your sub scale to aerodrome elevation..." or "...you are flying in the standard pressure region..." then you'd be doing a non standard pressure calculation as well...from high to low lookout below.

[aileron]

I recall that Natural Resources, from which the terrain on aviation maps in Canada are derived, is rounded to the nearest 100m... so answer B (1500') does make sense even though your calculation comes to 1832'

[Trematode]

Guys -- this question is meant to be solved with an E6B.

You do not need nor do you want to use the cold weather correction charts in the CAP GEN for the CPL exam -- they won't even be provided to you.

The original poster is correct, it is simply a matter of finding your TRUE altitude with your E6B and then figuring out the difference between that and your ridge height above sea level. For that you need the altimeter setting to calculate your pressure altitude.

start with that:

29.92 - 29.42 = .5 x 1000 = +500'

so pressure altitude is 7,500'

On your E6B you take your pressure altitude and line it up with the outside air temp in the altitude correction window. Calibrated/Indicated Altitude on the inner scale (7000'), gives a TRUE altitude of ~6,500'.

Therefore the clearance over the ridge @ 5000' AMSL is 1,500'. Answer is B.


You don't really get into the cold weather correction charts until the INRAT.

[skateosiris]

Thanks for all the help I think I got it now, just another quick question:

An Aircraft is loaded to its max certified takeoff weight of 7,450 lbs. It is found that the CG is 2" too far aft of the datum. In an attempt to rectify the situation, it is decided to move some cargo with a mass of 260 lbs forward. How far forward must the cargo be moved in order to bring the CG within acceptable limits?

a.)56"
b.)57"
c.)58"
d.)59"


So I've been using the weight shift formula:

260/7450=2/x

And the answer I keep getting is 57.3 so I chose B, but they say the answer is C, am I missing anything?


Thanks for the help

[Aeros]

In the question the C of G is 2" too far aft. In order to fix the problem you have to move it forward at least 2". In this case, the weight that is moved is fixed. I agree with your math -- it has to be moved 57.3" forward to bring the C of G into limits. I think that your problem comes in how you interpret this number and turn it into a "whole" number. In your case you rounded down, to 57". If you moved the weight only 57" you wouldn't actually move the C of G the full 2" forward. It would remain slightly too far aft.

The weight has to be moved 57.3". Since that answer isn't in the list the weight would have to moved slightly further forward (ie Round UP) to ensure that the C of G is moved sufficiently far forward.

It looks like another pesky "examism".

[Teamflyer]

The formula to find height above a ridge is as follows:
((Indicated Alt - Ground Elevation) / 1000) X Difference between ICAO and actual temp X 4 = amount to be subtracted from indicated altitude.

This gives true altitude. Take your true altitude, subtract it from the height of the ridge and you have your clearance.

Is that a formula i can use cold temp correction as well?

[loopa]

The formula to find height above a ridge is as follows:
((Indicated Alt - Ground Elevation) / 1000) X Difference between ICAO and actual temp X 4 = amount to be subtracted from indicated altitude.

This gives true altitude. Take your true altitude, subtract it from the height of the ridge and you have your clearance.

Out of curiosity, when do you find yourself using this formula in a plane?

[trey kule]

Edited, as I really dont understand the difficulty

[Rowdy Burns]

Guys -- this question is meant to be solved with an E6B.
[...]
Calibrated/Indicated Altitude on the inner scale (7000'), gives a TRUE altitude of ~6,500'.

Therefore the clearance over the ridge @ 5000' AMSL is 1,500'. Answer is B.

Kudos (props, applause, etc.) to Trematode for getting to the right start, although he/she is wrong with one respect:

You don't just put the Calibrated/Indicated Altitude on the inner scale, although in this one particular example you can get away with it. You put the Calibrated/Indicated Altitude above the altimeter setting source, which happens to be at Sea Level:

Your Altimeter indicates an altitude of 7000 feet, with an altimeter setting of 29.42 (based on a station at sea level)

See the attachment from the E6B manual. I print-screened, copied and pasted it so it's all there (including the answer to three practice problems).

Cheers,

Rowdy

edited to add he/she (instead of just he)

[Trematode]

100% correct. Thanks for the clarification.

I think the important take away is that you are only applying corrected values to the non-standard air mass above the station elevation. This goes for finding True Alt. as in the example, and also when calculating your cold weather corrected minimums on your IFR approach plates.

And I'm a dude!

[Gear Jerker]

Another ridge crossing question from the sharper edge cpl written prep:

Navigation question 3:

A VOR station is located along a ridge 1500' above the height of the surrounding terrain. How far from the station will an aircraft, flying at 4500' AGL be able to receive a signal from the station?

a) 100nm
b) 120nm
c) 130nm
d) 150nm

My thinking: Assumption that there is no other terrain which interferes with line of sight, and therefore 1500 "AGL" can be added to 4500' AGL in this case, for 6000' "vhf station elevation". I realize that there is really only a difference of 3000' between the VOR and the aircraft, but I'm adding the two because in terms of how far away the aircraft can pick up a signal, the result is equivalent to if the VOR was at 0' and the aircraft was at 6000'. Or am I way off...?

Now applying this to the formula 1.23*square root of station altitude

1.23*square root of 6000
1.23*77.4596
95.3nm

Nowhere near the choices of answers given, no idea why. The answer they give is c) 130nm

[metal]

I think something is wrong with either the question or the answers. The correct formula first off, is 1.23 x the square root of the height above the station . So it would be 1.23 x sqrt3000 (station is at 1500', and you're flying at 4500')

The answer should be 67 miles..

[Gear Jerker]

@Metal, thanks for the reply.

I agree, I think the answers are way off, just wanted to double check.

Assuming that future questions aren't as messed up, though:

Are you sure that you're using 3000' as height above station in this particular case? I thought so at first, but if the station itself is 1500' above surrounding terrain, that increases reception distance in itself. Then, if the a/c is flying at 4500AGL, that increases the distance again.

[metal]

I just did a quick google search and found this formula, which gives 130 miles as the answer.

1.23√Receiver Ht (in feet) + 1.23√Transmitter Ht (in feet)

So..

1.23√1500 + 1.23√4500 = 130 miles