Finding Clearance above a ridge
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Finding Clearance above a ridge
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
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
Re: Finding Clearance above a ridge
Of the top of my head, I would say A, based on the cold weather correction factor.
Success in life is when the cognac that you drink is older than the women you drink it with.

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Re: Finding Clearance above a ridge
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.
((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
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Re:
You do not need to clear the rige, but pass the ...*&^ exam...Beefitarian wrote: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.
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Re: Finding Clearance above a ridge
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, 1514=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.
TDFS (or temp diff from standard) X 4 (a constant) X altitude (in thousands) = altitude correction, in this case
21 degrees (7x2=14, 1514=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.
Re: Finding Clearance above a ridge
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'
Re: Finding Clearance above a ridge
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.
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.

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Re: Finding Clearance above a ridge
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
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
Re: Finding Clearance above a ridge
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".
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".
Re: Finding Clearance above a ridge
Is that a formula i can use cold temp correction as well?DHC1 Jockey wrote: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.
Re: Finding Clearance above a ridge
Out of curiosity, when do you find yourself using this formula in a plane?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.
Re: Finding Clearance above a ridge
Edited, as I really dont understand the difficulty

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 Joined: Wed Oct 19, 2011 1:07 pm
Re: Finding Clearance above a ridge
Kudos (props, applause, etc.) to Trematode for getting to the right start, although he/she is wrong with one respect:Trematode wrote: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.
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:
See the attachment from the E6B manual. I printscreened, copied and pasted it so it's all there (including the answer to three practice problems).skateosiris, OR QUESTION wrote:Your Altimeter indicates an altitude of 7000 feet, with an altimeter setting of 29.42 (based on a station at sea level)
Cheers,
Rowdy
edited to add he/she (instead of just he)
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Re: Finding Clearance above a ridge
100% correct. Thanks for the clarification.
I think the important take away is that you are only applying corrected values to the nonstandard 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!
I think the important take away is that you are only applying corrected values to the nonstandard 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
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 Joined: Tue Sep 27, 2011 12:48 am
Re: Finding Clearance above a ridge
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
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
Re: Finding Clearance above a ridge
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..
The answer should be 67 miles..
 Gear Jerker
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Re: Finding Clearance above a ridge
@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.
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.
Re: Finding Clearance above a ridge
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
1.23√Receiver Ht (in feet) + 1.23√Transmitter Ht (in feet)
So..
1.23√1500 + 1.23√4500 = 130 miles
 Gear Jerker
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 Posts: 188
 Joined: Tue Sep 27, 2011 12:48 am
Re: Finding Clearance above a ridge
The idea behind the formula is line of sight. Between the antenna and the horizon, then between the horizon and the aircraft.
Re: Finding Clearance above a ridge
Do yourself a favour  buy a @*lhanes for the test bank. The author of the Sharper Edge is a nice guy, and the study portion of the book was well done, but the test questions did not prep me for the written.
Re: Finding Clearance above a ridge
Hello all I am stuck on this part of the formula  Difference between ICAO and actual temp? I know standard temp is 15 however how do they get an answer of 42 when finding the difference between Indicated OAT 40 and ISA temp. ?
how exactly do i calculate this difference with accuracy? my particular question is:
ground elevation of airport: 3000 ft
altimeter setting: 30.43
indicated OAT at 6,500 ft: 40C
indiated alt: 6,500 ft
height of ridge: 5,850 ft
ans: 62 feet
any help would be greatly appreciated.
how exactly do i calculate this difference with accuracy? my particular question is:
ground elevation of airport: 3000 ft
altimeter setting: 30.43
indicated OAT at 6,500 ft: 40C
indiated alt: 6,500 ft
height of ridge: 5,850 ft
ans: 62 feet
any help would be greatly appreciated.
AP
Re: Finding Clearance above a ridge
Easy!
So you need to find the difference between the actual temperature and the ICAO temp.
ICAO temp = 15 C at MSL. You deduct 2 degrees per thousand from there as you go up. So at 6,500 feet you take 6.5 X 2 = 13, then deduct 13 from 15 = 2 at 6,500.
So if the ICAO temp at 6,500 is 2, and the indicated is 40, we have a differential of 42. That is to say, from 40 to 0 = 40, then from 0 to 2 is another 2, for a total difference of 42. Then you plug this into the true altitude formula. Piece of cake.
So you need to find the difference between the actual temperature and the ICAO temp.
ICAO temp = 15 C at MSL. You deduct 2 degrees per thousand from there as you go up. So at 6,500 feet you take 6.5 X 2 = 13, then deduct 13 from 15 = 2 at 6,500.
So if the ICAO temp at 6,500 is 2, and the indicated is 40, we have a differential of 42. That is to say, from 40 to 0 = 40, then from 0 to 2 is another 2, for a total difference of 42. Then you plug this into the true altitude formula. Piece of cake.