# Old Q that never seem to die or fade away.

Thread Starter

Join Date: Jul 2012

Location: Africa

Posts: 6

**Old Q that never seem to die or fade away.**

Good day to all.

I am apologizing in advance for asking the following questions over and over. I've read a lott about them and for some reason I can not understand where some guys get the figures they work on. The question I seek some help with we all would recognize out of which and book it comes and for which carrier they are normally associated with.

1) Given airfield elevation of 1500 ft, altimeter 1000 millibars, calculate pressure altitude:

2)Given an indicated altitude of 10,000 ft, and an actual OAT of -20°C, you set your altimeter to a local station setting of 29.62". If the station elevation is 2,500 ft what is your actual altitude'

Here's a explanation on what to do where ( no disrespect to the individual who posted it, its just I am not understanding it 100%).

The temperature correction formula is:

True Alt = Ind Alt + (ISA dev x 4/1000 x Ind Alt)

In this case, substitude 7500 for 10000 feet in the second "Ind Alt" in the above.

So True Alt = 10000 + (-15 x 4/1000 x 7500) = 9550

But why the 4/1000 ? is it something I am missing or what. I see some blogs say they use 6/1000 for the same Q ? Where does one get a solid A for the formula?

3) There are two runways, one is 8,000' and one is 10,000' long. With respect to second segment climb, which runway will allow for greater 2nd segment performance.

4)On a 3° glide path, OAT -25°, OM crossing height of 1760', the aircraft altitude will indicate

All information and explanations are greatly appreciated.

I am apologizing in advance for asking the following questions over and over. I've read a lott about them and for some reason I can not understand where some guys get the figures they work on. The question I seek some help with we all would recognize out of which and book it comes and for which carrier they are normally associated with.

1) Given airfield elevation of 1500 ft, altimeter 1000 millibars, calculate pressure altitude:

*This Q has been posted over and over yet I can not figure out what the hell to do. For I know I am not the sharpest tool in the shed, all I am asking is some guidance from all.*2)Given an indicated altitude of 10,000 ft, and an actual OAT of -20°C, you set your altimeter to a local station setting of 29.62". If the station elevation is 2,500 ft what is your actual altitude'

Here's a explanation on what to do where ( no disrespect to the individual who posted it, its just I am not understanding it 100%).

The temperature correction formula is:

True Alt = Ind Alt + (ISA dev x 4/1000 x Ind Alt)

In this case, substitude 7500 for 10000 feet in the second "Ind Alt" in the above.

So True Alt = 10000 + (-15 x 4/1000 x 7500) = 9550

But why the 4/1000 ? is it something I am missing or what. I see some blogs say they use 6/1000 for the same Q ? Where does one get a solid A for the formula?

3) There are two runways, one is 8,000' and one is 10,000' long. With respect to second segment climb, which runway will allow for greater 2nd segment performance.

4)On a 3° glide path, OAT -25°, OM crossing height of 1760', the aircraft altitude will indicate

All information and explanations are greatly appreciated.

Join Date: Jan 2008

Location: Reading, UK

Posts: 10,684

1) Given airfield elevation of 1500 ft, altimeter 1000 millibars, calculate pressure altitude:

*This Q has been posted over and over yet I can not figure out what the hell to do. For I know I am not the sharpest tool in the shed, all I am asking is some guidance from all.*The answer will differ in either case.

Join Date: Oct 2009

Location: UK

Posts: 1,271

Hi Diesel and Dust,

The Altimeter indicates the Pressure Altitude when set to 1013

Since QNH = 1,000 mbs, you'd simply wind on another 13mbs on the sub scale.

The Altimeter would then indicate a Pressure Altitude of 1500 + (13*27ft = 351 ft), i.e. 1851 ft.

The temperature correction is proportional to the Absolute Temperature scale (measured in degrees Kelvin) where ISA at sea level = 288° K. (273+15)

If the temperature deviation is ISA -15 degs C, then apply a correction of 15/288 * Height of the column of air (7,500) = 390 ft.

So I get 10,000 - 390 = 9610 ft.

I'm not certain what the examiner is asking here, but assuming all Obstacles are measured from the far end of the runways, then starting 2,000 feet further back with the same aircraft performance will give you more clearance during the second segment climb.

The 3° glide slope will place you physically at the correct height at the OM. Assume airfield is at sea level.

Since surface temp = -25 (ISA-40), then apply 40/288 * 1760 correction = 244ft

Therefore the Altimeter would over read by 244 feet, and show 2004 ft at the OM.

1) Given airfield elevation of 1500 ft, altimeter 1000 millibars, calculate pressure altitude:

Since QNH = 1,000 mbs, you'd simply wind on another 13mbs on the sub scale.

The Altimeter would then indicate a Pressure Altitude of 1500 + (13*27ft = 351 ft), i.e. 1851 ft.

But why the 4/1000 ? is it something I am missing or what.

If the temperature deviation is ISA -15 degs C, then apply a correction of 15/288 * Height of the column of air (7,500) = 390 ft.

So I get 10,000 - 390 = 9610 ft.

3) There are two runways, one is 8,000' and one is 10,000' long. With respect to second segment climb, which runway will allow for greater 2nd segment performance.

4)On a 3° glide path, OAT -25°, OM crossing height of 1760', the aircraft altitude will indicate

Since surface temp = -25 (ISA-40), then apply 40/288 * 1760 correction = 244ft

Therefore the Altimeter would over read by 244 feet, and show 2004 ft at the OM.

*Last edited by rudderrudderrat; 27th Aug 2012 at 04:02. Reason: typos & credit to Lord Kelvin*

Thread Starter

Join Date: Jul 2012

Location: Africa

Posts: 6

**Hi, RudderRudderrat**

Thanks for all the information regarding the calculations and runway performance. I've never worked from 288deg K @ MSL.Been using your method on various other examples, and it seems to be working correct.

''Since surface temp = -25 (ISA-40), then apply 40/288 * 1760 correction = 244ft'' . Just want to make 100% sure.

Isn't it ISA = 15 deg @ MSL.

Surface temp : -25 deg C = ISA-10 deg Cel. Then using the formula : 10/288deg K * 1760 = 61 feet.

= 1760+61= 1821 Feet @ the FAN marker/OM.

''Since surface temp = -25 (ISA-40), then apply 40/288 * 1760 correction = 244ft'' . Just want to make 100% sure.

Isn't it ISA = 15 deg @ MSL.

Surface temp : -25 deg C = ISA-10 deg Cel. Then using the formula : 10/288deg K * 1760 = 61 feet.

= 1760+61= 1821 Feet @ the FAN marker/OM.

Thread Starter

Join Date: Jul 2012

Location: Africa

Posts: 6

Hi Rudderrudderrat

Just verify this calculation. Does it make sense ?

An aircraft at FL230, temperature -41º C, QNH 983 HPA, is at a true altitude of:

OK, so the a/c is flying on 1013 and the QNH at the airfield is 983Hpa. : 1013 -983 = 30Hpa x 30feet/1 Hpa = 900 feet dif in elevation en 1013.25.

ISA @ FL 230 : ISA-10 deg Cel.

: @ MSL ISA is 288 deg Kelvin.

: kelvin is 273 +15 = 288 deg K

: 10 / 288 * 23900 feet = 830 feet.

= 23000 - 830 = 22170 feet (True .Alt)

Just verify this calculation. Does it make sense ?

An aircraft at FL230, temperature -41º C, QNH 983 HPA, is at a true altitude of:

OK, so the a/c is flying on 1013 and the QNH at the airfield is 983Hpa. : 1013 -983 = 30Hpa x 30feet/1 Hpa = 900 feet dif in elevation en 1013.25.

ISA @ FL 230 : ISA-10 deg Cel.

: @ MSL ISA is 288 deg Kelvin.

: kelvin is 273 +15 = 288 deg K

: 10 / 288 * 23900 feet = 830 feet.

= 23000 - 830 = 22170 feet (True .Alt)

*Last edited by Diesel and Dust; 27th Aug 2012 at 08:49.*