# ATPL theory questions

Join Date: Apr 2004

Location: London

Posts: 465

I agree with Alex. Common standard parallels are used in some regional sets. The publisher in question addresses this in its FAQ:

While this may be true for the chart in question it obviously depends on the actual latitude range to be charted.

Ignore it. The pertinent information was in the text and was adequately summarised in Deetz and Adams.

In practice because the scale on these projections is almost constant it is easier to assess distances by referring to a meridian graduated in minutes of latitude.

Q. Why does the scale of the VFR+GPS charts differ slightly from 1:500.000?

A. In order to support the European wide standard, a Lambert Conformal Conic Projection with Standard Parallels at N37° and N65° is used for production of all VFR+GPS charts. This enables the seamless combination of all charts and the use for continuous flight planning. A scale distortion is a consequence of the projection and varies with the distance from the Standard Parallels. The true scale is depicted on every VFR+GPS chart on the Cover panel (e.g. EB/EH: True Scale at N51° - 1:515.000) and near the Scale Bar. In addition, a factor is given to multiply measured distances with every parallel (e.g. 1.03).

A. In order to support the European wide standard, a Lambert Conformal Conic Projection with Standard Parallels at N37° and N65° is used for production of all VFR+GPS charts. This enables the seamless combination of all charts and the use for continuous flight planning. A scale distortion is a consequence of the projection and varies with the distance from the Standard Parallels. The true scale is depicted on every VFR+GPS chart on the Cover panel (e.g. EB/EH: True Scale at N51° - 1:515.000) and near the Scale Bar. In addition, a factor is given to multiply measured distances with every parallel (e.g. 1.03).

greater accuracy can be achieved by selecting two SPs at a range closer to 16 [deg]

I also must admit the math level ... is beyond my current knowledge.

In practice because the scale on these projections is almost constant it is easier to assess distances by referring to a meridian graduated in minutes of latitude.

Join Date: Aug 2005

Location: France

Posts: 10

**True Altitude Calculation**Hello

Could you please provide some assistance with the following question?

*Altimeter reading 4500 FT, OAT 20°C, calculate the true altitude*... has it got anything to do with pressure or density altitudes?

Join Date: Apr 2014

Location: Mexico

Age: 24

Posts: 5

I agree with Alex. Common standard parallels are used in some regional sets. The publisher in question addresses this in its FAQ

Quote:

Q. Why does the scale of the VFR+GPS charts differ slightly from 1:500.000?

A. In order to support the European wide standard, a Lambert Conformal Conic Projection with Standard Parallels at N37° and N65° is used for production of all VFR+GPS charts. This enables the seamless combination of all charts and the use for continuous flight planning. A scale distortion is a consequence of the projection and varies with the distance from the Standard Parallels. The true scale is depicted on every VFR+GPS chart on the Cover panel (e.g. EB/EH: True Scale at N51° - 1:515.000) and near the Scale Bar. In addition, a factor is given to multiply measured distances with every parallel (e.g. 1.03).

Quote:

Q. Why does the scale of the VFR+GPS charts differ slightly from 1:500.000?

A. In order to support the European wide standard, a Lambert Conformal Conic Projection with Standard Parallels at N37° and N65° is used for production of all VFR+GPS charts. This enables the seamless combination of all charts and the use for continuous flight planning. A scale distortion is a consequence of the projection and varies with the distance from the Standard Parallels. The true scale is depicted on every VFR+GPS chart on the Cover panel (e.g. EB/EH: True Scale at N51° - 1:515.000) and near the Scale Bar. In addition, a factor is given to multiply measured distances with every parallel (e.g. 1.03).

@

**dook ,**@

**Alex Whittingham**and

**@selfin**thank you very much for your help. I appreciate all the help and input you gave.

Join Date: Feb 2000

Location: It's a secret

Posts: 238

They've given you the OAT at 4500ft which means it's an ISA deviation question. You know what the ISA temp is at sea Level (15C) so you work out the ISA deviation then use the formula 4 x Indicated Alt in 1000's of feet x the ISA Deviation to calculate the height error.

Join Date: Jan 2017

Location: EDDH

Posts: 4

Hello everyone,

I am really struggling with two types of M&B questions, someone can help with a scheme?

First one:

"The loaded weight of an aeroplane is 100.000 kg.

The CG of the loaded aeroplane is at 20% MAC = Sta. 15,5 m.

The CG should be shifted to Sta. 16 m by moving cargo from the fwd. hold (sta. 10m) to the aft. hold (sta. 25m).

Initially, fwd. cargo load is 5.000kg and the aft. cargo load is 3.000kg.

How much cargo load is in the aft hold after the load shift to obtain the new CG?"

Second one:

"The loaded weight of an aeroplane is 13.000kg.

The CG of the loaded aeroplane is at Sta. 105,5in.

The aft. CG limit is at Sta. 102in.

How many seat rows (seat pitch 33in) must four passengers (75kg each) move forward from the last row (Sta. 224in), to bring the CG at least to the aft. limit?"

I am really struggling with two types of M&B questions, someone can help with a scheme?

First one:

"The loaded weight of an aeroplane is 100.000 kg.

The CG of the loaded aeroplane is at 20% MAC = Sta. 15,5 m.

The CG should be shifted to Sta. 16 m by moving cargo from the fwd. hold (sta. 10m) to the aft. hold (sta. 25m).

Initially, fwd. cargo load is 5.000kg and the aft. cargo load is 3.000kg.

How much cargo load is in the aft hold after the load shift to obtain the new CG?"

Second one:

"The loaded weight of an aeroplane is 13.000kg.

The CG of the loaded aeroplane is at Sta. 105,5in.

The aft. CG limit is at Sta. 102in.

How many seat rows (seat pitch 33in) must four passengers (75kg each) move forward from the last row (Sta. 224in), to bring the CG at least to the aft. limit?"

Join Date: Jan 2011

Location: England

Posts: 638

To solve any problem we need to begin by forming a clear picture of the situation.

In these two questions we are attempting to achieve a certain moment change by moving part of the load.

The moment change which we are trying to achieve is equal to the total mass multiplied by the distance we want to move the CofG. We could express this as M x S where M is the total mass and S is the required CofG shift.

We will achieve this by shifting a small part of the mass a certain distance. We could express this moment change which we are going to cause as m x s where m is the mass we move and s is the distance we shift it.

If we do the job properly the moment change we are trying to achieve (M xS) will be equal the moment change we cause by shifting the smaller mass (m x s)

So we will have M x S equals m x s

In the first question we are trying to calculate the mass to be moved, (s), so we rearrange the equation to give

(M x S) / m equals s (sorry my tablet does not appear to have an equals sign)

M is 100000, S is (16 - 15.5) which is 0.5, s is (25 - 10) which is 15

Inserting these number into our equation gives (100000 x 0.5) / 15 equals the mass to be moved to the aft hold.

Adding this to the initial mass of 3000 kg already in the hold gives us the answer.

The second question can be solved in a similar way, but this time we are trying to find the distance we need to move the 300 kg of the 4 passengers. Dividing this by the seat pitch will then give us the number of rows.

In these two questions we are attempting to achieve a certain moment change by moving part of the load.

The moment change which we are trying to achieve is equal to the total mass multiplied by the distance we want to move the CofG. We could express this as M x S where M is the total mass and S is the required CofG shift.

We will achieve this by shifting a small part of the mass a certain distance. We could express this moment change which we are going to cause as m x s where m is the mass we move and s is the distance we shift it.

If we do the job properly the moment change we are trying to achieve (M xS) will be equal the moment change we cause by shifting the smaller mass (m x s)

So we will have M x S equals m x s

In the first question we are trying to calculate the mass to be moved, (s), so we rearrange the equation to give

(M x S) / m equals s (sorry my tablet does not appear to have an equals sign)

M is 100000, S is (16 - 15.5) which is 0.5, s is (25 - 10) which is 15

Inserting these number into our equation gives (100000 x 0.5) / 15 equals the mass to be moved to the aft hold.

Adding this to the initial mass of 3000 kg already in the hold gives us the answer.

The second question can be solved in a similar way, but this time we are trying to find the distance we need to move the 300 kg of the 4 passengers. Dividing this by the seat pitch will then give us the number of rows.

*Last edited by keith williams; 7th Oct 2018 at 22:05.*

Join Date: Nov 2018

Location: New Delhi

Posts: 1

Hi everyone, A newbie here in the search of some help. I can't figure out how to solve this question of altimetry.

**Question :**

An aircraft takes off from A (elevation 600ft) aerodrome pressure 1008

An aircraft takes off from A (elevation 600ft) aerodrome pressure 1008

**mbs**

**. The altimeter on QNH reads 630 ft on**

**ground**

**has to clear a 7210 ft high hill midway by of margin 1500 ft before landing at B (elevation 330 ft) QFE = 1005**

**mbs**

**. If A/C**

**maintain**

**QNH of A throughout, find**

**Minimun altimeter reading to clear the hill.****Aircraft altimeter reading on landing.**

Join Date: Jul 2003

Location: UK

Posts: 446

If using 1Mb as 30ft (rounded up from 27ft) Then the QNH allowing for the over read of 30ft (630ft Alt) at A should be 1029. In theory, using 1029 and maintaining 8710 ft Altitude should give you the 1500ft clearance.

However! The pressure at B is lower (QNH + 11Mb = 330ft elev = 1016) Therefore, flying from a high pressure area to a lower pressure area, the adage "High to low, beware below" means that the altimeter reading a constant altitude but the aircraft would in fact descend in relation to the MSL. So you would need to fly higher. My suggestion would be the difference between 1029Mb and 1016 Mb ( 13Mb = 390ft) minimum (Altitude of 9100ft).

If you require more accurate, then use 27ft per Mb

However! The pressure at B is lower (QNH + 11Mb = 330ft elev = 1016) Therefore, flying from a high pressure area to a lower pressure area, the adage "High to low, beware below" means that the altimeter reading a constant altitude but the aircraft would in fact descend in relation to the MSL. So you would need to fly higher. My suggestion would be the difference between 1029Mb and 1016 Mb ( 13Mb = 390ft) minimum (Altitude of 9100ft).

If you require more accurate, then use 27ft per Mb

*Last edited by WASALOADIE; 13th Nov 2018 at 15:51. Reason: Addition of text*

Join Date: Apr 2014

Location: Mexico

Age: 24

Posts: 5

Hi everyone!

Quick and probably dumb question regarding ATS Routes: What does

I cannot find, in either the Jeppesen or anywhere else, any explanation beyond that they are used by international traffic. Does it mean that a "ULXXX" RNAV airway for example will not cross borders between states (countries)? Why are certain Juliet airways that do cross borders then?

Quick and probably dumb question regarding ATS Routes: What does

**ATS****regional routes**actually mean ?I cannot find, in either the Jeppesen or anywhere else, any explanation beyond that they are used by international traffic. Does it mean that a "ULXXX" RNAV airway for example will not cross borders between states (countries)? Why are certain Juliet airways that do cross borders then?

*Last edited by erikfj; 15th Nov 2018 at 00:47.*