Basic maths for ATP [Archive]

Keith.Williams.

13th Sep 2009, 11:32

Is the material below the kind of thing that you require?

CIRCLES 2.
Given: Position A 45° N, ?° E, Position B 45° N, 45° 15'E Distance A-B = 280 NM, B is to the East of A, What is the longitude of position A?

a. 38° 39'E.
b. 49° 57'E.
c. 51° 51'E.
d. 40° 33'E.

The distance in nm between any two meridians along a specified parallel of latitude is called the departure. This can calculated using the departure equation:

Departure (in nm) = Change in longitude x Cos latitude

This can be rearranged to give:

Change in longitude = Departure (in nm) / Cos latitude.

In this question the latitude is constant at 45degrees N, and the departure is 280 nm.

Inserting these figures into the above equation gives:

Change in longitude = 280 nm/Cos 45degrees = 395.98’ which divided by 60 = is 60 36’.

The question states that point B is to the east of point A so subtracting the above figure from the longitude of b gives: B (45° 15'E ) – (60 36’) = 38deg 39’ (option a).

CIRCLES 3.
A Lambert conformal conic chart has a constant of the cone of 0.75. The initial course of a straight-line track drawn on this chart from A (40° N 050° W) to B is 043° (T) at A; course at B is 055° (T). What is the longitude of B?

a. 34°W.
b. 36°W.
c. 38°W.
d. 41°W.

To solve this type of problem the following points must be noted:

a. A straight line on a Lambert conformal chart is a great circle.
b. The direction of great circles changes along their lengths by an amount
equal to the convergence of the meridians between any two points on
the great circle.
c. For a Lambert conformal chart the equation for the convergence of
meridians is:

Convergence = Change in longitude x constant of the cone

The initial course at point A (40° N 050° W) is 043° (T) and at point B is 055° (T). This means that the difference in course is 12 degrees, which is also the convergence between the meridians at A and B. So the convergence between points A and B is 12 degrees.

The question states that the constant of the cone is 0.75.

Inserting these values into the convergence equation above gives:

12 degrees = Change in longitude x 0.75

12 degrees / 0.75 = 16 degrees.

The initial longitude at point A is 050° W so the longitude at point B is 050° W - 16 degrees = 34 degrees (option a).

DESCENTS 18.
If there is a 15 knot decrease in headwind by what amount must the rate of descent be changed in order to maintain a 3° glideslope?

a. It must be increased by 50 ft/min.
b. It must be decreased by 50 ft/min.
c. It must be decreased by 75 ft/min.
d. It must be increased by 75 ft/min.

An approximate value for the change in ROD required to maintain a 30 glideslope for any given change in headwind or tailwind can be calculated using the following equation:

Change required (in ft/min) = 5 x Headwind or tailwind change (in kts).

So for a 15 kts decrease in headwind ROD must change by 5 x 15 = 75 ft/min.

Headwind increase glideslope angle, so to maintain a 3 degree slope the ROD must be decreased in a headwind and increased in a tailwind.

So to maintain a 3 degree glideslope in a 15 kt decrease in headwind, the ROD must be increased by 75 ft/min (option d).

DESCENTS 19.
An aircraft at FL290 is required to commence descent when 50 NM from a VOR and to cross that VOR at FL80. Mean GS during descent is 271kt. What is the minimum rate of descent required?

a. 1800 FT/MIN.
b. 1900 FT/MIN.
c. 2000 FT/MIN.
d. 1700 FT/MIN.

The aircraft is to descend from FL290 to FL80.

The required height change can be calculated using the equation:

Required change = (New FL – Initial FL) x 100 ft

Which is (80 - 290) x 100 = -21000 ft.

The ground distance available is 50 nm and the ground speed is 271 Kts.

So the time taken to descend will be 50 nm / 271 nm/h = 0.1845 hours, which is 11.1 minutes.

ROD can now be calculated using the equation: ROD = height change / time taken.

Inserting the figures calculated above gives:

ROD = 21000 ft / 11.1 m = 1892 ft/min or approximately 1900 ft/min(option b).

PSR & PET 70.
Given: Distance 'Q' to 'R' 1760 NM, Groundspeed 'out' 475 kt, Groundspeed 'back' 365 kt. The time from 'Q' to the Point of Equal Time
(PET) between 'Q' and 'R' is?

a. 91 MIN.
b. 106 MIN.
c. 92 MIN.
d. 97 MIN.

This type of problem can be solved using the standard equation below:

PET = DH / (O + H) where:

D = distance of total leg, H = homebound groundspeed, O = outbound groundspeed.

The distance of total leg is 1760 nm

Inserting these figures into the equation gives:

PET = (1760 nm x 365 nm/h) / (475 nm/h + 365 nm/h) = 765 nm.

The time taken to reach this point can be calculated using the equation:

Time taken = distance / groundspeed out = 765 nm / 475 kts = 1.61hour or 97 mins (option d).

PSR & PET 71.
Given: Distance 'A' to 'B' 3623 NM, Groundspeed 'out' 390 kt, Groundspeed 'back' 330 kt. The time from 'A' to the Point of Equal Time (PET) between 'A' and 'B' is?

a. 238 MIN.
b. 255 MIN.
c. 288 MIN.
d. 323 MIN.

This type of problem can be solved using the standard equation:

PET = DH / (O + H) where:

D = distance of total leg, H = homebound groundspeed, O = outbound groundspeed.

The distance of total leg is 3623 nm

Inserting these figures into the equation gives:

PET = (3623 nm x 330 nm/h) / (390 nm/h + 330 nm/h) = 1660 nm.

The time taken to reach this point can be calculated using the equation:

Time taken = distance / groundspeed out = 1660 nm / 390 kts = 4.26 hours or 255 mins (option b).

SCALE 25.
The total length of the 75°N parallel of latitude on a direct Mercator chart is 140 cm. What is the approximate scale of the chart at latitude 30°S?

a. 1 : 25 000 000.
b. 1 : 2 900 000.
c. 1:125000.
d. 1 : 6 000 000.

This type of problem can be solved using the equation:

Scale factor = Chart distance / Earth distance

The problem can be simplified by noting that on a Mercator chart, the lines of longitude are parallel, so the chart distance is the same at all latitudes.

The Earth distance at 30 degrees can be calculated using the departure equation:

Departure in nm = change of longitude in min x cos latitude

A full circle of the earth is 360 degrees which is 360 degrees x 60 = 21600 min

So at 30 degrees N this would be a departure of 21600 x cos 30 degrees = 18706.15nm

This can be converted into cm by multiplying by the factors 185200 cm/nm to give earth distance at 30 degrees N 18706.15 x 185200 = 3 464 378 743 cm

Inserting this and the chart distance of 140 cm into the scale equation gives:

Scale factor at 30 degrees N = 140 cm / 3 464 378 743 cm = 1 / 24745563 or approximately 1 : 25 000 000 (option a)

SCALE 26.
The total length of the 70°N parallel of latitude on a direct Mercator chart is 120 cm. What is the approximate scale of the chart at latitude 40°S?

a. 1 : 16 050 000.
b. 1 : 2 600 000.
c. 1:7 250 000.
d. 1 : 25 500 000.

This type of problem can be solved using the equation:

Scale factor = Chart distance / Earth distance

The problem can be simplified by noting that on a Mercator chart, the lines of longitude are parallel, so the chart distance is the same at all latitudes.

The Earth distance at 40 degrees can be calculated using the departure equation:

Departure in nm = change of longitude in min x cos latitude

A full circle of the earth is 360 degrees which is 360 degrees x 60 = 21600 min

So at 40 degrees N this would be a departure of 21600 x cos 40 degrees = 16546.6 nm

This can be converted into cm by multiplying by the factors 185200 cm/nm to give earth distance at 40 degrees N = 16546.6 x 185200 = 3064422907 cm

Inserting this and the chart distance of 120 cm into the scale equation gives:

Scale factor at 40 degrees N = 120 cm / 3064422907 cm = 1 / 25 536 857 or approximately 1 : 25 500 000 (option d)

SOLAR SYSTEM &TIME 43.
The sun rises at 55 º N and 050ºE on the 25th January at 0254 UTC. On the same day what time does it rise at 55 º N and 040ºW?

a. 0854 UTC.
b. 2154 UTC.
c. 0714 UTC.
d. 0514 UTC.

This type of problem can be solved using the standard equation:

Time difference (in hours) UTC = Change in longitude West (in degrees) /15

The change in longitude in this question = 40 degrees W + 50 degrees E = 90 degrees W

Inserting this into the above equation gives:

Time difference = 90 / 15 = 6 hours.

Adding this to the sunrise time at 50 degrees E gives:

Sunrise at 40 degrees 0254 UTC + 6hours = 0854 (option a).

SOLAR SYSTEM &TIME 44
The sun sets at 45 º N and 025ºE on the 21st March at 1633 UTC. On the same day what time does it rise at 45 º N and 060ºW?

a. 2254 UTC.
b. 2154 UTC.
c. 1714 UTC.
d. 2213 UTC.

This type of problem can be solved using the standard equation:

Time difference (in hours) UTC = Change in longitude West (in degrees) /15

The change in longitude in this question = 60 degrees W + 25 degrees E = 85 degrees W

Inserting this into the above equation gives:

Time difference = 85 / 15 = 5.67 hours, which is 5 h 40 mins.

Adding this to the sunrise time at 25 degrees E gives:

Sunset at 60 degrees W 1633 UTC + 5 h 40 min = 2213 UTC (option d).

If it is what you are looking for then you can get in my book entitled "1000 Question Answers And Explanations For JAR ATPL General Navigation" from POOLEYS or AFE.