Simplified Flight Planning Chart ATPL

[npocock]

This may seem basic to many, but please help me overcome my lack of insight.

I am having troubling understanding the useage between the dashed and solid lines on the simplified flight planning charts. After you have reached the ref line, which line should you follow? I realise that the instructions say interpolate, but I am not seeing between which lines to interpolate. Do I find a difference between the dashed lines or is it the solid lines or both?

This is driving me nuts and seriously holding me up.:ugh:

Thanks for your help in advance.

[Keith.Williams.]

They are all for pressure altitudes in 1000s of feet.

If you look at Figure 4.3.1b you will see that one of the solid line is marked 10. This is the 10000 ft line.

The solid line below it is for 14000 ft.

Now go up to the top of the grid and find the highest dotted line. This is for 18000 ft, next one down is 22000 ft, next one down is 25000 ft, next one down is 29000 ft, next one down is 33000 ft and the next one down (which is marked 37) is for 37000 ft.

Having reached the reference line you move parallel to these slopes until you are above the Landing weight, then go horizonatlly to the right to the edge of the graph to read off the Trip Fuel.

Try the question below and see if it helps.

(For this question use The CAP 697 Flight Planning Manual MRJT 1 Figure 4.3.1.b)

Given:
Estimated zero fuel mass 50 t;
Estimated landing mass at alternate 52 t;
Final reserve fuel 2 t; alternate fuel 1 t;
Flight to destination, distance 720 NM,
True course (TC) 030, W/V 340/30;
Cruise: long range FL 330,
Outside air temperature -30 ° C.

Find: estimated trip fuel and time

A 4 600 kg; 02: 05
B 4 800 kg; 01: 45
C 4 400 kg; 02: 05
D 4 750 kg; 02: 00

ISA temperature at 33000ft is 15 – 66 = -51°C.

The actual temperature is -30°C.

So the ISA deviation is (-30 actual temp) – (-51 ISA temp) = ISA +21°C.

Use CRP-5 to find headwind component, 20kt

Enter the bottom of the graph at trip distance 720NM.

Move vertically up to the reference line.

Move parallel to the curved lines to headwind 20kt.

Draw a line vertically up to the line marked “29 & above” close to the top of the page.

Move horizontally left to the reference line.

Move parallel to the sloping lines to the ISA +20°C line at the left edge of the graph.

Read off the time, 1.9h = 1h 54m.

At the lower set of pressure altitude lines find the intersection of the vertical line previously drawn and the 33000ft line.

Move right to the reference line.

Move parallel to the 33000ft sloping line to a point directly above the landing weight of 52000kg.

Move horizontally to the right edge of the graph.

Read off the fuel required, 4700kg.

[npocock]

Thank you Keith. The value of each line is what had been plaguing. Now that you have kindly pointed that out, I can move on with the right calculations. Thank you!

[npocock]

With that explanation in mind, how would you set about solving a questions like the one below?:

(Refer to CAP 697 figure 4.3.1)
Trip distance: 1900 NM
Fuel on board: 15 000 kg
Landing weight: 50 000kg
What is the minimum pressure altitude for this flight?


a FL370

b 10 000 ft

c 17 000 ft

d FL250

[Keith.Williams.]

Having looked at the matter at more length I can see that my earlier comments were incorrect. My interpretation appears to work for Figure 4.3.1b, but not for the others.

From item f) in the instructions in paragraph 3.2, I can see that each solid line represents one fixed pressure altitude and each dotted line represents another.

In Figure 4.3.1a for example the dotted lines all represent 33000 ft and above, while all of the solid lines represent 10000 ft. Looking at the example on Figure 4.3.1a, it is clear that the author has interpolated between the two lines to get the slope for the arrow from the reference line down to the 30000 kg landing weight.

In the case of this example, the actual pressure altitude of 25000 ft is about two thirds of the way up from the 10 line to the 33 line. If you draw this in your book you will see that it is parallel to the sloping arrow in the example.

I can suggest two ways of answering your new question. Both ways use Figure 4.3.1c.

Method 1.
There is no wind, so draw a straight line vertically upward from 1900 nm to the 10000 ft line in the centre of the table.

Draw a straight line vertically upward from 50000 kg landing weight to the top of the grid.

From the intersection of the 1900 nm vertical line and the 10000 ft pressure altitude line, draw a horizontal line to the right to the reference line.

Draw a sloping line parallel to the solid sloping (10000 ft) line, to intersect the vertical line from 50000 kg.

From the intersection of these two lines draw a horizontal line to the right edge of the table. This should give you about 16650 kg of fuel used.

Repeat the process for 17000 ft, 25000 ft and 37000 ft, but in these cases interpolate between the solid and dotted sloping lines.

From this process I get the following figures for fuel required.

10000 ft 16650 kg.
17000 ft 14800 kg.
25000 ft 12700 kg.
37000 ft 10800 kg.

From these figures it can be seen that 17000 ft with 14800 kg of fuel used, is the lowest altitude to achieve 1900 nm with 15000 kg of fuel.

Method 2.
The second method is to draw the two vertical lines as in method one.

From 15000 kg fuel at the right edge of the graph, draw a horizontal line to the left to interest the vertical line above 50000 kg landing weight.

We now need to slope down to the reference line, interpolating between the solid and dotted lines.

From the reference line we draw a horizontal line to intersect the vertical line above 1900 nm. This intersection gives us the required pressure altitude.

But without knowing the pressure altitude we cannot do the interpolation. In effect we need to know the answer before we can get the answer!

But the dotted and sloping lines are very close in this part of the table so it doesn’t make much difference. Whether we use the dotted line, the solid line or a line somewhere between the two, we get a figure of about 17000 ft for the pressure altitude.

The above methods are of course based on the assumption that we will use all or very nearly all of the 15000 kg of fuel. This means that we have little or no fuel left when we arrive overhead our destination. Flying at 17000 ft for example leaves only 200 kg for landing!

If this question were to appear in a JAR ATPL exam we would be left second guessing the examiner...did he/she want to catch us out overlooking reserves, or did he/she simply want us to demonstrate that we can use the graphs?

[npocock]

Thank you very much again Keith. I am due to sit tomorrow (TUES) in Florida, so we will see in about a months time what the examiner was looking for, if it comes up.

The explanation was just what I was looking for. Thank you!