The Lambert's Chart Flashcards

1
Q

On a simple conic projection, scale is only correct at the

A

standard parallel

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2
Q

Meridians on a simple conic projection are -

And the convergency is -

A

straight lines

Constant

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3
Q

On the simple conic projection convergency is constant, and the convergency factor is called -

A

constant of the cone or (n)

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4
Q

Constant of the cone =

Thus chart convergency =

A

Angular extent of the chart / Change of longitude

I.E, 231.4 / 360 = 0.6428, which is also the sine of the standard parallel.

D’ long x constant of the cone( in this case the constant = 0.6428) or sine parallel of origin

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5
Q

On the conformal conic projection, convergency(chart convergency) =

A

D’long x sin latitude of parallel of origin or constant of the cone

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6
Q

On the lambert chart, great circle track:

Is concave to -

But, this curve is so small, for practical reasons great circles on a lambert chart is considered -

A

the parallel of origin

straight

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7
Q

On the lambert chart, rhumb lines are -

Towards the -

A

curved

Equator

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8
Q

Convergence on a lambert chart is -

Chart convergence =

A

constant across the chart. But not zero as with the mercator chart!

ch.long x sin parallel of origin

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9
Q

On a lambert chart, scale expand outside -

and is correct at -

the scale contracts -

A

the standard parallels

The standard parallels

Between, within, the standard parallels

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10
Q

On a lambert chart, convergency/earth convergency is replaced by -

The formula =

Thus, chart conversion angle =

A

chart convergency

ch.long x sin parallel of origin

1/2 ch.long x sin parallel of origin

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11
Q

On the lambert chart, the convergence between 2 given meridians remains constant with -

A

Latitude, because the formula utilises the sine of the origin.

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12
Q

How to solve radio bearings plotting problems on a lambert chart:

A

Draw the lines that represent the meridians, according to the corresponding hemisphere.

Remember that we are looking for the difference between the NDB meridian at the aircraft meridian.

Once we have established which meridian is where, and after calculating the QUJ or QTE, simply take the aircraft’s meridian and place it over the meridian of the NDB. This will indicate whether we need to add or subtract the conversion angle.

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13
Q

If the CCF is 0.8, what is the parallel of origin latitude?

A

inverse of sine(0.8)

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14
Q

what is different between the earth convergency and chart convergency formulas?

A

Earth convergency = ch’long x sine of latitude

Chart convergency = ch’long x sine of parallel of origin

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15
Q

Normally the bearing required to plot on a mercator chart is not a GC bearing but a -

A

RL bearing

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16
Q

How to solve:

An aircraft heading 153° (T) obtains a relative bearing from an NDB of 293°. The convergency between the aircraft and the station is 9°. The bearing to plot on a lamberts chart in the northern hemisphere is:

A

Remember that on a lamberts chart we require the GC track, and not the RL track as with the mercator.

Also, since the bearing is already a great circle track, we need not convert it to rhumb line as we would with the mercator. Thus, we are only required to add convergency and not conversion angle.

153° (T) + 293 - 360 = 086(QTE GC)

Draw a sketch, and overlay the aircraft meridian over the NDB meridian. In order convert the NDB’s meridian to the aircraft’s meridian.

You will see that we need to add convergency.

086 + 9 = 93

93 + 180 = 275

17
Q

What is the constant of the cone for a Lambert conic projection whose Standard Parallels are at 50°N and 70°N?

A

50 + 70 / 2 = 60N.

sine of 60 = 0.8666

18
Q

Convergency on the lambert chart is -

But is only correct at the -

The formula is -

A

constant throughout latitude

parallel of origin

ch’long x constant of the cone

19
Q

For practical purposes a great circle on the lambert chart is a -

But has a slight curve concave to the -

A

straight line

parallel of origin

20
Q

Great circle distances on a lambert must be measured at the -

A

mid latitude