AM
od
ello
oN
|
Slide
Graphic
Computer
Slide
Graphic
Computer
Manual
life;
however,
some
components
are
made
of
plastic
materials.
Since
all
plastic
materials
are
sensitive
to
extreme
heat,
do
not
store
the
computer
in
direct
CAUTION:
This
computer
is
designed
for
along
service
|
sunlight
or
in
an
area
subject
to
high
temperatures.
|
|
|
©
Jeppesen
Sanderson,
nc,
1981,
1987
All
Rights
Reserves
89143050
55
Inverness
Driva
East,
Engiewood,
CO
00112-5490
rt
TABLE
OF
CONTENTS
Calculator
Side
of
Flight
Computer
..--..--
Time
and
Distance
.....
Finding
Time...
-
Finding
Distance
.
.
-
Finding
Speed...
.
Short
Time
and
Distance
.
Finding
Short
Time
.
Fuel
Consumption
.
Finding
Time.
.....
‘3
Fuel
Burned
(Gallons)
.
..
.
Fuel
Consumption
(g.p.h.)
-
True
Airspeed
.......-5-2-0
00+
a
Converting
Mach
Number
to
True
Airspeed..........
14
Changing
Nautical
Values
to
Statute
Equivalents,
StatutetoNautical,
and
StatutetoKilometers.
.
14
Multi-Part
Problems
.....
20...
os.ece
ns
16
Finding
Drift
Angle
(Off-Course
Problem)
.......
SAT
Time
and
Distance
toa
VOR
Station...............
20
BE
CUe
AIL
CUCG
acer
coteet
sarecwspent
entoremae
aoe
eer?
<n
23
PICISIC
VALE
TUCO
ei
fea
net
el
eestehas
Crates
=
cleeeicne
ale
caer
25
Multiplication
and
Division
...................-44
26
Multiplication
A
pr
yaZo)
oss,
OrRO
Mtns
RRL
26
DIWIslon
ver
aoncres
sc
Pais
Wins
Aeias:
wine
eee
27
Converting
Feet
Per
Nautical
Mile
to
ReatsReriminute
ts
«dest
ons
guicecn
eee
ecegee
28
Converting
Fahrenheit
to
Celsius
(centigrade)
........29
Wind
Side
of
Computer
.....................
al
Determining
Ground
Speed
and
True
Heading........
33
Finding
Unknown
Wind................0-.e-000
35
Finding
Altitude
for
Most
Favorable
Winds..........
38
RadlusiofvAction
sc
i's
fee
ace
ee
40
Metric\Conversions.....<-<:
oe
ee
46
Ottian
Conversions
<<;
...
<<...
as
Soren
-
48
Appendix
|
—
Answers
to
Practice
Problems........A-1
Appendix
Il
—
Glossary
of:Terms’....2..,.)
sae
A-5
INTRODUCTION
CALCULATOR
SIDE
OF
FLIGHT
:
ae
COMPUTER
Congratulations,
you
have
just
purchased
one
of
the
finest
computers
on
the
market
today.
It
is
well
designed
Cu
to
assist
you
in
solving
the
planning
and
navigating
problems
associated
with
flying.
It
is
simple
to
operate
;
and
adequate
instructions
and
formulas
are
printed
on
the
computer
so
that
the
user
need
not
worry
about
forgetting
how
to
work
the
computer
FLIGHT
COMPUTER
The
flight
computer
has
two
sides,
the
calculator
side
Cae
conversions,
al
e
era
division.
7
©
Unit
index
for
Multipli
@
“8”
Scale
{time
in
min
FLIGHT
COMPUTER
cation,
Division,
and
Rate
utes,
calibrated
altitude,
z
of
Climb/Descent.
calibrated
airspeed).
The
main
portion
of
the
calculator
side
of
the
flight
Imperia!
Gallon
Conver
"C"
Scale
(time
in
hours
computer
consists
of
three
separate
scales.
The
outer
mon
Arows
and
minutes)
scale
(10,fig.1)
is
fixed
to
the
computer,
The
second
Kilometer
Conversion
Lerpereure
Conversion
(11,
fig.
1)
and
third
(12,
fig.1)
scales,
inward,
are
printed
Arrows.
ee
on
a
disc
that
pivots
in
the
center
of
the
computer,
thus
©
Us.
Gallon
Conversion
Pounds
CON
a
oD
Permitting
them
to
be
rotated
within
the
outer
fixed
Arrows.
scale.
To
simplify
the
explanation,
the
scales
on
the
Seconds
Arrow.
Foot
Conversion
Arrow.
calculator
side
of
the
computer
are
referred
to
as
Meters
Conversion
©
GO
OO®O
®
Preacaltcee
ea
A”,
“B",
and
“C"
scales.
(See
fig.
1.)
Window.
Liters
Conversion
{
na
i
@
Kulogans
Contec
Rerdaa
The
“A”
scale
is
used
to
represent
miles,
gallons,
or
Arrow.
Speed
nde
true
airspeed.
When
used
as
miles,
the
scale
provides
©
Density
“altitude
®
Nauta
Cameron
the
distance
traveled
or
speed
of
the
aircraft
in
miles
Window.
aro
|
per
hour.
Gallons
on
this
scale
can
represent
two
values:
©
Bien
neti
Statirta’
Conversion
|
fuel
consumption
in
gallons
per
hour,
or
total
quantity
Wicdiaen
Arrows.
of
fuel
used
by
the
aircraft.
““T.A.S.""
on
the
“A”
scale
isan
i
@
“AY
Scale
(mites,
miles
uel
founes
Conversion
abbreviation
for
true
airspeed.
per
hour,
gallons,
gallons
per
hour,
true
airspeed,
@)
Oil/Pounds
Conversion
The
graduations
on
the
“‘B’
scale
are
used
for
time
in
I i
Arrow.
i
8
es
true
altitude)
rrow.
Minutes
or
for
“‘l.A.S.""
(Indicated
Air
Speed)
in
either
Figure
1
miles
per
hour
or
knots.
The
“C’’
scale
graduations
re-
Present
hours
and
minutes
only.
enemies
CHANGING
VALUES
In
order
to
accurately
read
the
scales
on
the
calculator
side
of
the
computer,
it
is
necessary
that
the
chang-
ing
values
of
the
graduations
of
these
scales
be
under-
stood.
When
using
the
“A”, “B”,
and
“C”
scales
to
solve
prob-
lems,
common
sense
must
be
used
to
determine
the
value
of
the
numbers.
If
a
short
distance
is
involved,
“25"
on
the
“A”
scale
might
be
read
as
“2.5
miles”.
If
a
long
distance
is
involved,
zeros
are
added
to
the
“25”
to
get
the
proper
answer.
For
example,
“25”
might
be
“250”
or
“2500
miles”.
First,
let’s
examine
the
changing
values
of
the
“A”
scale.
(See
fig.
2.)
If
the
number
“14”
is
used
as
14,
each
gradu-
ation
between
“14”
and
“15”
is
equal
to
.1.
If
these
num-
bers
are
used
as
140
and
150,
each
graduation
is
equal
to
1;
if
used
as
1400
and
1500,
they
represent
10;
and
if
used
as
14000
and
15000,
they
are
equal
to
100.
Between
the
numbers
“15”
and
“16,”
there
are
only
five
graduationsascompared
to
the
10
graduations
between
“14”
and “15
“.(See
fig.
2.)
When
these
num-
bers
are
used
as
15 and 16
respectively.
each
gradu-
ationisequalto.2;
when
used
as
150
and
160,
each
unit
represents2;when
visualizedas1500
and
1600,
each
unitisequalto20;
and
when
usedas15000
and
16000,
each
graduation
equals
200.
NOTE
The
graduationsonthe “B”
scale
are
identical
with
those
on
the
“A”
scale.
CHANGING
VALUES
Figure
2
CHANGING
VALUES
Figure
3
The
changing
values
on
the
“C”
scale
are
somewhat
different
than
those
on
the
“A”
and
“B”
scales.
(See
fig.
3.)
These
graduations
always
represent
minutes
and
are
equaltofiveorten
minutes
as
shown
by
the
arrows.
For
example:
between
1:50
and
2:00,
the
gradu-
ations
are
equaltofive
minutes;
and
between
2:00
and
2:30,
the
graduations
represent
ten minutes.
The
graduationsonthe
“C’”
scale
are
very
largeincom-
Parison
with
those
on
the
“B”
scale.
As
a
result,
there
are
times
when
finer
graduations
are
needed
than
are
Provided
on
the
“C”
scale.
When
this
happens,
the
smaller
graduations
on
the
“B”
scale
can
be
used
to
supplement
the
measurements
on
the
“C”
scale.
For
example:
in
figure
4,
the
graduations
on
the
“C”
scale
are
equal
to
ten
minutes,and
the
graduations
on
the
“B”
scale,
immediately
above,
are
equal
to
two
min-
utes.
Starting
at
the
left,
the
values
are
3:30
on
the
HOURS
ON
C—SCALE
MINUTES
ON
B—SCALE
36
3:38
340
%
rey
2
eletes
»
e
2
Scie
Ce
a0
os
L
insta
Figure
4
"C"
scale;
then,
moving
up
to
the
“B”
scale
for
3:32,
3:34,
3:36,
and
3:38;
and
then
back
to
the
“C”
scale
for
3:40.
Notice
that
3:40
on
the
“C”
scale
is
equal
to
220
minutes
on
the
“B”
scale.
SPEED
INDEX
The
speed
index
is
a
large
triangular
symbol
on
the
“B™
and
“C”
scales
and
is
used
as
a
reference
in
time
and
distance
problems.
(See
fig.
5.)
The
speed
index
always
represents
60
minutes
or
one
hour.
The
graduations
covered
by
this
symbol
are
also
used
as
6,
60,
600,
or
6000
on
the
“B”
scale.
TIME
AND
DISTANCE
these items
mustbeknown
t
are
three items
id
speed.
Two
of
the
problem.
FINDING
TIME
Ifanaircraftisflyingata
speedof120
miles
per
hour,
how
long
willittaketofly
140
miles?
The
steps
involvedinsolving
this
problem
areasfollows:
(See
fig.
6.)
1.
Turntothe
calculator
side of
the
computer
and
rotate
the
computer
disc
until
the
speed
index
is
located
directly
under
“12”
which
represents
120
miles
per
hou!
PRACTICE
PROBLEMS
—
FINDING
TIME
The
following
practice
problems
should
be
worked
to
gain
experience
in
working
time
problems
on
the
com-
puter.
Answers
to
these
problems
are
given
in
the
Ap-
endix
of
this
manual.
1.
Speed
—
180
m.p.h.;
distance
—
240
miles;
how
long
will
it
take
to
make
the
trip?
2.
Speed
—
142
m.p.h.;
distance
—
370
miles;
how
long
will
it
take
to
make
the
trip?
3.
Speed
—
110
mop.h.;
distance
—
33
miles;
whatisthe
time?
4.
Speed
—
136
m.p.h.;
distance
—
86
miles;
whatisthe
time?
FINDING
DISTANCE
an
aircraft
flies
at
110
m.p
r
2 .
for
a
two-hour
period,
ow
many
miles
will
it
fly’
The
steps
involved
in
solving
this
problem
are
as
follows:
(See
fig.
7.)
1.
Place
speed
index
under
“11”
which
represents
110
miles
per
hour.
0
ae)
v
iver
Ss
x)
TUT
PA
Ne
el
A
©
1592:
SS
os
FON
?
al
PRACTICE
PROBLEMS
—
FINDING
DISTANCE
The
answers
to
the
following
practice
problems
are
given
in
the
Appendix
of
this
manual.
1,
Speed
—
100
m.p.h.,
time
—
1:30;
what
dis-
tance
will
be
flown?
2.
Speed
—
126
mop.h.,
time
—
2:05;
what
is
the
distance
traveled?
3.
Speed
—
175
mp.h.;
time
—
4:00;
what
is
the
distance?
4,
Speed
—
133
m.ph.;
time
—
3:32;
what
is
the
distance?
FINDING
SPEED
lf an
aircraft
flies
21
miles
in
1:30,
what
ts'the
spe
“To
solve
this
problem,
i
8)
1.
Position
1:30
on
the
“C”
scale
directly
under
210
milesonthe
“A”
scale.
2.
Directly
over
the
speed
indexisthe
answer,
140
miles
per
hour.
Figure
8
PRACTICE
PROBLEMS
—
FINDING
SPEED
The
answers
to
the
following
practice
problems
are
given
in
the
Appendix
of
this
manual.
1.
Distance
—
90
miles;
time
—
0:43;
what
is
the
speed?
2.
Distance
—
320
miles;
time
—
2:00;
what
is
the
speed?
3.
Distance—35
miles;
time
—
0:19;
what
is
the
speed?
4
Distance
—
182
miles;
time
—
1:54;
what
is
the
speed?
SHORT
TIME
AND
DISTANCE
A
procedure
called
“short
time
and
distance”
is
used
on
the
computer
solve
problems
involving
short
dis-
tances,
such
as
;
In
these
problems,
a
very
“smail
amount
of
time
is
involved
in
checking
speed.
For
the
short
time
and
distance
procedure,
“36”
re;
the
speed
index.
This
"36"
is
equal
to
3600
seconds
in
one
hour.
Figure
9
shows
the
“36”
position
on
the
rotating
“B”
scale
of
the
computer,
When
using
the
short
time
and
distance
procedure,
all
minutes
on
the
“B”
scale
represent
seconds.
For
example,
figure
9
shows
that
at
101
m.p.h..
40
sec-
Onds
are
required
to
travel
1.12
miles.
Similarly,
the
“C"
scale
is
changed
from
hours
to
minutes.
Figure
9
shows
that
at
101
mph.,
it
will
take
5
minutes
to
fly
8.4
miles.
the
ircraft
is
always
placed
on
the
“
“AY
3600
index
on
the
“B”
scale.
ways
fi
ind
on
the
“A”
scale
directly
over
econds
on
the
“B”
scale,
or
over
time
in
min-
‘the
“C”
scale.
FINDING
SHORT
TIME
|
‘Flying
at
120
mop-h.,
for
a
distance
of
1.5
miles,
how
much
time
will
it
take?
This
problem
is
soived
as
follows:
(See
fig.
10.)
|
1.
Rotate
the
“B”
scale
until
the
“36”
is
directly
at
under
120
m.p.h.,
on
the
“A”
scale.
}
2.7
-
locate
the
number
“15”
on
the
“A”
scale.
his
problem,
this
number
is
actually
“1.5”
‘rather
than
“15”.
i
Figure
10
The
answers
to
the
following
practice
problems
are
‘Fuel
consumption
problems
are
solved
in
the
same
man-
PRACTICE
PROBLEMS
—
SHORT
TIME
given
in
the
Appendix
of
this
manual.
1.
Speed
—
140
m.p.h.;
time
—
3
minutes;
how
far
would
an
aircraft
fly?
2.
Distance
—
2.5
miles;
time
—
one
minute
and
30
seconds;
how
fast
is
the
aircraft
traveling?
FUEL
CONSUMPTION
ner
:
ime
and
distance
problems,
except
that
gallons
ier
hour
and
gallons
are
used
in
lieu
of
miles
per
hour
and
miles.
FINDING
TIME
If
an
aircraft
burns
fuel at
the
rate
of
nine
gallons
per
hour
and
has
45
gallons
of
useable
fuel
on
board,
how
fong
can
the
aircraft
fly?
The
steps
involved
in
solving
the
problem
are
as
follows:
(See
fig.
11.)
1.
Rotate
the
computer
disc
until
the
speed
index
is
directly
under
“90,
which
represents9gal-
lons
per
hourinthis
problem.
hen.
move
clockwise
on
the
“A”
scale
and
cate
“45,
which
represents
the
45
gallons
of
useable
fuel.
Directly
under
thisonthe
“C”
scaleisfound
5:00.
The
answer
to
our
prob-
lemis5
hours.
Figure
11
1
PRACTICE
FUEL
CONSUMPTION
PROBLEMS
—
PRACTICE
PROBLEMS
—
FUEL
BURNED
FINDING
TIME
The
answer
to
the
following
problems
are
given
in
the
Appendix
of
this
manual.
1,
Fuel
consumption
—
14
g.p.h.;
time
—
15
min-
utes;
how
much
fuel
was
burned?
The
answers
to
the
following
practice
problems
are
given
in
the
Appendix
of
this
manual.
1.
Fuel
consumption
—
12
g.p.h.,
useable
fuel
y
Note:
Don't
forget
to
use
care
in
determining
the
—
30
gallons;
how
long
can
the
aircraft
stay
decimal
point.
in
the
air?
2.
Fuel
consumption
—
21
g.p.h.;
time
—
2:40;
2.
Fuel
consumption
—
18
g.p-h.;
fuel
burned
how
much
fuel
was
burned?
—
68
gallons;
what
was
the
time?
3.
Fuel
consumption
—
12%
g.p.h.;
time
—
1:35;
how
much
fuel
was
burned?
3.
Fuel
consumption
—
11
gp-h.;
fuel
burned
—
24
gallons,
what
was
the
time?
4.
Fue!
consumption
—
11
g.p.h.;
time
—
0:28;
how
much
fuel
was
burned?
4.
Fuel
consumption
—
15
g.p.h.,
useable
fuel
1
a
ee
how
long
can
the
aircraft
stay
FUEL
CONSUMPTION
Ifan
aircraft
burns
80
gallons
of
fuel
in
2:30,
how
many
gallons
is
it
burning
per
hour?
The
problem
is
solved
as
follows:
(See
fig.
13.)
a
period
of
1.
Rotate
the
calculator
disc
until
2:30
on
the
FUEL
BURNED
|
“If
an
airera’
t
burns
8%
gallons
per
ho
2:00.
how
many
gallons
of
fuel
wer
“C”
scale
is
directly
under
"80°
on
the
“A”
scale.
The
problem
is
solved
as
follows:
(See
fig.
12.)
2)
Rotate
the
complete
computer
and
find
the
“90”
represent
“8”
and
“9”
re-
answer
on
the
“A”
scale
above
th
ic
gros
pa
as.
oat
e
problem
is
32
gallons
per
t
“Halfway
between
The
answer
tort
2
gallons
Place
the
hour
set
PRACTICE
PROBLEMS
—
FUEL
CONSUMPTION
The
answers
to
the
following
problems
are
given
in
the
Appendix
of
this
manual.
‘
1.
Fuel
burned
—
7
gallons;
time
—
0:40;
what
is
the
fuel
consumption?
Figure
12
11
.
Fuel
burned
—
47
gallons;
time
—
2:10;
what
is
the
fuel
consumption?
3.
Fuel
burned
—
75
gallons;
time
—
3:15;
what
is
the
fuel
consumption?
4.
Fuel
burned
—
36
gallons;
time
—
4:11;
what
is
the
fuel
consumption?
|
TRUE
AIRSPEED
Hi
The
true
airspeed
problem
is
solved
on
the
calculator
|
side
of
the
computer
as
follows:
|
|
scal
locate
the
indicated
“B”
scale.
The
“B”
scale
is
label-
/A.S.""
(calibrated
airspeed)
which,
for
|
all
practical
purposes,
is
equivalent
to
“indicated
i
airspeed”.
(See
fig.
15.)
|
i
Figure
15
3.fioe
the
true
«
os
:
i
ae
For
example:
what
is
the
T.A.S.
under
the
following
conditions:
altitude,
10,000
feet;
temperature,
-10°
C.;
and
L.A.S.,
130
m.p.h.?
The
problem
is
solved
as
follows:
(See
fig.
16.)
1.
Rotate
the
computer
disc
until
-10°
C.
is
located
directly
over
10,000
feet.
2.
Then
refer
to
the
“B”
scale
and
locate
“13”
which
represents
130
miles
per
hour
indicated
air-
speed
for
this
problem.
3.
Look
directly
over
130
m.p.h.
on
the
“B“
and
find
that
the
true
airspeed
is
150
mil
hour.
This
is
the
answer
to
this
problem.
scale
per
PRACTICE
PROBLEMS
—
FINDING
T.A.S.
The
answers
to
the
following
problems
are
given
in
the
Appendix
of
this
manual.
1.
Altitude
—
5,000
feet;
temperature
—
+15°
C.;
LAS.
—
125
m.p.h;
what
is
the
T.A.S.?
2.
Altitude
—
8,000
feet;
temperature
—
O°
C.;
1.A.S.
—
110
m.p.h.;
what
is
the
T.A.S
?
3.
Altitude
—
12,000
feet:
temperature
—
minus
6°
C.,
LA.S.
—
149
m.p.h.;
what
is
the
T.A.S.?
4.
Altitude
—
3,000
feet;
temperature
—
+15°
C.;
1.A.S.
—
105
m.p.h.;
what
is
the
T.A.S.?
13
Figure
17
CONVERTING
MACH
NUMBER
TO
TRUE
AIRSPEED
To
convert
mach
number
to
true
airspeed
in
knots,
use
the
following
procedure:
1.
Align
the
outside
air
temperature,
in
degrees
Celsius
(centigrade),
with
the
mach
number
index
as
shown
in
figure
17,
item
1.
2,
Read
the
true
airspeed,
in
knots,
on
the
“A”
scale
opposite
the
mach
number
on
the
“B”
scale.
Figure
17
points
out
several
mach
num-
bers
and
the
corresponding
true
airspeeds
at
an
air
temperature
of
+15°
C.
The
readings
are:
Mach
Number
True
Airspeed
0.8
(item
2)
528
knots
(item
5)
1.0
(item
3)
661
knots
(item
6)
1.36
(item
4)
898
knats
(item
7)
CHANGING
NAUTICAL
VALUES
TO
STATUTE
EQUIVALENTS,
STATUTE
TO
NAUTICAL,
AND
STATUTE
TO
KILOMETERS
Another
use
for
the
calculator
side
of
the
computer
is
for
converting
nautical
miles
to
statute
miles.
or
knots
to
miles
per
hour.
This
type
of
problem
is
made
very
simple
by
a
small
conversion
scale
consisting
of
two
arrows
labeled
“NAUT”
(nautical)
and
“STAT”
(statute)
14
aero
Figure
18
respectively.
These
arrows
are
located
on
the
“A”
scale
and
point
toward
the
“B”
scale.
NOTE
Knots
(nautical
miles
per
hour)
are
changed
to
statute
miles
per
hour
in
exactly
the
same
manner
For
example:
to
change
20
nautical
miles
to
statute
miles,
the
problem
is
solved
as
follows:
(See
fig.
18.)
1.
First,
rotate
the
calculator
disc
until
20
nau-
tical
miles
is
lined
up
with
the
nautical
arrow.
2.
Then,
read
the
equivalent
value,
in
statute
miles,
directly
under
the
statute
arrow.
The
answer
is
23
statute
miles
The
conversion
scale
can
be used
to
change
statute
to
nautical
miles
by
placing
the
statute
value
under
the
statute
arrow
and
reading
the
nautical
equivalent
under
the
nautical
arrow.
In
addition,
another
statute
index
arrow
is
included
on
the
“B”
scale.
It
can
be
used
to
convert
statute
miles
on
the
“B”
scale
to nautical
miles
or
kilometers
on
the
“A*
scale.
(See
fig.
19
and
20.)
Figure
19
Figure
20
15
To
convert
statute
to
nautical
equivalents:
(See
fig.
19.)
1.
Align
statute
index
on
the
“B”
scale
directly
under
the
nautical
index
on
the
“A”
scale.
2.
Directly
over
the
statute
value
on
“B”
scale,
read
the
equivalent
nautical
value
on
the
“A”
scale.
For
example,
90
statute
miles
is
equal
to
78.2
nautical
miles.
To
convert
statute
to
kilometer
equivalents:
(See
fig.
20.)
1.
Align
statute
index
on
the
“B”
scale
directly
under
the
kilometer
index
on
the
“A”
scale.
2.
Directly
over
the
statute
value
on
the
“B”
scale,
read
the
equivalent
kilometer
value
on
the
“A”
scale.
For
example,
90
statute
miles
is
equal
to
144.5
kilometers.
PRACTICE
PROBLEMS
—
NAUTICAL
TO
STATUTE,
STATUTE
TO
NAUTICAL,
STATUTE
TO
KILOMETERS
1.
140
nautical
milesisequal
to______statute
miles.
2
25knots,
wind
velocity,isequal
to.
m.p.h
3.77statute
milesisequal
to
nautical
miles
4.17knotsisequal
to.
m.p.h.
5.
133
statute
milesisequal
to.
kilometers
6.40kilometersisequal
to.
statute
miles
MULTI-PART
PROBLEMS
In
practice,
the
pilot
solves
a
series
of
interrelated
prob-
lems
leading
to
a
final
solution,
which
may
be
consider:
ed
one
multi-part
problem.
A
good
example
of
a
multi-
part
problem
is
finding
fuel
consumption
for
an
antici-
pated
flight.
The
following
multi-part
problems
are
provided
to
em-
phasize
the
order
of
information
required
to
determine
the
total
fuel
consumption.
In
each
case,
the
ground
Speed
is
determined
first,
then
the
time
en
route,
and
finally,
the
fuel
consumed.
In
each
case,
assume
that
LAS.
is
equal
to
C.A.S.
The
answers
to
the
problems
are
given
in
the
Appendix
of
this
manual.
16
.
Wind
—
O
m.p.h.;
ground
speed
—
120
m.p.h.;
distance
—
320
miles;
fuel
consumption
—
9
g.p.h.;
how
much
fuel
will
be
burned?
.
Altitude
—
7,500
feet;
|.AS.
—
105
m.p.h.
temperature
—
+15°
C.;
distance
—
256
mile:
fuel
consumption
—
11.5
g.p.h.;
wind
—
O
m.p.h.;
how
much
fuel
will
be
burned?
.
Altitude
—
9,000
feet;
|A.S.
—
115
m.p.h.;
temperature
—
-10°
C.;
distance
—
335
miles;
fuel
consumption
—
8.5g.p.h.;
wind
—
Om.p.h.;
how
much
fuel
will
be
burned?
.
Ground
speed
—
135
m.p.h.;
wind
—
O
m.p.h.;
temperature
—
-20°
C.;
altitude
—
9,000
feet;
distance
—
425
miles;
fuel
consumption
—
12
g.p.h.;
how
much
fuel
will
be
burned?
FINDING
DRIFT
ANGLE
(OFF-COURSE
PROBLEM)
The
computer
canbeusedtofind
the
drift
angle
when
the
aircraft
drifts off
course
duetoa
wind
shift,
errone-
ous
wind
information, or
navigation
error.
Thisisoften
known
asanoff-course
problem.
Figure
21
shows
a
typical
problem:
120
miles
from
departure
and
18
miles
off
course
to
the
right;
what
is
the
drift
angle?
1.Onthe
calculator
side,
set
the
number
of
miles
the
aircraft
has
flown,
120,
on
the
“B”
scale
Destination
OFF
COURSE
Figure
21
172
Figure
22
directly
under
the
number
of
miles
off
course,
18,
on
the
"A”
scale.
(See
fig.
22.)
Above
the
speed
index
is
the
drift
angle,
9
degrees,
(See
fig.
22),
which
is
also
the
number
of
degrees
to
change
heading
to
the
left
in
order
to
parallel
the
intended
true
course
between
Point
A
and
Point
B,
as
shown
in
figure
23.
To
find
the
additional
angle
needed
to
further
cor-
rect
the
course
from
Point
X
to
Point
B,
set
the
170
miles
to
go
on
the
“B"
scale,
under
18,
the
number
of
miles
off
course,
on
the
“A”
scale
(See
fig.
24).
(The
number
of
“miles
to
go”
is
measured
on
the
chart.)
Above
the
speed
indexisthe
additional
angle
neededtoarriveatthe
destination
(See
fig.
24.)
Use
the
nearest
whole
number.6,as
the
answer
Add
the
9
degrees
needed
to
parallel
the
intend-
ed
true
course
and
the
additional
6
degrees
Figure
23
for a
total
of
15
degrees
heading
correction
to
the
left,
necessary
to
fly
from
Point
X
to
the
destination,
Point
B.
(See
fig.
25.)
PRACTICE
PROBLEMS
—
OFF-COURSE
Refer
to
the
Appendix
for
answers
to
these
problems
1.
GIVEN
a.
Distance
out
from
departure
point
—
110
stat.
miles;
.
Distance
to
left
of
intended
true
course
—
13
stat.
miles;
.
Distance
to
destination
from
off
course
posi-
tion
—
200
stat.
miles.
PROBLEM:
a.
How
many
degrees
should
the
aircraft
be
turned
to
parallel
the
intended
course?.
b.
How
many
total
degrees
should
the
aircraft
be
turned
to
converge
on
the
destination?
c.
Which
direction
should
the
aircraft
be
turn-
ed
(right
or
left)?
2.
GIVEN:
a.
Distance
out
from
departure
point
—
150
stat.
miles,
b
Distance
to
right
of
intended
true
course
—
20
stat.
miles,
c.
Distance
to
destination
from
off
course
posi-
tion
—
140
stat.
miles.
PROBLEM:
a.
How
many
degrees
should
the
aircraft
be
turned
to
parallel
the
intended
true
course?
b.
How
many
degrees
should
the
aircraft
be
turned
to
converge
on
the
destination?
c.
Which
direction
should
the
aircraft
be
turn-
ed
(right
or
left)?
3.
GIVEN:
a.
Distance
out
from
departure
point
—
90
stat.
miles;
b.
Distance
to
left
of
intended
true
course
—
6
stat.
miles;
c.
Distance
to
destination
from
off
course
posi-
tion
—
180
stat.
miles.
PROBLEM
a.
How
many
degrees
must
the
aircraft
be
turnedtoparallel
the
intended
true
course?
b
How
many
degrees
must
the
aircraft
be
turned
to
converge
on
the
destination?
¢.
Which
direction
should
the
aircraftbeturn-
ed
(rightorleft)?
TIME
AND
DISTANCE
TO
A
VOR
STATION
\
The
time
and/or
distance
from
the
airplane
to
a
VOR
station
can
be
computed
easily
on
the
calculator
side
of
the
computer.
These
problems
are
based
on
the
pilot
20
ae
taking
a
time
check
to
make
a
measured
bearing
change
with
a
VOR
radio.
In
this
procedure,
the
pilot
flies
per-
pendicular
to
the
VOR
radials
involved
with
the
time
check.
TIME
TO
A
VOR
STATION
To
find
the
time
to a
VOR
station:
1.
Set
the
time
to
make
the
bearing
change,
in
minutes,
on
the
“C”
scale
directly
under
the
number
of
degrees
of
bearing
change
on
the
"A"
scale
2.
Find the
answeronthe
“B”
scale
directly
under
the
unit
index
(10)onthe
“A”
scale.
For
example:
what
ts
the
time
to
the
station
when
it
takes2minutes30secondstoaccomplish
five
degrees
of
bearing
change?
(See
fig.
26.)
This
problem
is
solved
as
follows
1.
Place
the
time
(two
minutes
and
thirty
seconds
—
2:30)
on
the
“C”
scale
directly
under
the
degrees
of
bearing
change
(5)
on
the
“A”
scale.
2.
Read
the
timetothe
stationonthe
“B”
scale
directly
under
the
unit
index
(10)
located
on
the
“A”
scale.
The
answeris30
minutestothe
station.
DISTANCE
TO
A
VOR
STATION
To
find
distancetoa
VOR
station.
ground
speed
must
alsobeknown
in
additiontothe
timetomakeabear-
ing
change.
The
solution
to
this
problem
involves
the
following
steps
1.
First,
solve
for
timetothe
station
GHT
COMPUTER
—.
el
Figure
26
-
Then,
set
the
speed
index
under
the
ground
speed.
.
Next,
locate
the
time
to
the
station,
in
minutes,
‘on
the
“B”
scale.
Look
directly
over
this
value
and
find
the
distance
to
the
station.
For
example:
Using
the
time
to
the
station
found
in
the
previous
problem
(ref.
fig.
26)
and
a
ground
speed
of
142
m.p.h.,
what
is
the
distance
to
the
station?
The
solution
is
determined
as
follows:
(See
fig.
27.)
1.
Place
speed
index
under
ground
speed
of
142
m.p.h.
on
the
“A”
scale.
2.
Locate
the
time
to
the
station,
30
minutes
on
the
“B”
scale
3.
Read
distance
to
station
on
the
“A”
scale
direct-
ly
above
30
minutes
on
the
“B”
scale.
The
answer
is
71
miles
to
the
station.
If
the
time
to
the
station
is
less
than
10
minutes,
the
problem
is
worked
in
the
same
way,
except
that
the
short
time
and
distance
method
is
used.
The
short
time
speed
index
“36”
is
used
in
lieu
of
the
triangular
speed
index,and
the
time
in
minutes
and
seconds
are
placed
on
the
“C”
scale.
For
example:
Time
to
station
—
4
min-
utes,
30
seconds;
ground
speed
—
127
m.p-h.;
what
is
the
distance
to
the
station?
(See
fig.
28.)
1.
Set
“36”
under
127
m.p.h
2.
Locate
“4:30”
on
“C”
scale
and
read
distance
directly
above
on
the
“A”
scale,
9.5
miles.
Figure
27
Figure
28
PRACTICE
PROBLEMS
—
TIME
AND
DISTANCE
TO
A
VOR
STATION
Refer
to
the
Appendix
for
answers
to
these
problems.
Degrees
of
Time
Aircraft
Time
Distance
Bearing
Between
Ground
To
To
Change
Bearings
Speed
Station
Station
i?
i
110
mph
10°
95
mph
be
-
135
mph
15~
in.
75
mph
20°
1min.
50sec.
140
mph
TRUE
ALTITUDE
True
altitude
is
obtained
by
using
the
window
labeled
“FOR
ALTITUDE
COMPUTATIONS”
which
incorpo-
rates
temperature
and
pressure
altitude
scales.
Pressure
altitude
is
indicated
on
the
altimeter
when
the
baromet-
ric
scale
is
set
to
the
atmospheric
standard
pressure
of
29.92
inches
of
mercury.
True
altitude
problems
can
be
solved
by
following
these
steps:
If
an
aircraft
is
flying
at
12,500
feet
with
an
outside
air
temperature
of
-20°C.
and
the
altimeter
set
on
30.42
inches
of
mercury,
what
is
the
true
altitude?
This
problem
is
solved
as
follows:
1.
Referring
to
the
altimeter,
note
the
altimeter
barometric
reading
of
30.42
and
record
it’on
a
work
sheet.
2.
Now,
rotate
the
barometric
scale
on
the
alti-
meter
to
29.92.
Note
that
the
pressure
altitude
is
12,000
feet.
3.
Next,
refer
to
the
altitude
computation
window
and
place
-20°C.
on
the
temperature
scale
directly
over
the
pressure
altitude
of
12,000
feet.
(See
fig.
29.)
4.
Move
to
the
B”
scale
of
the
computer
and
locate
the
original
indicated
or
calibrated
alti-
tude,
12,500
feet.
Look
directly
over
this
value
and
find
the
true
altitude
is
12,000
feet.
(See
fig.
30.)
This
is
the
answer
to
the
problem.
The
{
aircraft
is
actually
flying
500
feet
lower
than
the
altitude
originally
indicated
on
the
altimeter.
PRACTICE
PROBLEMS
—
TRUE
ALTITUDE
Refertothe
Appendix
for
answerstothese
problems:
1.
Pressure
altitude,
7,000
feet;
indicated
altitude,
6,500
feet;
temperature, -10°C.
Whatisthe
true
altitude?
2.
Pressure
altitude,
9,000
feet;
indicated
altitude,
10,000
feet;
temperature
-20°C.
Whatisthe
true
altitude?
Figure
30
3,
Pressure
altitude,
6,000
feet;
indicated
altitude,
5,500
feet;
temperature
-10°C.
What
is
the
true
altitude?
NOTE
Figure
answers
to
closest
50-foot
increment.
DENSITY
ALTITUDE
Density
altitude
problems
are
solved
on
the
calculator
side
of
the
computer
through
the
use
of
the
window
labeled
“FOR
TRUE
AIRSPEED
AND
DENSITY
ALT.”
The
procedure
for
solving
this
type
of
problem
is
shown
in
the
following
example:
Flying
at
10,000
feet
pressure
altitude
with
outside
air
temperature
at
-20°C.,
what
is
the
density
altitude?
This
problem
is
solved
as
follows:
1.
Place
-20°C
directly
over
10,000
feet
pressure
altitude
in
the
airspeed
and
density
altitude
window.
(See
fig.
31.)
2.
Refer
to
the
density
altitude
window
and
find
that
the
density
altitude
is
8,000
feet.
PRACTICE
PROBLEMS
—
DENSITY
ALTITUDE
Refer
to
the
Appendix
for
answers
to
these
problems.
1.
Pressure
altitude,
10,000
feet;
temperature,
-20°C.;
what
is
the
density
altitude?
2.
Pressure
altitude,
15,000
feet;
temperature,
-30°C.;
what
is
the
density
altitude?
3.
Pressure
altitude,
4,000
feet;
temperature,
-25°C.;
what
is
the
density
altitude?
Figure
31
MULTIPLICATION
AND
DIVISION
The
computer
can
also
be
used
for
multiplication
and
division.
The
index
for
these
problems
is
the
“10”
in
the
small
box.
(1,
fig.
1.)
To
Multiply:
1.
Rotate
the
“B”
scale
until
the
index
"10"
is
direct-
ly
under
the
number
to
be
multiplied.
2.
Without
moving
the
scales,
locate
the
multi-
plier
on
the
“B”
scale.
3.
Look
directly
over
the
multiplier
and
find
the
answer
on
the
“A”
scale.
To
Divide:
1.
Locate
the
number
to
be
divided
on
the
“A”
scale.
2.
Rotate
the
“B”
scale
until
the
divisor
is
located
directly
under
the
number
to
be
divided.
3.
Without
moving
the
scales,
find
the
answer
on
the
“A”
scale
directly
over
the
index
“10”.
MULTIPLICATION
Climbing
at
450
feet
per
minute
for
8
minutes,
how
much
altitude
would
be
gained?
This
problem
is
solved
by
multiplying
450
f.p.m.
x
8
minutes.
(See
fig.
32.)
1.
Place
450
on
the
“A”
scale
directly
over
“10”
in
the
box
on
the
“B”
scale.
2.
Find
8
minutes
on
the
“B”
scale
and,
looking
directly
over
this,
find
that
we
would
climb
3,600
feet
in
the
8
minutes.
Figure
32
DIVISION
An
aircraft
hastolose
8,000
feetin19
minutes.
What
is
the
rate-of-descent?
(See
fig.
33.)
1.
Place
8,000
feet
on
the
“A”
scale
directly
over
19
minutes
on
the
“B”
scale.
2.
Look
directly
over
the
“10”
and
find
that
the
aircraft
should
descend
at
420
f.p.m.
to
lose
8,000
feetin19
minutes.
PRACTICE
PROBLEMS
—
MULTIPLICATION
AND
DIVISION
Refer
to
the
Appendix
for
answers
to
these
problems:
1.
Rate-of-climb,
450
f.p.m.;
time
in
climb,
18
minutes;
whatisthe
altitude
gained?
.
Rate-of-descent,
600
f.p.m.;
time
in
descent,
4'2
minutes,
what
is
the
altitude
lost?
Altitudetolose.
6,500
feet;
timetolose
alti-
tude,9minutes;
whatisthe
rate-of-descent?
Altitude to
gain,
9,000
feet;
timetogain
alti-
tude,21minutes;
whatisthe
rate-of-climb?
Aircraft
cargo
arm,
55
inches;
cargo
weight,
226
pounds;
what
is
the
approximate
cargo
moment?
Total
aircraft
moment.
162,000
pound-inches;
total
aircraft
weight,
3,000
pounds;
what
is
the
center-of-gravity
arm?
CONVERTING
FEET
PER
NAUTICAL
MILE
TO
FEET
PER
MINUTE
Certain
IFR
departure
procedures
require
a
minimum
climb
rate
to
assure
proper
obstruction
clearance.
How-
ever,
the
minimum
climb
requirement
is
stated
in
terms
of
feet
to
be
gained
per
nautical
mile.
The
pilot
can
easily
convert
this
“feet
per
nautical
mile”
figure
to
a
“feet
per
minute”
figure
on
the
calculator
side
of
the
com-
puter.
The
following
example
outlines
the
procedure:
(See
fig.
34.)
1.
Set the
speed
index
at
the
appropriate
ground
speed
in
knots.
(Figure
34
shows
the
speed
index
set
at
a
ground
speed
of
120
knots.)
2.
All
figures
on
the
“B”
scale
will
represent
the
minimum
climb
requirement
in
feet
per
nau-
tical
mile.
All
figures
on
the
“A”
scale
will
repre.
sent
the
equivalent
vertical
velocity
in
feet
per
minute
and
are
read
directly
over
the
minimum
climb
requirement
figures.
For
example:
Ground
speed
—
120
knots;
climb
require-
ment
—
250
feet
per
nautical
mile;
what
is
the
rate of
climb
in
feet
per
minute?
To
solve
this
problem:
(See
fig.
34.)
1.
First
set
the
speed
index
at
120
knots.
2.
Then,
directly
above
the
250
on
the
“B”
scale,
find
the
answer,
500
feet
per
minute,
on
the
“A”
scale
FOR
TIME
a
For
Fi
IND
DISTANCE
CONSUMPTION
Figure
34
NOTE
At
the
same
ground
speed,
the
climb
require-
ment
of
350
feet
per
nautical
mile
is
equivalent
to
a
vertical
speed
of
700
feet
per
minute.
PRACTICE
PROBLEMS
—
CONVERTING
FEET
PER
NAUTICAL
MILE
TO
FEET
PER
MINUTE
Answers
to
these
problems
are
found
in
the
Appendix
of
this
manual.
Climb
Requirement
Required
in
Feet/Nautical
Minimum
Ground
Speed
Mile
Vertical
Speed
100
knots
240
f.p.m.
80
knots
300
f.p.m.
140
knots
300
f.p.m.
105
knots
400
f.p.m.
CONVERTING
FAHRENHEIT
TO
CELSIUS
(CENTIGRADE)
In
some
instances,
the
pilot
may
know
the
temperature
in
degrees
Fahrenheit
but
not
know
its
equivalent
in
the
Celsius
scale.
Since
the
computer
uses
Celsius
to
obtain
T.A.S.,
it
is
important
that
the
pilot
have
a
means
of
converting
Fahrenheit
to
Celsius.
The
flight
computer
described
in
this
manual
incor-
porates
a
temperature
conversion
scale
on
the
calculator
side
of
the
computer.
Temperature
conversions
can
be
read
directly
from
this
scale.
For
example:
Figure
35
shows
how
50°
F.
is
converted
to
10°
C.
Figure
35
WIND
SIDE
OF
COMPUTER
The
wind
side
of
the
computer
consists
of
a
rotating
azimuth
and
a
rectangular
grid
that
slides
up
and
down
through
the
azimuth.
(See
fig.
36.)
SLIDING
GRID
The
sliding
grid
is
nothing
more
than
a
section
taken
from
a
large
graduated
circle.
(See
fig.
37.)
The
lines,
Projecting
from
the
center
of
the
grid
and
radiating
outward,
represent
degrees
right
or
left
of
the
center
line.
The
lines
that
form
the
arcs
around
the
center
of
the
circle
represent
distance
from
the
center
and
are
labeled
in
miles.
The
computer
grid
has
two
sides,
a high
speed
side
and
a
low
speed
side.
(See
fig.
38.)
Since
most
private
aircraft
operate
in
the
speed
range
below
250
m.p.h.,
the
low
speed
side
is
generally
used
because
of
its
smaller
graduations
and
greater
accuracy.
The
low
speed
side
should
be
used
for
all
wind
problems
given
in
this
course,
AZIMUTH
The
azimuth
circle
rotates
freely
and
is
graduated
into
360°.
The
transparent
Portion
is
frosted
so
that
it
can
be
written
on
with
a
pencil.
The
center,
a
small
circle,
lies
directly
over
the
centerline
of
the
sliding
grid.
SLIDING
GRID
INSTRUCTIONS
FOR
FINDING
GROUND
SPEED
AND
TRUE
HEADING
©
INSTRUCTIONS
FOR
DETERMINING
MAGNETIC
COURSE,
MAGNETIC
HEADING,
AND
COMPASS
HEADING.
TRUE
INDEX.
WIND
CORRECTION
ANGLE
SCALE.
CENTER.
AZIMUTH.
eo
e000
Figure
36
DETERMINING
GROUND
SPEED
AND
TRUE
HEADING
SECTIONAL
AND
SECTIONAL
AND
WAC
SCALES
WAC
SCALES
To
maintain
a
specific
true
course,
the
pilot
must
deter-
(STATUTE)
(NAUTICAL)
mine
the
wind
correction
angle
and
adjust
the
aircraft
heading
accordingly.
The
time
en
route,
an
important
Cree
factor
in
any
flight.
is
influenced
by
wind
velocity
be-
pT
cause
it
affects
ground
speed.
To
determine
the
total
affect
of
wind
on
a
flight,
the
true
course,
true
airspeed,
and
wind
velocity
must
be
known.
The
operation
of
the
wind
side
of
the
computer
ts
des-
cribed
in
the
determination
of
true
heading
and
ground
speed
in
the
following
problem
GIVEN:
True
course
—030°
True
airspeed—
170
m.p.h.
Wind
—080?
at
20
m.p.h.
DETERMINE;
True
Heading
Ground
Speed.
The
solution
is
as
follows:
1.
Rotate
the
azimuth
until
the
wind
direction
of
080°
is
located
directly
under
the
true
index.
(See
fig.
40.)
2.
Next,
slide
the
grid
through
the
computer
un-
til
the
center
is
on
any
one
of
the
heavy
lines
extending
from
right
to
left
on
the
grid.
For
this
problem,
the
160-mile
grid
line
has
been
arbitrarily
chosen.
(See
fig.
40.)
HIGH
SPEED
SIDE
LOW
SPEED
SIDE
Figure
38
acre
Figure
40
Figure
41
.
Then,
place
the
wind
velocityonthe
azimuth
by
moving
up
from
the
center
20
miles
and
placing
a
pencil
mark
on
the
grid
centerline
(See
fig.
40.)
.
Now,
place
the
true
course
of
30°
under
the
true
index.
(See
fig.
41.)
.
With
the
true
course
under
the
true
index,
slide
the
grid
through
the
computer
until
the
pen-
cil
mark,
or
wind
dot,
is
positioned
on
the
true
airspeed
line
of
170
miles
per
hour
(See
fig
41.)
.
Find
the
wind
correction angle
by
checking
the
number
of
degrees
to
the
right
or
left
be-
tween
the
grid
centerline
and
the
wind
dot
In
this
problem,
the
wind
correction
angle
is
5°
to
the
right.
(See
fig.
42.)
_
To
find
true
heading,
the
scales
on
either
side
of
the
true
index
are
used.
Since
the
wind
correc
tion
angle
is
5°
right,
start
at
the
true
index
symbol,
count
five
units
to
the
right,
and
direct-
ly
below
this
graduation,
find
the
true
head-
ing
of
35°
on
the
azimuth
scale.
(See
fig.
42.)
If
wind
dot
is
to
left,
count
left
for
true
head-
ing,
if
wind
dot
is
to
right,
count
right.
NOTE
Another
method
is:
If
the
wind
correction
angle
is
to
the
right,
add
the
wind
correction
angle
to
the
true
course
value,
if
the
wind
correction
angle
is
to
the
left,
subtract
it
from
the
true
course.
8.
Without
moving
the
computer
setting,
read
the
ground
speed
under
the
center.
The
ground
speed
for
this
problem
is
156
miles
per
hour.
(See
fig.
42.)
PRACTICE
PROBLEMS
—
TRUE
HEADING
AND
GROUND
SPEED
For
answers,
refer
to the
Appendix
of
this
manual.
1.
True
course
—
310°,
T.A.S.
—
120
m.p.h.;
wind
—
180°
at
16
m.p.h.;
what
is
the
true
heading
and
ground
speed?
2.
True
course
—
178°,
T.A.S.
—
135
m.p.h.;
wind
—
045°
at
23
m.p.h.;
what
is
the
true
heading
and
ground
speed?
3.
True
course
—
050°,
T.A.S.
—
155
m.p.h.;
wind
—
165°
at
18
knots;
what
is
the
true
heading
and
ground
speed?
4.
True
course
—
270°,
T.A.S.
—
130
knots;
wind
—
344°
at
18
knots;
what
is
the
true
heading
and
ground
speed?
5.
True
course
—
095°,
I.A.S.
—
111
m.p.h.;
tem-
perature
—
+25°
C.;
altitude
—
7,500
feet;
wind
—
360°
at
10
knots;
what
is
the
true
heading
and
ground
speed?
FINDING
UNKNOWN
WIND
Another
problem
that
can
be
solved
on
the
wind
side
of
the
computer
is
finding
the
unknown
wind.
To
solve
this
problem,
four
factors
are
required:
true
course,
ground
speed,
true
heading.
and
true
airspeed.
To
find
the
unknown
wind
direction
and
speed,
use
the
wind
side
of
the
computer
and
proceed
as
foliows:
35
.
Place
true
course
under
true
index.
2.
Place
the
line
representing
the
ground
speed
under
the
center
of
the
azimuth.
3.
Determine
the
wind
correction
angle,
which
is
the
difference
between
the
true
course
and
the
true
heading,
and
note
whether
it
is
a
left
or
right
correction.
4.
Locate
the
true
airspeed
line
on
the
computer
and
count
the
number
of
degrees,
left
or
right,
necessary
to
correct
for
wind.
At
the
intersection
of
these
two
lines,
make
a
pencil
mark
to
repre-
sent
the
wind
dot.
5.
Rotate
the
azimuth
until
the
wind
dot
is
loca-
ted
on
the
centerline
directly
above
the
cen-
ter
of
the
azimuth.
The
wind
speed
is
the
differ-
ence
between
the
center
and
the
dot.
The
wind
direction,
in
degrees,
is
under
the
true
index.
EXAMPLE PROBLEM
—
FINDING
UNKNOWN
WIND
True
course
—
120°,
ground
speed
—
140
m.p.h.;
true
heading
—
115%;
true
airspeed
—
150
m.p.h.;
what
is
the
wind
direction
and
speed?
This
problem
is
solved
as
follows:
(See
fig.
43
and
44.)
1.
First,
place
the
true
course
of
120°
under
the
true
index.
(See
fig.
43.)
2.
Next,
position
the
grid
until
the
center
of
the
azimuth
is
over
the
line
representing
the
ground
speed
of
140
m.p.h.
(See
fig.
43.)
Figure
44
3.
Subtract
true
heading
from
true
course
and
find
that
the
true
heading
is
5°
less
than
the
true
course,
which
means
that
the
5°
is
a
left
wind
correction
angle
4.
With
the
center
on
140,
the
ground
speed,
move
up
the
grid
to
the
true
airspeed
line
of
150
m.p.h.
Then,
move
to
the
left
5°
and
make
a
pencil
mark,
or
wind
dot,
on
the
150
m.p.h.
line.
(See
fig.
44.)
5.
Rotate
the
azimuth
until
the
wind
dot
is
rest-
ing
directly
on
the
centerline.
(See
fig.
44.)
6.
By
checking
the
number
of
miles
between
the
center
of
the
azimuth
and
the
wind
dot,
find
that
the
wind
speed
is
16
m.p.h.
(See
fig.
44.)
7.
To
find
the
wind
direction.
look
under
the true
index.
In
this
problem,
the
wind
direction
is
065°
(See
fig.
45.)
Figure
45
TICE
PROBLEMS
—
FINDING
Dae
UNKNOWN
WIND
What
is
the
wind
direction
and
speed
for
the
follow-
ing
conditions?
Refer
to
the
Appendix
for
answers
to
these
problems.
_
True
course
—
190°;
ground
speed
—
110
m.p.h.;
true
heading
—
185°;
true
airspeed
—
115
mp.h.
True
course
—
355°;
ground
speed
—
130
m.p.h.;
true
heading
—
003°
;
true
airspeed
—
120
mph.
.
True
course
—
067°;
ground
speed
—
184
m.p.h.;
true
heading
—
O50°
;
true
airspeed
—
168
m.p.h.
FINDING
ALTITUDE
FOR
MOST
FAVORABLE
WINDS
The
wind
sideofthe
computer
can
be usedtodeter-
mine
the
best
altitudetoobtain the
highest
ground
speed
foragiven
course.
Thisisaccomplished
by
comparing
the
winds
aloft
with
the
coursetobe
flown,asfollows:
1.
The
wind
forecasts
for
each
altitude
are
plot-
ted
on
the
rotating
discinthe
same
manner
asinaground
speed-true
heading
problem.
The
only
difference,ofcourse,isthat
more
than
one
wind
is
plotted.
(See
fig.
46.)
.
Identify
each
wind
dot
with
the
appropriate
altitude.
.
Rotate
the
azimuth
to
place
the true
course
at
the
true
index
mark.
Figure
46
Figure
47
4.
To
examine
the
wind
effect
on
ground
speed,
position
the
curved
line
representing
the
appro-
Priate
true
airspeed
for
that
altitude
under
each
wind
dot.
Read
the
ground
speed
under
the
center
of
the
azimuth
for
each
wind
to
deter-
mine
the
highest
ground
speed.
For
example,
in
figure
47
is
shown
a
true
course
of
260°
and
winds:
3,000
ft.,
310°
at
22
m.p.h.;
6,000
ft.
340°
at
15
m.p.h.;
and
9,000
ft.,
030°
at
10
m.p.h.
With
a
true
airspeed
of
156
m.p.h.,
the
altitude
for
the
most
favorable
winds
is
9,000
ft.,
where
the
ground
speed
will
be
162
m.p.h.
Headwinds
would
be
realized
at
6,000
and
3,000
ft.
PRACTICE
PROBLEMS
—
FINDING
ALTITUDE
FOR
MOST
FAVORABLE
WINDS
Answers
to
these
problems
are
in
the
Appendix
of
this
manual.
WINDS
ALOFT:
3,000
ft.
—
from
300°
true,
at
21
knots
(24
m.p.h.)
6,000
ft.
—
from
330°
true,
at
12
knots
(14
m.p.h.)
9,000
ft.
—
from
030°
true,
at
12
knots
(14
m.p.h.)
Based
on
the
above
winds,
determine:
1.
Best
altitude
and
ground
speed
for:
T.C.
—
045°atTAS.
—
110
m.p.h.
ft.
Gs.
2.
Best
altitude
and
ground
speed
for
T.C.
—
330°atTA.S.
—
100
m.p-h
ft.
GS.
3.
Best
altitude
and
ground
speed
for:
T.C.
—
120°atTAS.
—
160
m.p-h.
ft
G.S,
4.
Best
altitude
and ground
speed
and
smallest
wind
correction
angle
for:
T.C.
—
170°
at
T.A.S.
—
130
m.p.h
ft.
GS.,
W.C.A,
5.
Best
altitude
and
ground
speed
for:
TC.
—
295°
atT.A.S
—
110
m.p.h
ft.
GS
RADIUS
OF
ACTION
Radius
of
action
means
the
time
or
distance
that
an
airplane
can
fly
out
on
a
given
course,
turn
around.
and
have
enough
fuel
to
return
to
the
departure
point.
A
radius
of
action
problem
is
solved
as
follows:
1.
First,
measure
the
true
course
outbound
from
the
departure
point
and
the
true
course
back
to
the
departure
point
NOTE
The
course
back
will
be
the
reciprocal
2.
Determine
the
ground
speed
out,
and
then
the
ground
speed
back.
NOTE
Wind
and
true
airspeed
must
be
known
for
this
step.
3.
Find
number
of
hours
of
fuel
on
board
the
air-
craft.
NOTE
Fuel
consumption
and
useable
fuel
must
be
known
for
this
step
4.
Add
the
ground
speed
out
to
the
ground
speed
back.
5.
Locate
the
total
of
ground
speed
out
and
ground
speed
back
on
the
“A”
scale
of
the
computer.
6.
Rotate
the
center
of
the
computer
until
the
fuel
duration,
in
hours
on
the
“C”
scale.
is
direct-
40
aE
ly
under
the
total
of
ground
speeds
out
and
back.
Move
along
the
“A”
scale
and
locate
the
ground
speed
back.
Look
directly
under
this
and
find
the
time
to
turn.
This
is
the
number
of
hours
and
minutes
that
the
aircraft
can
fly
outbound
‘on
the
course
before
it
has
to
turn
around
for
the
return
trip.
EXAMPLE
PROBLEM
—
RADIUS
OF
ACTION
True
course
—
060°;
true
airspeed
—
120
m.p-h.;
wind
—
052°
at
30
m.ph.;
useable
fuel
—
48
gallons:
and
fuel
consumption
—
6
g.p.h.
How
long
can
this
plane
fly
outbound
before
having
to
turn
and
come
back
to
the
departure
point?
This
problemisworked
as follows
1
Using
the
wind
side of
the
computer,
place
the
wind
directionof052°
under
the
true
index.
(See
fig.
48.)
Plot
up
from
the
center
the
wind
speed
of
30
m.p.h.
on
the
centerline
of
the
grid.
Rotate
the
azimuth
until
the
true
course
of
O60?
is
directly
under
the
true
index.
(See
fig.
49.)
To
find
the
ground
speed,
slide
the
grid
until
the
wind
dot
is
on
the
120
m.p.h.
line,
which
is
the true
airspeed.
Then,
check
the
center
of
the
azimuth
for
the
ground
speed.
which
41
Figure
49
is
90
m.p.h.
(See
fig.
49.)
Record
this
value
for
use
later
in
the
solution
to
this
problem.
.
Now,
reverse
the
azimuth
on
the
wind
side
of
the
computer
and
place
240°,
the
true
course
back
to
the
departure
point.
under
the
true
index.
(See
fig.
50.)
.
Slide
the
grid
until
the
wind
dot
rests
on
120
m.p-h.,
the
true
airspeed.
By
checking
the
cen-
ter
of
the
azimuth,
find
that
the
ground
speed
to
the
departure
point
will
be
150
m.p.h
(See
fig.
50.)
.
Add
ground
speed
outbound
to
ground
speed
back
to
the
departure
point
(90
+
150
=
240
m.p.h.).
The
total
is
240
m.p.h.
Next,
change
fuel
in
gallons
to
fuel
in
hours
as
follows:
a.
On
the
calculator
side
of
the
computer,
place
6
g.p.h.,
our
fuel
consumption,
on
the
“A’
scale
over
the
speed
index.
(See
fig.
51.)
itis)
id
Wie
250
|
Figure
51
b.
Look
directly
under
useable
fuel,
48
gals.
on
the
“A”
scale,
and
find,
on
the
“C”
scale,
that
the
total
useable
fuel
is
8
hours.
(See
fig.
51.)
9.
Now
place
the
240,
which
is
the
total
of
the
ground
speed
out
and
the
ground
speed
back,
directly
over
the
8
hours
of
fuel
available.
(See
fig.
52.)
Move
around
the
“A”
scale
to
the
ground
speed
back
of
150
m.p.h.
Next,
look
directly
under
150
and
find
that,
at
the
end
of
5
hours,
the
aircraft
should
start
back
to
its
departure
point
so
as
to
have
enough
fuel
to
make
the
return
trip.
10.
The
answer
to
the
problem
is:
time
to
turn
—
5
hours.
Time
is
converted
to
distance
as
follows:
a.
Place
the
ground
speed
out,
90
m.p.h.,
over
the
speed
index.
(See
fig.
53.)
Figure
52
11
Figure
53
b.
Look
directly
over5hours
on
the
“C”
scale
and
find
that
the
aircraft
will
travel
450
miles
in
this
length
of
time.
(See
fig.
53.)
450
miles
is
the
radius
of
action
in
miles.
5
hours
is
the
radius
of
action
in
time.
To
check
the
problem,
find
the
time
required
to
fly
back
over
the
450-mile
course
with
the
ground
speed
back
of
150
miles
per
hour
as
follows
a.
Place
the
speed
index
under
150,
our
ground
PRACTICE
PROBLEMS
—
RADIUS
OF
ACTION
Refer
to
the
Appendix
for
answers
to
these
problems:
1.
GIVEN:
a.
Wind,
230°
at
18
m.p.h.
b.
T.
C.
out,
270°
c.
T.
C.
back,
090°
.
d.
Fuel
available,
38
gal.
(not
counting
reserve.)
e.
Fuel
consumption,
8
g-p.h.
f.
T.A.S.,
124
m.p.h.
PROBLEM:
a.
What
is
the
time
to
turn?
b.
What
is
the
radius
of
action
in
statute
miles?
c.
What
is
the
total
time
out
and
back?
.
GIVEN:
a.
Ground
speed
out,
140
m.p.h.
b.
Ground
speed
back,
165
m.p.h.
c.
Fuel
available,
46
gal.
(not
counting
reserve.)
d.
Fuel
consumption,
11
g-p.h.
PROBLEM:
a.
What
is
the
time
to
turn?
b.
What
is
the
radius
of
action
in
statute
miles?
c.
What
is
the
total
time
out
and
back?
.
GIVEN:
speed
back.
(See
fig.
54.)
a
b.
Look
directly
under
450
miles
on
the
cay
b.
scale
and
find
that
it
will
take
3
hours
for
c.
Fuel
available,
68
gal.
(not
counting
reserve.)
d
e
.
Wind,
030°
at
14
knots.
.
T.C.
from
points
“A”
to
“B
“170°
.
|.
Fuel
consumption,
12
g.p.h.
.
T.
A.
S.,
164
m.p.h.
PROBLEM:
a.
What
is
the
time
to
turn?
b.
What
is
the
radius
of
action
in
statute
miles?
c.
What
is
the
total
time
out
and
back?
the
aircraft
to
make
the
return
trip
(See
fig.
54.)
This
is
the
final
check.
Add
time
out.
5
hours,
to
the
time
back
of
3
hours
for
a
total
of
8
hours.
This
time
is
equal
to
the
fuel
aboard
when
the
problem
solution
's
correct
METRIC
CONVERSIONS
KILOMETER
TO
STATUTE
OR
NAUTICAL
MILE
Kilometer-statute
mile
and
kilometer-nautical
mile
con-
versions
are
accomplished
by
aligning
the
Proper
arrows
on
the
B
and
A
scales.
The
conversion
arrow
for
kilometers,
labeled
KM,
is
located
near
12
on
the
B
scale.
When
this
arrow
is
aligned
with
the
STATUTE
arrow
on
the
scale
(at
76),
all
values
on
the
scale,
in
kilometers,
will
be
aligned
with
their
corresponding
values
on
the
scale,
in
statute
miles.
When
the
KM
arrow
on
the
scale
is
aligned
with
the
NAUTICAL
arrow
on
the
A
scale
(at
66),
all
of
the
kilometer
values
on
the
scale
will
be
aligned
with
their
corresponding
nautical
mile
values
on
the
scale.
EXAMPLE
PROBLEM
—
NAUTICAL
MILES
TO
KILOMETERS
GIVEN:
132
nautical
miles
DETERMINE:
Kilometers
The
problem
is
worked
as
follows:
1.
Set the
nautical
arrow
on
the
A
scale
opposite
the
KM
arrow
on
the
scale.
(See
fig.
55.)
2.
Read
244
kilometers
on
the
B
scale
opposite
132
NM
on
the
scale.
PROM
46
U.S.
GALLONS—IMPERIAL
(BRITISH)
GALLONS
Near
the
number
11
on
each
scale,
an
arrow
indicates
imperial
gallons,
and
near
the
number
13
another
arrow
indicates
U.
S.
gallons.
The
imperial
gallon,
used
in
Great
Britian
and
Canada,
is
equal
to
approximately
1.2
U.
S.
gallons.
To
convert
U.
S.
to
imperial
gallons
or
vice
versa,
line
up
the
imperial
gallon
arrow
on
one
scale
with
the
U.
S.
gallon
arrow
on
the
other
scale.
Read
imperial
gallons
on
the
same
scale
as
the
imperial
gallon
arrow
and
U.
S.
gallons
on
the
same
scale
as
the
U.
iS:
gallon
arrow.
Check
the
answer
to
see
that
the
number
of
U.
S.
gallons
is
larger
than
the
corresponding
number
of
imperial
gallons.
EXAMPLE
PROBLEM
—
U.S.
GALLONS
TO
IMPERIAL
GALLONS
GIVEN:
15
imperial
gallons
DETERMINE:
U.
S.
gallons
This
problem
is
solved
as
follows:
1.
Set
IMP.
GAL.
arrow
on
A
Scale
opposite
U.
S.
GAL.
arrow
on
B
scale.
(See
fig.
56.)
2.
Read
18
U.
S.
gallons
on
B
scale
opposite
15
on
A
scale.
PN
ciate
a
Figure
56
47
ee
OTHER
Co
NVER
GALLONS
—
LITERS;
PEEPS
torens:
POUNDs
—
KILOGRAMS
The
conversion
arrows
are
located
as
follows:
Liters—near
48.5
on
inside
and
outside
scales
Feet—near
the
figure
14
on
the
outside
scale
Meters—near
43.5
on
inside
scale
Pounds—near
36
on
outside
scale
Kilograms—near
the
figure
16
on
inside
scale
Note
that
the
arrows
for
feet,
meters,
pounds,
and
kilograms
each
appear
on
only
one
scale.
Hence,inthe
conversion
of
feet
to
meters
or
pounds
to
kilograms,
and
vice
versa,
there
is
only
one
way
of
matching
the
arrows
in
each
problem.
NOTE
1
imp.
gal.
=
1.2
U.
S.
gal.
4
liters
=
approx.
1
U.
S.
gal.
1
kg.
=
approx.
2
Ibs.
1
meter
=
approx.
3
ft.
m,
the
liter
is
the
i i
ic
syste!
ies
using
the
metric
sure
which
approximates
one
quart.
In
countr
liquid
mea:
FUEL—
POUNDS
AND
GALLONS
i
FUEL
LBS.
&
Le
ae
marked
:
at
128)
are
aligned,
e
B
scale
is
opposite
the
A
scale.
lance
in
this
hen
the
conver:
A
the
A
scale
(located
at
ae
US
GAL.
on
the
B
scale
(loca'
or
i
in
pounds
on
eae
responding
weight
in
P'
sat
ag
Ilons
can
be
converted
to
pound:
Gallo’
48
way
as
shown
in
Figure
57.
Similarly,
pounds
can
be
converted
to
gallons.
One
Gallon
of Fuel
weighs6pounds.
EXAMPLE
PROBLEM—
FUEL
IN
GALLONS
TO
POUNDS
GIVEN:
Fuel,
22.5
gallons
DETERMINE:
Fuel
weightinpounds
This
problemissolvedasfollows:
1.
Set
FUEL
LBS.
on
A
scale
opposite
U.S.
GAL.
on
scale.
(See
fig.
57.)
2.
Read
135
Ibs.
on
A
scale
opposite
22.5
GAL.
on
scale.
Figure
57
OIL—POUNDS
AND
GALLONS
The
procedure
for
obtaining
oil
weight
is
similar
to
that
for
fuel.
When
the
OIL
LBS.
arrow
on
the
A
scale
(located
at
96)
is
placed
over
the
U.S.
GAL.
arrow,
each
value
of
oil
in
gallons
is
set
on
the
B
scale
directly
under
its
corresponding
weight
in
Pounds
on
the
A
scale.
Figure
58
shows
an
example
of
the
conversion
of
oil
in
gallons
to
oil
weight
in
pounds.
Oil
weighs
7.5
pounds
per
gallon.
EXAMPLE
PROBLEM—
OIL
IN
QUARTS
TO
POUNDS
GIVEN:
Oil,
12
quarts
(3
gallons)
DETERMINE:
Oil
weight
in
pounds.
49
1.
Set
OIL
LBS.
on
A
scale
opposite
U.S.
GAL.
on
B
scale.
(See
fig.
58.)
2.
Read
Oil
Wt.
(22.5
Ibs.)
Opposite
3
Gal.
on
B
scale.
Figure
58
PRACTICE
PROBLEMS—
METRIC
AND
OTHER
CONVERSIONS
Fill
in
the
missing
values
and
check
your
answers
against
the
answers
in
the
Appendix.
Kilometers
Statute
Miles
Nautical
Miles
le
1,450
Ze
8
Se
54
Liters
U.S.
Gallons
Imperial
Gallons
4.
1,165
5.
550
6.
ig
7s
100
eee
Feet
Meters
8.
56
9
70
Pounds
Kilograms
10.
24
11.
1
12.
143
—_
:
Fuel,
Gallons
Pounds
Oil,
Gallons
13.
20
—
14.
150
—
4
15.
a,
7
16:
—
|
50
APPENDIX
I
ANSWERS
TO
PRACTICE
PROBLEMS
FINDING
TIME
(Page
5)
1:20
2:36
18
minutes
38
minutes
BoN=
FINDING
DISTANCE
(Page
6)
150
miles
262
miles
700
miles
470
miles
BW
FINDING
SPEED
(Page
7)
1
126
m.p.h
2
160
m.p.h.
3
110
m.p.h
a
96
m.p-h
SHORT
TIME
(Page
9)
1.
7
miles
2
100
m.p.h.
FUEL
CONSUMPTION
—
FINDING
TIME
(Page
10)
1
2:30
2
346
3
2:11
4
2:20
FUEL
BURNED
(Page
11)
1
3%
gal
2.
56
gal
e
19.8
gal
4.
5.15
gal
FUEL
CONSUMPTION
(Pages
11
and
12)
1
10'2
g.p.h
2
21.6
9.ph
3;
23
gph
4
86g.ph
A-l
EES
T.A.S.
(Page
13)
fe
137
m.p.h.
2
124
m.p.h.
SF
180
m.p.h.
4.
111
mph
NAUTICAL-STATUTE,
STATUTE-NAUTICAL,
AND
STATUTE-KILOMETER
(Page
16)
161
miles
28.8
m.p.h.
67
nautical
miles
19.6
m.p-h.
214
kilometers
24
8
statute
miles
OCI)
ON
MULTI-PART
PROBLEMS
(Page
17)
time
—
2:40;
24
gallons
TA.
121
m.p.h.;
time
—
2:07;
24.4
gallons
T.A.
130
m.p.h.;
time
—
2:34;
21.8
gallons
time
—
3:09;
37.8
gallons
ONS
OFFCOURSE
(Pages
19
and
20)
1
el
118
Right
g°
N72)
Left
4°
6°
Right
es2eg90c09
TIME
&
DISTANCE
TO
A
VOR
STATION
(Page
23)
36
min.,
66
miles
30
min.,
47.6
miles
4
min.,
9
miles
36
min.,
45
miles
5.5
min.,
12.85
miles
CSS
NS
TRUE
ALTITUDE
(Pages
24
and
25)
1.
6,250
feet
2.
9,400
feet
3.
5,250
feet
A-2
DENSITY
ALTITUDE
(Page
25)
1,
8,000
feet
2.
13,000
feet
3.
Sea
Level
MULTIPLICATION
AND
DIVISION
(Page 27)
8,100
feet
2,700
feet
720
f.p.m.
430
f.p.m.
12,400
pound-inches
54
inches
DOR
O
CONVERTING
FEET
PER
NAUTICAL
MILE
TO
FEET
PER
MINUTE
(Page
29)
1
400
f.p.m.
2.
400
f.p.m.
3.
700
f.p.m.
4
700
f.p.m.
TRUE
HEADING
AND
GROUND
SPEED
(Page
35)
1
True
heading—304°
Ground
speed—130
mp.h
2.
True
heading—171°
Ground
speed—150
mp.h
3
True
heading
—
057°
Ground
speed—163
mp_h.
(141
knots)
4
True
heading
—
278°
Ground
speed
—
124
knots
5
True
heading
—
090°
Ground
speed
—
130
mph
(113
knots)
FINDING
UNKNOWN
WIND
(Page
38)
1
125°
at11m.p.h
2
119°
at
20
m.p.h.
3
311°
at
55
m.p.h
A3
a
FINDING
ALTITUDE
FOR
MOST
FAVORABLE
WINDS
(Page
39
and
40)
1.
3,000
ft.,
115
m.p.h.
2.
9,000
ft.,92m.p.h.
3.
3,000
ft.,
183
m.p.h.
4.
6,000
ft.,
143
m.p.h.,2°right
5.
9,000
ft.,
110
m.p.h.
RADIUS
OF
ACTION
(Page
45)
Us
2.38
290
miles
4:45
2:15
315
miles
4:10
2:37
462
miles
5:40
epogeeope
oe
METRIC
AND
OTHER
CONVERSIONS
(Page
50)
1.
900
SM,
782
NM
2.
13
KM,
7
NM
3.
100
KM,
62
SM
4.
4,400
liters,
970
imp.
gal.
5.
2,500
liters,
660
U.
S,
gal.
6.
2.4
U.S.
gal.,
2
imp.
gal.
7.
378
liters,
83
imp.
gal.
8.
17
meters
9.
230
feet
10.
53
pounds
11.
5
kilograms
12.
65
kilograms
13.
120
pounds
14.
900
pounds
15.
30
pounds
16.
52.5
pounds
A4
———<——SSrtrcrclellmlrmmttt
APPENDIX
Il
GLOSSARY
OF
TERMS
AIRSPEED.
The
speed
of
an
aircraft relative
to
the
air.
CALIBRATED
(CAS).
Indicated
airspeed
corrected
for
pitot-static
instal-
lation
and
instrument
errors.
INDICATED
(IAS).
The
uncorrected
reading
obtained
from
the
air-
speed
indicator.
TRUE
(TAS).
Calibrated
airspeed
corrected
for
density
altitude
(pressure
and
temperature).
ALTITUDE.
The
height
of
an
aircraft
above
mean
sea
level
or
above
the
terrain.
ABSOLUTE
(AA).
True
altitude
corrected
for
terrain
elevation,
the
vertical
distance
of
the
aircraft
above
the
terrain.
CALIBRATED
(CA).
Indicated
pressure
altitude
corrected
for
instru-
ment
error.
Also
known
as
flight-level
pressure
altitude.
DENSITY
(DA).
Calibrated
altitude
corrected
for
temperature;
the
vertical
distance
of
the
aircraft
above
the
standard
datum
plane.
PRESSURE
ALTITUDE
(PA).
The
reading
of
the
pressure
altimeter
with
the
baro-
metric
window
set
at
29.92
TRUE.
The
density
altitude
corrected
for
pressure
alti-
tude
variation
(PAV);
the
vertical
distance
above
mean
sea
level
COURSE.
The
direction
of
the
intended
path
of
the
aircraft
over
the
earth;
or
the
direction
of
a
line
on
a
chart
representing
the
intended
aircraft
path,
ex-
pressed
as
the
angle
measured
from
a
specific
refer-
ence
datum
clockwise
from
O°
through
360°
to
the
line.
AS
EEE”
MAGNETIC
(MC).
The
course
of
an
aircraft
mea
q
to
the
north
magnetic
pola”
“*rerence
TRUE
(TC).
qapcoucse
of
an
aircraft
measured
with
reference
fue
north,
or
geographic,
pole.
(The
true
course
is
always
represented
b
of
the
flight
computer
grid.)
ice
SRoUND
SPEED.
The
speed
of
the
aircraft
relative
@
ground.
(Ground
speed
is
always
represented
by
the
center
of
the
azimuth
of
the
flight
computer.)
NOTES
HEADING.
The
angular
directionofthe
longitudinal
axisofan
aircraft
measured
clockwise
fromarefer-
ence
point.
COMPASS
(CH).
The
reading
taken
directly
from
the
compass.
MAGNETIC
(MH).
The
heading
of
an
aircraft
with
reference
to
magne-
tic
north.
TRUE
(TH).
The
heading
of
an
aircraft
with
reference
to
grid
north.
MACH
NUMBER.
The
speed
of
a
moving
object
com
pared
to
the
speed
of
sound
within
the
same
medium
of
movement.
A
speed
of
Mach
2.5
would
be
two
and
one-half
times
the
speed
of
sound
in
the
same
medium.
NAUTICAL
MILE
(NM).
A
unit
of
distance
used
In
navigation,
6080
feet;
the
mean
length
of
one
min-
ute
of
longitude
on
the
equator,
approximately
1
minute
of
latitude:
1.15
statute
miles.
STATUTE
MILE
(SM).
5.280
feet
or
827
nautical
miles.
WIND
COR
grees
that
t
placed,
to
the
right
or
I
true heading- (The
win
ber
of
de
RECTION
ANGLE.
The
num
:
he
aircraft
longitudinal
axis
must
be
dis:
left
of
the
true
course
and
the
ngle
Is
always
tween
the
centerline
of
the
grid
and
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