Beam and tape calculations
RESTRICTED
G.S. miles per hour = G.S. knots xl.15
G.S. feet per sec.= G.S. miles per
hrs.
times l.467
W.R.
ft.= G.S. in ft. per sec. times
A. T. F.
Trail in ft.= Alt. x Tin mils.
TO0O
A.R. In ft.= W.R. in ft. - Trail in ft.
Thus:
150 knots
1.15 = 172.5 MPH G.S.
172.5 X
1.467 = 253.06
ft/sec.
G. S.
253.06
sec. = 8857.1 ft.
W.R.
15,000
(alt.)
× 70 = 1050 ft.
trail.
1000
8857.1-1050=7807.1 ft.
A. R
1807ml=.5205 tan function A.R. angle
15000
D.
Same problem and mechanical Solution by Sight.
1.
Given:
Same as
above.
2. Find:
Position of rate quadrant (A.R.
angle) in tan function.
3. Solution:
a.
fot tape spe ta
Determine tape speed (analogous to
ground speed)
in mils per second.
To do this di-
vide G.S. ft/sec. by the number of feet subten-
ded by
one
mil at
that altitude.
At 15,000 ft.
one mil subtends 15 ft. --and
in one second 253.06
feet
are covered (see
above
) --and there
are
feet
in each
mil, so that tape speed mils/sec.
equals 16.88 mils.
For a more
accurate setting of the
tape,
multiply this amount by any
number of sec-
onds
so long as the answer
stays within the limits
of the steel
tape (1800 mils). Using a stop watch,
reguire that number of mils for the chosen number
of seconds
to pass
a given point.
b.
Set Disc Speed
into Sight. (divide
bombing constant by A.T.F.).
Aiso set trail in.
Set maximum
sighting angle,
turn on
telescope
the
tape.
motor
and synchronize
Synchrone until te
on some mark on
the
ma - k
÷ om
maX! num
Cominimum.
C .
Read
the position of te
*ate auad -
rant.
This can be done by referring to the scale
rate
oniy a rough
tan
quadrant,
but
is wiligive
function reading.
A more accur-
ate method of ascertaining the tan function posi-
in
2)
transverse
hair
216