LIST OF CONTENTS
CHAPTER 1- DEFINITIONS
SECTION – 1
SECTION – 2
INTRODUCTION
DEFINITIONS
CHAPTER 2 – INTERNAL BLSTS
SECTION – 3
SECTION – 4
SECTION – 5
SECTION – 6
GENERAL PRINCIPLES
CHARGE DESIGN
MUZZLE VELOCITY
ABNORMAL EFFECTS INTERNAL BLST
CHAPTER 3 – EXTERNAL BLSTS
SECTION – 7
SECTION – 8
SECTION – 9
SECTION – 10
SECTION – 11
SECTION – 12
SECTION – 13
SECTION – 14
SECTION – 15
SECTION – 16
GENERAL
MOTION IN VOCUO
AIR RESISTANCE AND PROJECTILE SHAPE
MOTION IN AIR
STABILITY OF A PROJECTILE
DRIFT
MAGNUS EFFECT
ROTATION OF THE EARTH
UNCONVENTIONAL SHELLS
METEOROLOGICAL CONDITIONS
CHAPTER 4 – MORTARS AND ROCKETS
SECTION – 17
SECTION – 18
MORTORS
ROCKETS
CHAPTER 5 – FIRING TABLES
SECTION – 19
SECTION – 20
COMPILATION OF FIRING TABLES
CONTENTS OF FIRING TABLES
CHAPTER 6 – NON STANDARD CONDITIONS
SECTION – 21
SECTION – 22
VARIATIONS AND EFFECTS
CORRECTIONS
SECTION – 23
DISTINCTIONS BETWEEN EFFECTS AND
CORRECTIONS
SECTION – 24
SECTION – 25
CORRN OF ERRORS IN DEF OR ELE
CREST CLEARANCE
CHAPTER 7 – PROBABILITY AND DISPERSION
SECTION – 26
SECTION – 27
SIMPLE PROBABILITY THEORY
CONSISTENCY AND ACCURACY
CHAPTER 8 – TERMINAL BALLISTICS
SECTION – 28
SECTION – 29
CLASSIFICATION OF PROJECTILES
HIGH EXPLOSIVE PROJECTILES
DATA BANK - BLSTS
PREP BY
MAJ MATEEN LAGHARI
SEC – 1
Question No 1.
Define the fol :-
a.
Blsts. The branch of science that deals with experimental and
theoretical studies and fd trials on the firing of proj from an
orthodox gun, how etc or an arty rkt, its subsequent motion and
its performance at a tgt.
b.
Internal Blsts. The branch of blsts that deals with the events
taking place in a gun from the time the prop chg is ignited until
the proj leaves the muzzle. Scope of internal blsts also covers
the blsts properties of props.
c.
Intermediate Blsts. The branch of blsts covering the pd from
the time the with the proj leaves the barrel until it ceases to be
affected by the muzzle outflow of prop gas.
d.
External Blsts. The branch of blsts dealing with the motion of
a proj from the time it leaves the influence of muzzle outflow of
the prop gas to the instant it strikes the tgt, or to the initiation of
the fuze.
e.
Terminal Blsts.
The branch of blsts that deals with the
performance of proj at a tgt and the tgt’s response to impact.
More gen, it deals with the events that fol the initiation of fuze.
The branch of blsts concerned with theoretical and
experimental studies of the behaviour of a proj on impact with a
tgt, the damage produced at the tgt, and the mech of causing
this damage.
f.
Gnry.
tgts.
It is the prac application of blsts to the engagement of
Question No 2.
What is the approx dist after the proj leaves the barrel
until it ceases to be affected by the muzzle outflow of prop gas?
Answer No 2.
Question No 3.
Answer No 3.
Approx dist is around 50 Ms.
What are the basic reqs of arty fire?
Fol are the three basic reqs :-
a.
Accy.
b.
Consistency.
c.
Eff.
Question No 4.
Although it is conventional to study and dev each
branch of blsts separately, yet all the branches are inter related and a
modification in the reqs of one branch on design factors effs other branches
as well. Your comments and prove your statement with an eg?
Answer No 4.
Yes, the statement is true and can be proved with the
help of fol eg. We know that in order to improve the terminal blsts
performance of a KE proj, its wt and impact vel are req to be enhanced.
The improvement of these factors would entail a change in internal,
intermediate and external blsts. The proj must be given a high MV thus req
the use o a higher chg within the gun safety limits and to retain its velocity it
must have an improved shell design and shape to reduce the air
resistance. Hence all branches although studied separately, yet are closely
related to each other and a change in one branch would entail a
modification in the other branches as well.
SEC – 2
Question No 1.
Define the fol :-
a.
Traj.
It is the path described by the CG of the proj form the
muzzle onwards.
b.
Tgt.
It is the specific pt at which the fire is dir. It is normally
on gr but may also be in the air.
c.
Wpn Axis. It is the axis of bore at the breech end, taken as a
straight line. In case of a rkt it is the longitudinal axis of the rkt
before launching taken as a straight line.
d.
Muzzle Axis.
straight line.
e.
Line of Sight (L/S).
It is a straight line passing through the
wpn or inst and the tgt.
f.
Line of Departure.
It is the tangent to the traj at the
commencement of proj’s flt.
g.
Line of Arrival.
or impact.
h.
Vertical Plane of Sight.
It is the vertical plane through the
wpn or inst containing the L/S.
j.
Lateral Plane of Sight.
It is the lateral plane through the
wpn or inst containing the L/S.
k.
Vertical Plane of Fire.
It is the vertical plane through the
wpn containing the wpn axis before firing.
l.
Vertical Plane of Departure.
It is the vertical plane through
the wpn containing the line of departure.
It is the axis of bore at the muzzle taken as a
It is the tangent to the traj at the pt of graze
m.
Horizontal Plane.
The horizontal plane at a given ref pt is
the tangent to the circumference of earth at that pt.
n.
Lvl Sfc.
The sfc of a sphere (the earth assuming a mean
radius) tangential to the horizontal plane at a ref pt, plus the
altitude of the ref pt.
o.
Slant Dist.
p.
Horizontal Dist.
plane.
q.
Map Range. The horizontal dist from a wpn to a pt vertically
above or below the tgt furnished by the grid in use.
r.
Ht.
The dist measured along the vertical line b/w the ref lvl
and the pt above or below it.
s.
Alt. The ht of a pt wrt mean sea lvl.
t.
Traj.
Traj is the curved path described by the centre of
gravity of proj from the muzzle onwards.
u.
Vertex.
It is the highest pt on the traj and on this pt the
vertical component of velocity is zero. It is also called the
culminating pt or the summit.
v.
Vertex Ht.
It is also known as the max ordinate and is the ht
of the vertex with respect to the lvl of the wpn.
w.
Pt of Graze.
The pt of intersection of traj and the lvl sfc on
which the wpn rests.
x.
Pt of Impact. The pt at which a proj first strikes an object.
y.
Pt of Burst.
z.
Blst Angles.
proj.
The dist b/w any two pts measured along the L/S.
The dist b/w any two pts on the horizontal
The pt at which the proj bursts in the air.
The angles which are associated with the traj of
aa.
Angle of Sight (A of S). The vertical acute angle measured
from the horizontal plane passing through the wpn or inst to the
L/S.
bb.
Ele. The vertical acute angle measured from the horizontal
plane to the wpn or inst axis.
cc.
Firing Table (FT) Ele.
The ele at which the gun is req to be
laid under std conditions to achieve the desired range.
dd.
Quadrant Ele (QE). The ele at which the gun is req to be laid
under the prevailing conditions to achieve the desired range. It
is also the vertical angle b/w the horizontal plane through
trunnions and the axis of bore when the gun is laid before firing.
ee.
Tangent Ele (TE).
The vertical acute angle measured from
the L/S and to the wpn or inst axis.
ff.
Angle of Projection (A of P). Vertical acute angle measured
from L/S to line of departure.
gg.
Angle of Departure (A of D).
The vertical acute angle
measured the horizontal plane passing through the wpn to the
line of departure.
hh.
Jump. The vertical acute angle measured from the muzzle
axis before firing to the line of departure.
jj.
Throw Off / Lateral Jump. The acute angle measured in the
horizontal plane from the muzzle axis before firing to the line of
departure.
kk.
Droop. The vertical acute angle measured from the wpn axis
to the muzzle axis.
ll.
Angle of Descent.
Angle b/w the line of arrival and the lvl
sfc at the pt of graze. In US terminology it is also known as the
angle of fall (A of F).
mm. Angle of Impact. The angle b/w the line of arrival and the sfc
of tgt at the pt of impact.
nn.
Angle of Incidence.
The acute angle b/w the line of arrival
and the line drawn at rt angles to the sfc of tgt (the normal) at
the pt of impact. It is the complement of angle of impact.
oo.
Angle of Drift.
The horizontal angle b/w the vertical plane of
departure and the vertical plane of impact or burst.
pp.
Time of Flt (TOF).
Time taken by the proj to travel b/w the
muzzle and a specified pt on the traj. When the pt is not
specified, the TOF ref to the pt of graze or burst.
qq.
MV.
The velocity with the proj leaves the muzzle of a gun. If
measured instrumentally, MV is deduced by the extrapolation
from the velocity from the velocity of proj measured at a
convenient pt on its traj.
rr.
Remaining Velocity.
impact.
ss.
Variation.
tt.
Eff. A change in blst performance due to one or more
variation.
uu.
Corrn.
A corrn is applied to basic data to cater for one or
more effs..
vv.
Low Angle Fire. Fire at gun ele below that at which the max
range is achieved.
Any diff b/w a non-std and a std condition.
ww. High Angle Fire.
range is achieved.
Question No 2.
range?
The velocity of proj at the pt of graze or
Fire at gun ele above that at which the max
What is the diff b/w the horizontal dist and the map
Answer No 2.
Horizontal dist is the dist b/w two pts in the horizontal
plane whereas the map range is the horizontal dist but furnished by the grid
in use.
Question No 3.
What is the diff b/w the alt and ht?
Answer No 3. Alt is the ht of a pt wrt mean sea lvl whereas ht ref to the
vertical dist b/w the ref lvl and the pt being considered above or below it.
Question No 4.
What is the diff b/w vertex and vertex ht?
Answer No 4.
Vertex is pt on the traj where the vertical component of
velocity becomes zero whereas vertex ht is the ht of that pt wrt to the lvl of
the wpn.
Question No 5.
What is the diff b/w jump and throw off?
Answer No 5.
Both the angles are made b/w axis of bore and line of
departure when the gun is laid before firing, with fol diffs :(a)
Jump is vertical angle where as throw off is the horizontal
angle.
(b)
Throw off is much smaller in magnitude than jump.
Question No 6.
positive A/S?
Answer No 6.
Diagrammatically show the angles at gun end for
Line of
Departure
Wpn Axis
Muzzle
Axis
A of P
Jump
A of D
L/S
Droop
TE Ele
Horizontal
Diagrammatically show the angles at gun end for
Plane
Line of
Wpn Axis
Departure
A/S
Question No 7.
negative A/S?
Answer No 7.
Muzzle
Axis
A of P
Jump
A of D
Droop
TE
Ele
Horizontal
Plane
A/S
L/S
Question No 8.
Ele and TE?
Answer No 8.
What is the eff of negative A/S on A of D and
It will have fol effs :-
a.
A of P would be more than the A of D.
b.
TE would more than ele.
Question No 9.
Find out A/S, ele and drift when fol data is aval :-
a.
TE
=
333 mils
b.
Jump
=
– 5 mils
c.
A of D
=
299 mils
A of P,
d.
Drift constant
=
17
Answer No 9.
A/S
=
– 29 mils
QE
=
304
Drift
=
5.67 = R 5.7
(17 x Tan A of P)
Question No 10. Draw all the angles formed at tgt end when tgt is at the
fwd slope and A/S is positive?
Answer No 10.
Normal
Angle of Incidence
Sfc of Tgt
900
L/S
Angle of Impact
Line of Arrival
Angle of Descent
Lvl Sfc
Question No 11.
Draw all the angles formed at tgt end when tgt is at
the fwd slope and A/S is negative?
Answer No 11.
Normal
Sfc of Tgt
Angle of Descent
Angle of Incidence
Angle of Impact
900
Lvl Sfc
L/S
Line of Arrival
Question No 12. Draw all the angles formed at tgt end when tgt is at the
reverse slope and A/S is positive?
Answer No 12.
Sfc of Tgt
Normal
L/S
Angle of
Incidence
Angle of Impact
Line of Arrival
Lvl Sfc
Angle of Descent
Question No 13. Draw all the angles formed at tgt end when tgt is at the
reverse slope and A/S is negative?
Answer No 13.
Normal
Angle of Impact
Angle of Incidence Angle of Arrival
Angle of Fall / Descent
Tgt Sfc
Question No 14.
Answer No 14.
Draw all the angles formed at tgt end when tgt is a tk?
Line of Fire
Gun
Angle of Impact
Tgt
Angle of Incidence
Normal
Pt of Impact
SEC – 3
Question No 1.
What is the purpose of a gun or a RL?
Answer No 1.
It is to project a proj with a sufficiently high velocity and
minimal initial disturbance to the traj, so as to inflict damage to a tgt at a
given range.
Question No 2.
Answer No 2.
What does a gun sys consist of?
It consists of :-
a.
The gun.
b.
The proj.
c.
The prop chg.
Question No 3.
Answer No 3.
What are the purposes of fitting a DB on a proj?
It is fitted for the fol reasons :-
a.
To prevent the escape of gases fwd past the proj.
b.
To offer certain initial resistance to mov which has the eff of
ensuring regularity of ignition of the prop chg.
c.
To engage with the rifling of the barrel and impart spin to the
proj as it moves up the bore.
d.
To asst in centering proj in bore.
e.
To hold proj in posn when loaded and prevent slip back when
the gun is elevated.
Question No 4.
Answer No 4.
What all factors of a proj eff the internal blsts?
Fol factors of proj affect the internal blsts :-
a.
Wt of the proj.
b.
Shape of the proj.
c.
Posn and shape of DB.
d.
Material of DB.
Question No 5.
Answer No 5.
a.
What are the laws of burning of props?
Laws of burning of props are :-
Prop burn parallel to their sfc throughout their length.
b.
Higher the pressure, greater the rate of burning. Rate of
burning is approx proportional to the pressure.
c.
Greater the area of burning sfcs higher the rate of burning and
pressure build up.
d.
Rate of burning depends upon :-
e.
(1)
Chemical composn of props.
(2)
Chg temp before ignition.
Props are bad conductors of heat.
Question No 6.
Answer No 6.
What factors of props affect the internal blsts?
The factors of props which affect the internal blsts are :-
a.
Wt of the prop.
b.
Nature of the prop.
c.
Shape of the prop.
d.
Size of the prop.
e.
Initial temp of the prop.
Question No 7.
Pressure(SSP)’?
What do you understand by the ‘Shot Start
Answer No 7.
The pressure produced inside the chamber of the gun
due to burning of props high enough to mov the proj fwd and engrave. The
proj then starts to accelerate.
Question No 8.
What do you understand by the ‘Peak Pressure (PP)’?
Answer No 8.
A stage comes inside the gun when the inc of pressure
due to evolution of gases is bal by the dec in pressure due to the inc of
space behind the proj. This pressure is known as the PP.
Question No 9.
Velocity is max at PP. Comment?
Answer No 9.
Statement is incorrect. Acceleration is max at PP,
whereas velocity will keep on inc after PP as gases will keep on expanding
but the acceleration will dec.
Question No 10.
(ABP)’?
What do you understand by the ‘All Burnt Posn
Answer No 10.
as the ‘ABP’.
The stage when the prop is completely burnt is known
Question No 11.
Why ABP is not a pt but a region?
Answer No 11.
It is because all grains are not of the same shape and
size, and also because of the lack of overall ignition in all grains the ‘ABP’
diffuses into a region.
Question No 12.
Why velocity is not max at ABP?
Answer No 12.
It is because after the ABP, the gasses cont to expand under
their own pressure, so the velocity is still inc but with a dec rate.
Question No 13. Draw Pressure & Velocity-Space curves for a gun when :a.
PP is 30 tons per sq inch and it occurs after one ft of shot travel, at this
moment the velocity of the proj is 400 m/s.
b.
ABP has reached after two ft of shot travel and at this moment the
velocity of the proj is 600 m/s and pressure is 9 tons per sq inch.
c.
When the proj is leaving the muzzle, MV is 800 m/s and pressure is 3
tons per sq inch (total shot travel is 8 ft).
Velocity m/s
Pressure Tons / Inch2
35
PP
800
Velocity space curve
30
600
25
20
400
Pressure space curve
200
1”
3”
4”
5”
6”
Answer No 13.
Question No 14.
Draw a velocity pressure space curve. Length of barrel is 10
ft, SSP is 3 TSI, PP is 14 TSI at 2 ft, ABP is at 4 ft where pressure is 8 TSI,
pressure at 6 ft is 6.5 TSI, pressure at muzzle is 2 TSI, velocity at 1 ft is 50 m/s, at
PP is 100 m/s, at 4 ft is 200 m/s, at 8 ft is 350 m/s and at muzzle is 400 m/s?
Answer No 14.
Pressure TSI
Velocity m/sec
15
10
5
2”
7”
8”
16
400
14
350
12
300
10
250
8
200
6
150
4
100
1
2
2
3
4
5
6
7
8
9
10
50
in ft
Question No 15.
Why it isLength
not advocated
to inc the length of barrel
beyond a certain length?
Answer No 15.
Because it is uneconomical to inc the barrel length
beyond a certain amount as the disadvantages like to silhouette, tac
limitations, mov etc outweigh the advantages of a small inc in velocity.
Question No 16.
barrel?
What do you understand by the concept of limiting
Answer No 16.
It is only a theoretical concept. It means that the length
of barrel is so adjusted that as the proj leaves the barrel the pressure of
gases behind it falls down to the atmospheric pressure. All projs fired on a
given chg would achieve same exactly the muzzle velocity and all velocity
variations would be automatically overcome.
Question No 17.
A limiting barrel if prac aval would be the most
consistent barrel. Comment?
Answer No 17.
Yes it is true, as all projs leave the muzzle with same
MV on a given chg and velocity variations are overcome.
Question No 18.
distr?
How the chemical energy produced by the prop is
Answer No 18.
Distr is as under :-
a.
Energy of proj, translational and rotational.
b.
Energy of recoil of the gun.
c.
Energy used in engraving of DB and in overcoming the
resistance during motion up the bore.
d.
Energy of unburnt chg.
e.
Energy lost as heat to gun.
f.
Energy that remains in the gas as heat.
Question No 19.
Why efficiency of gun as a heat engine is considered
to be only one third?
Answer No 19.
It is because only one-third of the energy is acquired by
the proj as KE whereas remaining two third energy remains in the gases
expelled from the barrel.
Question No 20.
Answer No 21.
How the efficiency of gun can be improved?
It can be done by fol :-
a.
By proper conditioning of barrel.
b.
By having the posn of ABP well back towards the breech end.
c.
Question No 22.
Answer No 22.
propelling the proj.
Define an efficient gun?
A gun which utilises 1/3 of the total prop energy in
Question No 23. Prove from the FT that diff chemical composn of props
are eff diff by varying atmospheric conditions. 105 mm How, chg 5?
Answer No 23.
‘Table E’ gives corrns to MV for variations in prop temp.
105 mm How uses two natures of props i.e dual granulation (US carts) and
NQ/R (POF carts). 70 F or 21 C is the std temp and both dual granulation
(US carts) and NQ/S (POF carts) do not req any corrn. However as the chg
temp changes both types of chgs are eff diff and req separate corrns for eg
at 100 F or 37.8 C a corrn to MV equal 3.0 m/s is req for US carts whereas
the POF carts req a corrn equal to 1.8 m/s only.
What do understand by the ‘Piobert’s Law’?
Question No 24.
Answer No 24.
As burning conts, the sfc of each grain recedes parallel
to itself. This law of burning by parallel layers is known as the Piobert’s
Law.
What do understand by the ‘Blsts Size’ of prop?
Question No 25.
Answer No 25.
sfcs.
The least dimension of a grain b/w two opposite burning
Question No 26.
Answer No 26.
What are the various types of props shapes?
Fol are the various types of prop shapes :-
a.
Cord.
b.
Long tube.
c.
Slotted tube.
d.
Multi-tube.
e.
Ribbons.
f.
Flakes.
Question No 27.
What is blsts size in case of a long tubes?
Answer No 27.
and the hole.
It is the wall thickness or the diff in the radii of cylinder
Question No 28.
Why slotted tubes were designed?
Answer No 28.
It is because the long tubes have not prove altogether
satisfactory in prac and tend to produce irregularities in pressure and
velocity. Ignition of internal sfc is not always satisfactory and the tubes tend
to split during burning due to the imprisonment of gases and consequent
dev of excess pressure in the hole.
Question No 29. The long tubes tend to split due to the imprisonment of
gases and consequent dev of excess pressure in the hole. Why the multitube does not split because of same reason?
Answer No 29.
It is because the cylinders forming the multi-tube are
comparatively short and before the excess pressure builds up theprop is
completely burnt.
Question No 30.
tube?
What do you understand by the web in case of multi-
Answer No 30.
The dist b/w any two holes and the dist of the outer
perforations from the outer sfc of the cylinder are all equal and this dist the
least dimension of the grain is called the web.
Question No 31.
Why the multi-tube pros are 85% progressive burning
and 15% digressive burning?
Answer No 31.
It is because until the web is consumed the burning sfc
is inc but afterwards the prop breaks up into 12 slivers and the burning sfc
starts to dec.
Question No 32.
for lower chgs?
In Hows, why is it essential to dec the blst size of props
Answer No 32.
The wt of prop in lower chs is less; they have less
burning sfcs causing low rate of burning and less PP, ABP movs towards
muzzle end thus causes inconsistency and low MV. To avoid this, blst size
of prop is reduced which means more burning sfcs, rapid build up of
pressure, more PP and ABP movs to breech end thus resulting in more MV
and consistency.
Question No 33.
Comment?
The blst size of props is req to be inc for lower chgs.
Answer No 33. The statement is incorrect. It is req to be dec because in
lower chgs less pressure is dev so rate of burning is also less, hence ABP
movs towards or at times outside the muzzle. To keep the ABP well back
towards breech, the blst size of the lower chgs is req to be dec.
Question No 34.
What is the eff of ABP on MV?
Answer No 34.
The velocity variations are greatest near the ABP and
the variations dec as the shot travel incs. Rd to rd regularity in MV is
improved if burning finishes i.e ABP occurs while the proj is still well back in
the bore.
Question No 35.
What is the eff of ABP on consistency?
Answer No 35.
The velocity variations are greatest near the ABP and
the variations dec as the shot travel incs. Rd to rd regularity in MV is
improved if burning finishes i.e ABP occurs while the proj is still well back in
the bore. Hence consistency will also improve.
Question No 36.
How the ABP effs the muzzle flash?
Answer No 36.
The temp of gases dec as the gases expand, an ABP
well back ensures that the tempo of the gases expelled from the gun is
lower than if burning conts until the proj is near the muzzle. There is,
therefore, less likelihood of muzzle flash occurring.
Question No 37.
What is the eff of fol on the pressure produced, ABP,
MV and consistency of a proj?
Answer No 37.
Ser
Variation Pressure
ABP
MV
Consistency
a.
Reduction
Later
Less
Less
Less
in chg wt
b.
Inc in chg
wt
More
Earlier
More
More
c.
Reduction
in
prop
size
More
Earlier
More
More
d.
Inc
in
prop size
Less
Later
Less
Less
e.
Reduction
in ECC
More
Earlier
More
More
f.
Inc
ECC
Less
Later
Less
Less
g.
Over
rammed
proj
Less
Later
Less
Less
h.
Under
rammed
proj
More
Earlier
Less
Less
j.
Reduction
in proj wt
Less
Later
More
More
k.
Inc in proj
wt
More
Earlier
Less
Less
l.
Reduction
in proj dia
More
Earlier
Less
Less
m.
Inc in proj
dia
Less
Later
More
More
n.
Reduction
in
chg
Less
Later
Less
Less
in
temp
o.
Inc in chg
temp
More
Earlier
More
More
Question No 38.
Prove with the help of FT 105 mm How that a
reduction in chg wt would produce lesser MV and lesser consistency?
Answer No 38.
We know that smaller chg means less chg wt while a
higher chg means more chg wt. Hence from FT 105 mm How, range 5,000
Ms:Ser
Chg
MV
Consistency
(Range PE Table G)
a.
7
464.8 m/s
11 Ms
b.
6
365.8 m/s
13 Ms
c.
5
301.8 m/s
22 Ms
d.
4
262.1 m/s
24 Ms
Question No 39.
Why smaller size props are used for reduced chgs?
Answer No 39.
Reduced chgs produce slower rate of burning, late
occurrence of ABP, lesser MV and poor consistency. Hence to compensate
for these effs smaller size of props are used.
Question No 40.
Which shape of prop is used in A/tk guns and why?
Answer No 40.
In the A/tk guns the req is to produce sufficiently high
MV with low PP to allow for the manufacture of a lt ord to give mob to the
gun. This req can be fulfilled by multi-tube shaped prop. The eff on
consistency due to lesser MV produced as compared to cord shape has no
eff due to the smaller ranges at which these guns are req to engage the
tgts.
Question No 41.
How the conflicting reqs of mob and range are met?
Answer No 41. By changing the shape of prop from cord to multi tube or
tube, the conflicting reqs of mob and range can be met. These props
produce lesser PP and the wpn can be made lt and hence more mob. The
MV produced is however less and this can be overcome by reducing the
size of the prop and using more wt of prop.
Question No 42.
For engagement of close tgts, the highest which is the
most consistent chg is used. Which gun is exception to it and why?
Answer No 42.
The exception is 25 Pdr Gun whose chg super is less
consistent than its chg 3. Chg super is having lesser consistency because it
is using the tubular shaped prop.
Question No 43.
(ECC)?
What do understand by the eff chamber capacity
Answer No 43.
It is the volume of chamber aval to the prop gases for
expansion after loading the rd. This ref to the chamber space not occupied
by the metal from the line of contact of DB with the forcing cone to the fwd
face of the means of obturation.
Question No 44.
Prove with the help of FT 105 mm How that an inc in
ECC would produce lesser MV?
Answer No 44.
We know that with the wear inside the gun chamber the
ECC incs and the MV produced will be less. Hence from FT 105 mm How,
page IX the table showing the approx losses in MV:Ser
Wear Measurement in mm
Loss in MV in m/s
a.
105.0036
0 m/s
b.
105.1052
2.7 m/s
c.
105.2322
4.0 m/s
d.
105.3084
4.3 m/s
Question No 45.
A badly rammed proj may produce an inc in pressure
and MV, yet it is inconsistent, why?
Answer No 45.
It is because of its poor external blsts performance.
Question No 46.
longer ranges?
Why a heavier proj despite its low MV ranges more at
Answer No 46.
It is because of its better carrying power and the ability
to retain its velocity better.
Question No 47.
Two projs of diff wt leave the muzzle at the same MV.
Which one will range farther?
Answer No 47.
If the MV is same the heavier one will range because of
its better carrying power and the ability to retain its velocity better.
Question No 48.
Two projs of diff dia but same wt leave the muzzle at
the same MV. Which one will range farther?
Answer No 48.
If the wt and MV is same the one with smaller diameter
will range farther because of its better carrying power and lesser air
resistance.
Question No 49.
Prove with the help of FT 105 mm How that a lighter
proj would range more at short ranges but a heavier projectile would range
more at longer ranges?
Answer No 49.
From FT 105 mm How, chg 7, table F, ranges 3,500,
7,500 and 10,500 Ms:Ser
Range
Corrn Req for Proj
Wt -1 Sq
Corrn Req for Proj
Wt +1 Sq
a.
3,500 Ms
-12 Ms
+13 Ms
b.
7,500 Ms
0 Ms
+3 Ms
c.
10,500 Ms
+19 Ms
-15 Ms
Question No 50.
Why HE and ill shells of 105 mm How dev diff MVs?
Answer No 50.
This is primarily because of wt diff. HE shell has a wt of
33 lbs where as ill shell weighs 36.6 lbs. Thus this diff of 3.6 lbs will have
eff on MV produced. Thus due to the inc in mass the velocity of ill shell is
less than HE shell.
Question No 51.
By using fatter and shorter projs we can achieve more
MV with less pressure, then why these shells are avoided?
Answer No 51.
Although these shells produce more MV and are more
suitable from internal blsts pt of view, yet their external blsts performance is
poor and they would achieve lesser range than an elongated proj.
Question No 52. Prove with the help of FT 105 mm How that a more the
chg temp more the MV produced and less the chg temp lesser the MV
produced?
Answer No 52.
From FT 105 mm How, chg 7, table E, chg temps 10.0
C, 21.1 C and 37.8 C :Ser
Chg Temp
Corrn To Vel
Corrn To Vel
US Carts
POF Carts
m/s
m/s
a.
10.0 C
-2.7 m/s
-8.2 m/s
b.
21.1 C
0 m/s
0 m/s
c.
37.8 C
+4.6 m/s
+12.2 m/s
Question No 53.
What is slip back and why does it occur?
Answer No 53.
Slip back occurs when a loaded fails to remain in its
seating when the gun is ele. It is caused by the failure of the DB to locate
correctly and tightly in the shot seating.
Question No 54.
When the slip back may occur?
Answer No 54.
Slip back may occur in the fol circumstances :-
a.
If there is a design fault which causes a mis-match b/w the front
taper on the DB and the taper of the forcing cone in the bore.
b.
If there is dirt on the DB or excessive fouling on the forcing
cone.
c.
If the mech rammer being used is not maint correctly, or timed
correctly, to ensure that the rammer head is applying the
correct force to the base of the proj as it enters its seating.
d.
If, during hand ramming, the force applied to the base opf the
proj is too weak or is severely misaligned.
Question No 55.
occurs?
Answer No 55.
Why it is not possible for the det to detect slip back if it
It is because of the noise of the ele mech.
Question No 56.
Why in case of the QF guns the proj would leave the
muzzle w/o any noticeable eff even if the slip back occurs?
Answer No 56.
It is because the cart case prevents the proj from falling
back more than a very short dist in the chamber.
Question No 57.
What would happen in case of the BL guns if after slip
back the chg and the proj are in the chamber and the gun is then fired?
Answer No 57.
One of the fol events may occur :-
a.
The proj will leave the bore and fall very short.
b.
The proj will break up as it leaves the muzzle.
c.
The proj will detonate in the bore as it comes violently into
contact with the area around the C of R.
d.
The burning gases can escape round the proj, leaving the proj
in the chamber.
Question No 58.
Why ‘spacer’ may be used?
Answer No 58.
It is loaded to prevent the proj from falling back if it
should become unseated.
Question No 59.
To avoid slip back ‘spoilers’ are used, Comment.
Answer No 59. Statement is incorrect. Spacer is used to avoid slip back.
Spoilers are instead fitted around the nose of the proj prior to loading. The
proj is fired with a chg high enough to ensure that the proj clears the gun.
The inc drag created by the spoiler, causes the proj to range shorter.
Question No 60.
avoided?
In gen, how the potential danger of slip can be
Answer No 61. It can be avoided by good design of the whole sys based
on rationale reqs. Careful attn to maint and drills at the gun is of great imp.
Question No 62.
Answer No 62.
Why crushed chg bags should not be used?
These should not be used for fol reasons :-
a.
Irregular burning sfcs and ignition.
b.
Irregular pressure.
c.
Irregular MV and inconsistency.
Question No 63.
than Hows?
Answer No 63.
Why guns are more sensitive to changes in rate of fire
It is bcause of fol reasons :-
a.
Guns have longer barrel than Hows.
b.
The no of chgs are less in case of guns.
c.
Wear is more in case of guns.
SEC – 4
Question No 1.
What factors must be considered while sel the
appropriate chg to be fired in a modern fd gun?
Answer No 1.
Fol factors may be considered :-
a.
Consistency and accy.
b.
Eqpt stability.
c.
Barrel wear.
d.
Crew fatigue.
Question No 2.
Upon what does the max wt of a chg that can be fired
from a gun depend?
Answer No 2.
It depends upon the fol :-
a.
Str of the gun.
b.
Wt of the gun.
c.
Stability of the gun.
Question No 3.
Answer No 3.
Question No 4.
occur?
What governs the min chg that can be fired from a gun?
It is governed by the phenomenon of ‘sticker’.
What is the phenomenon of ‘sticker’ and why does it
Answer No 4. In some hows it is possible for the proj to stick in the bore
when the lowest chg is fired, leaving hot gases trapped in the chamber
under pressure. This happens because the gas pressure dev by the lowest
chg is not always sufficient to over come the resistance of the DB and the
proj lodges in the bore, usually at, or very near, the C of R.
Question No 5.
What is the phenomenon of ‘cook off’?
Answer No 5.
It is the accidental firing of the proj due to the ignition of
chg because of the high temp inside the chamber. The temp which may
cause the phenomenon to occur may be as low as 170 C i.e the ignition
temp of most of the prop chgs.
Question No 6.
during firing?
Answer No 6.
Why chg should not be kept loaded for longer duration
Chg should not be kept loaded for fol reasons :-
a.
Chg temp will be inc.
b.
Premature ignition may occur if the gun is hot.
Question No 7.
How the chg sys are designed?
Answer No 7.
Chg sys are designed b/w the limitations imposed by the
str of the gun and the onset of ‘stickers’.
Question No 8.
What do you understand by the composite chg sys?
Answer No 8.
Composite chgs are chgs in which diff prop natures /
shapes / sizes are incl in the increments that constitute the normal chg.
This is a complex sys to cater for the wide range of chamber pressures.
Measurement of eff of chg temp on the dev MV is also complex. An eg of
an existing composite charge sys is British chg of 105 mm Abbot Mk-2.
Question No 9.
Why composite chgs are used?
Answer No 9.
In lower chg the wt of prop is very less. It results in
reduced pressure and reduced rate of burning. The PP decs and ABP
movs towards the muzzle. This incs the inconsistency and lesser MV.
Whereas the higher chgs produce unsafe chamber pressure. Therefore
change in shape and size of props is done or composite chgs are used in
order to avoid the a/m effs.
Question No 10.
What do you understand by the dual grain props?
Answer No 10.
The props used have two sizes i.e composite chg. In
lower chg the wt of prop is very less. It results in reduced pressure and
reduced rate of burning. The PP decs and ABP movs towards the muzzle.
This incs the inconsistency and lesser MV. To overcome this problem i.e to
inc the rate of burning props with diff blsts size are used for lower chgs.
Question No 11.
What do you understand by the zoned chg sys?
Answer No 11.
In these sys, only one prop nature is used in the
increments which constitute the full chg. This method reduces the
manufacturing and filling problems associated with composite chgs and is
the method most commonly used in US sys.
Question No 12.
Why green chg bags are used for the lower chgs?
Answer No 12.
The green chg bags are faster burning chgs. In order to
keep the ABP closer to the breech and ensure better consistency, the
green chg bags are used for the lower chgs.
Question No 13.
Why green chg bags produce lesser flash as
compared to the white chg bags?
Answer No 13.
The green chg bags are faster burning chgs and the
ABP in their case remains closer to the breech while the white chg bags
are cooler burning and so ABP the moves closer to the muzzle. Hence the
green chg bags produce lesser flash than the white bags.
Question No 14.
Prove with the help of FT 155 mm How (SB), range
5000 Ms, that the green bags despite low MV are more consistent than the
white bags?
Answer No 14.
Range PE is a measure of the consistency and lesser
the range PE more consistent is the chg and vice versa. From table G of
the FT, we find that the range PEs of green chg are less than the white chg
bags :Ser
Chg
MV
Consistency
(Range PE Table G)
a.
3G
464.8 m/s
11 Ms
b.
4G
365.8 m/s
13 Ms
c.
5G
301.8 m/s
22 Ms
d.
3W
262.1 m/s
24 Ms
e.
4W
f.
5W
Question No 15.
Why white chg bags should be avoided if a choice
exists b/w them and the green chg bags?
Answer No 15.
This is because the burning rate of the while bag is not
fast enough at the lower pressures to ensure that the ABP is well back in
the barrel. It is found that with white bag chgs 3 and 4, the ABP is often at,
or outside the muzzle, excessive muzzle flash occurs, and fall of shot is
erratic. When white bag chg 3 is fired, there is a danger of partially burnt
prop gases burning on exposure to air as the breech opens under its
automatic action. This can cause a flash inside the turret. For these
reasons, white bag chgs 3 and 4 should not be fired except in an
emergency.
Question No 16.
How M109 A2?
Why firing of chg 3 white bag is avoided for 155 mm
Answer No 16.
As breech opens under the auto action. There is a
danger of partially burnt gases burning on exposure to air because of slow
rate of burning of 3W. This may cause a flash inside turret.
Question No 17.
You have two chgs aval at gun posn for M109 A2 i.e
chg 5 white and chg 5 green. Which one will you use for engagement of tgt
and why?
Answer No 17.
Chg 5 green is preferred over chg 5 white because the
rate of burning of white bag is slow and ABP can not be well back in the
bore towards the breech, resultantly it is less consistent than green bag.
Question No 18.
What do you understand by ‘spoiler’?
Answer No 18. It a method used to solve the problem of stickers, and at
the same time to achieve a short min range. A spoiler, or disc, is fitted
around the nose of the proj prior to loading. The proj is fired with a chg high
enough to ensure that the proj clears the gun. The inc drag created by the
spoiler, causes the proj to range shorter.
SEC – 5
Question No 1.
What do you understand by ‘wear’?
Answer No 1.
Wear is the gradual removal of metal from the sfc of the
bore as a result of firing and is caused by the chemical action of hot gases
combined with the abrasive action of the DB.
Question No 2.
What is the eff of worn out gun on MV?
Answer No 2. In a worn out gun the initial resistance to the motion of the
proj is less. There is therefore a dec in the SSP. The proj can be rammed
slightly further fwd in a worn out gun, thus inc the initial space aval for
the expansion of the gases, resulting in a lowering of the pressure and a
further reduction in MV.
Question No 3.
Answer No 3.
How the wear of the gun effs its consistency?
The eff is as fol :-
a.
ECC of the gun incs, so less pressure, less PP and ABP
towards muzzle so less consistency.
b.
As the gun wears, the fwd seal of the gasses becomes less eff,
so the proj fired from it is more unsteady and the consistency
falls.
c.
Wear in trunnion bg, saddle and carriage union, ele and
traversing mechs and sight gears reduces consistency.
Question No 4. Prove with the help of FT 105 mm How that wear results
in reduction of MV?
Answer No 4.
From FT 105 mm How, page IX the table showing the
approx losses in MV :Ser
Wear Measurement in mm
Loss in MV in m/s
a.
105.0036
0 m/s
b.
105.1052
2.7 m/s
c.
105.2322
4.0 m/s
d.
105.3084
4.3 m/s
Question No 5. Prove with the help of FT 105 mm How, chg 7 that prop
lots manufactured by diff ctys may vary in performance and produce diff
MVs?
Answer No 5. From FT 105 mm How, chg 7, table E, two prop lots have
been given. These lots are eff diff by varying conditions producing diff MVs
at various chg temps, eg chg temps 10.00 C, 21.10 C and 37.80 C :Ser
Chg Temp
Corrn To Vel
Corrn To Vel
US Carts
POF Carts
m/s
m/s
a.
10.00 C
-2.7 m/s
-8.2 m/s
b.
21.10 C
0 m/s
0 m/s
c.
37.80 C
+4.6 m/s
+12.2 m/s
Question No 6.
compensated?
How the error due to ‘occasion to occasion eff’ can be
Answer No 6.
The method of compensating for this error is to take a
mean of MVs produced on as many occasions as possible and is being
done in cal firings.
SEC – 6
Question No1.
internal blsts?
What could be possible cause of abnormal effs of
Answer No 1.
It is thought that some abnormal effs are related to the
interaction of proj’s DB with the bore, since it has been observed that
abnormal variations are minimized when non-metallic DBs are used. But in
gen, it is true to say that the exact causes of these abnormal effs are not at
present understood.
Question No 2.
Answer No 2.
What are the abnormal effs of internal blsts?
Fol are the abnormal effs of internal blsts :-
a.
Cold gun eff / warmer eff.
b.
Order of fire eff.
c.
Hump eff.
Question No 3.
eff’?
Explain the phenomenon of ‘cold gun eff’ or ‘warmer
Answer No 3.
A gun fired after an interval will often dev a MV which is
not rep of the avg. This phenomenon is known as the ‘cold gun eff’,
sometimes ref to as the ‘warmer eff’ and can occur after quite short
intervals b/w firing, sometimes measured in mins.
Question No 4.
‘warmer eff’?
What could be possible cause of ‘cold gun eff’ or
Answer No 4.
The eff cannot be completely explained but is thought to
be concerned with a conditioning of the molecular structure of the metal
from which the barrel is made.
Question No 5. Cold gun eff occurs for first rd if the temp of barrel is less
than the surroundings. Comment?
Answer No 5.
Statement is incorrect. Cold gun eff occurs b/w the
intervals of firing for the first rd only. It happens even if the barrel is warm to
touch.
Question No 6.
Explain the phenomenon of ‘order of fire eff’?
Answer No 6. The order of the fire eff refs to the order in which chgs in a
multi-chg gun sys are fired at cal. The effs are confined to the lower chgs in
the sys. In fact, there appear to be two distinct effs:a.
One eff is related to barrel temp. If a warm gun fires a series of
rds at a high chg, immediately fol by a series at a low chg, the
mean MV of the low chg series may be several m/s higher than
normal.
b.
The other eff is related to a pd of inaction. If a gun which has
not been fired for some time fires a low chg first, then the mean
MV of a series of rds can be several m/s lower than that
expected from the FT. If after a pd of inactivity, a lower chg is
fired after a higher chg, the MV lvl for the lower chg is
unaffected and is normally as expected from the FT even
though the barrel may be cool.
Question No 7.
What is the suggested remedy for the ‘order of fire eff’?
Answer No 7.
Apart from always firing the highest chg first, there is no
certain way of eliminating this eff.
Question No 8.
firings?
Answer No 8.
Question No 9.
Why chgs are fired in descending order during cal
In order to eliminate the ‘order of fire eff’.
What do you understand by the ‘hump eff’?
Answer No 9.
Blst hump is used to describe the sit where during the
early life of a barrel, its MV rises to a peak as more rds are fired, reaches a
max value, and then falls to a lvl comparable with the state of wear.
Question No 10.
eff nowadays?
Why there is an inc in the freq of the occurrence hump
Answer No 10.
The use of cooler props and wear reducing additives
tend to inc the freq of occurrence, but hot props used previously caused
rapid wear and so masked the phenomenon.
Question No 11.
What could be possible cause of ‘hump eff’?
Answer No 11.
Hump is considered to be due to an inc in shot-start
resistance caused by the onset of craze cracking of the bore sfc with a
consequent inc in rate of burning of prop.
Question No 12.
What could be possible duration of ‘hump eff’?
Answer No 12.
Its occurrence is uncertain and its magnitude variable.
The duration of hump depends on the rate of wear and it can take anything
from 100 to 1,000 rds to return to new gun blsts. No exact figures can be
given for any particular eqpt because variations can occur depending on
the exact state of the bore when new.
Question No 13.
What is the suggested remedy for the ‘hump eff’?
Answer No 13.
The only solution is to measure the MV as often as
possible during the hump process and adopt an appropriate MV.
SEC – 7
Question No 1.
What are the main factors which govern the range and
dir to be achieved by the proj?
Answer No 1.
by the proj :a.
MV.
Fol main factors govern the range and dir to be achieved
b.
A of D.
c.
Gravity.
d.
Air resistance which is variable with :(1)
Met conditions.
(2)
Chars of proj itself eg the shape, wt etc.
(3)
Yaw of the proj.
e.
Drift.
f.
Magnus eff
g.
Rotation of Earth.
SEC – 8
Question No 1.
What is the Newton’s first law of motion?
Answer No 1.
Newton’s first law of motion states that a body at rest will
remain at rest and a body in motion conts to mov in a straight line at a
constant velocity unless acted upon by an external force.
Question No 2.
Prove diagrammatically that the value of ‘g’ is more at
the poles and less at the equator.
Answer No 2.
North Pole
Lesser Dist
Equator
Center of
the Earth
More Dist
South Pole
Question No 3.
What is a vector qty?
Answer No 3.
vector qty.
Anything that has a magnitude and a dir is termed as a
Question No 4.
resolved?
Answer No 4.
under :-
Into how many components the velocity can be
The velocity can be resolved into two components as
a.
Horizontal component of velocity i.e V Cos Ө.
b.
Vertical component of velocity i.e V Sin Ө.
V Sin Ө
V
Ө
V Cos Ө
Question No 5.
Work out the components of velocity various
Ds when the MV of gun is 1,000 m/s?
A of
Answer No 5
As of D
HORIZONTAL
COMPONENT
VERTICAL COMPONENT
00
1,000 X Cos 00 = 1,000 x 1 1,000 x Sin 00 = 1,000 x 0 =
= 1,000 m/s
0 m/s
300
1,000 x Cos 300 = 1,000 x 1,000 x Sin 300 = 1,000 x
.87 = 870 m/s
.50 = 500 m/s
450
1,000 x Cos 450 1= 1,000 x 1,000 x Sin 450 1= 1,000 x
.71 = 710 m/s
.71 = 710 m/s
600
1,000 x Cos 600 = 1,000 x 1,000 x Sin 600 = 1,000 x
.50 = 500 m/s
.87 = 870 m/s
900
1,000 x Cos 900 = 1,000 x 0 1,000 X Sin 900 = 1,000 x 1
= 0 m/s
= 1,000 m/s
Question No 6. Work out the components of velocity in-vacuo at various
TOFs for A of D 300 when the MV of gun is 196 m/s?
Answer No 6
TOFs
HORIZONTAL
COMPONENT
VERTICAL COMPONENT
0
196 X Cos 300 = 169.7 m/s
196 x Sin 300 = 98 m/s
1
169.7 m/s
98 – 9.8 = 88.2 m/s
2
169.7 m/s
98 – 2 x 9.8 = 78.4 m/s
10
169.7 m/s
98 – 10 x 9.8 = 0 m/s
Question No 7.
Why the horizontal component of velocity remains
constant in-vacuo throughout?
Answer No 7.
In-vacuo, the only factor which retards the velocity is
gravity and no component of gravity acts in the horizontal dir, therefore the
horizontal component of velocity remains constant throughout.
Question No 8.
Find out the remaining velocity of a proj after 4 secs
fired at an A of D of 300 mils with MV of 400 m/s?
Answer No 8. Remaining velocity =
SQR ((400 x Cos 300 mils)2 + (400 x Sin 300 mils – 4 x 9.8)2)
= 390.4 m/s
Question No 9.
Find out the horizontal dist and ordinate (vertical dist) of
a proj after 5 secs TOF fired at an A of D of 300 with MV of 400 m/s?
Answer No 9.
Horizontal dist = V Cos 300 x t
= 1732 Ms
Ordinate or vertical dist = V Sin 300 x t – ½ gt2
=
400 x Sin 300 x 5 – ½ x 9.8 x 52
=
877.5 Ms
Question No 10.
Find out the TOF ‘T’ for a given angle ‘Ө’?
Answer No 10. Time to vertex = t = (total time T) / 2
At vertex we know that gt = V Sin Ө
t = (V Sin Ө) / g
Therefore T = (2 V Sin Ө) / g
Question No 11.
Find out the range ‘R’ for a given angle ‘Ө’?
Answer No 11. Range ‘R’ = horizontal dist = V Cos Ө x T
But T = (2 V Sin Ө) / g
Therefore R = (V Cos Ө x 2 V Sin Ө) / g
=
(V2 2 Sin Ө Cos Ө) / g
But, 2 Sin Cos Ө = Sin 2 Ө
Hence R = (V2 Sin 2 Ө) / g
Question No 12.
Find out range and TOF of 155 mm Gun at ele 450
mils at chg Normal if the gun is fired at moon. Value of acceleration due to
gravity at moon is 1/6 of its value on Earth?
Answer No 12. R = (V2 Sin 2 Ө) / g
R = 193912.6744 Ms
R = 193913 Ms
Question No 13. Prove that the range achieved by two complimentary A
of Ds in-vacuo would be same?
Answer No 13. We know that R = (V2 Sin 2 Ө) / g. Range would thus
depend upon 2 Ө. If Ө = 300, then complimentary angle of 300 would be
600. So the value of 2 Ө for these two angles would be 600 and 1200. 600
and 1200 are suplimentary angles and the sin values of two suplimentary
are same. Hence proved that the range achieved by two complimentary A
of Ds in-vacuo would be same.
Question No 14.
Why the range achieved by two complimentary A of
Ds is same in-vacuo but not in air?
Answer No 14. We know that R = (V2 Sin 2 Ө) / g. Range would thus
depend upon 2 Ө. If Ө = 300, then complimentary angle of 300 would be
600. So the value of 2 Ө for these two angles would be 600 and 1200. 600
and 1200 are suplimentary angles and the sin values of two suplimentary
are same. Hence proved that the range achieved by two complimentary A
of Ds in-vacuo would be same. The a/m formula does not hold good in air.
The air resistance for two A of Ds is not same. The diff in the air resistance
results in diff ranges achieved by two A of Ds.
Find out the max ordinate (MO) ‘H’ for a TOF ‘T’?
Question No 15.
Answer No 15. Using the equation of motion, S = Vi t + ½ a t2
For a falling body at rest, S = ½ a t2
Substituting, we get H = ½ g t2
But t = T / 2, hence H = g T2 / 8
As g is 9.8 m/s2
H = 1.225 T2 approx
Question No 16.
a MV of 400 m/s?
Workout the MO of a proj fired at an A of D of 300 with
Answer No 16. TOF to pt of graze T = 2 V Sin Ө / g
=2 x 490 x Sin 300 / 9.8
= 50 secs
MO = 1.225 T2
= 1.225 x 502
= 3062.5 Ms
Question No 17.
If a proj is fired horizontally from a barrel that is
perfectly lvl at a higher gr, will the proj, at some pt, rise above the L/S?
Answer No 17.
No it can not rise above the L/S because the vertical
component of velocity is zero and this will start inc downward due to
gravitational pull. So it will keep moving down and further down from the
L/S throughout.
Question No 18.
We know that MO = 1.225 T2, the formula gives us an
impression that the MO is independent of the MV. Comment?
Answer No 18. No the statement it is incorrect
MO = 1.225 T2
But T = 2 V Sin Ө / g
Since T is dependent upon V Sin Ө whose value would
change with the value of MV, hence MO will also change
with a change in MV
Question No 19. We know that Cos 00 = 1 and horizontal component of
velocity is max at 00. The horizontal component of velocity gives range to
the proj but why the range at 00 A of D is 0 Ms?
Answer No 19. Horizontal dist = V Cos Ө x T
Which means that the horizontal dist is dependent on two
things i.e horizontal component of velocity and TOF
TOF = 2 V Sin Ө / g
And Sin 00 = 0
Hence Horizontal dist = V Cos 00 x 2 V Sin 00 / g
Horizontal dist = V x 0
Horizontal dist = 0
Question No 20.
Why a proj achieves max range at A of D of 450?
Answer No 20. We know that horizontal dist = (V2 Sin 2 Ө) / g
And Sin 900 = 1
Therefore for V Sin Ө to be max i.e equal to V, 0 should
be equal to 900
In the equation above (V2 Sin 2 Ө) / g, for vertical
component to be max, Ө should be equal to 450
Hence proj achieves max range at A of D 0f 450
Question No 21. Find out range and remaining velocity after 15 secs invacuo. 155 mm Gun, chg Normal, ele 525 mils?
Answer No 21. Range = 35972.89 Ms
Remaining velocity after 15 secs = 582.85 m/s
Question No 22.
How much time a shell will take to reach a tgt, if it is
fired in-vacuo when fol data is aval (suppose NR corrn is zero). 105 mm
How, chg 5, TE 140 mils, QE 180 mils, A of D 190 mils?
Answer No 22. First we find the A/S and A of P
A/S = QE – TE
= 180 – 150 = 40 mils
A of P = A of D – A/S
= 190 – 40 = 150 mils
TOF = 2 V Sin Ө / g
TOF = 2 x 301.8 x Sin 150 mils / 9.8
= 9.04 secs
Question No 23.
Ds beyond 450?
With high velocity guns why max range occurs at A of
Answer No 23. It is because of fol reasons :a.
Lesser air density at higher traj.
b.
With more A of D, the effs of Kσ are reduced.
c.
More range is achieved in lesser time.
Question No 24.
Answer No 24.
What are the chars of in-vacuo traj?
Fol are the chars of in-vacuo traj :-
a.
Traj is symmetrical about the vertical line through the vertex
and is parabola.
b.
The whole traj lies in the vertical plane containing the plane of
departure.
c.
Range depends upon MV and incs as the A of D incs upto 800
mils. For a given MV, max range is achieved at 800 mils.
d.
As Ө incs above 800 mils, range decs until it becomes zero
where Ө = 1600 mils. At this stage max vertex ht is reached
and the TOF is max.
e.
The angle of arrival equals the A of D, the remaining velocity at
the pt of impact equals the MV. Remaining vertical velocity ia a
min at the vertex.
f.
Traj is entirely independent of the nature of the proj, since air
resistance is assumed to be non-existent in-vacuo.
Question No 25.
For traj in air we have a separate vertical plane of
departure and a vertical plane of impact but for in-vacuo traj there is only
one vertical plane of departure. Comment?.
Answer No 25.
Yes, it is true. For traj in air, in order to ensure that the
proj remains stable and to ctr the air resistance, spin is imparted to the proj.
this spin causes the proj to drift away from the vertical plane of departure,
hence we have a vertical plane of impact. For in-vacuo traj since there is no
air resistance, spin is not req and hence there is no drift. Resultantly there
is only one vertical plane.
Question No 26.
traj?
Which would be the most eff shape of proj for in-vacuo
Answer No 26. The spherical shaped proj would be most eff for in-vacuo
traj as there would be no air resistance.
SEC – 9
Question No 1.
Define the fol :-
a.
Local Speed of Sound.
The speed of sound in the
atmosphere is normally taken to be 340 m/s. Actually it
changes with a change in the atmospheric temperature. The
actual speed of sound at any place in termed as the local speed
of sound.
b.
Subsonic Velocity. If the velocity of a body relative to the
surrounding atmosphere is less than the local speed of sound,
it is know as subsonic velocity (normally upto 340 m/s).
c.
Transonic Velocity.
If the velocity of a body relative to the
surrounding atmosphere is close to the local speed of sound, it
is known as transonic velocity (b/w 300 and 365 m/s).
d.
Supersonic Velocity.
If the velocity of a body relative to the
surrounding atmosphere is more than the local speed of sound,
it is known as supersonic velocity (normally above 365 m/s).
e.
Mach No. It is the ratio of the relative velocity of a proj to the
(local) speed of sound in the atmosphere. It is expressed as M
= V/a (where ‘M’ denotes mach no, ‘V’ denotes the velocity of a
proj and ‘a’ the local speed of sound.
f.
Fore body.
The part of the proj which has fwd facing sfc is
commonly termed the fore body.
Question No 2.
Which two fundamental properties of air affect the
nature and amount of air resistance on a proj?
Answer No 2.
Fol two properties affect the amount of resistance :-
a.
Viscosity.
b.
Compressibility.
Question No 3.
What is viscosity?
Answer No 3.
It is that property of fluids, semi fluids and gases by virtue
of which they resist to any instantaneous change in their shape and
disarrangement to their arrangement. In simple words it is the sticking
property of fluids.
Question No 4.
The viscosity is dependent upon what?
Answer No 4.
Viscosity is dependent upon density. More dense the air,
more would be the viscosity and the resistance to the movement of proj.
Question No 5.
Answer No 5.
Question No 6.
Answer No 6.
be reduced.
What types of drags are caused by viscosity of air?
Viscosity causes base and skin friction drags.
What is compressibility?
It is that property by which the size and volume of air can
Question No 7.
Why the density of air at higher alts is less as
compared to the density of air at mean sea lvl?
Answer No 7.
It is because of the wt of air. The air at higher alts
compresses the air at lower alts, hence inc the density.
Question No 8.
Why the amount of resistance changes for a
compressible fluid (air) with the inc / dec in volume?
Answer No 8.
To answer this question let us consider a proj mov
through the air. As the proj mov, it compresses the air in front of it and this
compression is transmitted to the surrounding air as a pressure wave of
disturbance. This incs the density and volume of the air and hence incs the
resistance.
Question No 9.
Why for mach no below 0.7, the air is considered to be
an incompressible fluid?
Answer No 9.
The pressure produced by the mov of a body in air mov
at the speed of sound. At velocities below mach no 0.7, it is possible for
these weak pressure waves to propagate fwd from a body and the air can
be pushed aside.
Question No 10.
Answer No 10.
a.
b.
The compressibility is dependent upon what?
Compressibility is dependent upon fol :-
Temp of air.
Density of air.
Question No 11.
Answer No 11.
What type of drag is caused by compressibility of air?
Compressibility causes wave drag.
Question No 12.
compared to vacuum?
Why a proj achieves lesser range less in air as
Answer No 12.
It is because as the proj movs through the air, the air is
req to be displaced and the further the proj movs, the more air is displaced.
Since the energy to displace this air can only come from the proj, there is a
continual drain on the energy of the ppoj, which shows itself in the form of a
drag. This drag causes a velocity decay and can easily halve the max
range of an arty shell to that which would occur in-vacuo.
Question No 13.
Prove from the FT 105 mm How, chg 5, A of D 300
mils that a proj fired in air would range less as compared to a proj fired in
vacuum ?
Answer No 13.
Ms at this A of D.
From FT we find that the proj achieves a range of 4500
For finding the range in-vacuo we shall use the formula as under :R = (V2 Sin 2 Ө) / g
R = (301.82 x Sin (2 x 300 mils) / 9.8
R = 5163.59 Ms
Question No 14.
Answer No 14.
What are the various forms of drag?
Fol are the various forms of drag :-
a.
Forebody form (head) drag also known as wave drag.
b.
Base drag.
c.
Skin friction drag.
d.
Excrescence drag.
Question No 15.
Why forebody form drag occurs?
Answer No 15.
As the proj movs through the air, it compresses the air
imed in front of it and this compression is transmitted to the surrounding air
as a pressure wave of disturbance. The wave travels through the air at the
speed of sound waves, since sound waves are themselves merely
disturbance of the air. The generation of these compression waves causes
forebody drag on the proj.
Question No 16.
Why compressibility effs of air are negligible at slow
speed below mach 0.7 forebody form drag occurs?
Answer No 16.
The effs are negligible as i is possible for the weak
pressure waves to propagate fwd from a body and the air is pushed aside.
Question No 17.
Why head resistance is considerably inc in the
supersonic zone and why the drag so produced is known as ‘wave drag’?
Answer No 17.
When a proj is moving faster than sound, no part of the
disturbance set up can escape dir in front of the proj since it is moving
faster than the disturbance. The result is the fmn of a shock wave at the
nose of the proj. The maint of this shock wave reqs considerable qty of
energy which can only come from the proj, and so the head resistance is
considerably inc. Since this inc is due to the fmn of shock waves, it is also
called ‘wave drag’.
Question No 18.
Why chgs that produce MVs only slightly above the
speed of sound are avoided?
Answer No 18.
Since it is common for the projs to be unsteady just
outside the muzzle of the gun due to initial yaw and the turning moment
which is strongest in the transonic region, it is undesirable to use chgs
which give MVs only slightly above the speed of sound. For this reason,
MVs in the region of the velocity of sound are avoided.
Question No 19.
Answer No 19.
What are the causes of initial yaw?
Initial yaw is caused due to fol :-
a.
Irregular dispersal of gases (because of CG and spin) at the
moment the proj leaves the muzzle.
b.
The traj is a curved path while axis of proj is a straight line.
c.
Transverse vibration produced inside the bore.
d.
Radial clearance b/w shell and bore.
e.
Unsymmetrical pressure behind the proj.
Question No 20.
The chgs that produce MVs in the transonic region are
avoided as much as possible as the projs becomes unsteady, yet we find
that chg 5 of 105 mm How produces MV equal to 301.8 m/s but still it is
one of the most consistent chg of the how?
Answer No 20.
Chg 5 of 105 mm How produces a MV equal to 301.8
m/s which falls in the transonic region. Yet the velocity of the proj drops into
the subsonic region i.e below 300 m/s within the first sec outside the
muzzle. With the velocity dropping down to the subsonic region the proj
becomes steady during its flight and the forebody drag reduces.
Question No 21.
What is ‘base drag’?
Answer No 21.
When a proj movs through the air, the air in front of it is
at a higher pressure than at the tail. The diff in pressure causes a vacuum
or wake behind the base of the proj due to the inability of the air flow to turn
quickly enough to fill the space. The vacuum sucks the proj backwards
causing retardation of the proj which is called the base or pressure drag.
Question No 22.
Why the base drag incs with velocity but tends to be
fairly steady in value at the velocity of sound?
Answer No 22.
This is because the base eff drag is due to the low
pressure behind the base of the proj. As the proj velocity incs, towards the
velocity of sound, so the air pressure behind the base of the proj tends to
zero. The drag eff behind the base cannot eff materially as the velocity incs
supersonically.
Question No 23.
How the base drag can be reduced?
Answer No 23.
It can be reduced by streamlining the base of the proj
and / or att a combustible material at the base of the proj (base bleed). The
combustible material generates fuel rich and oxygen deficient gases which
burn in the wake and locally inc in the pressure reduces the base drag upto
80%.
Question No 24.
Why little is gained by streamlining the base of a proj
which is to remain in the supersonic region for the most of its flight?
Answer No 24. At supersonic velocities the retardation experienced by a
proj is to a far greater extent due to the nose shock wave than to base
drag. Hence little is gained by streamlining the base of this proj.
Question No 25.
What is the cause of ‘skin friction drag’?
Answer No 25.
The air at the sfc of the shell movs at the same speed at
the proj. The next layer movs little more slowly and so on, outwards. There
are many of these layers of air and there s friction b/w each two. All these
packets of friction build up to form skin friction.
Question No 26.
How the skin friction drag can be reduced?
Answer No 26.
It can be reduced by smooth machining of the sfc of the
proj and by polishing with special paints.
Question No 27.
What is the cause of ‘excrescence drag’?
Answer No 27.
It is caused by eff of air on the excrescence from the
shell. It is minimised by eliminating all unnec protuberances from the shell.
The only one unavoidable is the driving band and this must be carefully
designed so as not to fan outwards after launch.
Question No 28.
How excrescence drag can be minimised?
Answer No 28. It is minimised by eliminating all unnec protuberances from
the shell. The only one unavoidable is the driving band and this must be
carefully designed so as not to fan outwards after launch.
Question No 29. During ascending portion of traj, find out the retardation
due to air resistance when retardation due to gravity is 9.8 m/sec2. 105
mm How chg 5, ele 300 mils?
Answer No 29.
= 301.8 m/s
From FT Table G, we can find MO = 363 Ms and velocity
Vertical component of velocity = 87.61 m/s
Final velocity at vertex = 0 m/s
Using the formula 2aS = Vf2 – Vi2
2 x a x 363 = 0 – (87.61)2
726 x a = – 7675.51
a = – 10.57 m/sec2
We know that g = – 9.8 m/sec2
Therefore eff of air resistance = a – g) = 0.77 m/sec 2
Question No 30.
Graphically show the the various types of air
resistances faced by the proj?
Total Air Resistance
Answer No 30.
Head Resistance
Base Drag
Resistance
Velocity o Sound
Skin Friction
Question No 31.
Answer No 31.
Question No 32.
(CRH)?
What does the proj shape ref to?
It ref only to the head of the proj.
What do you understand by 5/10 cal radius head
Answer No 32.
A head the length of which corresponds to that of a
simple ogive of radius 5 cals. The curvature is 10 cals i.e 10 proj diameters.
SEC – 10
Question No 1.
Why TOF for falling portion of traj in air is greater than
the rising portion of traj?
Answer No 1.
It is because during falling portion of traj the air
resistance acts to opposite to the gravity as against the rising portion of the
traj where both the air resistance and gravity retard the velocity of the proj.
Question No 2.
Answer No 2.
What are the chars of traj in air?
Fol are the chars of traj in air :-
a.
The traj is no longer symmetrical; the vertex is nearer to the pt
of impact than to the gun.
b.
The vertex ht is less than for the corresponding ‘in-vacuo’ traj.
However, as the eff on vertex ht is small, it is usually sufficiently
accurate to use the in-vacuo formula to calc vertex ht.
c.
The angle of descent is greater than the angle of departure.
This is so because as the proj drops after vertex, the horizontal
component of velocity is considerably reduced and the air
resistance causes the proj to fall more steeply.
d.
the terminal velocity is less than the MV.
e.
The larger the value of blsts coefficient the nearer pt where proj
falls in air is to where the proj would fall in-vacuo.
Question No 3.
Answer No 3.
Why traj is not symmetrical in air?
It is because o the fol :-
a.
Horizontal component of velocity is continuously reducing by
the amount of air resistance.
b.
Vertical component of velocity is reducing by the amount of
gravity and air resistance till vertex but after vertex it is inc by
the amount of gravity minus air resistance.
Question No 4.
Horizontal component of velocity remains constant invacuo throughout the flt of the proj. Prove from FT that it does not remain
constant in air. 105 mm How, chg 5, Range 5,000 Ms?
Answer No 4.
Using FT 105 mm How Table G, we know MV = 301.8
m/s, ele = 344.7 mils, terminal velocity = 248 m/s and A of F = 391 mils
Horizontal component of velocity at muzzle
= 301.8 x Cos 344.7 mils
= 284.68 m/s
Horizontal component of velocity at tgt
= 248 x Cos 391 mils
= 229.95 m/s
Hence proved that Horizontal component of velocity does not remain
constant in air.
Question No 5.
Prove from FT 105 mm How, chg 5, A of D 615.1 mils
that the vertex ht in air is less than the vertex ht in-vacuo?
Answer No 5.
Using FT 105 mm How Table G, we know that the proj
achieves a vertex ht 1326 Ms at A of D 615.1 mils. For in-vacuo we use the
formula as under :First we find the TOF for in-vacuo traj
T = 2 V Sin Ө / g
T = 2 x 301.8 x Sin 615.1 mils / 9.8
R = 34.974 secs
H = 1.225 T2
H = 1.225 x 34.9742
H = 1498.4 Ms
Hence proved that the proj would achieve more vertex ht in-vacuo as
compared to in air.
Question No 6.
Prove from FT 105 mm How, chg 5, A of D 615.1 mils
that the vertex ht in air is less than the vertex ht in-vacuo?
Answer No 6.
Using FT 105 mm How Table G, we know that the proj
achieves a vertex ht 1326 Ms at A of D 615.1 mils. For in-vacuo we use the
formula as under :First we find the TOF for in-vacuo traj
T = 2 V Sin Ө / g
T = 2 x 301.8 x Sin 615.1 mils / 9.8
R = 34.974 secs
H = 1.225 T2
H = 1.225 x 34.9742
H = 1498.4 Ms
Hence proved that the proj would achieve more vertex ht in-vacuo as
compared to in air.
Question No 7.
Prove from FT 105 mm How, chg 5, A of D 615.1 mils
that the formula for calc the vertex ht in-vacuo approx holds good for calc
the vertex ht in air also?
Answer No 7.
Using FT 105 mm How Table G, we know that the proj
achieves a vertex ht 1326 Ms at A of D 615.1 mils. Using the in-vacuo
formula we can find the vertex ht for air as under:First we find the TOF for in-vacuo traj
H = 1.225 T2
H = 1.225 x 32.92
H = 1325.95 Ms
Hence proved that the formula for calc the vertex ht in-vacuo approx
holds true for calc the vertex ht in air.
Question No 8.
Why formula for calc MO i.e MO = 4T2 holds good for
shorter ranges only and not for longer ranges?
Answer No 8.
It is because of the fol reasons :-
a.
At shorter ranges the traj remains in lower layers where density
although being more is almost constant, so it proportional eff on
MO and TOF is same.
b.
A longer ranges proj travels in higher layers, where density is
less but not constant, hence ratio of inc and dec in resistance is
more. Therefore eff on MO and TOF is diff.
Question No 9.
Prove from FT 105 mm How, chg 5, A of D 615.1 mils
that in air the angle of descent is more than the A of D?
Answer No 9.
Using FT 105 mm How Table G, we can find the angle of
descent of the proj as 709 mils.
Hence proved that in air the angle of
descent is more than the A of D.
Question No 10.
Prove from FT 105 mm How, chg 5, A of D 615.1 mils
that in air the terminal velocity is less than the MV?
Answer No 10.
Using FT 105 mm How Table G, we can find the MV of
proj is 301.8 m/s whereas for A of D 615.1 mils the terminal velocity is 240
m/s. Hence proved that in air terminal velocity is less than the MV.
Question No 11.
Prove from FT 105 mm How, chg 5, range 5,000 Ms
that TOF from muzzle to vertex is less than TOF from vertex to pt of graze?
Answer No 11. Using FT 105 mm How Table G, we can find the MV of
proj is 301.8 m/s, terminal velocity is 248 m/s and MO is 471 Ms
Vertical component of velocity at muzzle = 301.8 x Sin 344.7 =
100.19m/s
Vertical component of velocity at tgt = 248 x sin 391= 92.88 m/s
Vertical component of velocity at vertex = 0 m/s
Avg velocity for ascending portion = 50.095 m/s
Avg velocity for descending portion = 46.44 m/s
Dsit = MO = 471 M
S = Vxt
471 = 50.095 x t
or Upward t = 9.4 secs
471 = 46.44 x t
or Downward t = 10.14 secs
Question No 12.
Why the terminal velocity for high angle fire is more
than that of the low angle fire when considering common range?
Answer No 12.
During the falling portion of the traj the velocity of the
proj starts inc at an acceleration of 9.8 m/s i.e the value of g. We know that
the proj achieves more ht and hence more TOF in high angle fire. Hence
more is aval for inc in the vel of proj before it reaches the gr and so its
terminal in high ngle fire is more than that in the low angle fire. Secondly
the proj is falling steeply and less sfc is presented to face the air resistance,
therefore the force gravity can inc the vel more as compared to low angle
fire where more sfc is presented.
Question No 13.
A std proj and a heavier proj lt the muzzle at the same
MV. Which of the two projs would range farther?
Answer No 13.
If the two projs leave the muzzle at the same MV, the
heavier one would range farther because of its better carrying power.
Question No 14.
Answer No 14.
a.
Upon what does the carrying power of a proj depend?
It depends upon fol physical chars of the proj :-
Mass. The greater the mass, the greater the energy and the
greater the carrying power.
b.
Diameter.
The greater the size of the hole, which the proj
must bore through the air, the greater the resistance and the
less the carrying power. The size of the hole varies with d2.
c.
Shape.
This is deduced by K (kappa). In fact, it has been
found impossible to separate the eff of this factor from a
steadiness factor, σ (sigma). There product Kσ is used as
single factor and is determined by single factor. The greater the
value of Kσ, the less is the carrying power.
Question No 15.
Why Kσ is not considered as a constant for a given
proj wt and configuration?
Answer No 15.
of D.
It is because the value of Kσ varies with velocity and A
Question No 16.
Prove for FT 105 mm How, chg 7 that a heavier proj
despite lower MV would range farther at longer ranges than a lighter proj?
Answer No 16. From 105 mm How, chg 7, Table F, for a heavier proj of
3 sqs wt, from range 0 to 7,900 Ms a positive corrn for range is req. At
8,000 Ms the corrn becomes 0 whereas after range 8,200 Ms a negative
corrn is req meaning thereby that the heavier projs starts rg farther at
longer ranges than a lighter std proj.
Question No 17.
Two projs of diff wts leave the muzzle together at the
same MV. The heavier proj has a lesser diameter than the lighter proj.
Which proj will range farther at shorter ranges?
Answer No 17.
The heavier proj would range farther at shorter ranges
due to more carrying power and blsts coefficient as it has more mass and
lesser diameter.
SEC – 11
Question No 1.
Define the fol :-
a.
Centre of Gravity (CG).
It is a fixed pt in a body where the
wt of the body is conc or acting or the resultant of gravitations
pull is passing. It is also known as centre of mass. If an object
is suspended through this pt, it will remain in a state of
equilibrium.
b.
Centre of Pressure (CP). It is a pt in a body through which
the resultant of aerodynamic force is considered to be acting. It
is not a fixed pt and is liable to shift with the change is posn of
the body.
c.
Angle of Yaw.
The axis of a proj fired from a gun does not,
in gen lie exactly along the traj, which is the path of its CG. The
angle b/w the axis of the proj and the tangent to the traj is
known as the angle of yaw.
Tan to Traj
Axis Of Proj
Traj
Angle Of Yaw
d.
Static Stability Margin.
It is the dist b/w CG and CP.
e.
Statically Stable Proj.
If CG is ahead of CP, the static
margin is positive and the proj becomes statically stable.
CP
CG
f.
Statically Unstable Proj. If CG is behind CP, the static margin
is negative.
CG
CP
g.
Neutrally Stable Proj. If the static margin zero, the proj is
considered neutrally stable.
CG
CP
h.
Moment.
Product of a force and the dist from its line of
action to a pt, expressing the power of that force to cause
rotation about that pt.
D
C
B
A
j.
Gyroscopic Force.
The force possessed by a rotating body
by virtue of which it resists any disturbance caused to its
rotation and keeps the rotating body stable.
k.
Precession.
The circular path about the CG which takes the
shape of a dec spiral.
l.
Notation.
The rotational mov in a small circle which forms a
rosette pattern.
Question No 2.
small yaw?
Answer No 2.
What would happen if an un-rotated proj acquires a
Fol are the chars of traj in air :-
a.
Increased drag.
b.
A deviating force tending to alter the dir of motion.
c.
A turning moment which may inc or dec the yaw, according to
the design of the proj.
Question No 3.
Why modern arty shells have negative stability margin?
Answer No 3.
Because of pointed nose to bear least resistance and
strong base to withstand the firing stresses the CG is towards the base.
The CP is towards nose and ahead of CG therefore the stability margin is
negative.
Question No 4.
Answer No 4.
a.
What means of stabilization are being used in projs?
Fol means are being used :-
Fin stabilization.
b.
Spin stabilization.
c.
Combination of the two.
Question No 5.
How the sfc area of a proj is inc towards the rear to
make it statically stable?
Answer No 5.
the CG.
Question No 6.
Answer No 6.
It is done by attaching fins to the rear of the proj behind
What are the main causes of instability?
The main causes of instability are as fol :-
a.
Yaw.
b.
Air resitance incl drag and cross wind forces.
Question No 7.
Answer No 7.
What are the disadvantages of att fins to a proj?
Fins have fol disadvantages :-
a.
They tend to increase the drag and hence reduce the range.
b.
They also often create problems from the launching and payload carrying potential aspects.
c.
They give rise to large wind effs, particularly in the case of a rkt
being boosted by the burning of the prop at the beginning of its
flt thus causing inconsistency. (The combinations of low lvl wind
and motor thrust cause the rkt to turn into the wind. It is difficult
to allow for this adequately in a strong gusty wind and it
becomes a maj source of error in free flt rkts having a long
burning time).
Question No 8.
Why CG of a modern shell is near its base?
Answer No 8. It is because the shell has to face a large amount of firing
stresses due to the burning of prop chg at its base.
Question No 9.
Why a small amount of spin is nec for rkts?
Answer No 9.
It may be useful in decreasing the dispersion inherent in
fin stabilized projs caused by accidental asymmetric forces arising from fin
misalignment. The eff of this slow spin is to rotate such disturbing forces
and so virtually eliminate any resultant dispersion.
Question No 10.
proj?
Answer No 10.
What are the effs of an over stabilized (over spun)
Effs of an over stabilized proj are as under :-
a.
For a neutrally stable proj there is no aerodynamic overturning
moment, hence gyroscopic stabilization will not work correctly.
b.
Proj will tend to retain its initial attitude.
c.
Proj will not dip down.
d.
It would result in irregular blsts.
e.
Proj may land base first or on side depending the A of D.
Question No 11.
Answer No 11.
When projs may be over spun?
Projs may be over spun under fol cases :-
a.
New guns.
b.
Wt of chg is more than req.
Question No 12.
Answer No 12.
What are the effs of an under spun proj?
Effs of an under spun proj are as under :-
a.
The yaw would become too great.
b.
The nose would mov away from the traj.
c.
The proj may turn over in the air.
d.
This may also give irregularity in range and drift.
e.
Even if the proj does not topple over, it will make violent
precessional mov.
f.
Angle of yaw may be as great a 350.
Question No 13.
Answer No 13.
When projs may be under spun?
Projs may be under spun under fol cases :-
a.
Defective DB.
b.
Worn out guns.
SEC – 12
Question No 1.
Answer No 1.
What are two parts of the lateral deviation of a proj?
The lateral deviation consists of fol two parts :-
a.
Drift due to equilibrium yaw – for spun projs.
b.
Drift due to rotation of Earth.
Question No 2.
What do you understand by ‘equilibrium yaw’?
Answer No 2. A proj during its flight cannot travel with zero yaw. When a
correctly spun shells stabilities after leaving the muzzle, its axis lies along
the traj. Soon after being fired, the shell devs some vertical yaw as the traj
falls. Air pressure under the nose will tend to inc this yaw. But a spinning
shell is a gyroscope and the force applied to the nose will cause the shell to
‘precess’. For a shell spinning clockwise, the eff is that the nose will mov to
the rt and the base to the lt. The air pressure will now tend to inc the yaw to
the rt, but the gyroscopic eff now movs the nose downwards. Air pressure
attempts to inc the nose down attitude but the gyroscopic eff movs the nose
to the lt. The net result is that the nose of the shell tends to precess around
the traj. However, whilst this is happening, the traj is also dipping. The
combination of a sel clockwise spin rate, precession and a dipping traj
keeps the avg yaw more or less constant. The CG of the shell fols the traj
and the nose rosettes in trochoidal motion about the traj with an avg posn
off-set to the rt. This avg yaw to the rt is called ‘Equilibrium Yaw’.
Angle of Equilibrium Yaw
Shell pointing on avg to the rt of traj
Question No 3.
What would happen if the nose of proj is below the traj?
Answer No 3.
traj.
The Magnus eff and the drift will both be to the rt of the
Question No 4.
What is the main cause of drift?
Answer No 4.
We know that a shell movs in the air with an equilibrium
yaw. Thus the air stream coming towards the shell, on avg creates a high
pressure on the lt, thus pushing the shell to the rt.
Question No 5.
Why at very high angles of departure the nose may fail
to fol the traj at the vertex and may fall base first, at a reduced range and a
marked drift towards the lt?
Answer No 5.
It is because of the Magnus eff.
SEC – 13
Question No 1.
What do you understand by the Magnus eff?
Answer No 1.
It is the lifting upwards of rotating body moving in the
during its flt due to low air pressure created above and high air pressure
created below the body. Example is of a golf ball moving in the air.
Question No 2.
moving in the air?
Explain the phenomenon of Magnus eff for a shell
Answer No 2.
All shells travel with the nose on avg to the rt of the
trajectory and there is a cross flow of relative air flow, from lt to rt of the
shell body. In addn, air is being carried around the shell as it spins. If the
shell is viewed from the rear, with only the cross component of relative air
flow being shown. Thus a shell flying with an avg equilibrium yaw to the
right, will experience an avg Magnus Force upwards, which will tend to inc
the range.
Air Flow
Low
Pressure
Lifting Force
Air Being Carried Along
the Shell
Question No 3.
High
Pressure
Why Magnus moment is caused?
Answer No 3.
Since the Magnus force will not nec act through the CG,
there may be a Magnus moment in the vertical plane.
Question No 4.
What are the subsidiary effs which may also contribute
to the Magnus forces and Magnus moments?
Answer No 4.
The other subsidiary asymmetric effs can incl thickening
of the bdry layer on one side of the body and vortex shedding, resulting in
airflow breaking away and separating on one side to give a negative
Magnus force.
Question No 5.
What do you understand by vortex shedding?
Answer No 5.
Vortex shedding is an unsteady flow that takes place in
special flow velocities (according to the size and shape of the cylindrical
body). In this flow, vortices are created at the back of the body and detach
periodically from either side of the body. It is caused when a fluid flows past
a blunt object. The fluid flow past the object creates alternating lowpressure vortices on the downstream side of the object. The object will tend
to move toward the low-pressure zone.
SEC – 14
Question No 1.
Define the fol :-
a.
Sphere. A sfc which has all the pts on it equidistant from its
centre is known as sphere.
b.
Great Circle.
It is a circle drawn on the sfc of any sphere
whose centre is the centre of the sphere. It divides the sphere
in two equal parts.
c.
Polar Axis.
It is the axis about which the Earth performs
its daily rotation. It is a diameter of the Earth and meets the sfc
of the earth in North and South poles.
d.
Equator. The great circle on the Earth’s sfc, that is 900 every
where from either pole. Its plane passes through the centre of
the Earth and is at rt angle to the polar axis.
e.
Latitude. It is the angle subtended at the centre of Earth b/w
the plane of equator and the plane made by the dir of plumb
line from the obsr. It is measured from 00-900 and designated as
north or south depending upon whether the place is north or
south of equator.
f.
Meridians. A great circle passing through north and south poles
and cutting the equator at rt angles. Meridian passing through
Greenwich is called Std Meridian.
g.
Longitude. It is the angle subtended at the centre of Earth
by the arc of equator intercepted b/w the plane of std meridian
and the plane of local meridian. It is measured from 00-1800 and
designated east and west depending upon whether the place is
east or west of std meridian.
h.
Parallels of Latitude.
These are small circles on the sfc of
Earth parallel to equator. All pts on a parallel of latitude have
same latitude value.
Question No 2.
the travel of proj?
Answer No 2.
travel of proj :-
What factors influence the eff of rotation of Earth upon
Fol factors influence the eff of rotation of Earth upon the
a.
Dir of fire.
b.
A/S.
c.
Velocity of proj.
d.
Range to the tgt.
e.
The latitude of the gun.
Question No 3.
of Earth?
Answer No 3.
Earth :-
What factors eff the performance of proj due to rotation
Fol factors eff the performance of proj due to rotation of
a.
Centrifugal force.
b.
Tangential velocity.
c.
Gravity.
d.
Longitudinal eff.
Question No 4.
How the dir of fire influences the eff of rotation of Earth
upon the travel of proj?
Answer No 4.
Since the Earth is rotating in a fixed dir i.e from West to
East therefore the change in the dir of fire will influence the eff of rotation of
Earth. If a proj is fired northwards in the Northern Hemisphere, there will be
a net displacement to the East, whereas for firing southwards the
displacement will be towards the West. Similarly, a diff displacement will be
there for firing westwards and eastwards. The corns for Southern
Hemisphere are opposite to those of the Northern Hemisphere. The range
in the similar manner is eff differently for diff dirs of fire.
Question No 5.
the travel of proj?
How the A/S influences the eff of rotation of Earth upon
Answer No 5.
The A/S effs the TOF and range to the tgt. The change in
these two factors will eff the time spent by the proj in the air and the
amount by which the tgt will be rotated with ref to the gun after the proj has
been fired.
Question No 6.
How the velocity of proj influences the eff of rotation of
Earth upon the travel of proj?
Answer No 6.
The velocity of proj effs the TOF and range to the tgt.
The change in these two factors will eff the time spent by the proj in the air
and the amount by which the tgt will be rotated with ref to the gun after the
proj has been fired. Even if the range is kept constant a proj with more
velocity would take less time to reach the tgt as compared to a proj with
less velocity, hence both projs will be eff diff.
Question No 7.
Prove from FT that the df corrn due to rotation of Earth
decs with inc in MV. 8 inch How, chg 5 and 6, range 8,000 Ms, dir of fire
400 mils, latitude 400 North?
Answer No 7.
Df corrn at chg 5 when MV is 420.6 m/s = L 1.0
Df corrn at chg 6 (when MV is 499.9 m/s) = L 0.9
Hence proved that the df corrn due to rotation of Earth decs with inc
in MV
Question No 8.
How the range to the tgt influences the eff of rotation of
Earth upon the travel of proj?
Answer No 8.
The range to the tgt effs the TOF which will eff the time
spent by the proj in the air and the amount by which the tgt will be rotated
with ref to the gun after the proj has been fired.
Question No 9.
How the latitude of the gun influences the eff of rotation
of Earth upon the travel of proj?
Answer No 9. Having a bulge at the equator, the Earth’s circumference
and speed of rotation at various latitudes is diff. Due to these factors the eff
varies with the change in the latitude.
Question No 10.
What are the diffs b/w latitude and longitude?
Answer No 10. Latitude
Longitude
a.
It is a vertical angle.
a.
It is a horizontal angle.
b.
It is measured as North
b.
It is measured from as
or South.
c.
Measured from the
East or West.
c.
Measured from the std
equator.
Meridian.
d.
Value is from 0-900.
d.
Value is from 0-1800.
e.
It is not a great circle.
e.
All longitudes are great
circles.
Question No 11.
The Earth is rotating in which dir?
Answer No 11.
axis.
The Earth is rotating from West to East about its polar
Question No 12.
about its polar axis?
In how much time the Earth’s complete one rotation
Answer No 12.
mins and 4 secs.
The Earth completes one rotation of 3600 in 23 hrs, 56
Question No 13.
What is the circumference of Earth along equator?
Answer No 13.
It is about 40,075 kms.
Question No 14.
What is the speed of rotation of Earth on the equator?
Answer No 14.
It is approx 465 m/s.
Question No 15.
What is the speed of rotation of Earth on the poles?
Answer No 15.
On the poles the velocity falls down to 0 m/s.
Question No 16.
and South?
Work out the rotational velocity at latitude 340 North
Answer No 16. Rotational velocity at = Rotational velocity at the
any latitude
equator x Cos latitude
= 465 x Cos 340
= 385.5 m/s
Question No 17.
What do you understand by the ‘Coriolis eff’?
Answer No 17.
It is an apparent df of moving objects when they are
viewed from a rotating ref frame.
Question No 18.
Diagrammatically show the eff of rotational velocity of
Earth on the line of fire when firing due North and South in Northern and
Southern Hemisphere?
Answer No 18.
Firing North
Latitude
T
T’
Equator
G
G’
Latitude
Northern Hemisphere
T’
T
G
G’
Southern Hemisphere
Firing South
Latitude
G
T
G’
T’
Equator
Latitude
G’
G
T
T’
Question No 19.
In which dir the eff of tangential (rotational) velocity
will be most and why?
Answer No 19.
It will be most in the dir of East as the Earth is rotating
from West to East so Eastward tangential velocity is always imparted to
proj.
Question No 20.
What corrns are req to the line of fire when firing in diff
dirs in the Northern and Southern Hemisphere?
Answer No 20.
Fol corrns are req in the Northern Hemisphere :-
a.
Firing West there will be a displacement to the North.
b.
Firing East there will be a displacement to the South.
c.
Firing North there will be a displacement to the East. This case
produces the smallest displacement.
d.
Firing South there will be a displacement to the West. This case
produces the largest displacement.
In each case in the Northern Hemisphere, the gun must be aimed lt of
the tgt. The converse is true for the Southern Hemisphere.
Question No 21.
Why df corrn is zero at bg 1,600 and 4,800 mils on
0
latitude 0 , whereas on latitudes other than 00 it is not zero?
Answer No 21.
Once the proj is fired it always travels in the dir of great
circle. If it is fired from any of the parallels of latitudes, which will curve
compared to the dir of motion of proj, so some line corrn is req. Whereas
along equator this curvature is not there because that is itself a great circle,
so line corrn is zero.
Question No 22.
Diagrammatically show the eff of rotational velocity of
Earth on the range when firing due East and West?
Answer No 22.
Firing East
Dir of Fire
P1
0
G
T
G1
T1
South
Pole
Tangent to
Earth’s Sfc
Firing West
Dir of Fire
Tangent to
T
G
T1
G1
Earth’s Sfc
South
Pole
Question No 23.
latitude?
Why the magnitude of range effs also decs with inc in
Answer No 23.
As the rotational velocity decs with inc in latitude, the
magnitude of range effs also decs.
Question No 24.
Why the curvature of Earth effs the travel of the proj?
Answer No 24.
It exists because of the use of map range which
assumes the sfc of Earth to be flat while the actual range is measured on a
sphere. The gun to tgt range is cpt for a plane tangent to the sfc of the
Earth at the wpn.
T’
G
T
In above fig the range to tgt T is cpt as GT’ which is some what lesser
than the range GT along the sphere. Once the proj reaches T’ it is still
above the sfc of earth and will cont to drop and land at P which is slightly
longer than the original cpt range. This eff is of little significance except at
very long ranges. It is disregarded when using FTs, since FT ranges incl
curvature eff.
Question No 25.
What do understand by the proj lag?
Answer No 25. It means that as the proj is fired, it lags behind the tgt and
lands west of it after returning to the sfc of Earth. To understand consider a
shell fired vertically from the Earth’s sfc at the Equator as shown in the fig
below. When the shell is fired it has horizontal velocity, VE, which is the
peripheral velocity of the Earth. After rising to a ht, X, it still has this
horizontal velocity. However, to remain vertically above pt B it would req a
horizontal velocity
VA = VE x (X + r)/r, where r is the radius of the
VA
VA = Vel (r + x) / r
Earth. Consequently, when the shell returns to Earth, it will always land
West of pt B. This eff is really caused by the Coriolis acceleration.
Question No 26.
Prove that dec in rotational velocity at latitude near
equator is less as compared to latitudes near the poles?
Answer No 26. We know that,
Rotational Velocity at latitude = Rotational Velocity at Equator x Cos
latitude
Rotational Velocity at 00
= 465 x Cos 00
= 465 m/s
Rotational Velocity at 10
= 465 x Cos 10
= 464.9 m/s
Dec
= 0.1 m/s
Rotational Velocity at 50
= 465 x Cos 50
= 463.2 m/s
Dec
= 1.8 m/s
Rotational Velocity at 850
= 465 x Cos 850
= 40.5 m/s
Rotational Velocity at 890
= 465 x Cos 890
= 8.1 m/s
Dec
= 32.4 m/s
Rotational Velocity at 900
= 465 x Cos 900
= 0 m/s
Dec
= 8.1 m/s
Question No 27.
Find out the value of range eff at dir of 3,000 mils,
range 6,000 Ms, chg 5?
Answer No 27. We know that,
Range Eff at any bg = Rotational Eff at 1600 mils x Sin Bg
Range Eff at any 3,000 = – 17 x Sin 3000
Range Eff at any 3,000 = – 3.3
Question No 28.
Work out the range corrn at 350 South for a dir of fire
1,400 mils when range corrn at same latitude is +22 Ms for dir of fire 4,400
mils?
Answer No 28. Range corrn for 1,600 mils = 22 / Sin 4400 mils
= – 23.8126284 Ms
Range corrn for 4,400 mils = –23.8126284 Ms x
Sin 1400 mils
= – 23.36 Ms
Question No 29.
If dir of fire is 4,100 and 3,100 mils, from rotation of
Earth pt of view, what woul be the eff on FS?
Answer No 29.
Range effs for dir of fire b/w 0 – 3,200 mils the range
incs req a lengthen corrn. For dir of fire 3,200 – 6,400 mils the range decs,
the FS would be req to be shortened.
Question No 30.
With what velocity a satellite should revolve around
the Earth to stay vertically above the equator all the times at one pt. Dist of
satellite from sfc of Earth is 300 kms, radius of Earth is 6400 kms and
velocity of equator is 465 m/sec?
Answer No 30. 465 X (6400 + 300)
6400
= 486.8 m/s
= 487 m/s
SEC – 15
To be covered in Ammo subj.
SEC – 16
Question No 1.
Define the fol :-
a.
Met.
It is the science dealing with the atmospheric
phenomenon.
b.
Atmosphere.
Earth.
c.
Air.
An invisible substance which is mixture of several
gasses. Main constituents are nitrogen and oxygen.
It is an envelop of air which is surrounding the
d.
Lapse Rate.
lapse rate.
The rate of dec of temp with alt is known as the
e.
Inversion.
inversion.
The rate of inc of temp with alt is known as
f.
Isothermal.
Temp remains constant with inc in alt.
g.
Conduction.
Process by which heat energy is transmitted
through contact with neighboring molecules.
h.
Convection. Heat is transmitted by transporting of molecules
from place to place within a substance. Convection occurs in
fluids such as water and air, which mov freely.
j.
Radiation. Transfer of heat energy w/o the involvement of a
physical substance in the transmission. Radiation can transmit
heat through a vacuum.
Question No 2.
Answer No 2.
a.
b.
What is the composn of air?
Air is composed of the fol :-
Gasses.
Pure dry air consists of fol:-
(1)
Nitrogen
-
78 %
(2)
Oxygen
-
21 %
(3)
Other gasses
-
1 %
(a)
Argon
-
0.93 %
(b)
Carbon dioxide -
0.03 %
(c)
Neon, Krypton and Helium- 0.93 % (approx)
Moisture / Water Vapours.
It is present in the air in varying
qty. In many respects it is most imp constituent of the air. It is
moisture in atmosphere which forms clouds, rain, snow and fog.
It is 4 % of the atmosphere.
c.
Impurities.
In addn to moisture and gasses, air contains a
no of solid particles in the form of seeds, bacteria, dust, smk
and salt. To give an indication of the size of such particles it has
been est that there are about 4,000,000,000 (four billion)
particles present in an ordinary cigarette puff.
Question No 3.
What are the various layers / spheres of atmosphere
surrounding the Earth?
Answer No 3.
as under :-
The atmosphere has been divided into five distinct layers
a.
Troposphere.
The troposphere starts at the Earth's sfc and
extends 8 to 14.5 kms high (5 to 9 miles). This part of the
atmosphere is the densest. As we climb higher in this layer, the
temp drops from about 17 to –52 degs Celsius. Almost all
weather is in this region. The Tropopause separates the
troposphere from the next layer. The tropopause and the
troposphere are known as the lower atmosphere.
b.
Stratosphere.
The stratosphere starts just above the
troposphere and extends to 50 kms (31 miles) high. Compared
to the troposphere, this part of the atmosphere is dry and less
dense. The temp in this region incs gradually to –3 degs
Celsius, due to the absorption of ultra violet radiation. The
ozone layer, which absorbs and scatters the solar ultraviolet
radiation is in this layer. Ninety-nine percent of "air" is located in
the troposphere and stratosphere. The stratopause separates
the stratosphere from the next layer.
c.
Mesosphere.
The mesosphere starts just above the
stratosphere and extends to 85 kms (53 miles) high. In this
region, the temp again falls as low as –93 degs Celsius as you
inc in alt. The chemicals are in an excited state, as they absorb
energy from the Sun. The mesopause separates the
mesosphere from the thermosphere. Scientists call the regions
of the stratosphere and the mesosphere, along with the
stratopause and mesopause, the middle atmosphere.
d.
Thermosphere.
The thermosphere starts just above the
mesosphere and extends to 600 kms (372 miles) high. The
temp go up as you inc in alt due to the Sun's energy. Temp in
this region can go as high as 1,727 degs Celsius. Chemical
reactions occur much faster here than on the sfc of the Earth.
This layer is known as the upper atmosphere.
e.
Ionosphere / Exosphere. The layer above the thermosohere
is called ionosphere / exosphere. Temp inc with inc in ht in this
layer upto a top temp is 10,000 degs Celsius.
Question No 4.
Why the firing units are provided with met msg?
Answer No 4.
To cater for the variations in atmospheric conditions the
firing units are provided with a met msg to enable them to ascertain the
departure from std values and accordingly apply corrns to offset the
existing weather conditions.
Question No 5.
met forecast?
Answer No 5.
Enumerate the types of errors which can be there in a
Fol types of errors can be there :-
a.
Instrumental error.
b.
Reduction error.
c.
Time error.
d.
Space error.
Question No 6.
Answer No 6.
a.
What elms of air eff the external blsts?
Fol elms of air eff the external blsts :-
Density. Dense air will offer greater resistance to a proj than
less dense air, i.e, the Co will be smaller for air of higher
density and the shell will not range so far. It is not practical to
measure density directly and so it is cpt from measurements of
pressure, temp and humidity. Dec barometric pressure or high
alts will result in a dec in drag, and inc barometric pressure or
low alts will result in a rise in drag. Humidity also has an impact
the opposite of the one that most people expect. Since water
vapour has a density of .8 gms per lit, while dry air avgs about
1.3 gms per lit, higher humidity actually decs the air density,
and therefore decs the drag.
b.
Wind.
A head wind will impede the passage of a shell and a
fol wind will assist it. A crosswind blowing from lt to rt will cause
the shell to deviate to the rt and vice versa.
c.
Elasticity. This is a measure of the compressibility of the air,
which is intimately connected with the velocity of sound and
affects the resistance to a shell. The velocity of sound is mainly
dependent on temp and humidity and is not measured directly.
Question No 7.
expected?
Why humidity has an opposite eff to what is normally
Answer No 7.
It is because the water vapour has a density of .8 gms
per lit, while dry air avgs about 1.3 gms per lit, higher humidity actually
decs the air density, and therefore decs the drag and the shell would range
farther in higher humidity.
Question No 8.
measured?
How the density of air is calc and why is it not directly
Answer No 8.
It is not practical to measure density directly and so it is
cpt from measurements of pressure, temp and humidity.
Question No 9.
alts?
Why the air at higher alts is less than the air at lower
Answer No 9.
The air has a mass because of which the air above is
compressing the air below. Resultantly the the air below becomes
compressed and its density incs. Therefore the air at lower alts is more
dense than the air at higher alts.
Question No 10.
despite less temp?
Why the proj achieves more range at higher alts
Answer No 10. It is because of lesser density. Lower temp has its eff but
the lesser density at higher alts outweighs the eff of lower temp and the
proj despite lesser temp achieves more range.
Question No 11.
decided?
How the std atmosphere for compilation of FTs is
Answer No 11.
The actual choice of a std atmosphere is a matter of
convenience provided the stds do not vary too greatly from the actual
conditions experienced in prac. Besides assuming certain std values on gr,
the values of temp and density at pre sel ht aloft are also assumed. This
constitutes the Std Blsts Atmosphere. It is imp that these values must be
close to avg of actual prevailing values, otherwise the magnitude of corrns
would become unmanageable. Also if the prevailing conditions are not
known the accy of fire will be adversely affected.
Question No 12.
What std atmospheric values are being used for
various eqpt / FTs held with Pak Army?
Answer No 12.
Fol std values are being used :-
British
US
Eastern
ICAO
Wind
Nil
Nil
Nil
Nil
Pressure
3”
29.4”
(1015.9mb)
(996 mb)
29.92”
29.92”
(998.5 mb)
(1013.25mb)
Temp
60 F/15.6 C 59 F/15 C
59 F/15 C
59 F/15 C
Humidity
50 %
Nil
Nil
Nil
Density
Vel of
Sound
1222.5
1203.4
1207.2
1225
gm/m3
gm/m3
gm/m3
gm/m3
1220 f/s
1220 f/s
Nil
1117 f/s
Question No 13.
What are the various line nos being used and what
are the hts of these line nos?
Answer No 13.
under :-
Line nos being used are numbered from 1 – 15 as
Zone Ht (Ms)
Line No ZZ
Sfc
00
200
01
500
02
1,000
03
1,500
04
2,000
05
3,000
06
4,000
07
5,000
08
6,000
09
8,000
10
10,000
11
12,000
12
14,000
13
16,000
14
18,000
15
Question No 14.
What are weighing factors?
Answer No 14.
These are factors worked out for each zone for top and
bottom weather values of that zone in order to determine the proportional
eff on diff values.
Question No 15.
Why avg weighing factors are used?
Answer No 15.
Weighing factors vary not only from layer to layer but
also from one traj to another. To achieve the greatest accy there should be
a weighing factor for each layer of each traj but this is only prac when cpts
are done by the cptr. During manual cpts, avg weighing factors are used
which can be applied with reasonable accy to all fd eqpt.
Question No 16.
manual cpts?
What info is provided by the blsts std met msg for
Answer No 16.
provides :-
For manual computations, the std blsts met msg
a.
Equivalent constant wind velocity.
b.
Blsts air temp (expressed as a percentage of stds).
c.
Ballistic air density (expressed as a percentage of stds).
Question No 17.
svcs?
Answer No 17.
a.
How the weighing factors are cl for various arms /
Weighing factors are cl as under :-
Type – 1.
Used by PAF for descending traj.
b.
Type – 2.
c.
Type – 3.
Used by fd arty against gr tgts for both ascending
and descending trajs.
d.
Type – 4.
Used by long range guns for both ascending and
descending trajs.
Question No 18.
Wind’?
Used by AD for ascending traj.
What do you understand by ‘Equivalent Constant
Answer No 18. This is the wind which, if traveled at constant speed and
dir at all pts of the traj, would produce the same changes in range and in
line to the pt of graze as does the actual wind structure. Its value is a
weighed mean of the actual winds from the sfc to the vertex, the weighing
factors used are applied to the winds at various lvls, to allow for the fact
that time spent by the proj in layers of given thickness varies at diff lvls.
Question No 19.
What do you understand by ‘Blsts Air Temp’?
Answer No 19. This is analogous to the equivalent constant wind, being
derived from a weighed mean of the variations from the std temp in each ht
layer. The weighing factors for temp are not the same as for wind. When
producing blsts air temp, allowance is made for humidity.
Question No 20.
Why no separate allce for elasticity is made in the met
msg although it varies with air temp?
Answer No 20. Although elasticity also varies with air temp, no separate
allowance is made for it in the met msg because the elasticity variations are
incl in the figs given in FTs for the eff on range, of a non-std blsts air temp.
Question No 21.
Answer No 21.
How ‘Blsts Density’ is cpt?
This is cpt from the blsts air temp and pressure.
SEC – 17
Question No 1.
What is a mor?
Answer No 1.
Question No 2.
Answer No 2.
Question No 3.
Answer No 3.
the bore.
Mor is a simple, mob wpn designed to give high traj fire.
Why mors are gravity fired?
Mors are gravity fired for simplicity.
What is windage?
It is the diff b/w the cross sectional area of the bomb and
Question No 4.
Why windage is nec in mors?
Answer No 4.
bomb descends.
It is nec to allow the air in the barrel to escape as the
Question No 5.
What are the internal blsts chars of mors in comparison
with similar cal guns?
Answer No 5.
Fol are the internal blsts chars of mors in comparison
with similar cal guns :a.
Small amount of prop chg.
b.
Low chamber pressure.
c.
Greater variations in MV.
d.
Less wear of barrel.
Question No 6.
Why there is not much SSP or in-bore resistance as the
mor bomb starts mov in the bore?
Answer No 6.
Question No 7.
that of the guns?
It is because of the absence of DB.
Why the overall pressure in mors is much lower than
Answer No 7.
Due to absence of DB the SSP and in-bore resistance in
mors is much less hence the overall is much lower than that of the guns.
Question No 8.
Why mor barrels can be made lighter?
Answer No 8.
Due to absence of DB the SSP and in-bore resistance in
mors is much less hence the overall is much lower than that of the guns,
which for the const of lighter barrels.
Question No 9. Why there are greater variations in MV in case of mors
as compared to the guns?
Answer No 9.
The amount of gas escaping past the bomb is not always
the same and thus the rd-to-rd pressure varies causing more variations in
MV as compared to the guns.
Question No 10.
Why mech wear in case of mors is negligible?
Answer No 10.
Since there is little friction b/w the bomb and the barrel
walls, therefore the mech wear is negligible.
Question No 11.
Answer No 11.
What is the main cause of wear in mors?
Wear in mors is almost wholly due to gas erosion.
Question No 12.
drag design?
Why supersonic and transonic mor bombs have a low
Answer No 12.
In case of these mor bombs, the drag becomes a
significant factor, therefore these bombs have a low drag design.
Question No 13.
of low drag design?
How supersonic and transonic mor bombs are made
Answer No 13.
These mor bombs have an ogival nose, a cylindrical
body tapering down to a tail boom with a fin unit in the end.
Question No 14.
How the amplitude of yawing motion can be reduced?
Answer No 14.
least 0.5 cals.
It is reduced by having the static stability margin of at
Question No 15.
What are the launch velocities of subsonic mors?
Answer No 15.
Mach 0.5 or less.
Sub-sonic mors have launch velocities in the region of
Question No 16.
boom?
Why subsonic mor bombs can be designed w/o a tail
Answer No 16.
These mor bombs have very low velocities in the range
of Mach 0.5 or less. At these velocities drag does not have a very marked
eff on range. This means that a small mor bomb can have relatively large
drag w/o seriously degrading its max range. Therefore these bombs are
often designed w/o a tail boom.
Question No 17.
with heavier bomb?
Answer No 17.
Why is drum tail used with lighter bomb and fin tail
Because fins have more sfc and minor temp variations.
SEC – 18
Question No 1. Why the launcher of a rkt can be extremely lt, simple and
designed as a multi barrel?
Answer No 1.
The launcher only pts the rkt at the req bg and ele and
provides a means of ignition. It has no obturation or recoil problems to
conten with. Hence it can be extremely lt, simple and designed as a multi
barrel.
Question No 2.
interval?
Why the firing switch fires successive rds at a slight
Answer No 2. The firing switch fires successive rds at a slight interval to
avoid mutual interference of the rkts in flt.
Question No 3.
Answer No 3.
What does a rd of rkt consist of?
A rd of rkt consist of fol :-
a.
Fuze.
b.
Warhead.
c.
Motor.
Question No 4.
Why warhead casing of rkt can be made thin as
compared to a conventional shell?
Answer No 4.
Since the rkt does not have to sustain high in-bore
stresses, therefore the warhead casing of rkt can be made thin as
compared to a conventional shell.
Question No 5.
What is venturi?
Answer No 5. The motor of a rkt is sealed and fixed to the base of the
warhead at one end and has an open throat on the other end known as
venturi.
Question No 6.
Why exposed metal parts are protected by a sacrificial
layer of insulation?
Answer No 6.
Question No 7.
Answer No 7.
It is because of the high combustion temp involved.
In what all stages the flt of rkt can be divided?
It can be divided into fol two stages :-
a.
The stage when the prop is burning.
b.
The stage when the prop has been completely burnt.
Question No 8.
Explain the rkt blsts when the prop is still burning?
Answer No 8. On ignition, high pressure prop gas is produced inside the
rkt motor. These gases escape through the convergent-divergent nozzle at
a rate which is con by motor design. The rearward momentum of the gases
is offset by the fwd motion of the rkt. The rkt conts to be accelerated till all
the prop is burnt. At this moment the thrust ceases and the rkt becomes a
free proj. This stage, incl both internal and intermed blsts phs.
Question No 9.
Why intermed blsts of rkts is more imp than the guns?
Answer No 9.
Most arty rkts leave the launcher while the prop is still
burning therefore the intermed blsts stage in rkts is more imp than the
guns.
Question No 10.
understanding?
What imp aspects of rkt blsts are nec for basic
Answer No 10.
understanding :-
Fol imp aspects of rkt blsts are nec for basic
a.
Prop.
It is desirable that during the burning stage the rkt is
accelerated at a uniform rate. Towards this end, special grain
shapes are used in rkt motors, which are diff from those for
guns and have constant burning sfcs.
b.
Effs of Cross Wind.
During the burning stage a cross wind
tends to turn rkt into the wind by changing the thrust vector. For
eg, if wind is blowing from rt of the axis of rkt, it will cause the
out flowing gas jet to mov to the lt of its flow path and
consequently, the thrust of jet will change towards rt.
c.
Stability. Rkts may be spin stabilized or fin stabilized. Spin is
imparted to the rkt by offsetting the vanes from the axis of the
motor. This causes the gas to emerge at an angle to this axis
and impart a spinning motion to the proj.
Question No 11.
Answer No 11.
How spin can be imparted to the rkt?
Fol methods are used to impart spin to the rkt :-
a.
Spin is imparted to the rkt by offsetting the vanes from the axis
of the motor. This causes the gas to emerge at an angle to this
axis and impart a spinning motion to the proj.
b.
By fixing a curved rail or helical cam fixed to the launcher.
c.
By machining slow helical grooves in the launch.
Question No 12.
Why rkts are considered useful for large area tgts?
Answer No 12.
large dispersion.
It is because of the inaccys of their del which cause
Question No 13.
What factors influence the accy of arty rkt?
Answer No 13. Fol factors influence the accy of arty rkt :a.
Length of the launcher.
b.
Std of launcher manufacture.
c.
Method of generating roll on the launcher.
d.
Exhaust gas (blow-by).
e.
Launcher motions.
f.
Thrust misalignment.
g.
The effs of rkt flexure.
Question No 14.
Why a launcher for the rkt is req?
Answer No 14.
It req to guide the rkt during the burning pd and this is
achieved by contact forces at the interfaces (or bourrelets) of the rkt.
Question No 15.
What max launcher length is gen used for the rkts?
Answer No 15.
Max guidance length of 5 – 6 Ms for present gr sp rkt
systems is gen used which corresponds to approx 10% of the pd of burn of
the motor.
Question No 16.
Why significant dispersions occur in rkt and what is
the solution to them?
Answer No 16.
It is impractical to design arty rkts which would be allburnt on the launcher and an unconstrained boost phase is req. Due to the
fact that the rockets are unrestrained except for aerodynamic forces,
significant dispersions occur due to thrust misalignment unless the rkt is
spun at a rate of approx 25 radians / sec.
Question No 17.
What do you understand by ‘blow-by’?
Answer No 17.
If the rearward exhaust gas in a rkt meets any
constriction, then the flow can be reversed and ‘blow’ fwd through the
annular area b/w the rkt and launch tube. This process is commonly called
‘blow-by’ and can severely eff the pitch angular velocity of the rkt during
launch.
Question No 18.
How the problems of accy resulting due to reduced
structural mass of the launcher to ensure mob can be overcome / reduced?
Answer No 18.
Locking the vehicle suspension and the use of a spade
reduces the problems of accy.
Question No 19.
How the response of a fin-stabilized free-flt rkt to
thrust misalignment and fin misalignment may be reduced?
Answer No 19.
The response of a fin-stabilized free-flt rkt to thrust
misalignment and fin misalignment may be reduced by intro slow spin
(typically 20 – 30 radians / sec).
Question No 20.
What all forces act on the rkt during launch ph?
Answer No 20.
Fol forces act on the rkt during launch ph and may
generate bending moments and df of the rkt :a.
Thrust.
b.
Inertia.
c.
Spin.
d.
Gravity.
Question No 21.
Why range PE of 122 mm MBRL is more at shorter
ranges and reduces at longer ranges?
Answer No 21.
At shorter ranges the rkt is fired at lower eles below 30
degs and has lesser TOF. On these eles the rkt passes close to gr features
whose turbulent air current deflects the rkt from its path. Secondly the rkt is
unsteady once the rkt motor is still burning. At shorter ranges due to lesser
TOFs the not much time is aval to overcome this unsteadiness while at
longer ranges the rkt stabilization improves. Hence range PE of 122 mm
MBRL is more at shorter ranges and reduces at longer ranges.
Question No 22.
How range can be varied in 122 mm MBRL?
Answer No 22.
The range is varied by use of small and large breaking
rings. These rings are fitted on the fuze and by altering the shape of the
proj, the air resistance is varied and resultantly the range is changed.
SEC – 19
Question No 1. Why FTs, GFTs and cptr progms are nec from blsts pt of
view?
Answer No 1.
These are nec for fire con purposes. The data is req to
relate the performance of the wpn to the measureable initial conditions
prevailing in order to allow the user to hit the tgt. Accy of the data in the
FTs, GFTs and cptr progms bears dir on the accy of the wpn system,
especially in predicted fire.
Question No 2. What is the aim of R&A trials?
Answer No 2. The aim of these trials is to record for analysis the launch
parameters and envmt of a firing and the resultant performance of the eqpt.
Question No 3.
How bias errors are removed during R&A trials?
Answer No 3.
These are removed by firing on diff occasions, using a no
of guns and several eles for each chg.
Question No 4.
How the data for worn out barrels is worked out during
the compilation of FTs?
Answer No 4.
After the R&A trials and production of FTs, when
sufficiently worn barrels become aval, firings are conducted to compare the
results obtained from barrels at various stages of wear. These trials,
together with the results of experimental firings and of cal by units, enable
the variation of MV with barrel wear to be determined.
SEC – 20
Question No 1. Why the layout of FTs held with Pak Arty is not uniform?
Answer No 1.
These are not uniform due to the large variety of eqpt
procured from diff sources of origin. At present except for FTs of 122 mm
How and 122 mm MBRL which use the Eastern FTs, all other FTs incl 130
mm Gun and 122 mm How D–30 have been converted onto the ICAO stds.
Question No 2. What info may be contained in the title pages of an ICAO
stds FT?
Answer No 2. Fol info may be incl :a.
Nomenclature of the gun to which it applies.
b.
List of appropriate ammo.
c.
Std conditions on which the table is based.
d.
Table of contents.
e.
Security mkg.
f.
Army code no.
Question No 3.
ICAO stds FT?
What info may be contained in the intro portion of an
Answer No 3. Fol info may be incl :a.
A list of symbols and abvns used in the FTs.
b.
Gen info on the method of compilation of FTs.
c.
Details of wpn – proj combination to which the FT applies along
with corrns req for that particular combination.
d.
Details of wpn chars.
e.
A table of chg composns.
f.
A table of wear data and EFC.
g.
A table of proj – fuze combination and mean wt.
h.
Explanation of tables.
j.
Details of ICAO stds met msg and method of cpt C of M.
k.
Probability tables with explanation.
l.
Chg sel table.
Question No 4.
Answer No 4.
What are the chars of gen layout of ICAO stds FTs?
The chars of gen layout of ICAO stds FTs are as fols :-
a.
Indexing is provided to give an easy access to each chg and to
other secs.
b.
Conventional algebraic signs are used throughout the tables.
c.
Negative signs and values are printed in red, inc and tail wind
corrn colms are shaded and high angle data starts after a line
of asteriks.
SEC – 21
Question No 1. What is the eff of an inc in the A of P in both the low and
the high angle fire?
Answer No 1.
S/No
Low Angle Fire
High Angle Fire
a.
Range
Inc
Dec
b.
TOF
Inc
Inc
c.
Vertex Ht
Inc
Inc
d.
Drift
Inc
Inc
Question No 2. What is the eff of a inc in the MV in both the low and the
high angle fire?
Answer No 2.
S/No
Low Angle Fire
High Angle Fire
a.
Range
Inc
Inc
b.
TOF
Inc
Inc
c.
Vertex Ht
Inc
Inc
d.
Drift
Inc
Inc
Question No 3. What is the eff of an inc in the Co (other than due to an
inc in wt of proj) in both the low and the high angle fire?
Answer No 3.
S/No
Low Angle Fire
High Angle Fire
a.
Range
Inc
Inc
b.
TOF
Inc
Inc
c.
Vertex Ht
Inc
Inc
d.
Drift
Inc
Inc
Question No 4.
achieved?
What is the eff of an inc wt of proj on the range
Answer No 4.
With the inc in the wt of proj, there is a dec in the MV but
an inc in the Co. At short ranges the dec in MV outweighs the inc in Co but
at longer ranges the reverse is true.
Question No 5. What is the eff of head wind in both the low and the high
angle fire?
Answer No 5.
S/No
Low Angle Fire
High Angle Fire
a.
Range
Inc
Inc
b.
TOF
Inc (very small)
Inc (very small)
c.
Vertex Ht
Inc (very small)
Inc (very small)
d.
Drift
Inc (very small)
Inc (very small)
Question No 6.
angle fire?
What is the eff of tail wind in both the low and the high
Answer No 6.
S/No
Low Angle Fire
High Angle Fire
a.
Range
Dec
Dec
b.
TOF
Dec (very small)
Dec (very small)
c.
Vertex Ht
Dec (very small)
Dec (very small)
d.
Drift
Dec (very small)
Dec (very small)
Question No 7.
Why drift is more in high angle and less in low angle?
Answer No 7.
Drift changes with TOF and we know that TOF is more in
high angle fire and less in low angle fire. At the same time we know that
drift is proportional to tan A of P and value of tan incs with inc in A of P.
hence value of drift is more in high angle and less in angle.
Question No 8.
Keeping the range constant, why drift is more for lower
chgs and less for higher chgs?
Answer No 8.
It is because of fol reasons :-
a.
Drift changes with TOF and we know that TOF for same range
is more for lower chgs as compared to higher chgs.
b.
At the same time we know that the A of P for lower chgs is
more as compared to higher chgs and drift is proportional to tan
A of P. Value of tan incs with inc in A of P, hence value of drift
is more for lower hgs than highe chgs.
c.
Drift also changes with the rate of spin. Lower chgs have less
MV and less rate of spin as compared t higher chgs hence they
have more drift.
Question No 9.
If you see the FT of 8” How, you will find out that drift
corrn at range of 7,000 Ms is 9.0 at chg 3 whereas at the same range it is
6.7 at chg 4. Why is it so?
Answer No 9.
It is because of fol reasons :-
a.
At chg 4, MV is more so more spin, hence less drift.
b.
At chg 4, TOF is less, hence less drift.
c.
At chg 4, A of P is less, hence less drift.
Question No 10.
What is the eff of cross wind on the proj’s flt?
Answer No 10.
Cross wind effs the line of fire. A cross wind blowing
from the rt of the line of fire causes a variation to the lt and vice versa.
Question No 11.
What do you understand by the rigidity of traj?
Answer No 11.
Let us suppose that in order to hit a tgt with low angle
fire at zero A/S, a certain A of P is req. In order to hit another tgt vertically
above (or below) the first one it is then only nec to inc (or dec) the A of D by
the A/S to the sec tgt, provided this A/S is small (less than 10 mils). Thus
the A of Ps in the two cases are the same. This principle is called the
“Rigidity of the Traj” i.e, the traj can be “swung” rigidly up or down w/o
appreciably altering its form. This is shown in figure below:-
All A of Ps are
Equal
Horizontal
Question No 13.
Why NR corrn is req and what are its components?
Answer No 13.
In fig below if the same A of P req to hit a tgt T1 at zero
A/S be applied for Tgt T2 at a positive A/S, the proj will usually range short
of T2, and pass through T3. There are three reasons for the traj not
passing through T2 :a.
Slant Range.
The dist along the L/S from gun to T2 is
greater than the (horizontal) range from gun to T1.
b.
Horizontal Component of Velocity. In the higher traj the
horizontal component of velocity is less than in the lower one.
Therefore, the range is dec.
c.
Density. In the higher traj, the proj passes through air of
lower density than in the lower traj. Therefore its range is inc.
NR Corrn
T2
T3
T1
Question No 14. What are the effs of the components of NR for positive
A/S low angle fire?
Answer No 14.
S/No
Eff Due to Corrns for Positive A/S
Positive A/S
in Terms of
Range
A of P
a.
Slant Range
Short (–)
Add (+)
Ele (+)
b.
Horizontal
Component
of Velocity
Short (–)
Add (+)
Ele (+)
c.
Density
Over (+)
Drop (–) Depression (–)
d.
Overall Eff
Short (–)
Add (+)
Ele (+)
Question No 15.
Why for high velocity guns, at low angle fire the NR for
positive A/S may be negative?
Answer No 15.
It is because at low angle fire with high velocity guns,
the density eff may outweigh the slant range and horizontal component of
velocity effs; hence the result is that the NR corrn for positive A/S may
become negative.
Question No 16.
In-vacuo at low angle fire the NR for positive A/S will
always be positive even for the high velocity guns. Comment?
Answer No 16.
Yes, the statement is true. It is because in-vacuo there
is no air, so the eff of air density will be nil and rest of the two components
i.e, horizontal component of velocity and slant range, both have the eff to
bring the rd short of the tgt so a positiveve NR will always be req.
Question No 17. What are the effs of the components of NR for negative
A/S low angle fire?
Answer No 17.
S/No
a.
Eff Due to Corrns for Positive A/S
Negative A/S in Terms of
Slant Range
Short (–)
Range
A of P
Add (+)
Depression (–)
b.
Horizontal
Component
of Velocity
Over (+)
Drop (–) Ele (+)
c.
Density
Short (–)
Add (+)
Depression (–)
d.
Overall Eff
Short (–)
Add (+)
Depression (–)
Question No 18. What are the effs of the components of NR for positive
A/S high angle fire?
Answer No 18.
S/No
Eff Due to Corrns for Positive A/S
Positive A/S
in Terms of
Range
A of P
Depression (–)
a.
Slant Range
Short (–)
Add (+)
b.
Horizontal
Component
of Velocity
Over (+)
Drop (–) Ele (+)
c.
Density
Short (–)
Add (+)
Depression (–)
d.
Overall Eff
Short (–)
Add (+)
Depression (–)
Question No 19. What are the effs of the components of NR for negative
A/S high angle fire?
Answer No 19.
S/No
Eff Due to Corrns for Positive A/S
Negative A/S in Terms of
Range
A of P
Ele (+)
a.
Slant Range
Short (–)
Add (+)
b.
Horizontal
Component
of Velocity
Over (+)
Drop (–) Depression (–)
c.
Density
Short (–)
Add (+)
Ele (+)
d.
Overall Eff
Over (+)
Add (+)
Ele (+)
Question No 20.
high angle fire?
What is the eff of positive A/S in both the low and the
Answer No 20.
S/No
Low Angle Fire
High Angle Fire
a.
TOF
Negligible eff
Negligible eff
b.
Vertex Ht
Inc
Inc
c.
Drift
No eff
No eff
Question No 21.
What do understand by ‘linearity of effs’?
Answer No 21.
It means that change in any non std condition will have
lnr effs for eg if a change of 3% in air density has a total eff of 70 Ms in
range, then a change of 6% in air density will have an eff of 140 Ms in
range.
Question No 22.
What do you understand by the first order variations
and what are the numerical limits fixed for these variations?
Answer No 22.
When the variations in initial conditions (MV, Co and A
of D) are small, their effs can be assumed to be lnr and proportional,
provided all changes are small. These are termed as first order variations.
The numerical limits usually fixed for first order variations are that a change
in :a.
MV must not exceed 25 m/s.
b.
Co must not exceed 10%.
c.
A of D must not exceed 30 mils.
Question No 23.
What do you understand by ‘cross term eff’?
Answer No 23.
The method of combining the effs of two non-std
conditions is based on the assumption that effs due to diff variations are
indep of one another. In calc the effs due to one variation, it is assumed
that all other conditions are std. When variations are small, this assumption
does not intro great errors, but when variations are large, errors may arise
through making this simple assumption. This phenomenon is known as
‘cross term eff’.
Question No 24.
A proj under the influence of 40 knots cross wind and
40 knots head wind is deflected by 15 mils to the rt. What will be eff on dir
(for same wind angle) if magnitude of head wind decs to 20 knots?
Answer No 24.
indep of rg effs.
Proj will still be deflected by 15 mils as df / line effs are
SEC – 22
Question No 1. What std conditions are used in Pak Arty?
Answer No 1.
Std conditions are used in Pak Arty are as fol :-
a.
A/S.
Zero.
b.
MV.
As laid down for eqpt and chg used.
c.
Wt and Shape of Proj.
d.
Equivalent Constant Wind.
As laid down.
Zero.
700 F / 210 C. When using old FTs, as given in
e.
Chg Temp.
FTs.
f.
Density at Mean Sea Lvl.
g.
Pressure at Mean Sea Lvl.
h.
Temp at Mean Sea Lvl.
j.
Temp Lapse Rate.
k.
Earth is considered stationary.
1225 gm/m3.
1013.25 mbs.
150 C.
6.50 C per 1000 Ms.
Question No 2. Why std conditions cannot be kept as zero?
Answer No 2. Std conditions are kept closer to prevailing conditions and
cannot be kept as zero for fol reasons :a.
If std conditions are closer to prevailing conditions, the corrns
worked out are small and manageable.
b.
If std conditions are kept as zero, the magnitude of error in calc
will be enhanced as corrns worked out will be large.
Question No 3.
FTs?
Why diff ctys use diff std conditions for compilation of
Answer No 3.
It is because atmospheric conditions in diff parts of the
world are diff. Therefore diff std conditions are used by diff ctys which are
close to the prevailing conditions in that cty. At the same time diff ctys use
diff units of measurement.
Question No 4. What are the components / gps in which the range corrns
are divided?
Answer No 4.
Range corrns are divided into fol components / gps:-
a.
Range C of M.
b.
Non-std proj corrn.
c.
Rotation of Earth corrn.
d.
NR corrn.
e.
MV corrn.
Question No 5.
What do you understand by the range C of M?
Answer No 5. It consists of corrns, normally common to all guns of a bty
and which hold good for a definite pd of time, to ctr act the effs associated
with variations in :a.
Head or fol component of equivalent constant wind.
b.
Blsts air temp.
c.
Air density.
d.
Chg temp.
Question No 6.
Answer No 6.
Question No 7.
divided?
Answer No 7.
Why chg temp corrns is incl in the C of M?
It is incl in the C of M for convenience only.
What are the components in which the df corrns are
Df corrns are divided into fol components :-
a.
Line or df C of M.
b.
Drift corrn.
c.
Rotation of Earth corrn.
d.
Lack of lvl of trunnions.
Question No 8.
What do you understand by the line C of M?
Answer No 8. It consists of corrns, normally common to all guns of a bty
and which hold good for a definite pd of time namely to ctr act cross
component of equivalent constant wind.
Question No 9.
Why drift corrn is determined at P/range?
Answer No 9.
Since the drift varies with traj and TOF, therefore it must
be determined at P/range.
Question No 10. What are the components of FS corrns?
Answer No 10.
FS corrns have fol components :-
a.
Corrn for MV and chg temp.
b.
Corrn for range wind.
c.
Corrn for air density.
d.
Corrn for proj’s wt.
e.
Corrn for blsts air temp.
f.
Corrn for A/S.
Question No 11. Prove that eff of non-std conditions on range does not
correspond to their eff on TOF. 8 inch How, chg 1, range 2,500 Ms?
Answer No 11.
We assume any non std condition which reduces rg by
50 Ms. So the dec in rg is 2 %. TOF for 2,450 Ms rg is 10.4. Reduction in
TOF is 1.89 %.
Question No 12. Prove from the FT that the effs of non-std conditions on
TOF does not correspond to their eff on range. 8 inch How, chg 3, A of D –
442 mils?
Answer No 12.
Range
TOF
a.
In-vacuo (using formula)
7233 Ms
26.2”
b.
In air (from FT)
6400 Ms
25.3”
c.
Diff
833 Ms
1.9”
d.
Dec %
11.5 %
3.4 %
Hence proved that range and TOF are eff diff by air.
Question No 13.
std conditions?
What corrns to FS would be req for inc in various non-
Answer No 13.
S/No
Non-Std Conditions
Corrn
a.
MV
Lengthen
b.
Chg temp
Lengthen
c.
Head Wind
Shorten
d.
Air density
Shorten (lengthen for
powder
burning
fuzes)
e.
Blsts air temp
Lengthen / shorten
f.
Proj’s wt
Lengthen
g.
A/S
Lengthen (negligible
for powder burning
fuzes)
Question No 14. Why a dec in air density despite an inc in range may
req a shorten corrn for powder burning fuzes?
Answer No 14.
A dec in air density would reduce the pressure on the
fuze and thus inc the burning time thereby causing delay in functioning.
This delay nec shortening of the fuze.
Question No 15.
Why no corrn is req for a change in A/S despite a
change in range in case of powder burning fuzes?
Answer No 15.
It is because an inc in A/S is ctr by the eff of dec in
density on rate of powder burning, hence the corrn req is negligible.
Question No 16.
An inc in proj wt incs the range at longer ranges and
reqs a negative corrn, yet we find that the fuze corrn is shorten at all
ranges?
Answer No 16.
It is because with an inc in proj wt the proj achieves
lesser range at shorter ranges and a shorten corrn for fuze is req. However
for longer ranges the proj achieves more range yet a shorten corrn for fuze
because the proj is able to retain its velocity and achieves more range in
less time. Hence a shorten corrn is req.
SEC – 23
Question No 1. Prove that at a given range, the corrn to ctr act a variation
is not numerically equal to that variation with its sign reversed?
Answer No 1.
In the fig below, consider what happens when all
conditions are std except that there is a head wind.
a.
If it is req to hit a tgt T1, at range R1, under std conditions the A
of P is A1 and the shell will fol traj 1, hitting T1.
b.
Because of a head wind, if the shell was fired at A of P A1, it
would adopt traj 1a and hti T2 at the range R2. The diff E, b/w
R1 and R2 is the eff at range R1.
c.
In order to hit T1 in the presence of a head wind, an
A of P
A2, greater than A1 must be applied. Under std conditions this
A of P would lead to traj 2 and achieve T3 at range R3. The diff
C b/w R1 and R3 is the corrn at range R1.
d.
The corrn C is opposite in sign but not numerically equal to the
eff E.
2a
1
2
1a
A1
G
A2
R2
T2 E
R1
T1
T3
C
R3
SEC – 24
Question No 1. What is meant by dispersion?
Answer No 1. The scattering of rds on a gr tgt in terms of range and line
and on a vertical tgt in terms of ht and width is known as dispersion.
Question No 2. What do you understand by Gaussian distr?
Answer No 2.
If we take a variable found in nature, eg, ht of men, we
find that they are distr throughout the population in a manner similar to that
shown in fig below where ht is shown on the horizontal (X) and frequency
density the no of men of the same ht) on the vertical (Y) axis. The graph
No of men with mean ht
shows that there are many men of mean or near mean ht, with few giants
and dwarfs. Such a pattern is known as a normal or Gaussian distr, with
the curve of the symmetrical to the mean.
Frequency
(no of men
with the same
ht)
Dwarfs
Giants
Mean Ht
Attribute (Ht)
Question No 3.
What do you understand by probability?
Answer No 3.
It is the ratio of chances favouring an event with the total
no of chances for and against the event i.e the no of times an event can
occur divided by the total no of times it can or cannot occur. Probability
cannot be more than 100% and less than zero. 100% probability is denoted
by 1.
Question No 4.
What is PE?
Answer No 4.
The measure of dispersion is expressed in terms of PE
and is statistically defined as the error plus or minus which will not be
exceeded in 50 of the cases.
Question No 5.
range on a tgt?
What is the gen distr (in percentage) of rds in terms of
Answer No 5.
The gen distr is shown as under :-
2%
16% 25% MPI 25% 16%
7%
7%
2%
50%
2PE
Line of fire
82%
4PE
96%
6PE
Question No 6.
100%
8PE
Define the fol :-
a.
Range PE. It is an error in range that a wpn may be expected
to exceed as often as not.
b.
Df PE. It is an error in df that a wpn may be expected to
exceed as often as not.
c.
Vertical PE.
It is the vertical expression of the range PE at a
given A of F (Vertical PE = Range PE x Tan
A of F).
d.
Ht of Burst PE. It is the error in ht of burst that a wpn may be
expected to exceed as often as not. It reflects the combined eff
of the dispersion caused by variations in functioning of fuze and
factors causing dispersion in range.
e.
Circular PE (CPE).
It is the radial error that a wpn ma be
expected to exceed as often as not. This term is used in rkts,
high lvl bombing and guided msl fds.
Question No 7.
Prove from FT that the vertical PE is more than range
PE if the angle of fall is more 800 mils. 155 mm Gun, chg Super, range
20,000 Ms?
Answer No 7.
From Table G
Range PE = 50 Ms
A of F = 829 mils
Ver PE = 50 x Tan 829 mils = 52.9 Ms
Question No 8.
Prove that the vertical PE may exceed the range PE
even at A of D less than 800 mils. 105 mm How, chg 7, range 11,000 Ms?
Answer No 8.
From Table G
A of D
= 745.1 mils
Range PE = 21 Ms
A of F
= 901 mils
Vertical PE = Range PE x Tan A of F
= 21 x Tan (901 mils)
= 25.6399 Ms
Question No 9.
fuzes?
Why the ht of burst PE cannot be determined for VT
Answer No 9.
The ht of burst PE for VT fuzes cannot be determined
because it depends upon the A of F and type of terrain.
Question No 10.
Diagrammatically show the distr of CPE?
Answer No 10.
93.7%
99.8%
50%
43.7%
6.1%
Question No 11.
What are the uses of probability table?
Answer No 11.
Given the size of a PE, the probability table enables the
solution of fol problems :a.
How much area corresponds to a given size of PE.
b.
What is the size of any other percentage zone.
c.
What percentage rds on the avg will fall within a given dist on
any one side from the MPI.
d.
An extension of ‘c’. What percentage rds on the avg will fall
within a given dist symmetrically about the MPI.
e.
Given the size of range PE and df PE, what percentage of rds
on the avg will fall in a given area, a known dist from the MPI.
Question No 12.
Locate the area under the normal curve which
corresponds to 1.4 PE?
Answer No 12.
or 32.75%.
From the probability table the answer comes as 0.3275
Question No 13.
Locate the area under the normal curve which
corresponds to 1.44 PE?
Answer No 13.
or 33.43%.
Question No 14.
Ms?
From the probability table the answer comes as 0.3343
Find the size of 35% zone when 1 PE is equal to 28
Answer No 14.
Entering the probability table at the percentage of one
closest to 35%, we get the value 0.3506.
The value of T = 1.54
T x 28 = 43.12 Ms
Size of 35% zone would be equal to 43 Ms
Question No 15.
MPI of an arty piece has been accurately adjusted
upon a sel pt S. The range PE is 30 Ms. 18 rds are fired under adjusted
conditions. Find the percentage and no of rds that will fall beyond S at a
dist not to exceed 40 Ms?
Answer No 15. 40 Ms = 40/30 = 1.33 range PE
From the probability table for T 1.33 we get 0.3151
Percentage of rds falling beyond S at a dist not to exceed 40Ms =
0.3151 x 100 = 31.51%
No of rds = 0.3151 x 18 = 5 (fraction has been discarded)
Question No 16.
MPI of an arty piece has been accurately adjusted
upon a sel pt S. The range PE is 30 Ms. 18 rds are fired under adjusted
conditions. Find the percentage and no of rds that will fall on either side of
S at a dist not to exceed 40 Ms?
Answer No 16. 40 Ms = 40/30 = 1.33 range PE
From the probability table for T 1.33 we get 0.3151
Percentage of rds falling beyond S at a dist not to exceed 40Ms =
0.3151 x 100 = 31.51%
31.51% is for one side, for both sides = 31.51 x 2 = 63.02%
No of rds = 0.3151 x 18 x 2 = 11 (fraction has been discarded)
Question No 17.
Determine the probability of rd falling b/w 26 and 35
Ms over the adjusting pt if the range PE is 26 Ms?
Answer No 17. 26 Ms = 1 range PE
35 Ms = 35 / 26 = 1.346 = 1.35 PE
From the probability table for T 1.35 we get 0.3187
Probaibilty of rds falling over adjusting pt at a dist of 26 Ms or less i.e
T 1.00 = 0.2500
Probaibilty of rds falling b/w 35 and 26 Ms = 0.3187 – 0.2500
= 0.0687 = 6.87 %
Question No 18.
What percentage of rds will fall within a gun pit of size
15 Ms x 10 Ms if the MPI is adjusted in the centre and the dir of fire is
parallel to longer side of the gun pit (range PE = 10 Ms and df PE = 5 Ms)?
Answer No 18. T for range = 7.5 / 10 = 0.75 and T for df = 5 / 5 = 1
(taking dist on one side of the MPI)
Total probability for range = 0.1935 x 2 = 0.3870 (38.7 %)
Total probability for df = 0.2500 x 2 = 0.5000 (50.0 %)
38.7 % rds will fall within the gun pit for range and out of these 50 %
will fall within the gun pit for df
Total percentage of rds faling within the gun pit = 38.7 x 50/100 = 19
% (rounding off is done towards the lower fig)
Question No 19.
What percentage of rds will fall within a gun pit of size
15 Ms x 10 Ms if the MPI is adjusted in the at the bottom centre of the pit
and the dir of fire is parallel to longer side of the gun pit (range PE = 10 Ms
and df PE = 5 Ms)?
Answer No 19. T for range = 15 / 10 = 1.5 and T for df = 5 / 5 = 1
Total probability for range = 0.3442 (34.42 %)
Total probability for df = 0.5000 (50.0 %)
34.42 % rds will fall within the gun pit for range and out of these 50 %
will fall within the gun pit for df
Total percentage of rds faling within the gun pit = 34.42 x 50/100 = 17
% (rounding off is done towards the lower fig)
Question No 20.
What percentage of rds will fall in tgt area of size 30
Ms x 40 Ms if the MPI is adjusted as shown in fig (range PE = 15 Ms and df
PE = 5 Ms)?
30 Ms
Answer No 20.
30 Ms
40 Ms
25 Ms
5 Ms
10 Ms
T for range 10 Ms = 10/15 = 0.67 and T for range 30 Ms = 30/15 =
2.0
Probability for these values of T is 0.1743 and 0.4113 respectively
Total probability for range = 0.1743 + 0.4113 = 0.5856 (58.56%)
T for df 5 Ms = 5/5 = 1.0 and T for df 25 Ms = 25/5 = 5.0
Probability for these values of T is 0.2500 and 0.4996 respectively
Total probability for df = 0.2500 + 0.4996 = 0.7496 (74.96%)
Total percentage of rds faling within the tgt area = 58.56 x 74.96 / 100
= 43 % (rounding off is done towards the lower fig)
Question No 21.
Find the probability that a rd will fall short within a dist
of 1.503 PE from the adjusting pt?
Answer No 21. Probability corresponding to 1.50 = 0.3442
Probability corresponding to 1.51 = 0.3458
Probabiity for 1.503 = 0.3447
Question No 22.
Find out probability for greater accuracy that a rd will
fall plus of MPI within a dist of 2.973 PE?
Answer No 22. Probability corresponding to 2.970 = 0.4776
Probability corresponding to 2.980 = 0.4778
Probabiity for 2.973 = 0.47772
Question No 23.
% zone?
If 90 % zone of a gun is 90 Ms, how much will be 30
Answer No 23. Towards one side of MPI 45 % zone is 45 Ms
T = 2.45 = 45 / Range PE
Range PE = 18.37 Ms
For 15 % zone value of T is 0.57
T = 0.57 = zone / 18.37
Zone = 10.47 Ms = 15 % zone
30% zone = 20.94 Ms
Question No 24.
zone?
80 % zone of the gun is 80 Ms, how much will be 40%
Answer No 24. T = 1.9 + 1.9 = 3.8
1 PE = 80 / 3.8
so 40 % = 0.78 x 2 = 1.56 x 80 / 3.8 = 32.84 Ms
Question No 25.
MPI of 155 mm Gun at 4,000 Ms chg Normal is
accurately adjusted at the centre of a wall of 5 Ms x 5 Ms. If 30 rds are
fired, how many rds are expected to hit the wall?
Answer No 25. Rg PE = 19 Ms, A of F =76 Ms, df PE = 3 Ms and Ver
PE = 1.42 Ms
Range
T = 2.5 / 1.42 = 1.76
= 38.25 x 2 = 76.5 %
Df
T = 2.5 / 3 = 0.83 = 21.23% for both sides = 42.46 %
Rds falling on wall = 76.5 x 42.46 / 100 = 32.48 %
No of rds 30 x 32.48 / 100 = 9.74 = 9 rds
Question No 26.
What will be the hit probability if the MPI is placed at
the centre of 3 Ms x 3 Ms wall in aslt fire. 105 mm How, chg 6, range to tgt
2,000 Ms?
Answer No 26.
Range PE = 8 Ms, df PE = 1 Ms, A of F = 67 mils so
Ver PE = 0.53 Ms
Hit probability for ht = 47.18 x 2 (where T = 2.83) = 94.38 %
Hit probability for Df = 34.42 x 2 (where T = 1.5) = 68.84 %
Total Hit probability = 8.62 x 38.7 / 100
= 64.94 %
Question No 27.
After a precision shoot ,MPI was adjusted in the
centre of an RR bunker ‘A’ of 10 Ms x 10Ms, another RR bunker ’B’ of
same dimensions is at 50 Ms from first bunker exactly in the line of fire.
How many rds should be fired at adjusted MPI so that one rd falls in RR
bunker B. 105 mm How, chg 7, range 7,000Ms?
Answer No 27.
Df PE 4
Range
Range PE 14
Farther corner ‘B’
T = 65 / 14 = 4.64 = 49.92 %
Near corner of ‘B’
T= 55 / 14 = 3.93 = 49.6 %
Inside ‘B’ for range
49.92 – 49.6 = 0.32 %
Rt side Df
T= 5/4 = 1.25 = 30.03 %
On either side = 60.06
For bunker B
(60.06 x 0.32) / 100 = 0.19 %
For one rd
100 / 0.19 = 526.32
Or 527 rds are to be fired
Question No 28.
MPI is adjusted on a wall as shown in fig. What is
single rd hit probability in percentage. 8 inch How, chg 6, range 12,000 Ms?
20 Ms
Answer No 28.
20 Ms
15 Ms
25 Ms
5 Ms
5 Ms
Range PE = 22 Ms
Df PE = 3 Ms
Ver PE = 16.18 Ms
Range / Ht
T = 20 / 16.18 = 1.24 = 29.84 %
T = 5 / 16.18 = 0.31 = 8.3 %
Total = 38.14 %
Df
T = 5/3 = 1.67 = 3.7 %
T = 15/3 = 5 = 49.96 %
Rds falling on wall = 33.17 %
Single rd hit probability = 33.17 %
Question No 29.
How many rds should be fired to obtain one rd landing
into a gun pit of 12 Ms x 12 Ms, when MPI is adjusted in the centre of the
gun pit. 155 mm Gun, chg Super, range 15,000 Ms?
Answer No 29.
Range PE = 37Ms
Df PE = 8 Ms
Hit probability for range = 4.31 x 2 (where T = 0.16) = 8.62 % Hit
probability for df = 19.35 x 2 (where T = 0.75) =
38.7 %
Total Hit probability = 8.62 x 38.7 / 100 = 3.34 %
So no of rds = 100 / 3.34 = 29.97 = 30 rds
Question No 30.
What is the procedure for calc PE?
Answer No 30. Fol procedure is used :a.
Sample of 10 rds, precede by a warmer are fired under
specified conditions o ch and ele. The series is rpeted on diff
days in order to take acct of any occasion-to-occasion variation
in consistency.
b.
The firings are repeated on three diff eles and intermediate
results are derived by interpolation.
c.
Range and deviation from a fixed dir are recorded for each rd.
d.
A convenient datum is sel which lies wholly on one side of all
variations and data recorded at ‘c’ is converted into deviations
from this datum. Sqs of these deviations are also taken.
e.
An est of std deviation(s) is derived using the formula s = (1/(n
– 1) x (Xs – (1/n x X2)))1/2; where n is the no of rds, X is the
sum of all deviations and Xs is the sum of sqs of these
deviations.
f.
Results of several rds under a set condition are then combined
by taking the avg of all the s2 and taking the sq root of this avg.
This std deviation is denoted by S.
g.
PE is then cpt from formula, PE = 0.67449 S.
SEC – 25
Question No 1. What is consistency?
Answer No 1. It is the measure of the dispersion of a gp of rds about the
MP fired at a give ele about the same tgt fired under same conditions.
Question No 2. What is accy?
Answer No 2. It is the measure of the deviation of fire from the pt of aim,
expressed in terms o the dist b/w the pt of aim and the MPI.
Question No 3.
What is the basic diff b/w consistency and accy?
Answer No 3.
The basic diff is that in the case of consistency, we
consider the distr of the rds about MPI and in the case of accy, we consider
the distr of the MPI about the tgt (pt of aim).
Question No 4.
egs?
Explain the diff b/w consistency and accy with help of
Answer No 4.
The basic diff is that in the case of consistency, we
consider the distr of the rds about MPI and in the case of accy, we consider
the distr of the MPI about the tgt (pt of aim). The diff can be understood by
the fol egs :a.
If the MPI coincides with the tgt but indl rds are widely
dispersed from the MPI, we can say that the accy is perfect but
the consistency is poor.
b.
If the dispersion of a no of rds about a MPI is small but the MPI
is well away from the tgt we can say that consistency is good
but accy is poor.
c.
When all the rds fired hit the tgt, both accy and consistency are
perfect.
Question No 5. What factors eff the range consistency of a single gun?
Answer No 5.
a.
Fol factors eff the range consistency of a single gun:-
Bore Conditioning.
It will affect the regularity of MV which
will in turn affect the consistency. It incls fol :(1)
Wear.
Due to the wear, the ABP movs towards muzzle
of the gun becomes inconsistent.
(2)
Prop.
It may incl fol :-
(3)
b.
(a)
Moisture contents.
(b)
Accy in measuring chg temp.
(c)
Slight diff in size of props.
(d)
Slight diff in wt of props.
(e)
Slight diff in shape of props.
Ramming.
Non uniformity of ramming affects fol:-
(a)
ECC.
Variation in the ECC due to ramming will
affect the MV and will contribute to inconsistency.
(b)
Initial SSP. Variation in the posn of DB as a result
of ramming will correspondingly vary the resistance
offered by the proj and hence variation in initial
SSP.
Carriage Conditioning.
(1)
It incls fol :-
Wear.
A little play, because of the wear in the fol will
affect the consistency :(a)
Ele gears.
(b)
Sight setting gears.
(c)
Trunnion bgs.
(2)
Sights.
Setting of the sights can be made similar to
certain extent due its simplicity and there may remain
some diff b/w rd to rd.
(3)
Laying. Because sight scales must be simple and easy
to read, no two layers will make exactly the same lay,
which is a natural human limitation.
(4)
Jump.
It varies rd to rd and contributes to
inconsistency.
c.
Flight Conditions.
The variations in the shape, size and wt
of proj will change the Co and hence will cause inconsistency.
The rapid changes in atmospheric conditions i.e density, temp
and wind may eff each rd diff.
Question No 6. What factors eff the range consistency of more than one
gun?
Answer No 6.
When more than one gun is firing the dispersion of rds is
usually larger than when only one gun is firing, because the MPIs or MPBs
relating to the separate guns may not be in the same place. This dispersion
arises from the fol two causes :a.
There may be consistent diffs b/w the guns themselves.
b.
There may be large dist b/w the guns, so that met conditions
prevailing b/w one gun and the tgt may not be the same as
those b/w another gun and the tgt.
Question No 7. What factors eff the df consistency of a single gun?
Answer No 7.
Fol are the sources of inconsistency in dir:-
a.
Rd to rd variations in laying.
b.
Throw off.
c.
Departure of actual drift from FT drift.
d.
Rd to rd variations in crosswind component of wind.
e.
Play in traversing gears.
Question No 8.
Answer No 8.
a.
What factors eff the accy in range of a single gun?
Factors eff accy in range are as under:-
Errors of Initial Data.
(1)
Error in loc of gun.
(2)
Error in ht of gun.
It comprises of fol:-
b.
c.
(3)
Error in loc of tgt.
(4)
Error in ht of tgt.
(5)
Plotting errors.
Errors of Prediction. It comprises of fol:(1)
Measurement of wind data.
(2)
Measurement of air temp.
(3)
Measurement of chg temp.
(4)
Measurement of data from GFTs and GSTs.
Errors of Gun / Ammo.
It incls of fol:-
(1)
Droop. Varies from gun to gun and is dependent
the length and constr of the barrel.
(2)
Sights.
Due to firing or travelling there may be some
maladjustment in the sight on the gun. These errors are
reduced by sight testing.
(3)
Laying Errors. The incorrect laying will definitely result
in an inaccurate fire.
(4)
Cal. The adopted MV becomes stale on firing and only
frequent MV measurement can prevent cal error.
(5)
Prop. Lot to lot variations exist due to mfr processes but
these are kept to a min by proof firing. Mixed lots are
used for cal
in order to mean these variations but
most engagements are carried out with a single
lot.
(6)
FTs. These are compiled from range and accy firings
and represent the performance of the std gun under std
conditions, which will seldom be found in fd svc.
Question No 9.
How will the sights eff the accy & consistency?
on
Answer No 9.
All sighting systems have inherent tolerances due to the
mtg of the sight on the gun. These errors are reduced by sight testing.
Question No 10. What are the effs on consistency and accy of fol. Only
mention good, bad or nil for both. (e.g. accuracy bad, consistency good
etc)?
Answer No 10.
a.
b.
c.
d.
Guns Not Cal
(1)
Accy
–
Bad
(2)
Consistency
–
Nil
Inaccurate Measurement of Chg Temp
(1)
Accy
–
Bad
(2)
Consistency
–
Nil
Increased Droop
(1)
Accy
–
Bad
(2)
Consistency
–
Nil
Small Variation in Jump
(1)
Accy
–
Nil
(2)
Consistency
–
Bad
Question No 11.
fired on the tgt?
What will be the eff on accy if greater no of guns are
Answer No 11.
If greater no of guns are used there are more chances
that MPI will be on the tgt. It is so because the slight diff b/w the indl guns
will cancel the error of one another.
Question No 12.
How the prac value of PE is determined?
Answer No 12.
The dispersions obtained in experimental firings are
given in FTs for every 500 Ms of range. It is imp to note that in particular
they relate to a single and a new gun firing a single lot of prop. In
considering limitations for peacetime safety, allowance, in PE, is made to
cover the use of mixed lots and worn guns. FT dispersion data is therefore
multiplied by a factor of 1.33 for mixed lot and by a factor of 1.5 for guns
worn beyond sec qtr of life.
Question No 13.
mix lot?
Why prac value of PE is req for worn guns and firing
Answer No 13.
The PEs given in FTs have been compiled for std MVs
of new guns and single lot of ammo. When worn guns are firing or mix lot is
being fired, the PEs given in FTs does not hold good. Hence the prac value
of PE is req.
Question No 14.
Find out value of one prac range PE for 155 mm Gun,
chg Super, range 20,000 Ms, gun is in sec qtr of life firing single lot?
Answer No 14.
50 Ms.
Question No 15.
Find out value of one prac range PE for 155 mm Gun,
chg Super, range 9,000 Ms, gun is in third qtr of life firing single lot?
Answer No 15.
Question No 16.
depend upon?
37.5 Ms.
Upon what does the size of jump and throw off
Answer No 16.
muzzle brake.
Their size is dependent upon the type of carriage and
Question No 17.
Upon what does the size of droop depend upon?
Answer No 17.
Question No 18.
Answer No 18.
It depends upon the length and constr of barrel.
How the error of jump can be compensated?
It is compensated in cal.
Question No 19.
How the error in laying can be reduced?
Answer No 19. It can be reduced by laying on one side of the A/pt or
A/post which is normally the lt edge of the A/pt and A/post.
SEC – 26
Question No 1. What is the purpose of a wpn system?
Answer No 1. The purpose of a wpn system is to inflict damage of some
desired nature and lvl on a tgt or gp of tgts with roughly similar vuln chars.
Question No 2. The projs can be cl into what all types?
Answer No 2.
The proj can be cl into fol two types :-
a.
KE Projs.
Projs that do not carry a warhead but use KE to
attk a tgt. These are normally used by tks.
b.
Chemical Energy Projs. Projs that are fitted with a warhead
and a fuze of one type or another to activate it.
Question No 3.
APFSDS rds?
What do you understand by a ‘sabot’ and APDS /
Answer No 3.
A sub-cal rd must be adequately sp in a full cal sheath
during its motion in a gun barrel, with the facility of discarding this parasitic
sheath soon after it leaves the muzzle. Such sheaths are called sabots and
projs fitted with them are called APDS and APFSDS rds.
Question No 4.
Answer No 4.
Upon what does the eff of KE rd depend upon?
The eff of KE depend upon fol :-
a.
Proj’s mass.
b.
Proj’s striking velocity.
c.
Proj’s diameter.
d.
The angle of incidence.
Question No 5. How a large angle of incidence will eff the pen of a proj?
Answer No 5.
It may eff the pen as fol :-
a.
The thickness of armr plate in an oblique plane is more. Hence
the thickness req to be pierced would inc.
b.
The rd may ricochet off.
c.
The rd may break as a result of uneven stresses.
Question No 6.
projs?
Answer No 6.
What are main warhead types used in chemical energy
Fol main warhead types used in chemical energy projs :-
a.
Fragmentation warheads.
b.
Shaped (or hollow) chg warheads.
c.
HESH.
d.
Payload carriers.
Question No 7. How fragmentation projs cause damage onto the tgt?
Answer No 7.
These consist of an expl chg surrounded by a solid metal
casing. On detonation the solid casing fragments into pieces o a wide
range o shapes and sizes.
Question No 8. How the force of an expl can be conc on a small area of
a tgt material in case of shaped chg projs?
Answer No 8.
It can be conc on a small area of a tgt material if a hollow
is first made in the contact face of the expl.
Question No 9. What is the eff of the hollow which is first made in the
contact face of the expl in order to conc the force on a small area of tgt
material?
Answer No 9.
The eff of the hollow is to produce a crater that is
narrower and deeper than that resulting from an unmodified block of expl.
Question No 10. Upon what does the depth of pen of jet depend in case
of shaped chg projs?
Answer No 10.
It depends upon the length of jet and ideally detonation
is arranged to occur at a dist b/w the tgt and the base of cone (stand-off)
dist. The pen is indep of range and striking velocity.
Question No 11.
chg proj?
Answer No 11.
proj :-
What factors eff the engagement of tgt with a shaped
Fol factors eff the engagement of tgt with a shaped chg
a.
Obsn.
Eff range is ltd by obsn as the engagement would
invariably be by dir laying.
b.
Angle of Incidence.
The greater the angle of incidence the
more will be the thickness of the tgt req to be pen.
c.
Spin.
The spin of the proj has a serious degrading eff on the
jet, which is dispersed by centrifugal force.
Question No 12.
HESH rd?
What could be the size of the scab produced by a
Answer No 12.
The size of the scab produced by a HESH rd can be
upto 1.25 – 1.5 times its diameter.
Question No 13.
What do you understand by payload carrier shells?
Answer No 13.
These incl all proj that carry some contents which are
ejected at the tgt and thereafter the efs are achieved by these contents for
eg all types of smk shells, ill, incendiary flare, propaganda shells, ICM /
DPICM, ADAM, RAAM etc.
Question No 14.
What methods of del of contents on the tgt are used
by the payload carriers?
Answer No 14. Fol methods of del of contents on the tgt are used by the
payload carriers :-
a.
Ejection on a pt on traj short of the tgt.
b.
Ejection on impact.
Question No 15.
How ejection is achieved on a pt on traj short of the tgt
in case of payload carriers?
Answer No 15. It is achieved by using a time fuze, which functions at a
prescribed time and causes the contents to be ejected through the base or
nose of shell.
Question No 16.
What is the remaining velocity of the contents relative
to the remaining velocity of the shell in case of a payload carrier after the
contents are ejected at a pt short of the tgt?
Answer No 16. The remaining velocity of the contents is less than the
velocity of the shell in case base ejection and more in case of nose
ejection.
Question No 17.
Why the contents of a payload carrier after their
ejection at a pt short of the tgt are unable to fol the original traj path and fall
short of the tgt?
Answer No 17. It is because these contents lose their velocity quickly
after ejection as compared to the loss in velocity of the original shell (had
there been no ejection).
Question No 18.
Upon what does the pattern and extent of dispersion
of the contents of a payload carrier ejected short of the tgt depend upon?
Answer No 18. It depends upon the fol :a.
A of F.
b.
Remaining velocity.
c.
Ht of burst.
Question No 19.
blsts pt of view?
Why BE projs are fired at the lowest possible chg from
Answer No 19. It is req to inc the A of F and reduce the remaining
velocity.
Question No 20. Which type of carrier shells are fired with percussion
fuzes?
Answer No 20. Percussion fuzes are used in case of those payload
carriers whose accurate del with less dispersion is of value eg smk WP and
smk FS.
SEC – 27
Question No 1.
Define the fol :-
a.
Fragmentation.
When the filling of an HE shell detonates,
the steel body of the shell is rapidly broken into large no of
small fragments which are thrown outward at a very high
velocity aprox 900-1000 m/s.
b.
Harassed.
An en that suffers cas rg from 0 – 20% is
considered to be harassed.
c.
Suppressed.
An en that suffers cas rg from 20 – 50% is
considered to be suppressed.
d.
Destroyed. An en that suffers cas over 50% is considered to
be destroyed meaning thereby that the en unit(s) has lost its
ability to op efficiently. This can vary depending on trg and psy
state of tps.
Question No 2. What is approx velocity with which the fragments of HE
are thrown outward after detonation?
Answer No 2.
Question No 3.
med HE shells?
The velocity is approx equal to 900 – 1000 m/s.
What is the primary and secondary role of small and
Answer No 3.
The primary and secondary roles of small and med HE
shells are as fol :a.
Primary role is to inflict cas and to cause damage to eqpt by
means of these fast moving very hot fragments.
b.
Secondary role is to damage bldgs by blast eff.
Question No 4.
What are the three different types of fragments
produced by three portions of the shell?
Answer No 4.
These three portions of the shell produce three diff types
of fragments as under :a.
Base Spray.
The base of the shell breaks into few large
fragments. Large fragments comprising the whole of the body,
in the rear of DB, are often found short of the tgt (only in air
burst).
b.
Side Spray.
This portion of the shell produces large
fragments of small size. The dir of motion of spray is
perpendicular to the dir of motion of the shell.
c.
Nose Spray.
Nose portion of the shell produces a few large
fragments moving in the dir of flt of the proj.
Question No 5. Upon what does the lethality of HE shell depend?
Answer No 5.
It depends upon the fol :-
a.
Size of fragments.
b.
Speed of fragments.
c.
Pattern of fragments.
Question No 6.
determined?
Answer No 6.
How the kill probability of a certain hit on the tgt is
It depends upon the calc of fol :-
a.
Size of fragments.
b.
Speed of fragments.
c.
Pattern of fragments.
d.
Accy.
e.
PE.
Question No 7. What can be the wt and size of fragments produced as a
result of good detonation?
Answer No 7. The wt and size of the fragments may vary 2 – 4 gms and
1 - 3 cms respectively.
Question No 8.
Answer No 8.
Upon what does the size of the fragments depend?
The size of the fragments depends upon fol :-
a.
Velocity of Detonation. If the velocity of detonation is more,
the size of the fragments will be small. It means that the more
violent expl will produce smaller fragments on the tgt.
b.
Charge Weight Ratio
c.
(1)
A shell with thinner metallic casing and correspondingly
greater HE contents will produce smaller fragments.
(2)
Better fragmentation will result out of higher chg ratio.
(3)
The ideal CWR is 25 % of the HE filling in the shell but no
HE shell presently in svc fulfils that req.
(4)
Mor bombs are more lethal than shells for this reason
because it can afford to have thinner walls due to its
exposure to
lesser stresses.
Composition of the Metal of the Shell
(1)
Size of fragments depends upon the grade of steel.
(2)
For efficient fragmentation the metal must be brittle.
(3)
Str of the metal depends upon the magnitude of the firing
stresses.
(4)
Mor bombs are normally made of low grade steel as they
are not subjected to high pressure in the barrel that is
another reason that the mor bombs have better
fragmentation.
Question No 9.
Answer No 9.
Upon what does the velocity of fragments depend?
The velocity of fragments depends upon fol :-
a.
Velocity of detonation.
b.
Remaining velocity of the proj.
c.
CWR.
d.
Air resistance.
Question No 10.
fragments?
How the velocity of detonation will eff the velocity of
Answer No 10.
If the velocity of the detonation is more the resultant
velocity of the fragments will be more and vice versa. But when the velocity
of the detonation is more the fragments will mov away from the path / traj
fol by the proj.
Question No 11.
fragments?
How the remaining velocity of proj will eff the velocity of
Answer No 11. If the remaining velocity of the proj is more, more will be
the velocity of the fragments and vice versa.
Question No 12.
fragments?
How the chg-wt ratio (CWR) will eff the velocity of
Answer No 12.
Projs having more CWR, will have faster fragments
compared to those projs whose CWR is lesser.
Question No 13.
How the air resistance will eff the velocity of fragments?
Answer No 13. If the air resistance is more the velocity of the fragments
will be less because the fragments will loose some of the energy in
overcoming the air resistance and if air resistance is less the velocity of the
fragments will be more.
Question No 14.
Answer No 14.
Upon what does the pattern of fragments depend?
It depends upon the fol :-
a.
Angle of impact.
b.
Velocity of detonation and remaining velocity.
c.
Ht of burst.
Question No 15.
How the angle of impact effs the pattern of fragments?
Answer No 15.
When the angle of impact is closer to 900, greater
portion of the side spray and nose spray is del on the tgt. As the angle
decs, inc proportion of side spray gets wasted.
Question No 16.
same wt?
Why a mor bomb will be more lethal than a shell of
Answer No 16.
It is because the mors have a higher angle of impact
which delivers more eff fragments on the tgt than lower angle of impact.
Question No 17.
given range?
Why lower chgs are more eff than the higher chgs at a
Answer No 17.
It is because the lower chgs for a given range have a
higher angle of impact which delivers more eff fragments on the tgt than
lower angle of impact.
Question No 18.
Diagrammatically show as to how the angle of impact
effs the spread of fragments and lethal zone?
Answer No 18.
Lethal Zone – Steep Angle of Fall
Lethal Zone At Small Angle Of Fall
Lethal Zone
Question No 19.
How the velocity of detonation and remaining velocity
eff the pattern of fragments?
Answer No 19.
The dir and velocity of fragments is the resultant of the
velocity of the detonation and remaining velocity of the proj. Greater the
remaining velocity smaller is the spread of the fragments.
Question No 20.
guns?
Why mors and hows have more lethal area than
Answer No 20. Since the mors and hows have lesser remaining velocity,
hence they have more lethal area than the guns.
Question No 21.
Why mors have more lethal zone than guns / hows?
Answer No 21. It is because of fol reasons :a.
CWR is more.
b.
Low grade steel is used.
c.
Low remaining velocity.
d.
Steep A of F.
Question No 22.
chgs?
Why lower chgs have more lethal area than higher
Answer No 22. Since the lower chgs have lesser remaining velocity,
hence they are more lethal than the higher chgs.
Question No 23.
Diagrammatically show as to how the remaining
velocity effs the spread of fragments and lethal zone?
Answer No 23.
Lethal Zone – High Remaining Velocity
Velocity of Detonation
Dir of Fragments
Remaining
Velocity
Lethal Zone – Low Remaining Velocity
Velocity of Detonation
Dir of Fragments
Question No 24.
How the ht of burst effs the pattern of fragments?
Answer No 24. Their are four types of hts of bursts being used in arty
projs. Their effs are enumerated below :(1)
Gr Burst. Nearly half of the fragments get buried into the gr.
To obtain a useful fragment eff the shell must burst as the fuze
touches the gr.
Effs of Gr Burst
Traj
(2)
Airburst. Useful fragments have a downward velocity in the
airburst and hence are eff against tgts behind lateral cover (e.g
folds in the gr and slit trenches).
Eff of Air Burst
Base Spray
Nose Spray
Side Spray
(3)
Ricochet Airburst.
This type of airburst can be achieved by
certain percussion fuzes at low A of Fs. It has almost same eff
as an airburst. Due to uncertainly of ricochet action in extent
and dir, this type of bust is considered less eff. Its use in
unobserved fires and against entrenched troops is, therefore,
not well advised.
(4)
Mine Action.
Percussion fuzes with delay setting, when
fired at large A of Fs, result in shell being buried before the bust
occurs. Fragments in this case get buried into the gr and there
are no lethal fragments. However, there is enhanced blast eff or
mine action.
Question No 25.
fragments eff?
Why soft gr (such as sand or mud) reduce the
Answer No 25. Soft gr will reduce fragments eff, since the shell will bury
slightly before detonating.
Question No 26.
What is the eff of high and low air bursts on the
lethality of fragments?
Answer No 26.
With higher bursts the lethal zone of the shell is inc but
lethality of fragments is dec due to loss in velocity of fragments as they
travel towards the tgt. With low bursts the lethal zone is reduced but the hit
probability is inc.
Question No 27.
Why ricochet airburst is considered less eff than
airburst and is not advised to be used in unobserved fire / against
entrenched tps?
Answer No 27.
It is due to the uncertainty of ricochet action in extent
and dir, this type of bust is considered less eff. Its use in unobserved fires
and against entrenched troops is, therefore, not well advised.
Question No 28.
Mine action is used against what types of tgts?
Answer No 28.
This type of burst is useful against fd defs and BUAs.
Blast eff primarily depends on charge weight ratio.
Question No 29.
Upon what does the blast eff primarily depend?
Answer No 29.
Blast eff primarily depends on CWR. A thin walled shell
such as ac bomb produces tremendous blast and cratering eff.
Question No 30.
blast eff?
Why most lt and med shells are not eff in producing
Answer No 30.
Most lt and med shells are designed to give good
fragment eff and greater proportion of their chemical energy is converted to
the KE of fragments, hence these are not very eff in producing blast eff.
Question No 31.
155 mm How M 198 firing DPICM at chg 7 at eles 300
mils and 500 mils, on which ele the spread of grens will be more and why?
Answer No 31.
Spread will be more at ele 500 mils as there would be
less remaining velocity and grens will have comparatively less fwd
component of velocity and spin will cause more spread.