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Again, in triangle API, sin A: sin P = PI: AI,
cos i sin (ei) = PI: †h';
and combining this analogy with [1] above,
cos2i: sin2 (e-i)=h:}h',

4 h sin2 (e—i)
cos2 i

4h sin2 (ei) sec2 i;


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2h' 8h

= sin2 (e-i) sec2 i, g g


t=2 sin (e-i) seci; and therefore LatLsin (e-i) + L sec i + 1 L2 h − (20+† L32.2)


and since h' =gt2 (190), t2 =



1. The obliquity of the plane is 6° 4′, the elevation 37° 38', and the impetus 4200; required the amplitude.

Âns. 7040.

2. The impetus being 4200 and the inclination of an ascending plane 6o 4', required the greatest range.

Ans. 7596.

The elevation for this range is known (211); hence, find as above the corresponding range.

3. Required the time of flight of a shell discharged with an impetus of 2304 feet at an elevation of 45°, the plane being ascending with an inclination of 8° 30'.

Ans. 14.4 seconds.

213. PROBLEM II.-Given the elevation, the range, and inclination, to find the impetus or velocity.


r cos2 i
cosesin (e—i

Or, L4h Lr+2 L cos i+L sece+L cosec (e-i)—40.


214. When the plane is descending, e+i must be used instead of e-i, that is, in both cases the elevation above the oblique plane is used.

EXAMPLE. The elevation of a piece of ordnance is 30° 10'; with what impetus must a ball be discharged, in order to strike an object situated on an oblique plane, its angle of

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acclivity being 12° 45', and the distance of the object 3256 feet?


Lr+2L cos i + L sece + L cosec (e-i)-40 =L3256+2 L cos 12° 45' + L sec 30° 10' + L cosec 17° 25′-40

=3·512684+ 19·978314 + 10·063201 +10·523867 — 40 =4.078066= L 11969, and h = 2992.

The formula is easily obtained from that of the preceding 4 h cos e sin (e—i) problem, for in it r ; hence, cos2 i

r cos2 i

4h =r cos2 i· sec e cosec (e—i), cos e⚫ sin (e—i) for the sine and cosine of an arc are the reciprocals of the cosecant and secant respectively (Pl. Geom. Trig. p. 223); and therefore logarithmically, L4h=Lr+2 L cos i + L sec L cosec (ei) — 40.

EXERCISE. Find the impetus with which a shell must be discharged, in order to strike an object on an ascending plane, the elevation being 58° 26', the inclination 6° 4′, and the distance of the object 7040 feet. Ans. 4198 feet.

215. PROBLEM III.-The inclination, the impetus or velocity, and the range being given, to find the elevation.

Sin (2 ei) sin i +


2 h

which will give 2 e―i; and hence 2 e or e will be known.

Or, if n = sin (2 e — i) — sin i,

r cos2 i


LnLr+2 L cos i-(20+ L2h),

and sin (2 e-i) = n + sin i.

When the plane is descending, use 2e+i instead of 2 e-i, and - sin i instead of + sin i.

EXAMPLE. The inclination of a descending plane is 9° 12', the distance of an object on it is 7650 feet, and the impetus with which a shell is discharged, in order to strike that object, is 3609 feet; required the elevation.

LnLr+2L cos i-(20+ L2h)
= L 7650+2 L cos 9° 12′-(20+ L 7218)
=3·883661+ 19.988754 23-858417
013998 L 1·0328,






and sin (2e+i)
= n sin i
=sin 60° 48', or 2 e + i — 60° 48'.
Hence, 2e60° 48′-i-60° 48′-9° 12′ 51° 36',
e= = 25° 48';
and for the higher elevation, e' = 90°—i—e=55°,
and for the greatest range, E45°-i 40° 24.


Draw CL (fig. p. 131) parallel to DP to meet the circle DIP in L; draw CI, CP, and also KI and sP perpendicular to CL.



or, Hence,


Since (p. 131) PDI-e-i, hence PCI 2e2 i, and LCP CPD i, for CPO, DPE, are right angles; therefore LCI = 2e −2i+i=2e· e- i, and


CI: IK1: sin ICK,

IK = CI⚫sin ICK = CI· sin (2e — i);
CP: Ps 1: sin PCs,




Ps CP sin PCs CI sin i.

Ms-Ps = IK-Ps=

CI{sin (2e-i)-sin i}.... [1]

and PA: PM=1: cos i; hence PMPA cos i = 1r cos i, also, CP: Cs 1: cos i, or CI: h = 1: cos i,




2 cos i;

and substituting in [1] these values of PM and CI,


r cos i =

r cos2 i


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2 cos

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{sin (2 e- -i) — sin i}

= : sin (2 ei) — sin i.


1. The impetus with which a ball is projected to strike an object 7040 feet distant on an ascending plane, whose inclination is 6° 4′, is 4200 feet; what is the elevation?

Ans. 37° 38', and 58° 26'.

2. If a ball is projected with a velocity of 439 feet, so as to strike an object at the distance of 6745 feet on a descending plane, whose inclination is 8° 15′; required the elevation.

Ans. 32° 46', and 48° 59′.


216. Although the parabolic theory of projectiles affords a tolerable approximation to fact in the case of smaller velocities, not exceeding 300 or 400 feet per second, at least for the larger kinds of shells, yet its results deviate so widely from truth for greater velocities, that ranges which, calculated by this theory, exceed 20 or 30 miles, are found in fact to be only 2 or 3 miles. The cause of so great a difference is, that when the velocity of a projectile exceeds 1200 or 1300 feet, there is a vacuum formed behind it, because air rushes into a vacuum with a velocity of only about 1300 feet in a second; and, therefore, there is not merely the ordinary resistance of the air retarding the motion in this case, but also the atmospheric pressure of the air on its anterior surface, with scarcely any pressure on its posterior surface to counteract it; and even with less velocities than this, the pressure of the rarefied air on the posterior surface is so small, that the unbalanced pressure on the anterior surface causes a great retardation, far exceeding that produced by the ordinary resistance, which is nearly proportional to the square of the velocity.

217. It has been found by experiment that the square of the initial velocity of a projectile varies as the charge of powder directly, and as the weight of the ball inversely. By experiments made by Dr Hutton and Sir Thomas Blomwhere the

2 c

field, it was found that v = 1600 ↓


initial velocity, c = the charge of powder, and the weight of the ball; but by more recent experiments performed by Dr Gregory, and a select committee of artillery officers, it has been found that the velocity is considerably greater on account of the improved manufacture of gun

powder, and that the formula v = 1600√

approximation to the initial velocity.

218. Experiments for determining the velocity of a projectile are performed by means of a ballistic pendulum, which was invented by Robins, and consists of a massy block of wood suspended by a horizontal axis, the ball being fired against which, causes it by its impact to oscillate; and the first or greatest deviation from its vertical position being measured, the velocity of the impinging ball can then be computed by dynamical principles.



Another method for determining the initial velocity was invented by Count Rumford; by this method the velocity is calculated from the extent of recoil caused by the discharge when the gun is suspended.


219. PROBLEM I.-Of the charge of powder, the weight of the projectile and the initial velocity, any two being given, to find the third.


v = the initial velocity,

c = the weight of the charge,
the weight of the ball;


3 c

v = 1600√




= 3 (1600

c =

3 c



b = 3c (1000) 3

affords a near

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220. Also, the velocities are proportional to the square roots of the charges directly, and of the weights of the projectiles inversely.


v ÷

(Algebra, art. 414),

√3c, when b is constant,



when c is constant.

That is, if v, c, b, are the velocity, charge, and weight of

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