261-

theory AND PRACTICE

TRACT 31.

1000, and the quotient '016 be sought in a table of naturaltangents, it will give 55', or almost 1°, for the angle the gunmust be elevated to. Or, if the log. of - 016 or j-^be foundin the log, tangents, it will give the same angle of elevation55'.

107. Or, to find the time by the formula in the 11th

problem, it will be 32-f X log of = 32! x log.

1131 - 231 891 , 900.891 „„ , 30891

891 - 231 ' 1131 “’ '■’’*** X °6’ 660.1131 — 32 Y X °^‘ *2.1131 —

AU £

32 i X lo S-^ = 32| X -031114

1*003426, or 1 second

nearly, the same as above found, by dividing the distanceby the medium velocity.

10s. Example. 2. To determine the elevation of a 12-pounder gun, charged with 4lb of powder, to hit an objectat 500 yards or 1500 feet distance.—Here, the weight ofpowder being y of that of the ball, by the table in prob. 13,the initial velocity is 1306 = v, the ball’s diameter r/=4*403,

the object’s distance d = 1500 = I338i/ X log. ~—= 5891

x lo s* St P rob - 11 ; hencc dh = IS? = * 25462 >

the number to which log. is 1*1973 = n, that is, n =s

—TV 37 i this gives v = -4- 231 ?=■ 329 = v the last ve-

*—231 & n 1

locity at the distance (1500) of the object. Then —~ =

2

1068 the middle velocity, and ~g- = 1*405" the time offlight by this method.

But, by the method in prob. 11,/= x log. J- ^. y

5891

X log.

1306-231

829 a -,

-=: 2 . 6 i

1306

, 1075.829

x loir.-= 25! x

231 " 829 - 231 'licjfi — 1 ~ ,v '8* 598.1306

•051319 = 1*4616" the time, more exact, but not much dif-ferent from the former.

Then, as l 2 : 1 *46 z : : 16 : 34*1 feet, the height to bepointed above the object. Or, — *- = *022737, the tangentcf 1° 18', the angle of elevation to hit the object,