THE ROYAL ARTILLERY INSTITUTION. 
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•through the derrick, main beam, and rear guy. If they are situated in a plane 
perpendicular to the plane of the paper, there will, of course, be no resolved 
parts in that plane. 
Prob. —To determine the drains upon the various parts of a derrick with 
main beam . (Fig. 14). 
Let W = weight of gun and lifting tackle, 
a — effective length of main beam, 
l— i, n derrick, 
t = tension of head tackle BC, 
T — tension of rear guy BH (including resolved portions, if any, of 
. side guys), 
Z 0 = inclination of derrick to the horizon, 
£ (f> = a rear guy u 
iY = normal pressure along main beam AB, 
R = resultant pressure along derrick AC. 
Now, if we suppose the directing guys upon the head of the derrick merely 
to steady it laterally, the point C may be considered in equilibrium through 
the action of the forces W, R, and t; and since the lines of application of 
these forces are parallel to the sides BA, AC, CB, of the triangle BAC, those 
sides may be taken as respectively proportional to the forces. Again—since 
the point B may be taken as kept at rest by the three forces T, t, and N —if 
through A we draw AK parallel to BC, it is evident that the three forces are 
proportional to the three sides BK, KA, AB, of the triangle ABK, taken in 
order. 
I. To find the above forces by construction. 
We can easily obtain a diagram of the strains upon the derrick. (See 
Pig. 15). 
Draw a line vertical and therefore parallel to AB, and set off from any 
point a in this line ab representing W on any convenient scale; through a 
and b draw lines parallel to AC and BC respectively; through b draw bh 
parallel to the direction of the rear guy BH, and draw ch vertical; then, 
upon the same scale, be and ac will represent t and R, while bh and he give 
us T and N. 
We observe from the diagram that the force R is invariable for the same 
value of W, that the tension t diminishes as the inclination (0) of the derrick 
increases; the tension T also varies inversely as 6, but directly as the inclina¬ 
tion ((f) of the back guy; N decreases as 0 is increased, but increases with (f>. 
II. Analytically. (Pig. 15). 
Prom the triangle abc, 
bc=- fa 2 + ft — %al sin 9, 
R _ ac ___ l 
db~ a 9 
a 
be W , _ 
t «= W. —■ — — \/ a? P — 2al sin 0. 
ab a 
( 1 ) 
(2) 
