WEIGHT OF MATERIAL IN THE CONSTRUCTION OF IRON-CLAD SHIPS. 479 
Now, 
Displacement in tons= Hg of mem level 5ec ' * drau g ht = t 8 ’ 56 * 1 * 24 ^= 13 -W. 
35 35 
And displacement is also equal to weight of hull -(-weight of armour and backing 
-(-weight of engines and coals-]- 1000 tons. Whence we have the equation 
13-26 2 =188-756+15L696-|-63-066-|-1000 ; (A) 
.-. 13-26 2 — 403-56=1000, 
and 
6=32-87. 
From this value of 6 we get the following dimensions and particulars of the new ship 
of the ‘ Minotaur ’ type with thick armour, viz. — 
Length between perpendiculars =14-96 =489*7 feet. 
489'7 
Breadth extreme = ~4qq~X59-^= 72 - 5 „ 
Displacement 
= 13-26 2 
Indicated horse-power 
=220-76 
Nominal horse-power 
220-7 7 
= —6 
Weight of hull 
= 188-756 
Weight of armour and backing 
= 151-696 
Weight of engines and coals 
= 63-066 
=14253 tons. 
= 7254 H.P. 
=10361 „ 
= 6194 tons. 
=4986 „ 
= 2073 „ 
Taking now the new ship of the ‘ Bellerophon ’ type, and calling the half breadth of 
her mean level section 6', we shall have, by a similar process to the preceding, 
Length along the side of the) 
— 305*5 y 
=11-756'. 
mean level section j 
26 w 
Area of mean level section 
=1> 
= 15-46' 2 . 
Length of armoured side 
320 
——6' 
26 0 
= 12-36'. 
Length between perpendiculars 
300 7 , 
= 26 6 
=11-546'. 
Area (approximate) of im-) 
-My 
= 43-36'. 
mersed mid. sec. / 
“26 
Area of armoured sides =2 X 12-36' X 20-3=499-386'. 
Area of surface for hull =2{1L756' X 221+12-36' X 14-3} = 889-546'. 
Weight of armour and backing =499-386' X ‘22 ton= 109-866'. 
Weight of hull =889-546' x *152 ton=135-216'. 
MDCCCLXVIII. 3 X 
