SHIP-BUILDING. 



operation for the Plane of the Keel, ^c. 

 To find the centre of gravity for the plane of the 

 keel, &c. 



The length on the upper fide or plane of the \ 



keel, from the aft-fide of the rudder, is - J 



Multiplied by its thicknefs ... 



'] 



Area of the plane 



Diftance of its centre of gravity from the' 



aft-fide of the rudder, being equal to half 



its length 



Now 265 feet 9 inches, multiplied by 88 feet 7 inches, 

 is equal to the momentum 23540 feet if inch. 



The centres of gravity of the fix planes having been 

 found, the'dillance of the centre of gravity of the whole 

 bottom of the ihip, from the aft-fide of the rudder, is ob- 

 tained as follows : 



From the principles already explained, the diftance of 

 the centre of gravity of the bottom, from the aft-fide of the 

 rudder, is equal to tlie fum of the momenta of an infinite 

 number of horizontal planes, divided by the fum of thefe 

 planes ; or, which is the fame, by the folidity of the bot- 

 tom. As, however, we have no more than fix planes, we 

 muft conceive their momenta as the ordinates of a curve, 

 whofe diftanccs may be the fame as that of the horizontal 

 planes. 



Now the fum of thefe ordinates, or planes, except the 

 firft and laft, of which take but half, being multiplied by 

 their diftance, gives the furfacc of the curve ; of which any 

 ordinate whatever reprefents the momentum of the hori- 

 zontal plane at the fame height as thefe ordinates ; and the 

 whole iurface will reprefent the fum of the momenta of all 

 the horizontal planes. 



Now 2553537 feet 45 inches, divided by 261 14 feet 

 5 of an inch, gives 97 feet 9^ inches, the diftance of the 

 centre of gravity of the bottom of the ftiip from the aft-fide 

 of the rudder. 



The height of the centre of gravity of the bottom may 

 be thus found. 



To half of the plane of the keel and half of the upper 

 horizontal plane, add all the intermediate planes, and mul- 

 tiply them progreflively as before, taking the upper fide of 

 the keel for the axis of the momenta ; then that fum being 

 multiplied by the diftance between the planes, and divided 

 by the fum of the planes, taking half of the firft and laft, 

 gives the height of the centre of gravity of the bottom 

 above the keel. 



Area of the Planes. Diftant 



Ft. In. 



1 Keel. 



Produfls. 

 Ft. In. 



Now 80068 feet 5 1 inches, divided by 261 14 feet 

 J of an inch, gives 3 feet |- of an inch ; which, multiplied by 

 3.6 feet, the diftance between the horizontal feitions, gives 

 II feet 5 of an inch, the height of the centre of gravity 

 of the bottom of the ihip above the under fide of the keel. 



The height of the centre of gravity of the bottom of the 

 (hip, and its diftance from the aft-fide of the rudder, being 

 found, the fhip being fuppofed in an upright pofition, the 

 centre of gravity will ueceftarily be in the perpendicular longi- 

 tudinal feftion, fuppofed to divide the (hip in two equal and 

 fimilar parts at 97 feet 9 J inches, the diftance of the centre 

 of gravity of the bottom of the (hip before the aft-fide of the 

 rudder, which comes between the frame 6 and 2 in the after- 

 body. It may now be afcertained whether the ftiip will be 

 in her natural pofition when floating at the upper hori- 

 zontal line, or conftrufted to fail on an even keel. Thus, 

 feparate the difplacemcnt of that part of the bottom before 

 the centre of gravity or fupport, and fee how it agrees with 

 that part of the bottom abaft it, as we may then exambc 

 the difference, if any, as in the following examples. 



3 X 2 



Find 



