CARPENTRY. 



517 



LV. 



OMMtrae-' planes ; so that in a solid angle there arc six parti, 

 any three of which being given, the other three may 

 '; be found. Solid angles are either right angled or 

 ^ oblique: when they are right angled, two of the pluct 

 form a right angle with each other ; and this is the 

 only case necessary to be considered, as all the cases 

 of oblique solid angles can be solved by the help of 

 those which are right angled. The planes contain- 

 ing the right angle are called legs ; the plane sub- 

 tending the right angle, the hypothenuse ; and the 

 three inclinations, angles. 



PKOB. II. In a right angled solid angle are given 

 the two planes ABC and ABD, containing the right 

 angle to find the hypothenuse, and the angle con- 

 tained by the hypothenuse and the leg ABD. 



In AB take any point A ; draw AC at right 

 angles with AB, and ADE at right angles with 

 BD ; make BE equal to BC, and BE will be the 

 hypothenuse ; from AB cut off AF equal to AD, 

 and join FC, and AFC will be the angle required. 



For if the plane ABD be raised upon AB, at a 

 right angle with the plane ABC, and the plane ACF 

 turned up on AC until AF coincide with AD, and, 

 lastly, the plane DBE turned round the line DB 

 until BE fall upon FC ; then BE will fall upon BC, 

 FA and FC will be at right angles to BD, and con- 

 sequently AFC will be the inclination of the planes. 



PKOB. III. Given one of the legs, and the angle 

 opposite, to find the other leg. 



Let ABC be the given leg ; make ADC equal to 

 the given angle ; from the point C draw CA at 

 right angles with AB ; from A, with the radius AD, 

 describe the arc DE, and draw BE a tangent to the 

 arc at E ; then will ABE be the other leg required. 



By these problems, the various levels in roofing 

 are ascertained, as the backing of the hips, and the 

 side joints of purloins and jack rafters ; and in hand- 

 railing, the spring of the plank, and the intersection 

 of the plane of the plank with a horizontal plane. 



PHOB. IV. In a right angled solid angle, consist- 

 ing of three plane angles, are given one of the legs 

 and the adjacent angle to find the hypothenuse. 



Let ABC be the given leg; from any point A 

 draw AD- perpendicular to AB ; make the angle 

 DAE equal to the adjacent angle ; from A, with any 

 radius AD, describe an arc DE, cutting AE at E ; 

 draw DF and EC parallel to AB ; and draw CF pa- 

 rallel to AD, and join F,B, and FBA will be the hy- 

 pothenuse. 



PROB. V. In an oblique angled solid angle, con- 

 sisting of three plane angles, are given one of the 

 sides, and the two adjacent angles, to find the other 

 tide. 



Let ABC be the given side ; in AB take any 

 point A, and draw AD perpendicular to AB ; make 

 the angle DAE equal to one of the given angles ; 

 draw EK and DL parallel to AB, the former cut- 

 ting AD at M, any where in CB, or in CB produ- 

 ced take any point F, draw FG perpendicular to 

 CF, and make t the angle GFI equal to the other 

 given angle ; produce CF to H, and make FH equal 

 to ME ; draw HI parallel to FG, iK parallel to 

 BC, ai.d KL parallel to AD ; join BL, and ABL 

 is the side required. 



The applications of these problems arc numerous, 



in culling timk-i & to lit at any angle agiinst aootttrr ; C '*ittst9 

 likewise in oblique itone arches, which have cylin 

 dric or cylindroidic intradoses, in cutting the sides ^ 

 of the bevels of the (tone to the given anglrt. 



< rics of bevcJs may be found, which will 

 have one side common, while the other side of each 

 is only varied in the most easy manner. If the arch 

 stand in a vertical wall, the constant angle will be a 

 right angle ; but if otherwise, it will be acute or ob- 

 tuse : Thus, suppose AB, Fig. 8. to be the line of PLATK 

 an erect wall upon the ground, and CD the direction CXV. 

 of the arch ; or, k-t AB be the intersection of any ft (' * 

 plane, and CD the direction of a prismatic piece of 

 timber, consisting of several sides, to be cut so 2s to 

 fit against the plane : Suppose ECF, ECG, ECH, 

 &c. to be the angles which the several beds make 

 with each other, in a plane perpendicular to the sides 

 of the arch, or the angles which a series of planes 

 would make at the same intersection, parallel to the 

 sides of the same prism ; then CD/", CDg, CD/i, 

 &c. are respectively the angles which two adjoining 

 sides of the beds make with each other, or the angles 

 which the several arrises of the prism make with the 

 .sides of the end to be cut. In like manner, Fig. 9- Fig. <J. 

 shows a series of angles when the plane inclines or 

 reclines ; each is to be found in the same manner as 

 in Fig. 7. the angles DC/J DCg, DCA, &c. corre- 

 sponding with EDF, EDG, EDH, Sec. In Fig. 8. 

 it will be only necessary to find the angles on one 

 side ; but in Fig. 9. they must be found on both 

 sides. 



PROB. VI. The base and one of the planes of a 

 prism being given, to find the section of the prism 

 oblique to the base, but at right angles to the plane 

 of the side given, the line of inclination being g 

 in position upon the given side of the prism. 



Place the base contiguous to the given side of the 

 prism, so as to join to their common side, or line of 

 concourse, or line of junction ; take as many ordi- 

 nates in the base as may be thought sufficient, and 

 produce them to the line of inclination ; from the in- 

 tersected points in the line of inclination draw per- 

 pendiculars, which make equal to their correspond- 

 ing ordinates in the base ; and if the points in the 

 base where the lines proceed are the junctions of 

 straight lines, join every two adjacent points by 

 straight lines, and the section will be formed ; or, if 

 the points proceed from a curve, a curve must be 

 traced through the points found in the section. 



Examples. In Fig. 1. the base A BCD being a PLAT* 

 rectangle, the two sides AE and BF of the given CXVi. 

 plane in a straight line with AD and BC, two of F'g ' 

 the sides of the base answer the purpose of ordinates, 

 and therefore we have only to complete the section 

 to the length of the inclination EF, and to the 

 breadth BC or AD of the base. 



In Fig. 2. the base is bounded by several straight p;- 2 . 

 lines, and therefore every two contiguous points in 

 the section are also joined by straight lin-.s. 



In Fig. 3. the base is the arc of a circle, therefore pig. s. 

 the remaining boundary of the section must be a 

 curve passing through the several points: The sec- 

 tion here is the segment of an ellipse. 



PROB. VII. To find the section of a prism, the 

 base and one of the adjoining planes being given, also 



