CARPENTRY. 



Construc- 

 tive 

 Carpentry, 



CXVH. 

 Fig. 5. 



1%. 6. 



-Pig. 7. 



Fig. 8. 



PLATE 

 CXVIII. 



intersection EF of a cutting or sectional plane in 

 position to one another, likewise the altitude of the 

 vertex above its seat, and the inclination of the cut- 

 ting plane; to find the line of section on the triangu- 

 lar plane, provided that the angle of the cutting 

 plane is small enough to admit of an intersection with 

 the triangular plane. 



Draw DG perpendicular to EF, cutting EF at 

 G, and draw DH perpendicular to DG, make the 

 angle DGH equal to the inclination of the cutting 

 plane; produce GD to I -; make DI equal to the 

 perpendicular altitude of the triangle ; produce HD 

 to meet AC in K; join KIj draw DB perpendicular 

 to AC ; from K, with the distance KI, describe an 

 arc cutting DB at B, and join BA, BC, and BK; 

 make DL equal to DH ; draw LM parallel to DK, 

 cutting KI at M; from KB cut off KN, equal to KM, 

 and produce CA to E ; draw ENO, cutting BC at 

 O, then OP is the section required. 



Pr?oe. XIV. The seat A of the vertex and base 

 BCDE of a rectangular pyramid, and the intersection 

 FG, of a sectional plane being given in position, and 

 the height of the pyramid in quantity, to determine 

 the lines of section made on the sides ofthe pyramid 

 by the sectional plane. 



Proceed as in the last problem, and find the sec- 

 tion of each triangle, which will be the same as the 

 sections of the sides of the pyramid required. It 

 may here be observed, that one triangle HKI will 

 serve for the whole instead of Dl K, and the two sides 

 BE and DC, which do not meet HI, must be pro- 

 duced till they meet it, and all the intersections must 

 be drawn to K, as in Fig. 5. 



PROD. XV. Given the base ABCDEFA of an 

 erect prism, and the intersection HI of a sectional 

 plane, and the inclination of this plane to find the 

 sectional lines on the sides of the prism. 



Draw lines GH, GL, and GI, parallel to the 

 sides of the prism, till they cut the intersecting line ; 

 draw GK perpendicular to HI, meeting it at K, and 

 GP perpendicular to GK; make the angle GKP 

 equal to the inclination ; make GM, GN, GO, each 

 equal to GP; draw MH, NL, OI, and GMH, 

 GNL, GOI, are the angles which the sectional lines 

 make with the lines of concourse of the angles. 



From what has now been said, Fig. 8. will be 

 plain to inspection. It only differs from the last, in 

 the base being a square. 



PROB. XVI. Any three points being given on 

 the surface of a cylinder, to find an envelope that 

 would coincide with a section passing through these 

 three points, and over the surface of the segment of 

 the cylinder, between the two straight lines drawn 

 on the surface, through the two most remote points 

 parallel to the axis, the seat of the three points being 

 given on the base. 



This will admit of three different cases. Let 

 Fig. 1. be the cylinder, and let the points be B, D, 

 F; draw BA, DC, and FE parallel to the axis cut- 

 ting the base at A, C, E ; then A, C, and E are the 

 seats of the three points, and ACEFDBA is the 

 portion which requires the envelope. 



Now, suppose the cylinder to be cut by a plane 

 ABFE, which will be parallel to the axis, besides 

 the plane of the section BDFGHI 5 then the plane 

 2 



ABFE may cut the plane of the section through Construe- 

 the three points at right angles, or the plane A BFE tive 

 may form an acute angle with the plane BDFA, or ^ 

 any obtuse angle with the said part according to the "^^ 

 situation of the point D. 



In the general description,, it will only be neces- 

 sary to have the seats of the three points given on 

 the base, and the heights of the points on the cylin- 

 der from their seats. Let ABC, Fig. 2, 3, 4, be p LATB 

 the part of the base of the cylinder, and A, B. C, CXVIIF. 

 the seats of the three points. Join AC ; draw CE Fig. 2,3,4 

 and AD perpendicular to AC ; make AD equal to 

 the height of the point above A, CE equal to the 

 height of the point above C, and CF equal to the 

 height of the point above the seat B ; join ED ; 

 draw FG parallel to CA, and GH parallel to EC 

 or DA, cutting AC at H, and join HB; produce 

 AC to I ; divide the curve A BC into any number 

 of equal parts, and extend these parts upon CI ; 

 through the points of division in the curve ABC, 

 draw lines parallel to BH intersecting AC ; from 

 the points of intersection in AC, draw lines parallel 

 to GH to meet DE ; from the points of meeting ia 

 DE, draw lines parallel to AC ; and from the divi 

 sions in CI, draw lines parallel to CE, so as to inter- 

 sect the other parallels last drawn, the one conti- 

 nually nearer to CI and more remote from CE than 

 the last, and draw a curve EK through the several 

 points, and CIKEC will be the envelope required. 



Fig. 1. is the case where the rectangular plane 

 makes a right angle with the elliptic section ; Fig. 2. 

 that where the angle made by the rectangular plane 

 and the elliptic section is acute; and Fig. 3. the case 

 where the angle formed by these two planes is obtuse. 



In Fig. 3. and 4. DME is the orthographical pro- 

 jection of the curve, with which the curve line EK 

 of the envelope would coincide : This is found by 

 drawing parallels to GH through the points of di- 

 vision in the curve ABC, to meet the parallels of 

 AC, as shown by the dotted lines. 



Coverings of Solids. 



PROB. XVII. To cover any portion of the eur- Coverings 

 face of a cylinder, cut by any two oblique planes at ot 

 right angles to an axal plane, given the inclination 

 of the cutting plane to the axis ; so that the enve- 

 lope, or covering, will exactly cover the surface of 

 the cylinder terminated by the sections. 



Let that part of the axal plane intercepted by the 

 two oblique sections, and one or both sides of the 

 cylinder, be denominated the seat of the envelope; 

 and suppose the cylinder, or portion of the cylinder, 

 to be cut by a plane at right angles to the axis, the 

 intersection of this and the axal plane will either fall 

 within the seat of the envelope or its plane produced ; 

 again, let the arc of the section at right angles to the 

 axis, be called the regulating arc ; the section itself, 

 the regulating section ; the base of the said section 

 on the axal plane, the primary abscissa ; and the re- 

 gulating arc extended in a straight line, the secondary 

 abscissa ; and all lines at right angles to the primary 

 abscissa, terminated by the sides of the seat of the 

 envelope, ordinates. 



The method of finding the envelope will then be 



