530 



CARPENTRY. 



CXXIII. 



&- 3 - 



Fig. 5. 



Construe- the form of the envelopes will be obtained, as in Pro- 



tive blem XX. page 521. 

 Carpentry. j n the3e j. wo i agt p ro blems, the reader will have ob- 



served the inconvenience of obtaining the centres of 

 the boards near the base, and the absolute impossi- 

 bility of obtaining that at the bottom. The methods 

 of finding the forms of the boards where the centres 

 are thus inaccessible, will be shown by the following 

 problem, to which the three following lemmas are a 

 preparation. 



Pr.ATE Lemma I. If, in a circle, there be taken any two 



equal chords AB, BC, and AE and BF any two pa- 

 rallels ; and if the angle CBF be bisected by BD, 

 meeting the circumference in D, and AD drawn, then 

 AD will also bisect the angle EAB. Fig. 3. 



Lemma 2. If, in a circle, there be taken any se- 

 ries of equal chords AB, BC, CD, DE, and if the 

 angle ABF be bisected by BK, meeting the circum- 

 ference in K, and KC, KD, KE be joined, and CG, 

 DH, El be drawn parallel to BF; then KC, KD, 

 KE will bisect the angles BCG, CDH, DEI. 



Lemma 3. If there be any two inclined lines AB 

 and CD and if AE be drawn parallel to CD, form- 

 ing an obtuse angle BAE, and if the angle BAE be 

 bisected, the bisecting line AC will make equal an- 

 gles BAG and DCA, with each inclined line AB 

 and CD. 



PROB. X. To find the form of the boards at the 

 bottom of a dome, considering the surface as the sur- 

 faces of as many cones as are rows of boards, when 

 the centres of the circular edges of the boards are in- 

 accessible. 



Let ABC be an axal section of the dome, and sup- 

 pose f g to be the place of the last board on this sec- 

 tion ; draw the representation of the axis DB ; pro- 

 duce DB and gy to meet in E ; from the centre E, 

 with the distance Ey describe the arcfl, the con- 

 cave edge of the board cutting DB at Z; joinyZ, 

 and produce fl, cutting the circumference ABC at 

 m ; let g k, h i, i k be the places of the following 

 boards nearer to the bottom ; draw gm t km, im, 

 km, cutting DB at t, u, v, tv ; draw g p, h q, ir, 

 k s parallel to AC the base, cutting DB again in the 

 points />, q, r, s : then will pg be half the length, and 

 pt the height of the first board, the centre of which 

 is inaccessible ; also h q half the length, and qu the 

 height of the next board ; ir half the length, and rv 

 the height of the next ; likewise sk half the length, 

 and stv the height of the last. 



For drawy parallel to DB, and the angle nfl 

 will be equal -to the alternate angle^YE ; but be- 

 cause E/ and Ey are equal, the angley/E is equal to 

 the angle IfE. ; therefore the angle // is equal to the 

 angle nfl, and consequently the angle /E is bisect- 

 ed byyz ; but, by Lemma 2, If there be taken any 

 number of equal chorda fg,g h, hi, ik,kC, and if 

 the angle n/E be bisected bjjm, meeting the cir- 

 cumference in m, and gm, hm, im, km be joined, and 

 gx, hy, iz, k, &c. drawn parallel to fn ; thengwz, km, 

 im, km, will bisect the angles^a:, ghy, hiz, ik, &c. 

 It may now be shown as follows, that if any angle 

 ghy be bisected, the angle ghm orghu will be equal 

 to the angle BwA ; therefore, if kg and u B were pro- 

 duced, their point of concourse would be the centre 

 -ef an arc that would pats through h and u. For, 



since hy is parallel to BD, the alternate angle yhu Construe* 

 is equal to the angle huB ; but the angle yhu is tive 

 equal to the angle ghu ; therefore the angle ghu it Carpentry, 

 equal to the angle huB, as was to be shown. > 



Fig. 1. represents a semicircular dome, 

 with the joints in vertical planes. Fi . 



Fig. 2. is a segment dome, boarded with the joints Fig. 2.' 

 in vertical planes. 



Fig. 3. is a bell roof, boarded with the joints in pio- <* 

 , t J & * 



vertical planes. 



Fig. 4. The manner of finding the form of the Fig- 4. 

 board or envelope, by considering the surface of the 

 dome, as consisting of as many cylindric parts as there 

 are boards. See Problem XVII. page 520. In 

 this, the axal section is considered as the section of 

 the cylinder. 



PUOB. XI. To find the envelope or form of the 

 board, having the axal section given, Fig. 5. Fig. 5. 



Let AB be the breadth of a board, or the breadth 

 of the envelope j bisect AB at C by the perpendicu- 

 lar DE ; make CE equal to the length of the arc 

 from the bottom of the dome to the summit, which 

 may be found by calculation ; divide the arc AD in- 

 to any number of equal parts, and draw the sines pa- 

 rallel to AB ; divide CE into the same number of 

 equal parts, and draw lines parallel to AB ; make 

 ordinatea on each side of CE, reversed to the order 

 of the sines, and the curve being drawn on both sides, 

 will form the board. 



Fig. 6. shows the manner of finding the envelope Fig. 6. 

 of Fig. 2, in the same manner as in Fig. 4. 



Fig. 7. shows the method of finding the envelope Fig. 7. 

 of Fig. 2, without the use of the axal section. In 

 this Figure, AB is the breadth of a board ; AEDFB 

 a circular zone, similar to the contour of Fig. 2 ; the 

 part EDF being flat, corresponding to ef Fig. 2; 

 the arc AE Fig. 7, being similar to ae Fig. 2 ; and 

 CG, Fig. 7, equal to ae Fig. 2. The operation is 

 the same as in Fig. 5, as will be obvious by inspection. 



Fig. 8. shews the manner of finding the covering of Fig. 8. 

 Fig. 3, in the same manner as in Figs. 4. and 6. 



Fig. 9. is an elliptical dome : No. 1. the plan; No. Fig. 9. 

 2. the elevation ; No. 3. the half section of the eleva- 

 tion ; No. 4. a section of the plan. 



Fig. 10. is a circular dome and elliptic elevation : pig. 10. 

 No. 1. the plan ; No. 2. the section. 



Fig. 11. shews the manner of covering the domes in pjg, u 4 

 Figs. 9 and 10, the same as in Figs. 4, 6, 8, by laying 

 down the rotative section, whether on the plan or on 

 the elevation. 



The principle of executing niches with spheric 

 heads, depends entirely upon the properties of the 

 sphere ; and therefore, since all the sections of the 

 sphere are circles, and those which pass through the 

 centre are equal, and those which are more remote 

 from the centre are less than those which are nearer 

 to it, if a hemisphere be cut at right angles to the 

 plane of its base, its section will be a semicircle of 

 greater or less diameter, as it is nearer or more re- 

 mote from the centre. 



First, let the section be made through the centre 

 at right angles to the plane of the base, then the 

 hemisphere will be divided into two equal por- 

 tions, each a quarter of the sphere. Each of these 

 quarters will have two planes at right angles, and 



