272 



THE POPULAR EDUCATOR. 



and 4 feet above it. Anywhere upon the HL mark the PS (point 

 of sight). From PS as a centre, and with the distance of 6 feet 

 in the compasses, draw the semicircle DE 1 , DE 2 . Before we go 

 any farther we will examine this. To assist in understanding 

 the position and meaning of this semicircle we refer back 

 to Tig. 21, Lesson IV., Vol. II., page 360. There it will be 

 seen that E represents the eye, and its distance from the PP 

 from E to PS. Of course PS is opposite the eye E, and a line 

 between the two would 



form right angles with sc 



the PP. Now it is neces- 

 sary to set off on the HL 

 the distance of the eye from 

 the PP, that is, the dis- 

 tance from E to PS, for a 

 reason to be explained 

 presently ; therefore, the 

 proper way to do that is 

 to draw a semicircle, and 

 mark the extremities meet- 

 ing the HL as DE 1 and 

 DE 2 . In the eidograph 

 (Fig. 21), the dotted semi- 

 circle through the eye E 

 (ending on one side at 

 DE 1 and the other at DE 2 ) 

 is in an horizontal posi- 

 tion ; it is afterwards 

 supposed to be turned up, 

 or rabatted, upon the 

 PP passing through E 2 (to 

 the same points). This 

 will be the position in which we shall place it for the future, 

 and as seen in the figures which immediately follow Fig. 21. To 

 proceed with Fig. 49 : draw a line x x tangential to the semi- 

 circle, and parallel to the HL or PP. Our problem states that 

 the inclination of the pole to be represented is at an angle o/40 

 with tJie PP. Therefore, from E draw a line at that angle with 

 x x, meeting the HL in VP. There will be no difficulty in compre- 

 hending this, if we consider that because a; a; is parallel with the 

 PP, therefore if the plan of an object is known to be at a certain 

 angle with the PP (as in the ground-plan method), it will form 

 the same angle with x x. 



This, then, is the way a X. 



VP is found without the 

 necessity of a plan. From 

 PS draw the perpendicular 

 PS a, and mark one foot 

 to the right of a, viz., a b, 

 because the nearest end 

 of the pole is 1 foot to 

 the right of the eye. 

 Draw b PS, and some- 

 where upon b PS will be 

 found the position of the 

 nearest end of the pole, 

 to be determined in the 

 following manner : From 

 6 set off b c equal to 2 

 feet, draw a line from c 

 to DE 2 , cutting 6 PS in d, 

 the point required. This, 

 with the exception of the 

 plan, is precisely the 

 same that was done with f 



the line A B in Fig. 23, 



that is, by making c D equal to c A, a was found to be the 

 nearest end of the line A B. We now come to a stage of 

 the proceedings which will demand the closest attention of 

 our pupils. It is that of cutting off a portion of a vanish- 

 ing or retiring line, to give the perspective length of the 

 object, in this case the pole. From d, the nearest end of the 

 pole, draw a line to the VP (the vanishing line of the pole) ; upon 

 this line will be cut off the perspective length, as follows : 

 From VP, with the radius VP E, draw the arc E DVP ; DVP is the 

 distance of the eye from the VP, and is set off upon the HL upon 

 the same principle as the other distance points are placed ; the 

 use of this distance point is to enable us to cut off upon the 



P2. 



vanishing line the length required, in the same way that we used 

 the DE 2 for cutting off the point d in the line b PS. From DVP 

 through d draw a line to meet the PP in e, make e f equal to the 

 length of the pole, 6 feet, and from / draw another line back to 

 the DVP, cutting the vanishing line from d in h; h d will then be 

 the perspective length and representation of the pole. We must 

 dwell upon this for a minute or two, as this cutting (as we term 

 it) a vanishing line is important. Our pupils will have observed 



that we drew a line, com- 



x mencing from the DVP, 



through d to e, and after 

 we had marked the given 

 length e f on the PP, we 

 drew from / back again to 

 DVP; in other words, in 

 order to determine the 

 space upon the PP which 

 is to contain the length 

 of the pole, and at the 

 same time secure the per- 

 spective position, we 

 brought the nearest end, 

 d, of the pole to the PP 

 at e, measured its length, 

 ef, and then ruled back 

 again to the distance 

 point. This may be 

 summed up in very few 

 words, in the form of a 

 rule to be remembered 

 that every vanishing line 

 is cut by its own distance 

 point. The ability to draw an object in perspective upon this 

 lineal system depends principally upon a clear understanding of 

 the above rule in the several ways in which it may be applied. 

 We advise our pupils to get this first problem well up, by doing 

 it again at other angles, and other given distances and propor- 

 tions. Afterwards they will have very little difficulty in under- 

 standing all that is to follow. 



PROBLEM XXVIII. (Fig. 50). Two lines, each 3 feet long, 

 form a right angle ; one of the lines is at an angle of 40 with the 

 PP, nearest point 2 feet to the left of the eye, and 1 foot within the 



picture ; height of eye, dis- 



X, tance, and scale as in the 



last problem. 



Draw the PP, horizon- 

 tal line, and semicircle 

 through E at the given 

 distance as before, make 

 E VP 1 at an angle of 40 

 with x x, and draw E VP 2 

 at a right angle with it. 

 From each of the vanish- 

 ing points draw arcs from 

 E to the HL for the respec- 

 tive distance points ; pro- 

 duce E PS to a, and make 

 a b equal to 2 feet ; join 

 b PS, make b c equal to 1 

 foot, and draw a line from 

 c to DE 1 ; where this last 

 lino cuts b PS in d will be 

 the position of the angle ; 

 draw a line from d to 

 VP 1 . Now we must remem- 

 ber the rule given in the 



last problem, every vanishing line is cut by its own distance point; 

 consequently, as DVP 1 is the distance point of VP 1 , we must draw 

 a line from DVP 1 through d to the PP at e, make e f equal to 3 

 feet, the- length of one of the lines forming the right angle, and 

 from / rule back again to DVP 1 , cutting d VP 1 in h; dh will be 

 the length of the line. The other line of the right angle must 

 be similarly treated ; as it vanishes at VP 2 , the distance point of 

 VP 2 must be used for cutting off its perspective length, by bring- 

 ing a line first from DVP 2 through d to the PP at m; make m n 

 equal to the length of the line, and draw from n back again to 

 determine o in the vanishing line ; h d o will be the perspective 

 representation of the right angle as required. 



Fig. SO. 



PP 



