GEOMETRICAL PERSPECTIVE. 



IH 



DVP 1 



which it retired bcforo it was rained out of iU horizontal posi- 

 tion in other words, the position of the new vanishing point is 

 n/ir iif tho inclination of tho lino or piano ; 

 thw brings us to our object, to show where to find tho vr, lj 



i-tin 1 tin- . 



PBOHI.I.M \\\1. (Fig. 53). fliiv tho pertpective repwenta- 

 d> the ground at an angle of 30. Th 

 plan i >f the pole is at an angle of 50 

 with the PP. Length of pol< 

 the end on the ground w 2 feet n-itliin 

 the picture. The distance of the eye 

 f,-tn the PP 8 feet ; its height from 

 the ground 4 feet. First draw tho 

 111., and upon it, from tho PS as a 

 centre, draw the Bemicirolo with a 

 radius equal to tho distance of the 

 eye from the PP; raise a perpeudicu- 

 l.ir Hue from PS to E, and through B 



tial to the semicircle draw a 



::-;illelto tho iiij. From E draw 

 ii lino (K VP) at an angle of 50 with 

 the tangential lino. Draw the BP 

 (base of tho picture) parallel to tho 

 HL at a distance of 4 feet. Draw 

 PS c, and make c d equal to 2 feet ; 

 draw a line from d to DE, cutting 

 PS c in o ; this will give the point where tho polo rests upon tho 

 ground. Now if tho pole were in an horizontal position, its 

 vanishing point would be at tho VP on tho HL, but being in- 

 clined, its true vanishing point is above it (if tho inclination had 

 been downwards, its vanishing point would have been below the 

 HL). Therefore through the VP on the HL draw an indefinite 

 perpendicular lino ; find the distance point of the VP by draw- 

 ing the arc E DVP from VP as a centre, and with the radius 

 VP E. From DVP draw a lino at an angle of 30, meeting the 

 perpendicular from VP in VP 2 ; the VP 2 will bo the vanishing 

 point for the inclined line. Through tho point a draw a lino 

 directed to VP and meeting the BP in / (the point of contact) ; 

 from / draw the perpendicular / g It, (the line of contact). 

 Again, the pupil must bo reminded of a rule we gave in our last 

 lesson, tJiat every vanishing line must be cut from its own distance 

 point. Now tho vanishing line in this case is of tho pole only 

 from a to vp 2 , and upon this line wo must cut off a portion 

 equal to the 

 length of the 

 pole, conse- 

 quently we 

 mnst first 

 find tho dis- 



point 



of VP": thus, 

 from VP 2 as 

 a centre, and 

 with the dis- 

 tance to DVP 

 on the HL, 

 draw an aro 

 from DVP to 

 DVP. With 

 tho use of 

 this dis- 

 tance point 

 we now cut 

 off the 

 length of tha 

 pole : draw 

 a line from 



DVP 8 , through a, to tho lino of contact at g; mark off g h 

 equal to tho length of tho pole, 6 feet; and from h draw 

 a lino back again to DVP 1 , cutting tho vanishing lino of 

 tho pole in b ; ab will be the required perspective represcn- 

 tatio:i of the pole. To prove this, draw anywhere upon BP 

 the line m n, 6 feet long, and at an angle of 30 ; the pupil 

 will see that this is the full length of the pole at the given 

 angle, consequently its height from the ground at n is shown ; 

 draw n o parallel to HL in other words, mark the height 

 of the polo from the ground upon the line of contact ; draw 

 a line from o to the VP, it -will be found to cut the top of 



the pole M previously found in 6. This U ooo of toe 

 we recommend our pupiLi to repeat several times, 

 the pole at other angles, and turning it the other way in 

 the picture. A thorough knowledge of the practice of cut- 

 ting vanishing lines from their distance poinU ia the key. 

 stono of the principle contained in this method of representing- 

 objects in perspective. We purpose now to show how this may 

 be applied to give the inclination of 

 a roof, and aa it will be necessary to 

 draw the whole fignre we will give 

 out the whole problem, and advise 

 that it should be done on a larger 

 scale: our diagram U drawn to a 

 scale of 60 feet to the inch to econo- 

 mise space; it should be drawn by 

 our pupils en a scale of about 10 or 

 12 feet to the inch. 



PKOBLEM XXXII. (Fig. 54). 

 Draw the pertpective view of a square 

 tower having wingt: the bate* of the 

 tower and the wing* are each a tquare 

 of 48 feet ride ; height of tower 96 

 feet, and of the walU of the wingt 

 48 feet : the inclination of the roof 

 30, HL 10 feet, nearest end 12 feet 

 u-ithin the PP; distance of the eye 



from the PP, 120 feet ; angle of the front of the building with 

 the PP, 50. 



Having repeated in the lost problem the process which waa 

 explained in the last lesson, of finding the PS, E, and HL, the 

 vanishing points and their distance points, we will commence by 

 finding tho position of the nearest corner of the building. Draw 

 from PS to a; make a b equal 12 feet ; draw from b to DB, the 

 intersection will give the point required, from which a lino must 

 bo drawn to vp 1 . The next part of the process is the stumbling- 

 point of most beginners in this branch of perspective, and we 

 therefore request their attention to it. Find the distance point of 

 vp 1 , viz., DVP 1 . From DVP I draw a lino through the nearest corner 

 already found to the BP at e ; measure from e to /, from fiog, 

 and from g to h, each distance equal to the lengths of tho bases 

 of the wings and tower ; rule from these points back again ia 

 DVP 1 , we shall then have cut the several proportions of the 

 front of the building off the vanishing line that is, from the 



nearest an- 

 gle below c 

 to vp 1 by 

 the help of 

 the distance 

 point of VP 1 . 

 We make no 

 excuse for 

 repeating 

 this, because 

 we know 

 from practi- 

 cal experi- 

 ence how 

 often this 

 is forgotten. 

 The end of 

 the build- 

 ing must be 

 treated in 

 the same- 

 way, begin- 

 ning with a 

 line from 



DVP S to p ; p r ib the width of the building ; VP* its vanish- 

 ing point ; tho heights on the line of contact aro at n and 

 o. We presume there will be no difficulty with the rest of the 

 perpendicular and horizontal lines, and we now proceed with 

 tho roof. Because the ridge of the roof is over the centre of the 

 body of the building, there is no necessity in this case for finding 

 more than ono vanishing point for the roof, viz., tho inclination 

 s t. Tho vanishing point for that inclination is VP* on the per- 

 pendicular from vp'-, found by making an angle of 30 from 

 DVP*. The centre of tho building is found by drawing the 

 diagonals at the end and a perpendicular through their interseo* 



