---------------------------------------- ---------------------------- ,transcriber's ,note ,in the original book the author used lowercase italic letters and lowercase small capitals to label the geometric diagrams4 ,these have been here transcribed as lowercase and uppercase italic letters respectively 7a,a74 ,this file should be read using a font that supports the following ,unicode characters3 angle1 ratio1 proportion1 therefore1 prime1 and double prime4 ---------------------------------------- ---------------------------- ---------------------------------- ,library ,edition ---------------------------------- ,,the ,,complete ,,works ,,of ,,john ,,ruskin ,,elements ,,of ,,drawing ,,and ,,perspective ,,the ,,two ,,paths ,,unto ,,this ,,last ,,munera ,,pulveris ,,sesame ,,and ,,lilies ,,ethics ,,of ,,the ,,dust ---------------------------------- ,,national ,,library ,,association ,,new ,,york ,,chicago ---------------------------------- ,,the ,,elements ,,of ,,perspective ,,arranged ,,for ,,the ,,use ,,of ,,schools ,,and ,,intended ,,to ,,be ,,read ,,in ,,connection ,,with ,,the ,,first ,,three ,,books ,,of ,,euclid4 ,,contents4 ,,page ,preface ix ,introduction #a ,,problem ,i4 ,to fix the ,position of a given ,point #aj ,,problem ,,ii4 ,to draw a ,right ,line between two given ,points #ac ,,problem ,,iii4 ,to find the ,vanishing-,point of a given ,horizontal ,line #ag ,,problem ,,iv4 ,to find the ,dividing-,points of a given ,horizontal ,line #bc ,,problem ,v4 ,to draw a ,horizontal ,line1 given in ,position and ,magnitude1 by means of its ,sight-,magnitude and ,dividing-,points #bd ,,problem ,,vi4 ,to draw any ,triangle1 given in ,position and ,magnitude1 in a ,horizontal ,plane #bg ,,problem ,,vii4 ,to draw any ,rectilinear ,quadrilateral ,figure1 given in ,position and ,magnitude1 in a ,horizontal ,plane #bi ,,problem ,,viii4 ,to draw a ,square1 given in ,position and ,magnitude1 in a ,horizontal ,plane #ca ,,problem ,,ix4 ,to draw a ,square ,pillar1 given in ,position and ,magnitude1 its ,base and ,top being in ,horizontal ,planes #cd ,,problem ,x4 ,to draw a ,pyramid1 given in ,position and ,magnitude1 on a ,square ,base in a ,horizontal ,plane #cf ,,problem ,,xi4 ,to draw any ,curve in a ,horizontal or ,vertical ,plane #ch ,,problem ,,xii4 ,to divide a ,circle drawn in ,perspective into any given ,number of ,equal ,parts #db ,,problem ,,xiii4 ,to draw a ,square1 given in ,magnitude1 within a larger ,square given in ,position and ,magnitude2 the ,sides of the two ,squares being ,parallel #de ,,problem ,,xiv4 ,to draw a ,truncated ,circular ,cone1 given in ,position and ,magnitude1 the ,truncations being in ,horizontal ,planes1 and the ,axis of the ,cone vertical #dg ,,problem ,,xv4 ,to draw an ,inclined ,line1 given in ,position and ,magnitude #ej ,,problem ,,xvi4 ,to find the ,vanishing-,point of a given ,inclined ,line #ec ,,problem ,,xvii4 ,to find the ,dividing-,points of a given ,inclined ,line #ee ,,problem ,,xviii4 ,to find the ,sight-,line of an ,inclined ,plane in which ,two ,lines are given in ,position #eg ,,problem ,,xix4 ,to find the ,vanishing-,point of ,steepest ,lines in an ,inclined ,plane whose ,sight-,line is given #ei ,,problem ,,xx4 ,to find the ,vanishing-,point of ,lines perpendicular to the ,surface of a given ,inclined ,plane #fa ,,appendix4 ,i4 ,practice and ,observations on the preceding ,problems #fi ,,ii4 ,demonstrations which could not conveniently be included in the ,text #ii ,,preface4 ,for some time back ,i have felt the want1 among ,students of ,drawing1 of a written code of accurate ,perspective ,law2 the modes of construction in common use being various1 and1 for some problems1 insufficient4 ,it would have been desirable to draw up such a code in popular language1 so as to do away with the most repulsive difficulties of the subject2 but finding this popularization would be impossible1 without elaborate figures and long explanations1 such as ,i had no leisure to prepare1 ,i have arranged the necessary rules in a short mathematical form1 which any schoolboy may read through in a few days1 after he has mastered the first three and the sixth books of ,euclid4 ,some awkward compromises have been admitted between the first-attempted popular explanation1 and the severer arrangement1 involving irregular lettering and redundant phraseology2 but ,i cannot for the present do more1 and leave the book therefore to its trial1 hoping that1 if it be found by masters of schools to answer its purpose1 ,i may hereafter bring it into better form4#a ,an account of practical methods1 sufficient for general purposes of sketching1 might indeed have been set down in much less space3 but if the student reads the following pages carefully1 he will not only find himself able1 on occasion1 to solve perspective problems of a complexity greater than the ordinary rules will reach1 but obtain a clue to many important laws of pictorial effect1 no less than of outline4 ,the subject thus examined becomes1 at least to my mind1 very curious and interesting2 but1 for students who are unable or unwilling to take it up in this abstract form1 ,i believe good help will be soon furnished1 in a series of illustrations of practical perspective now in preparation by ,mr4 ,le ,vengeur4 ,i have not seen this essay in an advanced state1 but the illustrations shown to me were very clear and good2 and1 as the author has devoted much thought to their arrangement1 ,i hope that his work will be precisely what is wanted by the general learner4 ,students wishing to pursue the subject into its more extended branches will find1 ,i believe1 ,cloquet's treatise the best hitherto published4#b #a ,some irregularities of arrangement have been admitted merely for the sake of convenient reference2 the eighth problem1 for instance1 ought to have been given as a case of the seventh1 but is separately enunciated on account of its importance4 ,several constructions1 which ought to have been given as problems1 are on the contrary given as corollaries1 in order to keep the more directly connected problems in closer sequence2 thus the construction of rectangles and polygons in vertical planes would appear by the ,table of ,contents to have been omitted1 being given in the corollary to ,problem ,,ix4 #b ,nouveau ,trait lmentaire de ,perspective4 ,bachelier1 #ahbc4 ,,the ,,elements ,,of ,,perspective4 ,,introduction4 ,when you begin to read this book1 sit down very near the window1 and shut the window4 ,i hope the view out of it is pretty2 but1 whatever the view may be1 we shall find enough in it for an illustration of the first principles of perspective 7or1 literally1 of 8looking through074 ,every pane of your window may be considered1 if you choose1 as a glass picture2 and what you see through it1 as painted on its surface4 ,and if1 holding your head still1 you extend your hand to the glass1 you may1 with a brush full of any thick color1 trace1 roughly1 the lines of the landscape on the glass4 ,but1 to do this1 you must hold your head very still4 ,not only you must not move it sideways1 nor up and down1 but it must not even move backwards or forwards2 for1 if you move your head forwards1 you will see more of the landscape through the pane2 and1 if you move it backwards1 you will see less3 or considering the pane of glass as a picture1 when you hold your head near it1 the objects are painted small1 and a great many of them go into a little space2 but1 when you hold your head some distance back1 the objects are painted larger upon the pane1 and fewer of them go into the field of it4 ,but1 besides holding your head still1 you must1 when you try to trace the picture on the glass1 shut one of your eyes4 ,if you do not1 the point of the brush appears double2 and1 on farther experiment1 you will observe that each of your eyes sees the object in a different place on the glass1 so that the tracing which is true to the sight of the right eye is a couple of inches 7or more1 according to your distance from the pane17 to the left of that which is true to the sight of the left4 ,thus1 it is only possible to draw what you see through the window rightly on the surface of the glass1 by fixing one eye at a given point1 and neither moving it to the right nor left1 nor up nor down1 nor backwards nor forwards4 ,every picture drawn in true perspective may be considered as an upright piece of glass1#c on which the objects seen through it have been thus drawn4 ,perspective can1 therefore1 only be quite right1 by being calculated for one fixed position of the eye of the observer2 nor will it ever appear deceptively right unless seen precisely from the point it is calculated for4 ,custom1 however1 enables us to feel the rightness of the work on using both our eyes1 and to be satisfied with it1 even when we stand at some distance from the point it is designed for4 ,supposing that1 instead of a window1 an unbroken plate of crystal extended itself to the right and left of you1 and high in front1 and that you had a brush as long as you wanted 7a mile long1 suppose71 and could paint with such a brush1 then the clouds high up1 nearly over your head1 and the landscape far away to the right and left1 might be traced1 and painted1 on this enormous crystal field4#d ,but if the field were so vast 7suppose a mile high and a mile wide71 certainly1 after the picture was done1 you would not stand as near to it1 to see it1 as you are now sitting near to your window4 ,in order to trace the upper clouds through your great glass1 you would have had to stretch your neck quite back1 and nobody likes to bend their neck back to see the top of a picture4 ,so you would walk a long way back to see the great picture--a quarter of a mile1 perhaps1--and then all the perspective would be wrong1 and would look quite distorted1 and you would discover that you ought to have painted it from the greater distance1 if you meant to look at it from that distance4 ,thus1 the distance at which you intend the observer to stand from a picture1 and for which you calculate the perspective1 ought to regulate to a certain degree the size of the picture4 ,if you place the point of observation near the canvas1 you should not make the picture very large3 vice vers1 if you place the point of observation far from the canvas1 you should not make it very small2 the fixing1 therefore1 of this point of observation determines1 as a matter of convenience1 within certain limits1 the size of your picture4 ,but it does not determine this size by any perspective law2 and it is a mistake made by many writers on perspective1 to connect some of their rules definitely with the size of the picture4 ,for1 suppose that you had what you now see through your window painted actually upon its surface1 it would be quite optional to cut out any piece you chose1 with the piece of the landscape that was painted on it4 ,you might have only half a pane1 with a single tree2 or a whole pane1 with two trees and a cottage2 or two panes1 with the whole farmyard and pond2 or four panes1 with farmyard1 pond1 and foreground4 ,and any of these pieces1 if the landscape upon them were1 as a scene1 pleasantly composed1 would be agreeable pictures1 though of quite different sizes2 and yet they would be all calculated for the same distance of observation4 ,in the following treatise1 therefore1 ,i keep the size of the picture entirely undetermined4 ,i consider the field of canvas as wholly unlimited1 and on that condition determine the perspective laws4 ,after we know how to apply those laws without limitation1 we shall see what limitations of the size of the picture their results may render advisable4 ,but although the size of the picture is thus independent of the observer's distance1 the size of the object represented in the picture is not4 ,on the contrary1 that size is fixed by absolute mathematical law2 that is to say1 supposing you have to draw a tower a hundred feet high1 and a quarter of a mile distant from you1 the height which you ought to give that tower on your paper depends1 with mathematical precision1 on the distance at which you intend your paper to be placed4 ,so1 also1 do all the rules for drawing the form of the tower1 whatever it may be4 ,hence1 the first thing to be done in beginning a drawing is to fix1 at your choice1 this distance of observation1 or the distance at which you mean to stand from your paper4 ,after that is determined1 all is determined1 except only the ultimate size of your picture1 which you may make greater1 or less1 not by altering the size of the things represented1 but by taking in more1 or fewer of them4 ,so1 then1 before proceeding to apply any practical perspective rule1 we must always have our distance of observation marked1 and the most convenient way of marking it is the following3 ,illustration3 ,fig4 #a4 ,,placing ,,of ,,the ,,sight-,,point1 ,,sight-,,line1 ,,station-,,point1 ,,and ,,station-,,line4 ,i4 ,,the ,,sight-,,point4--,let ,,abcd1 ,fig4 #a41 be your sheet of paper1 the larger the better1 though perhaps we may cut out of it at last only a small piece for our picture1 such as the dotted circle ,,nopq4 ,this circle is not intended to limit either the size or shape of our picture3 you may ultimately have it round or oval1 horizontal or upright1 small or large1 as you choose4 ,i only dot the line to give you an idea of whereabouts you will probably like to have it2 and1 as the operations of perspective are more conveniently performed upon paper underneath the picture than above it1 ,i put this conjectural circle at the top of the paper1 about the middle of it1 leaving plenty of paper on both sides and at the bottom4 ,now1 as an observer generally stands near the middle of a picture to look at it1 we had better at first1 and for simplicity's sake1 fix the point of observation opposite the middle of our conjectural picture4 ,so take the point ,s1 the center of the circle ,,nopq2--or1 which will be simpler for you in your own work1 take the point ,s at random near the top of your paper1 and strike the circle ,,nopq round it1 any size you like4 ,then the point ,s is to represent the point opposite which you wish the observer of your picture to place his eye1 in looking at it4 ,call this point the 8,sight-,point40 ,,ii4 ,,the ,,sight-,,line4--,through the ,sight-point1 ,s1 draw a horizontal line1 ,,gh1 right across your paper from side to side1 and call this line the 8,sight-,line40 ,this line is of great practical use1 representing the level of the eye of the observer all through the picture4 ,you will find hereafter that if there is a horizon to be represented in your picture1 as of distant sea or plain1 this line defines it4 ,,iii4 ,,the ,,station-,,line4--,from ,s let fall a perpendicular line1 ,,sr1 to the bottom of the paper1 and call this line the 8,station-,line40 ,this represents the line on which the observer stands1 at a greater or less distance from the picture2 and it ought to be imagined as drawn right out from the paper at the point s4 ,hold your paper upright in front of you1 and hold your pencil horizontally1 with its point against the point ,s1 as if you wanted to run it through the paper there1 and the pencil will represent the direction in which the line ,,sr ought to be drawn4 ,but as all the measurements which we have to set upon this line1 and operations which we have to perform with it1 are just the same when it is drawn on the paper itself1 below ,s1 as they would be if it were represented by a wire in the position of the leveled pencil1 and as they are much more easily performed when it is drawn on the paper1 it is always in practice1 so drawn4 ,,iv4 ,,the ,,station-,,point4--,on this line1 mark the distance ,,st at your pleasure1 for the distance at which you wish your picture to be seen1 and call the point ,t the 8,station-,point40 ,illustration3 ,fig4 #b4 ,in practice1 it is generally advisable to make the distance ,,st about as great as the diameter of your intended picture2 and it should1 for the most part1 be more rather than less2 but1 as ,i have just stated1 this is quite arbitrary4 ,however1 in this figure1 as an approximation to a generally advisable distance1 ,i make the distance ,,st equal to the diameter of the circle ,,nopq4 ,now1 having fixed this distance1 ,,st1 all the dimensions of the objects in our picture are fixed likewise1 and for this reason3-- ,let the upright line ,,ab1 ,fig4 #b41 represent a pane of glass placed where our picture is to be placed2 but seen at the side of it1 edgeways2 let ,s be the ,sight-point2 ,,st the ,station-line1 which1 in this figure1 observe1 is in its true position1 drawn out from the paper1 not down upon it2 and ,t the ,station-point4 ,suppose the ,station-line ,,st to be continued1 or in mathematical language 8produced10 through ,s1 far beyond the pane of glass1 and let ,,pq be a tower or other upright object situated on or above this line4 ,now the apparent height of the tower ,,pq is measured by the angle ,,qtp1 between the rays of light which come from the top and bottom of it to the eye of the observer4 ,but the actual height of the image of the tower on the pane of glass ,,ab1 between us and it1 is the distance ,p,q between the points where the rays traverse the glass4 ,evidently1 the farther from the point ,t we place the glass1 making ,,st longer1 the larger will be the image2 and the nearer we place it to ,t1 the smaller the image1 and that in a fixed ratio4 ,let the distance ,,dt be the direct distance from the ,station-point to the foot of the object4 ,then1 if we place the glass ,,ab at one-third of that whole distance1 ,p,q will be one-third of the real height of the object2 if we place the glass at two-thirds of the distance1 as at ,,ef1 ,p,q 7the height of the image at that point7 will be two-thirds the height#e of the object1 and so on4 ,therefore the mathematical law is that ,p,q will be to ,,pq as ,,st to ,,dt4 ,i put this ratio clearly by itself that you may remember it3 ,p,q ,,pq ,,st ,,dt or in words3 ,p dash ,q dash is to ,,pq as ,,st to ,,dt ,in which formula1 recollect that ,p,q is the height of the appearance of the object on the picture2 ,,pq the height of the object itself2 ,s the ,sight-point2 ,t the ,station-point2 ,d a point at the direct distance of the object2 though the object is seldom placed actually on the line ,,ts produced1 and may be far to the right or left of it1 the formula is still the same4 ,for let ,s1 ,fig4 #c41 be the ,sight-point1 and ,,ab the glass--here seen looking down on its upper edge1 not sideways2--then if the tower 7represented now1 as on a map1 by the dark square71 instead of being at ,d on the line ,,st produced1 be at ,e1 to the right 7or left7 of the spectator1 still the apparent height of the tower on ,,ab will be as ,s,t to ,,et1 which is the same ratio as that of ,,st to ,,dt4 ,illustration3 ,fig4 #c4 ,now in many perspective problems1 the position of an object is more conveniently expressed by the two measurements ,,dt and ,,de1 than by the single oblique measurement ,,et4 ,i shall call ,,dt the 8direct distance0 of the object at ,e1 and ,,de its 8lateral distance40 ,it is rather a license to call ,,dt its 8direct0 distance1 for ,,et is the more direct of the two2 but there is no other term which would not cause confusion4 ,lastly1 in order to complete our knowledge of the position of an object1 the vertical height of some point in it1 above or below the eye1 must be given2 that is to say1 either ,,dp or ,,dq in ,fig4 #b4f3 this ,i shall call the 8vertical distance0 of the point given4 ,in all perspective problems these three distances1 and the dimensions of the object1 must be stated1 otherwise the problem is imperfectly given4 ,it ought not to be required of us merely to draw a room or a church in perspective2 but to draw this room from this corner1 and that church on that spot1 in perspective4 ,for want of knowing how to base their drawings on the measurement and place of the object1 ,i have known practiced students represent a parish church1 certainly in true perspective1 but with a nave about two miles and a half long4 ,it is true that in drawing landscapes from nature the sizes and distances of the objects cannot be accurately known4 ,when1 however1 we know how to draw them rightly1 if their size were given1 we have only to assume a rational approximation to their size1 and the resulting drawing will be true enough for all intents and purposes4 ,it does not in the least matter that we represent a distant cottage as eighteen feet long1 when it is in reality only seventeen2 but it matters much that we do not represent it as eighty feet long1 as we easily might if we had not been accustomed to draw from measurement4 ,therefore1 in all the following problems the measurement of the object is given4 ,the student must observe1 however1 that in order to bring the diagrams into convenient compass1 the measurements assumed are generally very different from any likely to occur in practice4 ,thus1 in ,fig4 #c41 the distance ,,ds would be probably in practice half a mile or a mile1 and the distance ,,ts1 from the eye of the observer to the paper1 only two or three feet4 ,the mathematical law is however precisely the same1 whatever the proportions2 and ,i use such proportions as are best calculated to make the diagram clear4 ,now1 therefore1 the conditions of a perspective problem are the following3 ,the ,sight-line ,,gh given1 ,fig4 #a42 ,the ,sight-point ,s given2 ,the ,station-point ,t given2 and ,the three distances of the object1#g direct1 lateral1 and vertical1 with its dimensions1 given4 ,the size of the picture1 conjecturally limited by the dotted circle1 is to be determined afterwards at our pleasure4 ,on these conditions ,i proceed at once to construction4 #c ,if the glass were not upright1 but sloping1 the objects might still be drawn through it1 but their perspective would then be different4 ,perspective1 as commonly taught1 is always calculated for a vertical plane of picture4 #d ,supposing it to have no thickness2 otherwise the images would be distorted by refraction4 #e ,i say 8height0 instead of 8magnitude10 for a reason stated in ,appendix ,i41 to which you will soon be referred4 ,read on here at present4 #f ,p and ,q being points indicative of the place of the tower's base and top4 ,in this figure both are above the sight-line2 if the tower were below the spectator both would be below it1 and therefore measured below ,d4 #g ,more accurately1 8the three distances of any point1 either in the object itself1 or indicative of its distance40 ,,problem ,i4 ,,to ,,fix ,,the ,,position ,,of ,a ,,given ,,point4#h ,let ,p1 ,fig4 #d41 be the given point4 ,illustration3 ,fig4 #d4 ,let its direct distance be ,,dt2 its lateral distance to the left1 ,,dc2 and vertical distance beneath the eye of the observer1 ,,cp4 ,let ,,gh be the ,sight-line1 ,s the ,sight-point1 and ,t the ,station-point4#i ,it is required to fix on the plane of the picture the position of the point ,p4 ,arrange the three distances of the object on your paper1 as in ,fig4 #d4aj ,join ,,ct1 cutting ,,gh in ,q4 ,from ,q let fall the vertical line ,,qp4 ,join ,,pt1 cutting ,,qp in ,p4 ,p is the point required4 ,if the point ,p is above the eye of the observer instead of below it1 ,,cp is to be measured upwards from ,c1 and ,,qp drawn upwards from ,q4 ,the construction will be as in ,fig4 #e4 ,illustration3 ,fig4 #e4 ,and if the point ,p is to the right instead of the left of the observer1 ,,dc is to be measured to the right instead of the left4 ,the figures #d4 and #e41 looked at in a mirror1 will show the construction of each1 on that supposition4 ,now read very carefully the examples and notes to this problem in ,appendix ,i4 7page #fi74 ,i have put them in the ,appendix in order to keep the sequence of following problems more clearly traceable here in the text2 but you must read the first ,appendix before going on4 #h ,more accurately1 8,to fix on the plane of the picture the apparent position of a point given in actual position40 ,in the headings of all the following problems the words 8on the plane of the picture0 are to be understood after the words 8to draw40 ,the plane of the picture means a surface extended indefinitely in the direction of the picture4 #i ,the sentence within brackets will not be repeated in succeeding statements of problems4 ,it is always to be understood4 #aj ,in order to be able to do this1 you must assume the distances to be small2 as in the case of some object on the table3 how large distances are to be treated you will see presently2 the mathematical principle1 being the same for all1 is best illustrated first on a small scale4 ,suppose1 for instance1 ,p to be the corner of a book on the table1 seven inches below the eye1 five inches to the left of it1 and a foot and a half in advance of it1 and that you mean to hold your finished drawing at six inches from the eye2 then ,,ts will be six inches1 ,,td a foot and a half1 ,,dc five inches1 and ,,cp seven4 ,,problem ,,ii4 ,,to ,,draw ,a ,,right ,,line ,,between ,,two ,,given ,,points4 ,illustration3 ,fig4 #f4 ,let ,,ab1 ,fig4 #f41 be the given right line1 joining the given points ,a and ,b4 ,let the direct1 lateral1 and vertical distances of the point ,a be ,,td1 ,,dc1 and ,,ca4 ,let the direct1 lateral1 and vertical distances of the point ,b be ,,td1 ,,dc1 and ,c,b4 ,then1 by ,problem ,i41 the position of the point ,a on the plane of the picture is a4 ,and similarly1 the position of the point ,b on the plane of the picture is b4 ,join ab4 ,then ab is the line required4 ,,corollary ,i4 ,if the line ,,ab is in a plane parallel to that of the picture1 one end of the line ,,ab must be at the same direct distance from the eye of the observer as the other4 ,therefore1 in that case1 ,,dt is equal to ,d,t4 ,then the construction will be as in ,fig4 #g42 and the student will find experimentally that ab is now parallel to ,,ab4#aa ,illustration3 ,fig4 #g4 ,and that ab is to ,,ab as ,,ts is to ,,td4 ,therefore1 to draw any line in a plane parallel to that of the picture1 we have only to fix the position of one of its extremities1 a or b1 and then to draw from a or b a line parallel to the given line1 bearing the proportion to it that ,,ts bears to ,,td4 ,,corollary ,,ii4 ,if the line ,,ab is in a horizontal plane1 the vertical distance of one of its extremities must be the same as that of the other4 ,therefore1 in that case1 ,,ac equals ,,bc 7,fig4 #f474 ,and the construction is as in ,fig4 #h4 ,illustration3 ,fig4 #h4 ,in ,fig4 #h4 produce ab to the sight-line1 cutting the sight-line in ,v2 the point ,v1 thus determined1 is called the ,,vanishing-,,point of the line ,,ab4 ,join ,,tv4 ,then the student will find experimentally that ,,tv is parallel to ,,ab4#ab ,,corollary ,,iii4 ,if the line ,,ab produced would pass through some point beneath or above the station-point1 ,,cd is to ,,dt as ,c,d is to ,d,t2 in which case the point c coincides with the point c1 and the line ab is vertical4 ,therefore every vertical line in a picture is1 or may be1 the perspective representation of a horizontal one which1 produced1 would pass beneath the feet or above the head of the spectator4#ac #aa ,for by the construction ,,at a,t ,,bt b,t2 and therefore the two triangles ,,abt1 ab,t1 7having a common angle ,,atb17 are similar4 #ab ,the demonstration is in ,appendix ,,ii4 ,article ,i4 #ac ,the reflection in water of any luminous point or isolated object 7such as the sun or moon7 is therefore1 in perspective1 a vertical line2 since such reflection1 if produced1 would pass under the feet of the spectator4 ,many artists 7,claude among the rest7 knowing something of optics1 but nothing of perspective1 have been led occasionally to draw such reflections towards a point at the center of the base of the picture4 ,,problem ,,iii4 ,,to ,,find ,,the ,,vanishing-,,point ,,of ,a ,,given ,,horizontal ,,line4 ,illustration3 ,fig4 #i4 ,let ,,ab1 ,fig4 #i41 be the given line4 ,from ,t1 the station-point1 draw ,,tv parallel to ,,ab1 cutting the sight-line in ,v4 ,v is the ,vanishing-point required4#ad ,,corollary ,i4 ,as1 if the point b is first found1 ,v may be determined by it1 so1 if the point ,v is first found1 b may be determined by it4 ,for let ,,ab1 ,fig4 #aj41 be the given line1 constructed upon the paper as in ,fig4 #h42 and let it be required to draw the line ab without using the point ,c4 ,illustration3 ,fig4 #aj4 ,find the position of the point ,a in a4 7,problem ,i47 ,find the vanishing-point of ,,ab in ,v4 7,problem ,,iii47 ,join a,v4 ,join ,,bt1 cutting a,v in b4 ,then ab is the line required4#ae ,,corollary ,,ii4 ,we have hitherto proceeded on the supposition that the given line was small enough1 and near enough1 to be actually drawn on our paper of its real size2 as in the example given in ,appendix ,i4 ,we may1 however1 now deduce a construction available under all circumstances1 whatever may be the distance and length of the line given4 ,illustration3 ,fig4 #aa4 ,from ,fig4 #h4 remove1 for the sake of clearness1 the lines ,c,d1 b,v1 and ,,tv2 and1 taking the figure as here in ,fig4 #aa41 draw from a1 the line a,r parallel to ,,ab1 cutting ,,bt in ,r4 ,then a,r is to ,,ab as a,t is to ,,at4 ---- ---- as c,t is to ,,ct4 ---- ---- as ,,ts is to ,,td4 ,that is to say1 a,r is the sight-magnitude of ,,ab4#af ,illustration3 ,fig4 #ab4 ,therefore1 when the position of the point ,a is fixed in a1 as in ,fig4 #ab41 and a,v is drawn to the vanishing-point2 if we draw a line a,r from a1 parallel to ,,ab1 and make a,r equal to the sight-magnitude of ,,ab1 and then join ,,rt1 the line ,,rt will cut a,v in b4 ,so that1 in order to determine the length of ab1 we need not draw the long and distant line ,,ab1 but only a,r parallel to it1 and of its sight-magnitude2 which is a great gain1 for the line ,,ab may be two miles long1 and the line a,r perhaps only two inches4 ,,corollary ,,iii4 ,in ,fig4 #ab41 altering its proportions a little for the sake of clearness1 and putting it as here in ,fig4 #ac41 draw a horizontal line a,r and make a,r equal to a,r4 ,through the points ,r and b draw ,r,m1 cutting the sight-line in ,m4 ,join ,,tv4 ,now the reader will find experimentally that ,,vm is equal to ,,vt4#ag ,illustration3 ,fig4 #ac4 ,hence it follows that1 if from the vanishing-point ,v we lay off on the sight-line a distance1 ,,vm1 equal to ,,vt2 then draw through a a horizontal line a,r1 make a,r equal to the sight-magnitude of ,,ab1 and join ,r,m2 the line ,r,m will cut a,v in b4 ,and this is in practice generally the most convenient way of obtaining the length of ab4 ,,corollary ,,iv4 ,removing from the preceding figure the unnecessary lines1 and retaining only ,r,m and a,v1 as in ,fig4 #ad41 produce the line a,r to the other side of a1 and make a,x equal to a,r4 ,join ,xb1 and produce ,xb to cut the line of sight in ,n4 ,illustration3 ,fig4 #ad4 ,then as ,,xr is parallel to ,,mn1 and a,r is equal to a,x1 ,,vn must1 by similar triangles1 be equal to ,,vm 7equal to ,,vt in ,fig4 #ac474 ,therefore1 on whichever side of ,v we measure the distance ,,vt1 so as to obtain either the point ,m1 or the point ,n1 if we measure the sight-magnitude a,r or a,x on the opposite side of the line a,v1 the line joining ,r,m or ,,xn will equally cut a,v in b4 ,the points ,m and ,n are called the 8,,dividing-,,points0 of the original line ,,ab 7,fig4 #ab471 and we resume the results of these corollaries in the following three problems4 #ad ,the student will observe1 in practice1 that1 his paper lying flat on the table1 he has only to draw the line ,,tv on its horizontal surface1 parallel to the given horizontal line ,,ab4 ,in theory1 the paper should be vertical1 but the station-line ,,st horizontal 7see its definition above1 page #e72 in which case ,,tv1 being drawn parallel to ,,ab1 will be horizontal also1 and still cut the sight-line in ,v4 ,the construction will be seen to be founded on the second ,corollary of the preceding problem4 ,it is evident that if any other line1 as ,,mn in ,fig4 #i41 parallel to ,,ab1 occurs in the picture1 the line ,,tv1 drawn from ,t1 parallel to ,,mn1 to find the vanishing-point of ,,mn1 will coincide with the line drawn from ,t1 parallel to ,,ab1 to find the vanishing-point of ,,ab4 ,therefore ,,ab and ,,mn will have the same vanishing-point4 ,therefore all parallel horizontal lines have the same vanishing-point4 ,it will be shown hereafter that all parallel inclined lines also have the same vanishing-point2 the student may here accept the general conclusion--8,all parallel lines have the same vanishing-point40 ,it is also evident that if ,,ab is parallel to the plane of the picture1 ,,tv must be drawn parallel to ,,gh1 and will therefore never cut ,,gh4 ,the line ,,ab has in that case no vanishing-point3 it is to be drawn by the construction given in ,fig4 #g4 ,it is also evident that if ,,ab is at right angles with the plane of the picture1 ,,tv will coincide with ,,ts1 and the vanishing-point of ,,ab will be the sight-point4 #ae ,i spare the student the formality of the reductio ad absurdum1 which would be necessary to prove this4 #af ,for definition of ,sight-,magnitude1 see ,appendix ,i4 ,it ought to have been read before the student comes to this problem2 but ,i refer to it in case it has not4 #ag ,the demonstration is in ,appendix ,,ii4 ,article ,,ii4 p4 #aja4 ,,problem ,,iv4 ,,to ,,find ,,the ,,dividing-,,points ,,of ,a ,,given ,,horizontal ,,line4 ,illustration3 ,fig4 #ae4 ,let the horizontal line ,,ab 7,fig4 #ae47 be given in position and magnitude4 ,it is required to find its dividing-points4 ,find the vanishing-point ,v of the line ,,ab4 ,with center ,v and distance ,,vt1 describe circle cutting the sight-line in ,m and ,n4 ,then ,m and ,n are the dividing-points required4 ,in general1 only one dividing-point is needed for use with any vanishing-point1 namely1 the one nearest ,s 7in this case the point ,m74 ,but its opposite ,n1 or both1 may be needed under certain circumstances4 ,,problem ,v4 ,,to ,,draw ,a ,,horizontal ,,line1 ,,given ,,in ,,position ,,and ,,magnitude1 ,,by ,,means ,,of ,,its ,,sight-,,magnitude ,,and ,,dividing-,,points4 ,illustration3 ,fig4 #af4 ,let ,,ab 7,fig4 #af47 be the given line4 ,find the position of the point ,a in a4 ,find the vanishing-point ,v1 and most convenient dividing-point ,m1 of the line ,,ab4 ,join a,v4 ,through a draw a horizontal line ab and make ab equal to the sight-magnitude of ,,ab4 ,join b,m1 cutting a,v in b4 ,then ab is the line required4 ,,corollary ,i4 ,illustration3 ,fig4 #ag4 ,supposing it were now required to draw a line ,,ac 7,fig4 #ag47 twice as long as ,,ab1 it is evident that the sight-magnitude ac must be twice as long as the sight-magnitude ab2 we have1 therefore1 merely to continue the horizontal line ab1 make bc equal to ab1 join c,m1 cutting a,v in c1 and ac will be the line required4 ,similarly1 if we have to draw a line ,,ad1 three times the length of ,,ab1 ad must be three times the length of ab1 and1 joining d,m1 ad will be the line required4 ,the student will observe that the nearer the portions cut off1 bc1 cd1 etc41 approach the point ,v1 the smaller they become2 and1 whatever lengths may be added to the line ,,ad1 and successively cut off from a,v1 the line a,v will never be cut off entirely1 but the portions cut off will become infinitely small1 and apparently 8vanish0 as they approach the point ,v2 hence this point is called the 8vanishing0 point4 ,,corollary ,,ii4 ,it is evident that if the line ,,ad had been given originally1 and we had been required to draw it1 and divide it into three equal parts1 we should have had only to divide its sight-magnitude1 ad1 into the three equal parts1 ab1 bc1 and cd1 and then1 drawing to ,m from b and c1 the line ad would have been divided as required in b and c4 ,and supposing the original line ,,ad be divided irregularly into any number of parts1 if the line ad be divided into a similar number in the same proportions 7by the construction given in ,appendix ,i471 and1 from these points of division1 lines are drawn to ,m1 they will divide the line ad in true perspective into a similar number of proportionate parts4 ,the horizontal line drawn through a1 on which the sight-magnitudes are measured1 is called the 8,,measuring-,,line40 ,and the line ad1 when properly divided in b and c1 or any other required points1 is said to be divided 8,,in ,,perspective ,,ratio0 to the divisions of the original line ,,ad4 ,if the line a,v is above the sight-line instead of beneath it1 the measuring-line is to be drawn above also3 and the lines b,m1 c,m1 etc41 drawn down to the dividing-point4 ,turn ,fig4 #ag4 upside down1 and it will show the construction4 ,,problem ,,vi4 ,,to ,,draw ,,any ,,triangle1 ,,given ,,in ,,position ,,and ,,magnitude1 ,,in ,a ,,horizontal ,,plane4 ,illustration3 ,fig4 #ah4 ,let ,,abc 7,fig4 #ah47 be the triangle4 ,as it is given in position and magnitude1 one of its sides1 at least1 must be given in position and magnitude1 and the directions of the two other sides4 ,let ,,ab be the side given in position and magnitude4 ,then ,,ab is a horizontal line1 in a given position1 and of a given length4 ,draw the line ,,ab4 7,problem ,v47 ,let ab be the line so drawn4 ,find ,v and ,v1 the vanishing-points respectively of the lines ,,ac and ,,bc4 7,problem ,,iii47 ,from a draw a,v1 and from b1 draw b,v1 cutting each other in c4 ,then abc is the triangle required4 ,if ,,ac is the line originally given1 ac is the line which must be first drawn1 and the line ,vb must be drawn from ,v to c and produced to cut ab in b4 ,similarly1 if ,,bc is given1 ,vc must be drawn to c and produced1 and ab from its vanishing-point to b1 and produced to cut ac in a4 ,,problem ,,vii4 ,,to ,,draw ,,any ,,rectilinear ,,quadrilateral ,,figure1 ,,given ,,in ,,position ,,and ,,magnitude1 ,,in ,a ,,horizontal ,,plane4 ,illustration3 ,fig4 #ai4 ,let ,,abcd 7,fig4 #ai47 be the given figure4 ,join any two of its opposite angles by the line ,,bc4 ,draw first the triangle ,,abc4 7,problem ,,vi47 ,and then1 from the base ,,bc1 the two lines ,,bd1 ,,cd1 to their vanishing-points1 which will complete the figure4 ,it is unnecessary to give a diagram of the construction1 which is merely that of ,fig4 #ah4 duplicated2 another triangle being drawn on the line ,,ac or ,,bc4 ,,corollary4 ,it is evident that by this application of ,problem ,,vi4 any given rectilinear figure whatever in a horizontal plane may be drawn1 since any such figure may be divided into a number of triangles1 and the triangles then drawn in succession4 ,more convenient methods may1 however1 be generally found1 according to the form of the figure required1 by the use of succeeding problems2 and for the quadrilateral figure which occurs most frequently in practice1 namely1 the square1 the following construction is more convenient than that used in the present problem4 ,,problem ,,viii4 ,,to ,,draw ,a ,,square1 ,,given ,,in ,,position ,,and ,,magnitude1 ,,in ,a ,,horizontal ,,plane4 ,illustration3 ,fig4 #bj4 ,let ,,abcd1 ,fig4 #bj41 be the square4 ,as it is given in position and magnitude1 the position and magnitude of all its sides are given4 ,fix the position of the point ,a in a4 ,find ,v1 the vanishing-point of ,,ab2 and ,m1 the dividing-point of ,,ab1 nearest ,s4 ,find ,v1 the vanishing-point of ,,ac2 and ,n1 the dividing-point of ,,ac1 nearest ,s4 ,draw the measuring-line through a1 and make ab1 ac1 each equal to the sight-magnitude of ,,ab4 7,for since ,,abcd is a square1 ,,ac is equal to ,,ab47 ,draw a,v and c,n1 cutting each other in c4 ,draw a,v1 and b,m1 cutting each other in b4 ,then ac1 ab1 are the two nearest sides of the square4 ,now1 clearing the figure of superfluous lines1 we have ab1 ac1 drawn in position1 as in ,fig4 #ba4 ,illustration3 ,fig4 #ba4 ,and because ,,abcd is a square1 ,,cd 7,fig4 #bj47 is parallel to ,,ab4 ,and all parallel lines have the same vanishing-point4 7,note to ,problem ,,iii47 ,therefore1 ,v is the vanishing-point of ,,cd4 ,similarly1 ,v is the vanishing-point of ,,bd4 ,therefore1 from b and c 7,fig4 #bb47 draw b,v1 c,v1 cutting each other in d4 ,then abcd is the square required4 ,,corollary ,i4 ,it is obvious that any rectangle in a horizontal plane may be drawn by this problem1 merely making ab1 on the measuring-line1 ,fig4 #bj41 equal to the sight-magnitude of one of its sides1 and ac the sight-magnitude of the other4 ,,corollary ,,ii4 ,let abcd1 ,fig4 #bb41 be any square drawn in perspective4 ,draw the diagonals ad and bc1 cutting each other in ,c4 ,then ,c is the center of the square4 ,through ,c1 draw ef to the vanishing-point of ab1 and gh to the vanishing-point of ac1 and these lines will bisect the sides of the square1 so that ag is the perspective representation of half the side ab2 ae is half ac2 ch is half cd2 and bf is half bd4 ,illustration3 ,fig4 #bb4 ,,corollary ,,iii4 ,since ,,abcd1 ,fig4 #bj41 is a square1 ,,bac is a right angle2 and as ,,tv is parallel to ,,ab1 and ,,tv to ,,ac1 ,v,,tv must be a right angle also4 ,as the ground plan of most buildings is rectangular1 it constantly happens in practice that their angles 7as the corners of ordinary houses7 throw the lines to the vanishing-points thus at right angles2 and so that this law is observed1 and ,,vtv is kept a right angle1 it does not matter in general practice whether the vanishing-points are thrown a little more or a little less to the right or left of ,s3 but it matters much that the relation of the vanishing-points should be accurate4 ,their position with respect to ,s merely causes the spectator to see a little more or less on one side or other of the house1 which may be a matter of chance or choice2 but their rectangular relation determines the rectangular shape of the building1 which is an essential point4 ,,problem ,,ix4 ,,to ,,draw ,a ,,square ,,pillar1 ,,given ,,in ,,position ,,and ,,magnitude1 ,,its ,,base ,,and ,,top ,,being ,,in ,,horizontal ,,planes4 ,let ,,ah1 ,fig4 #bc41 be the square pillar4 ,then1 as it is given in position and magnitude1 the position and magnitude of the square it stands upon must be given 7that is1 the line ,,ab or ,,ac in position71 and the height of its side ,,ae4 ,illustration3 ,fig4 #bc4 ,illustration3 ,fig4 #bd4 ,find the sight-magnitudes of ,,ab and ,,ae4 ,draw the two sides ab1 ac1 of the square of the base1 by ,problem ,,viii41 as in ,fig4 #bd4 ,from the points a1 b1 and c1 raise vertical lines ae1 cf1 bg4 ,make ae equal to the sight-magnitude of ,,ae4 ,now because the top and base of the pillar are in horizontal planes1 the square of its top1 ,,fg1 is parallel to the square of its base1 ,,bc4 ,therefore the line ,,ef is parallel to ,,ac1 and ,,eg to ,,ab4 ,therefore ,,ef has the same vanishing-point as ,,ac1 and ,,eg the same vanishing-point as ,,ab4 ,from e draw ef to the vanishing-point of ac1 cutting cf in f4 ,similarly draw eg to the vanishing-point of ab1 cutting bg in g4 ,complete the square gf in h1 by drawing gh to the vanishing-point of ef1 and fh to the vanishing-point of eg1 cutting each other in h4 ,then aghf is the square pillar required4 ,,corollary4 ,it is obvious that if ,,ae is equal to ,,ac1 the whole figure will be a cube1 and each side1 aefc and aegb1 will be a square in a given vertical plane4 ,and by making ,,ab or ,,ac longer or shorter in any given proportion1 any form of rectangle may be given to either of the sides of the pillar4 ,no other rule is therefore needed for drawing squares or rectangles in vertical planes4 ,also any triangle may be thus drawn in a vertical plane1 by inclosing it in a rectangle and determining1 in perspective ratio1 on the sides of the rectangle1 the points of their contact with the angles of the triangle4 ,and if any triangle1 then any polygon4 ,a less complicated construction will1 however1 be given hereafter4#ah #ah ,see page #if 7note71 after you have read ,problem ,,xvi4 ,,problem ,x4 ,,to ,,draw ,a ,,pyramid1 ,,given ,,in ,,position ,,and ,,magnitude1 ,,on ,a ,,square ,,base ,,in ,a ,,horizontal ,,plane4 ,illustration3 ,fig4 #be4 ,let ,,ab1 ,fig4 #be41 be the four-sided pyramid4 ,as it is given in position and magnitude1 the square base on which it stands must be given in position and magnitude1 and its vertical height1 ,,cd4#ai ,illustration3 ,fig4 #bf4 ,draw a square pillar1 ,,abge1 ,fig4 #bf41 on the square base of the pyramid1 and make the height of the pillar ,,af equal to the vertical height of the pyramid ,,cd 7,problem ,,ix474 ,draw the diagonals ,,gf1 ,,hi1 on the top of the square pillar1 cutting each other in ,c4 ,therefore ,c is the center of the square ,,fghi4 7,prob4 ,,viii4 ,cor4 ,,ii47 ,illustration3 ,fig4 #bg4 ,join ,,ce1 ,,ca1 ,,cb4 ,then ,,abce is the pyramid required4 ,if the base of the pyramid is above the eye1 as when a square spire is seen on the top of a church-tower1 the construction will be as in ,fig4 #bg4 #ai ,if1 instead of the vertical height1 the length of ,,ad is given1 the vertical must be deduced from it4 ,see the ,exercises on this ,problem in the ,appendix1 p4 #gi4 ,,problem ,,xi4 ,,to ,,draw ,,any ,,curve ,,in ,a ,,horizontal ,,or ,,vertical ,,plane4 ,illustration3 ,fig4 #bh4 ,let ,,ab1 ,fig4 #bh41 be the curve4 ,inclose it in a rectangle1 ,,cdef4 ,fix the position of the point ,c or ,d1 and draw the rectangle4 7,problem ,,viii4 ,coroll4 ,i47#bj ,let ,,cdef1 ,fig4 #bi41 be the rectangle so drawn4 ,illustration3 ,fig4 #bi4 ,if an extremity of the curve1 as ,a1 is in a side of the rectangle1 divide the side ,,ce1 ,fig4 #bi41 so that ,,ac shall be 7in perspective ratio7 to ,,ae as ,,ac is to ,,ae in ,fig4 #bh4 7,prob4 ,v4 ,cor4 ,,ii47 ,similarly determine the points of contact of the curve and rectangle e1 f1 g4 ,if an extremity of the curve1 as ,b1 is not in a side of the rectangle1 let fall the perpendiculars ,ba1 ,bb on the rectangle sides4 ,determine the correspondent points a and b in ,fig4 #bi41 as you have already determined ,a1 ,b1 e1 and f4 ,from b1 ,fig4 #bi41 draw b,b parallel to ,,cd1#ba and from a draw a,b to the vanishing-point of ,,df1 cutting each other in ,b4 ,then ,b is the extremity of the curve4 ,determine any other important point in the curve1 as ,p1 in the same way1 by letting fall ,pq and ,pr on the rectangle's sides4 ,any number of points in the curve may be thus determined1 and the curve drawn through the series2 in most cases1 three or four will be enough4 ,practically1 complicated curves may be better drawn in perspective by an experienced eye than by rule1 as the fixing of the various points in haste involves too many chances of error2 but it is well to draw a good many by rule first1 in order to give the eye its experience4#bb ,,corollary4 ,if the curve required be a circle1 ,fig4 #cj41 the rectangle which incloses it will become a square1 and the curve will have four points of contact1 ,,abcd1 in the middle of the sides of the square4 ,illustration3 ,fig4 #cj4 ,draw the square1 and as a square may be drawn about a circle in any position1 draw it with its nearest side1 ,,eg1 parallel to the sight-line4 ,let ,,ef1 ,fig4 #ca41 be the square so drawn4 ,draw its diagonals ,,ef1 ,,gh2 and through the center of the square 7determined by their intersection7 draw ,,ab to the vanishing-point of ,,gf1 and ,,cd parallel to ,,eg4 ,then the points ,,abcd are the four points of the circle's contact4 ,illustration3 ,fig4 #ca4 ,on ,,eg describe a half square1 ,,el2 draw the semicircle ,,kal2 and from its center1 ,r1 the diagonals ,,re1 ,,rg1 cutting the circle in x1 y4 ,from the points x y1 where the circle cuts the diagonals1 raise perpendiculars1 ,px1 ,qy1 to ,,eg4 ,from ,p and ,q draw ,,pp1 ,,qq1 to the vanishing-point of ,,gf1 cutting the diagonals in m1 n1 and o1 p4 ,then m1 n1 o1 p are four other points in the circle4 ,through these eight points the circle may be drawn by the hand accurately enough for general purposes2 but any number of points required may1 of course1 be determined1 as in ,problem ,,xi4 ,the distance ,,ep is approximately one-seventh of ,,eg1 and may be assumed to be so in quick practice1 as the error involved is not greater than would be incurred in the hasty operation of drawing the circle and diagonals4 ,it may frequently happen that1 in consequence of associated constructions1 it may be inconvenient to draw ,,eg parallel to the sight-line1 the square being perhaps first constructed in some oblique direction4 ,in such cases1 ,,qg and ,,ep must be determined in perspective ratio by the dividing-point1 the line ,,eg being used as a measuring-line4 ,obs4 ,in drawing ,fig4 #ca4 the station-point has been taken much nearer the paper than is usually advisable1 in order to show the character of the curve in a very distinct form4 ,if the student turns the book so that ,,eg may be vertical1 ,fig4 #ca4 will represent the construction for drawing a circle in a vertical plane1 the sight-line being then of course parallel to ,,gl2 and the semicircles ,,adb1 ,,acb1 on each side of the diameter ,,ab1 will represent ordinary semicircular arches seen in perspective4 ,in that case1 if the book be held so that the line ,,eh is the top of the square1 the upper semicircle will represent a semicircular arch1 above the eye1 drawn in perspective4 ,but if the book be held so that the line ,,gf is the top of the square1 the upper semicircle will represent a semicircular arch1 below the eye1 drawn in perspective4 ,if the book be turned upside down1 the figure will represent a circle drawn on the ceiling1 or any other horizontal plane above the eye2 and the construction is1 of course1 accurate in every case4 #bj ,or if the curve is in a vertical plane1 ,coroll4 to ,problem ,,ix4 ,as a rectangle may be drawn in any position round any given curve1 its position with respect to the curve will in either case be regulated by convenience4 ,see the ,exercises on this ,problem1 in the ,appendix1 p4 #he4 #ba ,or to its vanishing-point1 if ,,cd has one4 #bb ,of course1 by dividing the original rectangle into any number of equal rectangles1 and dividing the perspective rectangle similarly1 the curve may be approximately drawn without any trouble2 but1 when accuracy is required1 the points should be fixed1 as in the problem4 ,,problem ,,xii4 ,,to ,,divide ,a ,,circle ,,drawn ,,in ,,perspective ,,into ,,any ,,given ,,number ,,of ,,equal ,,parts4 ,let ,,ab1 ,fig4 #cb41 be the circle drawn in perspective4 ,it is required to divide it into a given number of equal parts2 in this case1 #bj4 ,let ,,kal be the semicircle used in the construction4 ,divide the semicircle ,,kal into half the number of parts required2 in this case1 #aj4 ,produce the line ,,eg laterally1 as far as may be necessary4 ,from ,o1 the center of the semicircle ,,kal1 draw radii through the points of division of the semicircle1 p1 q1 r1 etc41 and produce them to cut the line ,,eg in ,p1 ,q1 ,r1 etc4 ,from the points ,,pqr draw the lines ,,pp1 ,,qq1 ,,rr1 etc41 through the center of the circle ,,ab1 each cutting the circle in two points of its circumference4 ,then these points divide the perspective circle as required4 ,if from each of the points p1 q1 r1 a vertical were raised to the line ,,eg1 as in ,fig4 #ca41 and from the point where it cut ,,eg a line were drawn to the vanishing-point1 as ,,qq in ,fig4 #ca41 this line would also determine two of the points of division4 ,illustration3 ,fig4 #cb4 ,if it is required to divide a circle into any number of given unequal parts 7as in the points ,a1 ,b1 and ,c1 ,fig4 #cc471 the shortest way is thus to raise vertical lines from ,a and ,b to the side of the perspective square ,,xy1 and then draw to the vanishing-point1 cutting the perspective circle in a and b1 the points required4 ,only notice that if any point1 as ,a1 is on the nearer side of the circle ,,abc1 its representative point1 a1 must be on the nearer side of the circle abc2 and if the point ,b is on the farther side of the circle ,,abc1 b must be on the farther side of abc4 ,if any point1 as ,c1 is so much in the lateral arc of the circle as not to be easily determinable by the vertical line1 draw the horizontal ,,cp1 find the correspondent p in the side of the perspective square1 and draw pc parallel to ,,xy1 cutting the perspective circle in c4 ,illustration3 ,fig4 #cc4 ,,corollary4 ,it is obvious that if the points ,p1 ,q1 ,r1 etc41 by which the circle is divided in ,fig4 #cb41 be joined by right lines1 the resulting figure will be a regular equilateral figure of twenty sides inscribed in the circle4 ,and if the circle be divided into given unequal parts1 and the points of division joined by right lines1 the resulting figure will be an irregular polygon inscribed in the circle with sides of given length4 ,thus any polygon1 regular or irregular1 inscribed in a circle1 may be inscribed in position in a perspective circle4 ,,problem ,,xiii4 ,,to ,,draw ,a ,,square1 ,,given ,,in ,,magnitude1 ,,within ,a ,,larger ,,square ,,given ,,in ,,position ,,and ,,magnitude2 ,,the ,,sides ,,of ,,the ,,two ,,squares ,,being ,,parallel4 ,illustration3 ,fig4 #cd4 ,let ,,ab1 ,fig4 #cd41 be the sight-magnitude of the side of the smaller square1 and ,,ac that of the side of the larger square4 ,draw the larger square4 ,let ,,defg be the square so drawn4 ,join ,,eg and ,,df4 ,on either ,,de or ,,dg set off1 in perspective ratio1 ,,dh equal to one half of ,,bc4 ,through ,h draw ,,hk to the vanishing-point of ,,de1 cutting ,,df in ,i and ,,eg in ,k4 ,through ,i and ,k draw ,,im1 ,,kl1 to vanishing-point of ,,dg1 cutting ,,df in ,l and ,,eg in ,m4 ,join ,,lm4 ,then ,,iklm is the smaller square1 inscribed as required4#bc ,,corollary4 ,illustration3 ,fig4 #cf4 ,if1 instead of one square within another1 it be required to draw one circle within another1 the dimensions of both being given1 inclose each circle in a square4 ,draw the squares first1 and then the circles within1 as in ,fig4 #cf4 #bc ,illustration3 ,fig4 #ce4 ,if either of the sides of the greater square is parallel to the plane of the picture1 as ,,dg in ,fig4 #ce41 ,,dg of course must be equal to ,,ac1 and ,,dh equal to ,,bc_/#b1 and the construction is as in ,fig4 #ce4 ,,problem ,,xiv4 ,,to ,,draw ,a ,,truncated ,,circular ,,cone1 ,,given ,,in ,,position ,,and ,,magnitude1 ,,the ,,truncations ,,being ,,in ,,horizontal ,,planes1 ,,and ,,the ,,axis ,,of ,,the ,,cone ,,vertical4 ,let ,,abcd1 ,fig4 #cg41 be the portion of the cone required4 ,illustration3 ,fig4 #cg4 ,as it is given in magnitude1 its diameters must be given at the base and summit1 ,,ab and ,,cd2 and its vertical height1 ,,ce4#bd ,and as it is given in position1 the center of its base must be given4 ,illustration3 ,fig4 #ch4 ,draw in position1 about this center1#be the square pillar afd1 ,fig4 #ch41 making its height1 bg1 equal to ,,ce2 and its side1 ab1 equal to ,,ab4 ,in the square of its base1 abcd1 inscribe a circle1 which therefore is of the diameter of the base of the cone1 ,,ab4 ,in the square of its top1 efgh1 inscribe concentrically a circle whose diameter shall equal ,,cd4 7,coroll4 ,prob4 ,,xiii47 ,join the extremities of the circles by the right lines kl1 nm4 ,then klnm is the portion of cone required4 ,,corollary ,i4 ,if similar polygons be inscribed in similar positions in the circles kn and lm 7,coroll4 ,prob4 ,,xii471 and the corresponding angles of the polygons joined by right lines1 the resulting figure will be a portion of a polygonal pyramid4 7,the dotted lines in ,fig4 #ch41 connecting the extremities of two diameters and one diagonal in the respective circles1 occupy the position of the three nearest angles of a regular octagonal pyramid1 having its angles set on the diagonals and diameters of the square ad1 inclosing its base47 ,if the cone or polygonal pyramid is not truncated1 its apex will be the center of the upper square1 as in ,fig4 #bf4 ,,corollary ,,ii4 ,if equal circles1 or equal and similar polygons1 be inscribed in the upper and lower squares in ,fig4 #ch41 the resulting figure will be a vertical cylinder1 or a vertical polygonal pillar1 of given height and diameter1 drawn in position4 ,,corollary ,,iii4 ,if the circles in ,fig4 #ch41 instead of being inscribed in the squares bc and fg1 be inscribed in the sides of the solid figure be and df1 those sides being made square1 and the line bd of any given length1 the resulting figure will be1 according to the constructions employed1 a cone1 polygonal pyramid1 cylinder1 or polygonal pillar1 drawn in position about a horizontal axis parallel to bd4 ,similarly1 if the circles are drawn in the sides gd and ec1 the resulting figures will be described about a horizontal axis parallel to ab4 #bd ,or if the length of its side1 ,,ac1 is given instead1 take ae1 ,fig4 #cg41 equal to half the excess of ,,ab over ,,cd2 from the point e raise the perpendicular ce4 ,with center a1 and distance ,,ac1 describe a circle cutting ce in c4 ,then ce is the vertical height of the portion of cone required1 or ,,ce4 #be ,the direction of the side of the square will of course be regulated by convenience4 ,,problem ,,xv4 ,,to ,,draw ,,an ,,inclined ,,line1 ,,given ,,in ,,position ,,and ,,magnitude4 ,we have hitherto been examining the conditions of horizontal and vertical lines only1 or of curves inclosed in rectangles4 ,illustration3 ,fig4 #ci4 ,illustration3 ,fig4 #dj4 ,we must1 in conclusion1 investigate the perspective of inclined lines1 beginning with a single one given in position4 ,for the sake of completeness of system1 ,i give in ,appendix ,,ii4 ,article ,,iii4 the development of this problem from the second4 ,but1 in practice1 the position of an inclined line may be most conveniently defined by considering it as the diagonal of a rectangle1 as ,,ab in ,fig4 #ci41 and ,i shall therefore1 though at some sacrifice of system1 examine it here under that condition4 ,if the sides of the rectangle ,,ac and ,,ad are given1 the slope of the line ,,ab is determined2 and then its position will depend on that of the rectangle4 ,if1 as in ,fig4 #ci41 the rectangle is parallel to the picture plane1 the line ,,ab must be so also4 ,if1 as in ,fig4 #dj41 the rectangle is inclined to the picture plane1 the line ,,ab will be so also4 ,so that1 to fix the position of ,,ab1 the line ,,ac must be given in position and magnitude1 and the height ,,ad4 ,illustration3 ,fig4 #da4 ,if these are given1 and it is only required to draw the single line ,,ab in perspective1 the construction is entirely simple2 thus3-- ,draw the line ,,ac by ,problem ,i4 ,let ,,ac1 ,fig4 #da41 be the line so drawn4 ,from a and c raise the vertical lines ad1 cb4 ,make ad equal to the sight-magnitude of ,,ad4 ,from d draw db to the vanishing-point of ac1 cutting bc in b4 ,join ab4 ,then ab is the inclined line required4 ,illustration3 ,fig4 #db4 ,if the line is inclined in the opposite direction1 as ,,dc in ,fig4 #db41 we have only to join dc instead of ab in ,fig4 #da41 and dc will be the line required4 ,i shall hereafter call the line ,,ac1 when used to define the position of an inclined line ,,ab 7,fig4 #dj471 the 8relative horizontal0 of the line ,,ab4 ,,observation4 ,illustration3 ,fig4 #dc4 ,in general1 inclined lines are most needed for gable roofs1 in which1 when the conditions are properly stated1 the vertical height of the gable1 ,,xy1 ,fig4 #dc41 is given1 and the base line1 ,,ac1 in position4 ,when these are given1 draw ,,ac2 raise vertical ,,ad2 make ,,ad equal to sight-magnitude of ,,xy2 complete the perspective-rectangle ,,adbc2 join ,,ab and ,,dc 7as by dotted lines in figure72 and through the intersection of the dotted lines draw vertical ,,xy1 cutting ,,db in ,y4 ,join ,,ay1 ,,cy2 and these lines are the sides of the gable4 ,if the length of the roof ,,aa is also given1 draw in perspective the complete parallelopiped ,a,d,,bc1 and from ,y draw ,,yy to the vanishing-point of ,,aa1 cutting ,d,b in ,y4 ,join ,a,y1 and you have the slope of the farther side of the roof4 ,illustration3 ,fig4 #dd4 ,the construction above the eye is as in ,fig4 #dd42 the roof is reversed in direction merely to familiarize the student with the different aspects of its lines4 ,,problem ,,xvi4 ,,to ,,find ,,the ,,vanishing-,,point ,,of ,a ,,given ,,inclined ,,line4 ,if1 in ,fig4 #dc4 or ,fig4 #dd41 the lines ,,ay and ,a,y be produced1 the student will find that they meet4 ,let ,p1 ,fig4 #de41 be the point at which they meet4 ,from ,p let fall the vertical ,,pv on the sight-line1 cutting the sight-line in ,v4 ,then the student will find experimentally that ,v is the vanishing-point of the line ,,ac4#bf ,complete the rectangle of the base ,,ac1 by drawing ,a,c to ,v1 and ,,cc to the vanishing-point of ,,aa4 ,join ,y,c4 ,now if ,,yc and ,y,c be produced downwards1 the student will find that they meet4 ,let them be produced1 and meet in ,p4 ,produce ,,pv1 and it will be found to pass through the point ,p4 ,therefore if ,,ay 7or ,,cy71 ,fig4 #de41 be any inclined line drawn in perspective by ,problem ,,xv41 and ,,ac the relative horizontal 7,,ac in ,figs4 #ci1 #dj471 also drawn in perspective4 ,through ,v1 the vanishing-point of ,,av1 draw the vertical ,,pp upwards and downwards4 ,produce ,,ay 7or ,,cy71 cutting ,,pp in ,p 7or ,p74 ,then ,p is the vanishing-point of ,,ay 7or ,p of ,,cy74 ,illustration3 ,fig4 #de4 ,the student will observe that1 in order to find the point ,p by this method1 it is necessary first to draw a portion of the given inclined line by ,problem ,,xv4 ,practically1 it is always necessary to do so1 and1 therefore1 ,i give the problem in this form4 ,theoretically1 as will be shown in the analysis of the problem1 the point ,p should be found by drawing a line from the station-point parallel to the given inclined line3 but there is no practical means of drawing such a line2 so that in whatever terms the problem may be given1 a portion of the inclined line 7,,ay or ,,cy7 must always be drawn in perspective before ,p can be found4 #bf ,the demonstration is in ,appendix ,,ii4 ,article ,,iii4 ,,problem ,,xvii4 ,,to ,,find ,,the ,,dividing-,,points ,,of ,a ,,given ,,inclined ,,line4 ,illustration3 ,fig4 #df4 ,let ,p1 ,fig4 #df41 be the vanishing-point of the inclined line1 and ,v the vanishing-point of the relative horizontal4 ,find the dividing-points of the relative horizontal1 ,d and ,d4 ,through ,p draw the horizontal line ,,xy4 ,with center ,p and distance ,,dp describe the two arcs ,,dx and ,d,y1 cutting the line ,,xy in ,x and ,y4 ,then ,x and ,y are the dividing-points of the inclined line4#bg ,obs4 ,the dividing-points found by the above rule1 used with the ordinary measuring-line1 will lay off distances on the retiring inclined line1 as the ordinary dividing-points lay them off on the retiring horizontal line4 ,another dividing-point1 peculiar in its application1 is sometimes useful1 and is to be found as follows3-- ,illustration3 ,fig4 #dg4 ,let ,,ab1 ,fig4 #dg41 be the given inclined line drawn in perspective1 and ,ac the relative horizontal4 ,find the vanishing-points1 ,v and ,e1 of ,ac and ,,ab2 ,d1 the dividing-point of ,ac2 and the sight-magnitude of ,ac on the measuring-line1 or ,,ac4 ,from ,d erect the perpendicular ,,df4 ,join ,,cb1 and produce it to cut ,,de in ,f4 ,join ,,ef4 ,then1 by similar triangles1 ,,df is equal to ,,ev1 and ,,ef is parallel to ,,dv4 ,hence it follows that if from ,d1 the dividing-point of ,ac1 we raise a perpendicular and make ,,df equal to ,,ev1 a line ,,cf1 drawn from any point ,c on the measuring-line to ,f1 will mark the distance ,,ab on the inclined line1 ,,ab being the portion of the given inclined line which forms the diagonal of the vertical rectangle of which ,,ac is the base4 #bg ,the demonstration is in ,appendix ,,ii41 p4 #ajd4 ,,problem ,,xviii4 ,,to ,,find ,,the ,,sight-,,line ,,of ,,an ,,inclined ,,plane ,,in ,,which ,,two ,,lines ,,are ,,given ,,in ,,position4#bh ,as in order to fix the position of a line two points in it must be given1 so in order to fix the position of a plane1 two lines in it must be given4 ,illustration3 ,fig4 #dh ,let the two lines be ,,ab and ,,cd1 ,fig4 #dh4 ,as they are given in position1 the relative horizontals ,,ae and ,,cf must be given4 ,then by ,problem ,,xvi4 the vanishing-point of ,,ab is ,v1 and of ,,cd1 ,v4 ,join ,,vv and produce it to cut the sight-line in ,x4 ,then ,,vx is the sight-line of the inclined plane4 ,like the horizontal sight-line1 it is of indefinite length2 and may be produced in either direction as occasion requires1 crossing the horizontal line of sight1 if the plane continues downward in that direction4 ,x is the vanishing-point of all horizontal lines in the inclined plane4 #bh ,read the ,article on this problem in the ,appendix1 p4 #ig1 before investigating the problem itself4 ,,problem ,,xix4 ,,to ,,find ,,the ,,vanishing-,,point ,,of ,,steepest ,,lines ,,in ,,an ,,inclined ,,plane ,,whose ,,sight-,,line ,,is ,,given4 ,illustration3 ,fig4 #di4 ,let ,,vx1 ,fig4 #di41 be the given sight-line4 ,produce it to cut the horizontal sight-line in ,x4 ,therefore ,x is the vanishing-point of horizontal lines in the given inclined plane4 7,problem ,,xviii47 ,join ,,tx1 and draw ,,ty at right angles to ,,tx4 ,therefore ,y is the rectangular vanishing-point corresponding to ,x4#bi ,from ,y erect the vertical ,,yp1 cutting the sight-line of the inclined plane in ,p4 ,then ,p is the vanishing-point of steepest lines in the plane4 ,all lines drawn to it1 as ,,qp1 ,,rp1 ,,np1 etc41 are the steepest possible in the plane2 and all lines drawn to ,x1 as ,,qx1 ,,ox1 etc41 are horizontal1 and at right angles to the lines ,,pq1 ,,pr1 etc4 #bi ,that is to say1 the vanishing-point of horizontal lines drawn at right angles to the lines whose vanishing-point is ,x4 ,,problem ,,xx4 ,,to ,,find ,,the ,,vanishing-,,point ,,of ,,lines ,,perpendicular ,,to ,,the ,,surface ,,of ,a ,,given ,,inclined ,,plane4 ,illustration3 ,fig4 #ej4 ,as the inclined plane is given1 one of its steepest lines must be given1 or may be ascertained4 ,let ,,ab1 ,fig4 #ej41 be a portion of a steepest line in the given plane1 and ,v the vanishing-point of its relative horizontal4 ,through ,v draw the vertical ,,gf upwards and downwards4 ,from ,a set off any portion of the relative horizontal ,,ac1 and on ,,ac describe a semicircle in a vertical plane1 ,,adc1 cutting ,,ab in ,e4 ,join ,,ec1 and produce it to cut ,,gf in ,f4 ,then ,f is the vanishing-point required4 ,for1 because ,,aec is an angle in a semicircle1 it is a right angle2 and therefore the line ,,ef is at right angles to the line ,,ab2 and similarly all lines drawn to ,f1 and therefore parallel to ,,ef1 are at right angles with any line which cuts them1 drawn to the vanishing-point of ,,ab4 ,and because the semicircle ,,adc is in a vertical plane1 and its diameter ,,ac is at right angles to the horizontal lines traversing the surface of the inclined plane1 the line ,,ec1 being in this semicircle1 is also at right angles to such traversing lines4 ,and therefore the line ,,ec1 being at right angles to the steepest lines in the plane1 and to the horizontal lines in it1 is perpendicular to its surface4 99 99 99 99 99 ,the preceding series of constructions1 with the examples in the first ,article of the ,appendix1 put it in the power of the student to draw any form1 however complicated1#cj which does not involve intersection of curved surfaces4 ,i shall not proceed to the analysis of any of these more complex problems1 as they are entirely useless in the ordinary practice of artists4 ,for a few words only ,i must ask the reader's further patience1 respecting the general placing and scale of the picture4 ,as the horizontal sight-line is drawn through the sight-point1 and the sight-point is opposite the eye1 the sight-line is always on a level with the eye4 ,above and below the sight-line1 the eye comprehends1 as it is raised or depressed while the head is held upright1 about an equal space2 and1 on each side of the sight-point1 about the same space is easily seen without turning the head2 so that if a picture represented the true field of easy vision1 it ought to be circular1 and have the sight-point in its center4 ,but because some parts of any given view are usually more interesting than others1 either the uninteresting parts are left out1 or somewhat more than would generally be seen of the interesting parts is included1 by moving the field of the picture a little upwards or downwards1 so as to throw the sight-point low or high4 ,the operation will be understood in a moment by cutting an aperture in a piece of pasteboard1 and moving it up and down in front of the eye1 without moving the eye4 ,it will be seen to embrace sometimes the low1 sometimes the high objects1 without altering their perspective1 only the eye will be opposite the lower part of the aperture when it sees the higher objects1 and vice vers4 ,there is no reason1 in the laws of perspective1 why the picture should not be moved to the right or left of the sight-point1 as well as up or down4 ,but there is this practical reason4 ,the moment the spectator sees the horizon in a picture high1 he tries to hold his head high1 that is1 in its right place4 ,when he sees the horizon in a picture low1 he similarly tries to put his head low4 ,but1 if the sight-point is thrown to the left hand or right hand1 he does not understand that he is to step a little to the right or left2 and if he places himself1 as usual1 in the middle1 all the perspective is distorted4 ,hence it is generally unadvisable to remove the sight-point laterally1 from the center of the picture4 ,the ,dutch painters1 however1 fearlessly take the license of placing it to the right or left2 and often with good effect4 ,the rectilinear limitation of the sides1 top1 and base of the picture is of course quite arbitrary1 as the space of a landscape would be which was seen through a window2 less or more being seen at the spectator's pleasure1 as he retires or advances4 ,the distance of the station-point is not so arbitrary4 ,in ordinary cases it should not be less than the intended greatest dimension 7height or breadth7 of the picture4 ,in most works by the great masters it is more2 they not only calculate on their pictures being seen at considerable distances1 but they like breadth of mass in buildings1 and dislike the sharp angles which always result from station-points at short distances4#ca ,whenever perspective1 done by true rule1 looks wrong1 it is always because the station-point is too near4 ,determine1 in the outset1 at what distance the spectator is likely to examine the work1 and never use a station-point within a less distance4 ,there is yet another and a very important reason1 not only for care in placing the station-point1 but for that accurate calculation of distance and observance of measurement which have been insisted on throughout this work4 ,all drawings of objects on a reduced scale are1 if rightly executed1 drawings of the appearance of the object at the distance which in true perspective reduces it to that scale4 ,they are not small drawings of the object seen near1 but drawings the real size of the object seen far off4 ,thus if you draw a mountain in a landscape1 three inches high1 you do not reduce all the features of the near mountain so as to come into three inches of paper4 ,you could not do that4 ,all that you can do is to give the appearance of the mountain1 when it is so far off that three inches of paper would really hide it from you4 ,it is precisely the same in drawing any other object4 ,a face can no more be reduced in scale than a mountain can4 ,it is infinitely delicate already2 it can only be quite rightly rendered on its own scale1 or at least on the slightly diminished scale which would be fixed by placing the plate of glass1 supposed to represent the field of the picture1 close to the figures4 ,correggio and ,raphael were both fond of this slightly subdued magnitude of figure4 ,colossal painting1 in which ,correggio excelled all others1 is usually the enlargement of a small picture 7as a colossal sculpture is of a small statue71 in order to permit the subject of it to be discerned at a distance4 ,the treatment of colossal 7as distinguished from ordinary7 paintings will depend therefore1 in general1 on the principles of optics more than on those of perspective1 though1 occasionally1 portions may be represented as if they were the projection of near objects on a plane behind them4 ,in all points the subject is one of great difficulty and subtlety2 and its examination does not fall within the compass of this essay4 ,lastly1 it will follow from these considerations1 and the conclusion is one of great practical importance1 that1 though pictures may be enlarged1 they cannot be reduced1 in copying them4 ,all attempts to engrave pictures completely on a reduced scale are1 for this reason1 nugatory4 ,the best that can be done is to give the aspect of the picture at the distance which reduces it in perspective to the size required2 or1 in other words1 to make a drawing of the distant effect of the picture4 ,good painting1 like nature's own work1 is infinite1 and unreduceable4 ,i wish this book had less tendency towards the infinite and unreduceable4 ,it has so far exceeded the limits ,i hoped to give it1 that ,i doubt not the reader will pardon an abruptness of conclusion1 and be thankful1 as ,i am myself1 to get to an end on any terms4 #cj ,as in algebraic science1 much depends1 in complicated perspective1 on the student's ready invention of expedients1 and on his quick sight of the shortest way in which the solution may be accomplished1 when there are several ways4 #ca ,the greatest masters are also fond of parallel perspective1 that is to say1 of having one side of their buildings fronting them full1 and therefore parallel to the picture plane1 while the other side vanishes to the sight-point4 ,this is almost always done in figure backgrounds1 securing simple and balanced lines4 ,,appendix4 ,i4 ,,practice ,,and ,,observations4 ,,ii4 ,,demonstrations4 ,i4 ,,practice ,,and ,,observations ,,on ,,the ,,preceding ,,problems4 ,,problem ,i4 ,an example will be necessary to make this problem clear to the general student4 ,the nearest corner of a piece of pattern on the carpet is #d feet beneath the eye1 #b feet to our right and #c feet in direct distance from us4 ,we intend to make a drawing of the pattern which shall be seen properly when held #a foot from the eye4 ,it is required to fix the position of the corner of the piece of pattern4 ,illustration3 ,fig4 #ea4 ,let ,,ab1 ,fig4 #ea41 be our sheet of paper1 some #c feet wide4 ,make ,,st equal to #a foot4 ,draw the line of sight through ,s4 ,produce ,,ts1 and make ,,ds equal to #b feet1 therefore ,,td equal to #c feet4 ,draw ,,dc1 equal to #b feet2 ,,cp1 equal to #d feet4 ,join ,,tc 7cutting the sight-line in ,q7 and ,,tp4 ,let fall the vertical ,,qp1 then ,p is the point required4 ,if the lines1 as in the figure1 fall outside of your sheet of paper1 in order to draw them1 it is necessary to attach other sheets of paper to its edges4 ,this is inconvenient1 but must be done at first that you may see your way clearly2 and sometimes afterwards1 though there are expedients for doing without such extension in fast sketching4 ,it is evident1 however1 that no extension of surface could be of any use to us1 if the distance ,,td1 instead of being #c feet1 were #ajj feet1 or a mile1 as it might easily be in a landscape4 ,it is necessary1 therefore1 to obtain some other means of construction2 to do which we must examine the principle of the problem4 ,in the analysis of ,fig4 #b41 in the introductory remarks1 ,i used the word 8height0 only of the tower1 ,,qp1 because it was only to its vertical height that the law deduced from the figure could be applied4 ,for suppose it had been a pyramid1 as ,,oqp1 ,fig4 #eb41 then the image of its side1 ,,qp1 being1 like every other magnitude1 limited on the glass ,,ab by the lines coming from its extremities1 would appear only of the length ,q,s2 and it is not true that ,q,s is to ,,qp as ,,ts is to ,,tp4 ,but if we let fall a vertical ,,qd from ,q1 so as to get the vertical height of the pyramid1 then it is true that ,q,s is to ,,qd as ,,ts is to ,,td4 ,illustration3 ,fig4 #eb4 ,supposing this figure represented1 not a pyramid1 but a triangle on the ground1 and that ,,qd and ,,qp are horizontal lines1 expressing lateral distance from the line ,,td1 still the rule would be false for ,,qp and true for ,,qd4 ,and1 similarly1 it is true for all lines which are parallel1 like ,,qd1 to the plane of the picture ,,ab1 and false for all lines which are inclined to it at an angle4 ,hence generally4 ,let ,,pq 7,fig4 #b4 in ,introduction1 p4 #f7 be any magnitude parallel to the plane of the picture2 and ,p,q its image on the picture4 ,then always the formula is true which you learned in the ,introduction3 ,p,q is to ,,pq as ,,st is to ,,dt4 ,now the magnitude ,p dash ,q dash in this formula ,i call the 8,,sight-,,magnitude0 of the line ,,pq4 ,the student must fix this term1 and the meaning of it1 well in his mind4 ,the 8sight-magnitude0 of a line is the magnitude which bears to the real line the same proportion that the distance of the picture bears to the distance of the object4 ,thus1 if a tower be a hundred feet high1 and a hundred yards off2 and the picture1 or piece of glass1 is one yard from the spectator1 between him and the tower2 the distance of picture being then to distance of tower as #a to #ajj1 the sight-magnitude of the tower's height will be as #a to #ajj2 that is to say1 one foot4 ,if the tower is two hundred yards distant1 the sight-magnitude of its height will be half a foot1 and so on4 ,but farther4 ,it is constantly necessary1 in perspective operations1 to measure the other dimensions of objects by the sight-magnitude of their vertical lines4 ,thus1 if the tower1 which is a hundred feet high1 is square1 and twenty-five feet broad on each side2 if the sight-magnitude of the height is one foot1 the measurement of the side1 reduced to the same scale1 will be the hundredth part of twenty-five feet1 or three inches3 and1 accordingly1 ,i use in this treatise the term 8sight-magnitude0 indiscriminately for all lines reduced in the same proportion as the vertical lines of the object4 ,if ,i tell you to find the 8sight-magnitude0 of any line1 ,i mean1 always1 find the magnitude which bears to that line the proportion of ,,st to ,,dt2 or1 in simpler terms1 reduce the line to the scale which you have fixed by the first determination of the length ,,st4 ,therefore1 you must learn to draw quickly to scale before you do anything else2 for all the measurements of your object must be reduced to the scale fixed by ,,st before you can use them in your diagram4 ,if the object is fifty feet from you1 and your paper one foot1 all the lines of the object must be reduced to a scale of one fiftieth before you can use them2 if the object is two thousand feet from you1 and your paper one foot1 all your lines must be reduced to the scale of one two-thousandth before you can use them1 and so on4 ,only in ultimate practice1 the reduction never need be tiresome1 for1 in the case of large distances1 accuracy is never required4 ,if a building is three or four miles distant1 a hairbreadth of accidental variation in a touch makes a difference of ten or twenty feet in height or breadth1 if estimated by accurate perspective law4 ,hence it is never attempted to apply measurements with precision at such distances4 ,measurements are only required within distances of1 at the most1 two or three hundred feet4 ,thus it may be necessary to represent a cathedral nave precisely as seen from a spot seventy feet in front of a given pillar2 but we shall hardly be required to draw a cathedral three miles distant precisely as seen from seventy feet in advance of a given milestone4 ,of course1 if such a thing be required1 it can be done2 only the reductions are somewhat long and complicated3 in ordinary cases it is easy to assume the distance ,,st so as to get at the reduced dimensions in a moment4 ,thus1 let the pillar of the nave1 in the case supposed1 be #db feet high1 and we are required to stand #gj feet from it3 assume ,,st to be equal to #e feet4 ,then1 as #e is to #gj so will the sight-magnitude required be to #db2 that is to say1 the sight-magnitude of the pillar's height will be #c feet4 ,if we make ,,st equal to #b feet1 the pillar's height will be #a foot1 and so on4 ,and for fine divisions into irregular parts which cannot be measured1 the ninth and tenth problems of the sixth book of ,euclid will serve you3 the following construction is1 however1 ,i think1 more practically convenient3-- ,the line ,,ab 7,fig4 #ec47 is divided by given points1 a1 b1 c1 into a given number of irregularly unequal parts2 it is required to divide any other line1 ,,cd1 into an equal number of parts1 bearing to each other the same proportions as the parts of ,,ab1 and arranged in the same order4 ,draw the two lines parallel to each other1 as in the figure4 ,join ,,ac and ,,bd1 and produce the lines ,,ac1 ,,bd1 till they meet in ,p4 ,join a,p1 b,p1 c,p1 cutting c,d in f1 g1 h4 ,then the line ,,cd is divided as required1 in f1 g1 h4 ,in the figure the lines ,,ab and ,,cd are accidentally perpendicular to ,,ap4 ,there is no need for their being so4 ,illustration3 ,fig4 #ec4 ,now1 to return to our first problem4 ,the construction given in the figure is only the quickest mathematical way of obtaining1 on the picture1 the sight-magnitudes of ,,dc and ,,pc1 which are both magnitudes parallel with the picture plane4 ,but if these magnitudes are too great to be thus put on the paper1 you have only to obtain the reduction by scale4 ,thus1 if ,,ts be one foot1 ,,td eighty feet1 ,,dc forty feet1 and ,,cp ninety feet1 the distance ,,qs must be made equal to one eightieth of ,,dc1 or half a foot2 and the distance ,,qp1 one eightieth of ,,cp1 or one eightieth of ninety feet2 that is to say1 nine eighths of a foot1 or thirteen and a half inches4 ,the lines ,,ct and ,,pt are thus practically useless1 it being only necessary to measure ,,qs and ,,qp1 on your paper1 of the due sight-magnitudes4 ,but the mathematical construction1 given in ,problem ,i41 is the basis of all succeeding problems1 and1 if it is once thoroughly understood and practiced 7it can only be thoroughly understood by practice71 all the other problems will follow easily4 ,lastly4 ,observe that any perspective operation whatever may be performed with reduced dimensions of every line employed1 so as to bring it conveniently within the limits of your paper4 ,when the required figure is thus constructed on a small scale1 you have only to enlarge it accurately in the same proportion in which you reduced the lines of construction1 and you will have the figure constructed in perspective on the scale required for use4 ,,problem ,,ix4 ,the drawing of most buildings occurring in ordinary practice will resolve itself into applications of this problem4 ,in general1 any house1 or block of houses1 presents itself under the main conditions assumed here in ,fig4 #ed4 ,there will be an angle or corner somewhere near the spectator1 as ,,ab2 and the level of the eye will usually be above the base of the building1 of which1 therefore1 the horizontal upper lines will slope down to the vanishing-points1 and the base lines rise to them4 ,the following practical directions will1 however1 meet nearly all cases3-- ,illustration3 ,fig4 #ed4 ,let ,,ab1 ,fig4 #ed41 be any important vertical line in the block of buildings2 if it is the side of a street1 you may fix upon such a line at the division between two houses4 ,if its real height1 distance1 etc41 are given1 you will proceed with the accurate construction of the problem2 but usually you will neither know1 nor care1 exactly how high the building is1 or how far off4 ,in such case draw the line ,,ab1 as nearly as you can guess1 about the part of the picture it ought to occupy1 and on such a scale as you choose4 ,divide it into any convenient number of equal parts1 according to the height you presume it to be4 ,if you suppose it to be twenty feet high1 you may divide it into twenty parts1 and let each part stand for a foot2 if thirty feet high1 you may divide it into ten parts1 and let each part stand for three feet2 if seventy feet high1 into fourteen parts1 and let each part stand for five feet2 and so on1 avoiding thus very minute divisions till you come to details4 ,then observe how high your eye reaches upon this vertical line2 suppose1 for instance1 that it is thirty feet high and divided into ten parts1 and you are standing so as to raise your head to about six feet above its base1 then the sight-line may be drawn1 as in the figure1 through the second division from the ground4 ,if you are standing above the house1 draw the sight-line above ,b2 if below the house1 below ,a2 at such height or depth as you suppose may be accurate 7a yard or two more or less matters little at ordinary distances1 while at great distances perspective rules become nearly useless1 the eye serving you better than the necessarily imperfect calculation74 ,then fix your sight-point and station-point1 the latter with proper reference to the scale of the line ,,ab4 ,as you cannot1 in all probability1 ascertain the exact direction of the line ,,av or ,,bv1 draw the slope ,,bv as it appears to you1 cutting the sight-line in ,v4 ,thus having fixed one vanishing-point1 the other1 and the dividing-points1 must be accurately found by rule2 for1 as before stated1 whether your entire group of points 7vanishing and dividing7 falls a little more or less to the right or left of ,s does not signify1 but the relation of the points to each other does signify4 ,then draw the measuring-line ,,bg1 either through ,a or ,b1 choosing always the steeper slope of the two2 divide the measuring-line into parts of the same length as those used on ,,ab1 and let them stand for the same magnitudes4 ,thus1 suppose there are two rows of windows in the house front1 each window six feet high by three wide1 and separated by intervals of three feet1 both between window and window and between tier and tier2 each of the divisions here standing for three feet1 the lines drawn from ,,bg to the dividing-point ,d fix the lateral dimensions1 and the divisions on ,,ab the vertical ones4 ,for other magnitudes it would be necessary to subdivide the parts on the measuring-line1 or on ,,ab1 as required4 ,the lines which regulate the inner sides or returns of the windows 7a1 b1 c1 etc47 of course are drawn to the vanishing-point of ,,bf 7the other side of the house71 if ,,fbv represents a right angle2 if not1 their own vanishing-point must be found separately for these returns4 ,but see ,practice on ,problem ,,xi4 ,illustration3 ,fig4 #ee4 ,interior angles1 such as ,,ebc1 ,fig4 #ee4 7suppose the corner of a room71 are to be treated in the same way1 each side of the room having its measurements separately carried to it from the measuring-line4 ,it may sometimes happen in such cases that we have to carry the measurement up from the corner ,b1 and that the sight-magnitudes are given us from the length of the line ,,ab4 ,for instance1 suppose the room is eighteen feet high1 and therefore ,,ab is eighteen feet2 and we have to lay off lengths of six feet on the top of the room wall1 ,,bc4 ,find ,d1 the dividing-point of ,,bc4 ,draw a measuring-line1 ,,bf1 from ,b2 and another1 g,c1 anywhere above4 ,on ,,bf lay off ,,bg equal to one third of ,,ab1 or six feet2 and draw from ,d1 through ,g and ,b1 the lines ,gg1 ,bb1 to the upper measuring-line4 ,then gb is six feet on that measuring-line4 ,make bc1 ch1 etc41 equal to bg2 and draw ce1 hf1 etc41 to ,d1 cutting ,,bc in e and f1 which mark the required lengths of six feet each at the top of the wall4 ,,problem ,x4 ,this is one of the most important foundational problems in perspective1 and it is necessary that the student should entirely familiarize himself with its conditions4 ,in order to do so1 he must first observe these general relations of magnitude in any pyramid on a square base4 ,let ,,agh1 ,fig4 #ef41 be any pyramid on a square base4 ,illustration3 ,fig4 #ef4 ,the best terms in which its magnitude can be given1 are the length of one side of its base1 ,,ah1 and its vertical altitude 7,,cd in ,fig4 #be472 for1 knowing these1 we know all the other magnitudes4 ,but these are not the terms in which its size will be usually ascertainable4 ,generally1 we shall have given us1 and be able to ascertain by measurement1 one side of its base ,,ah1 and either ,,ag the length of one of the lines of its angles1 or ,,bg 7or ,b,g7 the length of a line drawn from its vertex1 ,g1 to the middle of the side of its base4 ,in measuring a real pyramid1 ,,ag will usually be the line most easily found2 but in many architectural problems ,,bg is given1 or is most easily ascertainable4 ,observe therefore this general construction4 ,illustration3 ,fig4 #eg4 ,let ,,abde1 ,fig4 #eg41 be the square base of any pyramid4 ,draw its diagonals1 ,,ae1 ,,bd1 cutting each other in its center1 ,c4 ,bisect any side1 ,,ab1 in ,f4 ,from ,f erect vertical ,,fg4 ,produce ,,fb to ,h1 and make ,,fh equal to ,,ac4 ,now if the vertical altitude of the pyramid 7,,cd in ,fig4 #be47 be given1 make ,,fg equal to this vertical altitude4 ,join ,,gb and ,,gh4 ,then ,,gb and ,,gh are the true magnitudes of ,,gb and ,,gh in ,fig4 #ef4 ,if ,,gb is given1 and not the vertical altitude1 with center ,b1 and distance ,,gb1 describe circle cutting ,,fg in ,g1 and ,,fg is the vertical altitude4 ,if ,,gh is given1 describe the circle from ,h1 with distance ,,gh1 and it will similarly cut ,,fg in ,g4 ,it is especially necessary for the student to examine this construction thoroughly1 because in many complicated forms of ornaments1 capitals of columns1 etc41 the lines ,,bg and ,,gh become the limits or bases of curves1 which are elongated on the longer 7or angle7 profile ,,gh1 and shortened on the shorter 7or lateral7 profile ,,bg4 ,we will take a simple instance1 but must previously note another construction4 ,it is often necessary1 when pyramids are the roots of some ornamental form1 to divide them horizontally at a given vertical height4 ,the shortest way of doing so is in general the following4 ,illustration3 ,fig4 #eh4 ,let ,,aec1 ,fig4 #eh41 be any pyramid on a square base ,,abc1 and ,,adc the square pillar used in its construction4 ,then by construction 7,problem ,x47 ,,bd and ,,af are both of the vertical height of the pyramid4 ,of the diagonals1 ,,fe1 ,,de1 choose the shortest 7in this case ,,de71 and produce it to cut the sight-line in ,v4 ,therefore ,v is the vanishing-point of ,,de4 ,divide ,,db1 as may be required1 into the sight-magnitudes of the given vertical heights at which the pyramid is to be divided4 ,illustration3 ,fig4 #ei4 ,illustration3 ,fig4 #fj4 ,from the points of division1 #a1 #b1 #c1 etc41 draw to the vanishing-point ,v4 ,the lines so drawn cut the angle line of the pyramid1 ,,be1 at the required elevations4 ,thus1 in the figure1 it is required to draw a horizontal black band on the pyramid at three fifths of its height1 and in breadth one twentieth of its height4 ,the line ,,bd is divided into five parts1 of which three are counted from ,b upwards4 ,then the line drawn to ,v marks the base of the black band4 ,then one fourth of one of the five parts is measured1 which similarly gives the breadth of the band4 ,the terminal lines of the band are then drawn on the sides of the pyramid parallel to ,,ab 7or to its vanishing-point if it has one71 and to the vanishing-point of ,,bc4 ,if it happens that the vanishing-points of the diagonals are awkwardly placed for use1 bisect the nearest base line of the pyramid in ,b1 as in ,fig4 #ei4 ,erect the vertical ,,db and join ,,gb and ,,dg 7,g being the apex of pyramid74 ,find the vanishing-point of ,,dg1 and use ,,db for division1 carrying the measurements to the line ,,gb4 ,in ,fig4 #ei41 if we join ,,ad and ,,dc1 ,,adc is the vertical profile of the whole pyramid1 and ,,bdc of the half pyramid1 corresponding to ,,fgb in ,fig4 #eg4 ,illustration3 ,fig4 #fa4 ,we may now proceed to an architectural example4 ,let ,,ah1 ,fig4 #fj41 be the vertical profile of the capital of a pillar1 ,,ab the semi-diameter of its head or abacus1 and ,,fd the semi-diameter of its shaft4 ,let the shaft be circular1 and the abacus square1 down to the level ,e4 ,join ,,bd1 ,,ef1 and produce them to meet in ,g4 ,therefore ,,ecg is the semi-profile of a reversed pyramid containing the capital4 ,construct this pyramid1 with the square of the abacus1 in the required perspective1 as in ,fig4 #fa42 making ,,ae equal to ,,ae in ,fig4 #fj41 and ,,ak1 the side of the square1 equal to twice ,,ab in ,fig4 #fj4 ,make ,,eg equal to ,,cg1 and ,,ed equal to ,,cd4 ,draw ,,df to the vanishing-point of the diagonal ,,dv 7the figure is too small to include this vanishing-point71 and ,f is the level of the point ,f in ,fig4 #fj41 on the side of the pyramid4 ,draw ,fm1 ,fn1 to the vanishing-points of ,,ah and ,,ak4 ,then ,fn and ,fm are horizontal lines across the pyramid at the level ,f1 forming at that level two sides of a square4 ,illustration3 ,fig4 #fb4 ,complete the square1 and within it inscribe a circle1 as in ,fig4 #fb41 which is left unlettered that its construction may be clear4 ,at the extremities of this draw vertical lines1 which will be the sides of the shaft in its right place4 ,it will be found to be somewhat smaller in diameter than the entire shaft in ,fig4 #fj41 because at the center of the square it is more distant than the nearest edge of the square abacus4 ,the curves of the capital may then be drawn approximately by the eye4 ,they are not quite accurate in ,fig4 #fb41 there being a subtlety in their junction with the shaft which could not be shown on so small a scale without confusing the student2 the curve on the left springing from a point a little way round the circle behind the shaft1 and that on the right from a point on this side of the circle a little way within the edge of the shaft4 ,but for their more accurate construction see ,notes on ,problem ,,xiv4 ,,problem ,,xi4 ,it is seldom that any complicated curve1 except occasionally a spiral1 needs to be drawn in perspective2 but the student will do well to practice for some time any fantastic shapes which he can find drawn on flat surfaces1 as on wall-papers1 carpets1 etc41 in order to accustom himself to the strange and great changes which perspective causes in them4 ,illustration3 ,fig4 #fc4 ,the curves most required in architectural drawing1 after the circle1 are those of pointed arches2 in which1 however1 all that will be generally needed is to fix the apex1 and two points in the sides4 ,thus if we have to draw a range of pointed arches1 such as ,,apb1 ,fig4 #fc41 draw the measured arch to its sight-magnitude first neatly in a rectangle1 ,,abcd2 then draw the diagonals ,,ad and ,,bc2 where they cut the curve draw a horizontal line 7as at the level ,e in the figure71 and carry it along the range to the vanishing-point1 fixing the points where the arches cut their diagonals all along4 ,if the arch is cusped1 a line should be drawn1 at ,f to mark the height of the cusps1 and verticals raised at ,g and ,h1 to determine the interval between them4 ,any other points may be similarly determined1 but these will usually be enough4 ,figure #fc4 shows the perspective construction of a square niche of good ,veronese ,gothic1 with an uncusped arch of similar size and curve beyond4 ,illustration3 ,fig4 #fd4 ,in ,fig4 #fd4 the more distant arch only is lettered1 as the construction of the nearest explains itself more clearly to the eye without letters4 ,the more distant arch shows the general construction for all arches seen underneath1 as of bridges1 cathedral aisles1 etc4 ,the rectangle ,,abcd is first drawn to contain the outside arch2 then the depth of the arch1 ,aa1 is determined by the measuring-line1 and the rectangle1 abcd1 drawn for the inner arch4 ,aa1 ,bb1 etc41 go to one vanishing-point2 ,,ab1 ab1 etc41 to the opposite one4 ,in the nearer arch another narrow rectangle is drawn to determine the cusp4 ,the parts which would actually come into sight are slightly shaded4 ,,problem ,,xiv4 ,several exercises will be required on this important problem4 ,i4 ,it is required to draw a circular flat-bottomed dish narrower at the bottom than the top2 the vertical depth being given1 and the diameter at the top and bottom4 ,illustration3 ,fig4 #fe4 ,let ab1 ,fig4 #fe41 be the diameter of the bottom1 ac the diameter of the top1 and ad its vertical depth4 ,take ,,ad in position equal to ac4 ,on ,,ad draw the square ,,abcd1 and inscribe in it a circle4 ,therefore1 the circle so inscribed has the diameter of the top of the dish4 ,from ,a and ,d let fall verticals1 ,,ae1 ,,dh1 each equal to ad4 ,join ,,eh1 and describe square ,,efgh1 which accordingly will be equal to the square ,,abcd1 and be at the depth ad beneath it4 ,within the square ,,efgh describe a square ,,ik1 whose diameter shall be equal to ab4 ,describe a circle within the square ,,ik4 ,therefore the circle so inscribed has its diameter equal to ab2 and it is in the center of the square ,,efgh1 which is vertically beneath the square ,,abcd4 ,therefore the circle in the square ,,ik represents the bottom of the dish4 ,now the two circles thus drawn will either intersect one another1 or they will not4 ,if they intersect one another1 as in the figure1 and they are below the eye1 part of the bottom of the dish is seen within it4 ,illustration3 ,fig4 #ff4 ,to avoid confusion1 let us take then two intersecting circles without the inclosing squares1 as in ,fig4 #ff4 ,draw right lines1 ab1 cd1 touching both circles externally4 ,then the parts of these lines which connect the circles are the sides of the dish4 ,they are drawn in ,fig4 #fe4 without any prolongations1 but the best way to construct them is as in ,fig4 #ff4 ,if the circles do not intersect each other1 the smaller must either be within the larger or not within it4 ,if within the larger1 the whole of the bottom of the dish is seen from above1 ,fig4 #fg4 a4 ,illustration3 ,fig4 #fg4 ,if the smaller circle is not within the larger1 none of the bottom is seen inside the dish1 b4 ,if the circles are above instead of beneath the eye1 the bottom of the dish is seen beneath it1 c4 ,if one circle is above and another beneath the eye1 neither the bottom nor top of the dish is seen1 d4 ,unless the object be very large1 the circles in this case will have little apparent curvature4 ,,ii4 ,the preceding problem is simple1 because the lines of the profile of the object 7ab and cd1 ,fig4 #ff47 are straight4 ,but if these lines of profile are curved1 the problem becomes much more complex3 once mastered1 however1 it leaves no farther difficulty in perspective4 ,let it be required to draw a flattish circular cup or vase1 with a given curve of profile4 ,the basis of construction is given in ,fig4 #fh41 half of it only being drawn1 in order that the eye may seize its lines easily4 ,illustration3 ,fig4 #fh4 ,two squares 7of the required size7 are first drawn1 one above the other1 with a given vertical interval1 ,,ac1 between them1 and each is divided into eight parts by its diameters and diagonals4 ,in these squares two circles are drawn2 which are1 therefore1 of equal size1 and one above the other4 ,two smaller circles1 also of equal size1 are drawn within these larger circles in the construction of the present problem2 more may be necessary in some1 none at all in others4 ,it will be seen that the portions of the diagonals and diameters of squares which are cut off between the circles represent radiating planes1 occupying the position of the spokes of a wheel4 ,now let the line ,,aeb1 ,fig4 #fi41 be the profile of the vase or cup to be drawn4 ,inclose it in the rectangle ,,cd1 and if any portion of it is not curved1 as ,,ae1 cut off the curved portion by the vertical line ,,ef1 so as to include it in the smaller rectangle ,,fd4 ,draw the rectangle ,,acbd in position1 and upon it construct two squares1 as they are constructed on the rectangle ,,acd in ,fig4 #fh42 and complete the construction of ,fig4 #fh41 making the radius of its large outer circles equal to ,,ad1 and of its small inner circles equal to ,,ae4 ,the planes which occupy the position of the wheel spokes will then each represent a rectangle of the size of ,,fd4 ,the construction is shown by the dotted lines in ,fig4 #fi42 c being the center of the uppermost circle4 ,illustration3 ,fig4 #fi4 ,within each of the smaller rectangles between the circles1 draw the curve ,,eb in perspective1 as in ,fig4 #fi4 ,draw the curve xy1 touching and inclosing the curves in the rectangles1 and meeting the upper circle at y4#cb ,then xy is the contour of the surface of the cup1 and the upper circle is its lip4 ,if the line xy is long1 it may be necessary to draw other rectangles between the eight principal ones2 and1 if the curve of profile ,,ab is complex or retorted1 there may be several lines corresponding to ,,xy1 inclosing the successive waves of the profile2 and the outer curve will then be an undulating or broken one4 ,illustration3 ,fig4 #gj4 ,,iii4 ,all branched ornamentation1 forms of flowers1 capitals of columns1 machicolations of round towers1 and other such arrangements of radiating curve1 are resolvable by this problem1 using more or fewer interior circles according to the conditions of the curves4 ,fig4 #gj4 is an example of the construction of a circular group of eight trefoils with curved stems4 ,one outer or limiting circle is drawn within the square ,,edcf1 and the extremities of the trefoils touch it at the extremities of its diagonals and diameters4 ,a smaller circle is at the vertical distance ,,bc below the larger1 and ,a is the angle of the square within which the smaller circle is drawn2 but the square is not given1 to avoid confusion4 ,the stems of the trefoils form drooping curves1 arranged on the diagonals and diameters of the smaller circle1 which are dotted4 ,but no perspective laws will do work of this intricate kind so well as the hand and eye of a painter4 ,,iv4 ,there is one common construction1 however1 in which1 singularly1 the hand and eye of the painter almost always fail1 and that is the fillet of any ordinary capital or base of a circular pillar 7or any similar form74 ,it is rarely necessary in practice to draw such minor details in perspective2 yet the perspective laws which regulate them should be understood1 else the eye does not see their contours rightly until it is very highly cultivated4 ,illustration3 ,fig4 #ga4 ,fig4 #ga4 will show the law with sufficient clearness2 it represents the perspective construction of a fillet whose profile is a semicircle1 such as ,,fh in ,fig4 #fj41 seen above the eye4 ,only half the pillar with half the fillet is drawn1 to avoid confusion4 ,q is the center of the shaft4 ,,pq the thickness of the fillet1 sight-magnitude at the shaft's center4 ,round ,p a horizontal semicircle is drawn on the diameter of the shaft ab4 ,round ,q another horizontal semicircle is drawn on diameter cd4 ,these two semicircles are the upper and lower edges of the fillet4 ,then diagonals and diameters are drawn as in ,fig4 #fh41 and1 at their extremities1 semicircles in perspective1 as in ,fig4 #fi4 ,the letters ,a1 ,b1 ,c1 ,d1 and ,e1 indicate the upper and exterior angles of the rectangles in which these semicircles are to be drawn2 but the inner vertical line is not dotted in the rectangle at ,c1 as it would have confused itself with other lines4 ,then the visible contour of the fillet is the line which incloses and touches#cc all the semicircles4 ,it disappears behind the shaft at the point ,h1 but ,i have drawn it through to the opposite extremity of the diameter at d4 ,turned upside down the figure shows the construction of a basic fillet4 ,the capital of a ,greek ,doric pillar should be drawn frequently for exercise on this fourteenth problem1 the curve of its echinus being exquisitely subtle1 while the general contour is simple4 #cb ,this point coincides in the figure with the extremity of the horizontal diameter1 but only accidentally4 #cc ,the engraving is a little inaccurate2 the inclosing line should touch the dotted semicircles at ,a and ,b4 ,the student should draw it on a large scale4 ,,problem ,,xvi4 ,it is often possible to shorten other perspective operations considerably1 by finding the vanishing-points of the inclined lines of the object4 ,thus1 in drawing the gabled roof in ,fig4 #dc41 if the gable ,,ayc be drawn in perspective1 and the vanishing-point of ,,ay determined1 it is not necessary to draw the two sides of the rectangle1 ,a,d and ,d,b1 in order to determine the point ,y2 but merely to draw ,,yy to the vanishing-point of ,,aa and ,a,y to the vanishing-point of ,,ay1 meeting in ,y1 the point required4 ,again1 if there be a series of gables1 or other figures produced by parallel inclined lines1 and retiring to the point ,v1 as in ,fig4 #gb41cd it is not necessary to draw each separately1 but merely to determine their breadths on the line ,,av1 and draw the slopes of each to their vanishing-points1 as shown in ,fig4 #gb4 ,or if the gables are equal in height1 and a line be drawn from ,y to ,v1 the construction resolves itself into a zigzag drawn alternately to ,p and ,q1 between the lines ,,yv and ,,av4 ,the student must be very cautious1 in finding the vanishing-points of inclined lines1 to notice their relations to the horizontals beneath them1 else he may easily mistake the horizontal to which they belong4 ,thus1 let ,,abcd1 ,fig4 #gc41 be a rectangular inclined plane1 and let it be required to find the vanishing-point of its diagonal ,,bd4 ,find ,v1 the vanishing-point of ,,ad and ,,bc4 ,draw ,,ae to the opposite vanishing-point1 so that ,,dae may represent a right angle4 ,let fall from ,b the vertical ,,be1 cutting ,,ae in ,e4 ,join ,,ed1 and produce it to cut the sight-line in ,v4 ,illustration3 ,fig4 #gb4 ,then1 since the point ,e is vertically under the point ,b1 the horizontal line ,,ed is vertically under the inclined line ,,bd4 ,illustration3 ,fig4 #gc4 ,so that if we now let fall the vertical ,v,p from ,v1 and produce ,,bd to cut ,v,p in ,p1 the point ,p will be the vanishing-point of ,,bd1 and of all lines parallel to it4#ce #cd ,the diagram is inaccurately cut4 ,,yv should be a right line4 #ce ,the student may perhaps understand this construction better by completing the rectangle ,,adfe1 drawing ,,df to the vanishing-point of ,,ae1 and ,,ef to ,v4 ,the whole figure1 ,,bf1 may then be conceived as representing half the gable roof of a house1 ,,af the rectangle of its base1 and ,,ac the rectangle of its sloping side4 ,in nearly all picturesque buildings1 especially on the ,continent1 the slopes of gables are much varied 7frequently unequal on the two sides71 and the vanishing-points of their inclined lines become very important1 if accuracy is required in the intersections of tiling1 sides of dormer windows1 etc4 ,obviously1 also1 irregular triangles and polygons in vertical planes may be more easily constructed by finding the vanishing-points of their sides1 than by the construction given in the corollary to ,problem ,,ix42 and if such triangles or polygons have others concentrically inscribed within them1 as often in ,byzantine mosaics1 etc41 the use of the vanishing-points will become essential4 ,,problem ,,xviii4 ,before examining the last three problems it is necessary that you should understand accurately what is meant by the position of an inclined plane4 ,cut a piece of strong white pasteboard into any irregular shape1 and dip it in a sloped position into water4 ,however you hold it1 the edge of the water1 of course1 will always draw a horizontal line across its surface4 ,the direction of this horizontal line is the direction of the inclined plane4 7,in beds of rock geologists call it their 8strike407 ,illustration3 ,fig4 #gd4 ,next1 draw a semicircle on the piece of pasteboard2 draw its diameter1 ,,ab1 ,fig4 #gd41 and a vertical line from its center1 ,,cd2 and draw some other lines1 ,,ce1 ,,cf1 etc41 from the center to any points in the circumference4 ,now dip the piece of pasteboard again into water1 and1 holding it at any inclination and in any direction you choose1 bring the surface of the water to the line ,,ab4 ,then the line ,,cd will be the most steeply inclined of all the lines drawn to the circumference of the circle2 ,,gc and ,,hc will be less steep2 and ,,ec and ,,fc less steep still4 ,the nearer the lines to ,,cd1 the steeper they will be2 and the nearer to ,,ab1 the more nearly horizontal4 ,when1 therefore1 the line ,,ab is horizontal 7or marks the water surface71 its direction is the direction of the inclined plane1 and the inclination of the line ,,dc is the inclination of the inclined plane4 ,in beds of rock geologists call the inclination of the line ,,dc their 8dip40 ,to fix the position of an inclined plane1 therefore1 is to determine the direction of any two lines in the plane1 ,,ab and ,,cd1 of which one shall be horizontal and the other at right angles to it4 ,then any lines drawn in the inclined plane1 parallel to ,,ab1 will be horizontal2 and lines drawn parallel to ,,cd will be as steep as ,,cd1 and are spoken of in the text as the 8steepest lines0 in the plane4 ,but farther1 whatever the direction of a plane may be1 if it be extended indefinitely1 it will be terminated1 to the eye of the observer1 by a boundary line1 which1 in a horizontal plane1 is horizontal 7coinciding nearly with the visible horizon72--in a vertical plane1 is vertical2--and1 in an inclined plane1 is inclined4 ,this line is properly1 in each case1 called the 8sight-line0 of such plane2 but it is only properly called the 8horizon0 in the case of a horizontal plane3 and ,i have preferred using always the term 8sight-line10 not only because more comprehensive1 but more accurate2 for though the curvature of the earth's surface is so slight that practically its visible limit always coincides with the sight-line of a horizontal plane1 it does not mathematically coincide with it1 and the two lines ought not to be considered as theoretically identical1 though they are so in practice4 ,it is evident that all vanishing-points of lines in any plane must be found on its sight-line1 and1 therefore1 that the sight-line of any plane may be found by joining any two of such vanishing-points4 ,hence the construction of ,problem ,,xviii4 ,,ii4 ,,demonstrations ,,which ,,could ,,not ,,conveniently ,,be ,,included ,,in ,,the ,,text4 ,i4 ,,the ,,second ,,corollary1 ,,problem ,,ii4 ,in ,fig4 #h4 omit the lines ,,cd1 ,c,d1 and ,,ds2 and1 as here in ,fig4 #ge41 from a draw ad parallel to ,,ab1 cutting ,,bt in d2 and from d draw de parallel to ,,bc4 ,illustration3 ,fig4 #ge4 ,now as ad is parallel to ,,ab-- ,,ac ac ,,bc de2 but ,,ac is equal to ,,bc-- ac de4 ,now because the triangles ac,v1 bc,v1 are similar-- ac bc a,v b,v2 and because the triangles de,t1 bc,t are similar-- de bc d,t b,t4 ,but ac is equal to de-- a,v b,v d,t b,t2 the two triangles abd1 b,t,v1 are similar1 and their angles are alternate2 ,,tv is parallel to ad4 ,but ad is parallel to ,,ab-- ,,tv is parallel to ,,ab4 ,,ii4 ,,the ,,third ,,corollary1 ,,problem ,,iii4 ,in ,fig4 #ac41 since a,r is by construction parallel to ,,ab in ,fig4 #ab41 and ,,tv is by construction in ,problem ,,iii4 also parallel to ,,ab-- a,r is parallel to ,,tv1 ab,r and ,tb,v are alternate triangles1 a,r ,,tv ab b,v4 ,again1 by the construction of ,fig4 #ac41 a,r is parallel to ,,mv-- ab,r and ,mb,v are alternate triangles1 a,r ,,mv ab b,v4 ,and it has just been shown that also a,r ,,tv ab b,v-- a,r ,,mv a,r ,,tv4 ,but by construction1 a,r a,r-- ,,mv ,,tv4 ,,iii4 ,,analysis ,,of ,,problem ,,xv4 ,we proceed to take up the general condition of the second problem1 before left unexamined1 namely1 that in which the vertical distances ,,bc and ,,ac 7,fig4 #f4 page #ac71 as well as the direct distances ,,td and ,,td are unequal4 ,in ,fig4 #f41 here repeated 7,fig4 #gf471 produce ,c,b downwards1 and make ,c,e equal to ,,ca4 ,illustration3 ,fig4 #gf4 ,join ,,ae4 ,then1 by the second ,corollary of ,problem ,,ii41 ,,ae is a horizontal line4 ,draw ,,tv parallel to ,,ae1 cutting the sight-line in ,v4 ,v is the vanishing-point of ,,ae4 ,complete the constructions of ,problem ,,ii4 and its second ,corollary4 ,then by ,problem ,,ii4 ab is the line ,,ab drawn in perspective2 and by its ,corollary ae is the line ,,ae drawn in perspective4 ,from ,v erect perpendicular ,,vp1 and produce ab to cut it in ,p4 ,join ,,tp1 and from e draw ef parallel to ,,ae1 and cutting ,,at in f4 ,now in triangles ,,ebt and ,,aet1 as eb is parallel to ,,eb and ef to ,,ae2--eb ef ,,eb ,,ae4 ,but ,,tv is also parallel to ,,ae and ,,pv to eb4 ,therefore also in the triangles a,p,v and a,v,t1 eb ef ,,pv ,,vt4 ,therefore ,,pv ,,vt ,,eb ,,ae4 ,and1 by construction1 angle ,,tpv ,,aeb4 ,therefore the triangles ,,tvp1 ,,aeb1 are similar2 and ,,tp is parallel to ,,ab4 ,now the construction in this problem is entirely general for any inclined line ,,ab1 and a horizontal line ,,ae in the same vertical plane with it4 ,so that if we find the vanishing-point of ,,ae in ,v1 and from ,v erect a vertical ,,vp1 and from ,t draw ,,tp parallel to ,,ab1 cutting ,,vp in ,p1 ,p will be the vanishing-point of ,,ab1 and 7by the same proof as that given at page #ag7 of all lines parallel to it4 ,illustration3 ,fig4 #gg4 ,next1 to find the dividing-point of the inclined line4 ,i remove some unnecessary lines from the last figure and repeat it here1 ,fig4 #gg41 adding the measuring-line a,m1 that the student may observe its position with respect to the other lines before ,i remove any more of them4 ,now if the line ,,ab in this diagram represented the length of the line ,,ab in reality 7as ,,ab does in ,figs4 #aj4 and #aa471 we should only have to proceed to modify ,corollary ,,iii4 of ,problem ,,ii4 to this new construction4 ,we shall see presently that ,,ab does not represent the actual length of the inclined line ,,ab in nature1 nevertheless we shall first proceed as if it did1 and modify our result afterwards4 ,in ,fig4 #gg4 draw ad parallel to ,,ab1 cutting ,,bt in d4 ,therefore ad is the sight-magnitude of ,,ab1 as a,r is of ,,ab in ,fig4 #aa4 ,illustration3 ,fig4 #gh4 ,remove again from the figure all lines except ,,pv1 ,,vt1 ,,pt1 ab1 ad1 and the measuring-line4 ,set off on the measuring-line am equal to ad4 ,draw ,,pq parallel to am1 and through b draw m,q1 cutting ,,pq in ,q4 ,then1 by the proof already given in page #bj1 ,,pq ,,pt4 ,therefore if ,p is the vanishing-point of an inclined line ,,ab1 and ,,qp is a horizontal line drawn through it1 make ,,pq equal to ,,pt1 and am on the measuring-line equal to the sight-magnitude of the line ,,ab in the diagram1 and the line joining m,q will cut a,p in b4 ,we have now1 therefore1 to consider what relation the length of the line ,,ab in this diagram1 ,fig4 #gg41 has to the length of the line ,,ab in reality4 ,now the line ,,ae in ,fig4 #gg4 represents the length of ,,ae in reality4 ,but the angle ,,aeb1 ,fig4 #gg41 and the corresponding angle in all the constructions of the earlier problems1 is in reality a right angle1 though in the diagram necessarily represented as obtuse4 ,illustration3 ,fig4 #gi4 ,therefore1 if from ,e we draw ,,ec1 as in ,fig4 #gi41 at right angles to ,,ae1 make ,,ec ,,eb1 and join ,,ac1 ,,ac will be the real length of the line ,,ab4 ,now1 therefore1 if instead of am in ,fig4 #gh41 we take the real length of ,,ab1 that real length will be to am as ,,ac to ,,ab in ,fig4 #gi4 ,and then1 if the line drawn to the measuring-line ,,pq is still to cut a,p in b1 it is evident that the line ,,pq must be shortened in the same ratio that am was shortened2 and the true dividing-point will be ,q in ,fig4 #hj41 fixed so that ,q,p shall be to ,,qp as am is to am2 am representing the real length of ,,ab4 ,but amis therefore to am as ,,ac is to ,,ab in ,fig4 #gi4 ,therefore ,,pq must be to ,,pq as ,,ac is to ,,ab4 ,but ,,pq equals ,,pt 7,fig4 #gh472 and ,,pv is to ,,vt 7in ,fig4 #gh47 as ,,be is to ,,ae 7,fig4 #gi474 ,hence we have only to substitute ,,pv for ,,ec1 and ,,vt for ,,ae1 in ,fig4 #gi41 and the resulting diagonal ,,ac will be the required length of ,,pq4 ,illustration3 ,fig4 #hj4 ,it will be seen that the construction given in the text 7,fig4 #df47 is the simplest means of obtaining this magnitude1 for ,,vd in ,fig4 #df4 7or ,,vm in ,fig4 #ae47 ,,vt by construction in ,problem ,,iv4 ,it should1 however1 be observed1 that the distance ,,pq or ,,px1 in ,fig4 #df41 may be laid on the sight-line of the inclined plane itself1 if the measuring-line be drawn parallel to that sight-line4 ,and thus any form may be drawn on an inclined plane as conveniently as on a horizontal one1 with the single exception of the radiation of the verticals1 which have a vanishing-point1 as shown in ,problem ,,xx4 ,,the ,,end4 ,transcriber's ,note ,a handful of unequivocal typographical errors has been corrected4 ,for increased clarity1 a few diagrams have been shifted from their original position in the text4