[p38]PROBLEM XI.

[p38]PROBLEM XI.TO DRAW ANY CURVE IN A HORIZONTAL ORVERTICAL PLANE.[Geometric diagram]Fig. 28.LetA B,Fig. 28., be the curve.Inclose it in a rectangle,C D E F.Fix the position of the pointCorD, and draw the rectangle. (Problem VIII.Coroll. I.)[Footnote20]LetC D E F,Fig. 29., be the rectangle so drawn.[Geometric diagram]Fig. 29.If an extremity of the curve, asA, is in a side of the rectangle, divide the sideC E,Fig. 29., so thatA Cshall be (in perspective ratio) toA EasA Cis toA EinFig. 28.(Prob. V.Cor. II.)Similarly determine the points of contact of the curve and rectanglee,f,g.If an extremity of the curve, asB, is not in a side of the rectangle, let[p39]fall the perpendicularsBa,Bbon the rectangle sides. Determine the correspondent pointsaandbinFig. 29., as you have already determinedA,B,e, andf.Fromb,Fig. 29., drawbBparallel toC D,[Footnote21]and fromadrawaBto the vanishing-point ofD F, cutting each other inB. ThenBis the extremity of the curve.Determine any other important point in the curve, asP, in the same way, by letting fallPqandPron the rectangle’s sides.Any number of points in the curve may be thus determined, and the curve drawn through the series; in most cases, three or four will be enough. Practically, complicated curves may be better drawn in perspective by an experienced eye than by rule, as the fixing of the various points in haste involves too many chances of error; but it is well to draw a good many by rule first, in order to give the eye its experience.[Footnote22]COROLLARY.If the curve required be a circle,Fig. 30., the rectangle which incloses it will become a square, and the curve will have four points of contact,A B C D, in the middle of the sides of the square.[Geometric diagram]Fig. 30.Draw the square, and as a square may be drawn about a circle in any position, draw it with its nearest side,E G, parallel to the sight-line.LetE F,Fig. 31., be the square so drawn.[p40]Draw its diagonalsE F,G H; and through the center of the square (determined by their intersection) drawA Bto the vanishing-point ofG F, andC Dparallel toE G. Then the pointsA B C Dare the four points of the circle’s contact.[Geometric diagram]Fig. 31.OnE Gdescribe a half square,E L; draw the semicircleK A L; and from its center,R, the diagonalsR E,R G, cutting the circle inx,y.From the pointsxy, where the circle cuts the diagonals, raise perpendiculars,Px,Qy, toE G.FromPandQdrawP P′,Q Q′, to the vanishing-point ofG F, cutting the diagonals inm,n, ando,p.Thenm,n,o,pare four other points in the circle.Through these eight points the circle may be drawn by the hand accurately enough for general purposes; but any number of points required may, of course, be determined, as inProblem XI.The distanceE Pis approximately one-seventh ofE G, and may be assumed to be so in quick practice, as the error involved is not greater than would be incurred in the hasty operation of drawing the circle and diagonals.It may frequently happen that, in consequence of associated[p41]constructions, it may be inconvenient to drawE Gparallel to the sight-line, the square being perhaps first constructed in some oblique direction. In such cases,Q GandE Pmust be determined in perspective ratio by the dividing-point, the lineE Gbeing used as a measuring-line.[Obs.In drawingFig. 31.the station-point has been taken much nearer the paper than is usually advisable, in order to show the character of the curve in a very distinct form.If the student turns the book so thatE Gmay be vertical,Fig. 31.will represent the construction for drawing a circle in a vertical plane, the sight-line being then of course parallel toG L; and the semicirclesA D B,A C B, on each side of the diameterA B, will represent ordinary semicircular arches seen in perspective. In that case, if the book be held so that the lineE His the top of the square, the upper semicircle will represent a semicircular arch,abovethe eye, drawn in perspective. But if the book be held so that the lineG Fis the top of the square, the upper semicircle will represent a semicircular arch,belowthe eye, drawn in perspective.If the book be turned upside down, the figure will represent a circle drawn on the ceiling, or any other horizontal plane above the eye; and the construction is, of course, accurate in every case.][Footnote20:Or if the curve is in a vertical plane,Coroll. to Problem IX. As a rectangle may be drawn in any position round any given curve, its position with respect to the curve will in either case be regulated by convenience. See the Exercises on this Problem, in the Appendix,p. 85.]Return to text[Footnote21:Or to its vanishing-point, ifC Dhas one.]Return to text[Footnote22:Of course, by dividing the original rectangle into any number of equal rectangles, and dividing the perspective rectangle similarly, the curve may be approximately drawn without any trouble; but, when accuracy is required, the points should be fixed, as in the problem.]Return to text

[Geometric diagram]Fig. 28.

LetA B,Fig. 28., be the curve.

Inclose it in a rectangle,C D E F.

Fix the position of the pointCorD, and draw the rectangle. (Problem VIII.Coroll. I.)[Footnote20]

LetC D E F,Fig. 29., be the rectangle so drawn.

[Geometric diagram]Fig. 29.

If an extremity of the curve, asA, is in a side of the rectangle, divide the sideC E,Fig. 29., so thatA Cshall be (in perspective ratio) toA EasA Cis toA EinFig. 28.(Prob. V.Cor. II.)

Similarly determine the points of contact of the curve and rectanglee,f,g.

If an extremity of the curve, asB, is not in a side of the rectangle, let[p39]fall the perpendicularsBa,Bbon the rectangle sides. Determine the correspondent pointsaandbinFig. 29., as you have already determinedA,B,e, andf.

Fromb,Fig. 29., drawbBparallel toC D,[Footnote21]and fromadrawaBto the vanishing-point ofD F, cutting each other inB. ThenBis the extremity of the curve.

Determine any other important point in the curve, asP, in the same way, by letting fallPqandPron the rectangle’s sides.

Any number of points in the curve may be thus determined, and the curve drawn through the series; in most cases, three or four will be enough. Practically, complicated curves may be better drawn in perspective by an experienced eye than by rule, as the fixing of the various points in haste involves too many chances of error; but it is well to draw a good many by rule first, in order to give the eye its experience.[Footnote22]

If the curve required be a circle,Fig. 30., the rectangle which incloses it will become a square, and the curve will have four points of contact,A B C D, in the middle of the sides of the square.

[Geometric diagram]Fig. 30.

Draw the square, and as a square may be drawn about a circle in any position, draw it with its nearest side,E G, parallel to the sight-line.

LetE F,Fig. 31., be the square so drawn.

[p40]Draw its diagonalsE F,G H; and through the center of the square (determined by their intersection) drawA Bto the vanishing-point ofG F, andC Dparallel toE G. Then the pointsA B C Dare the four points of the circle’s contact.

[Geometric diagram]Fig. 31.

OnE Gdescribe a half square,E L; draw the semicircleK A L; and from its center,R, the diagonalsR E,R G, cutting the circle inx,y.

From the pointsxy, where the circle cuts the diagonals, raise perpendiculars,Px,Qy, toE G.

FromPandQdrawP P′,Q Q′, to the vanishing-point ofG F, cutting the diagonals inm,n, ando,p.

Thenm,n,o,pare four other points in the circle.

Through these eight points the circle may be drawn by the hand accurately enough for general purposes; but any number of points required may, of course, be determined, as inProblem XI.

The distanceE Pis approximately one-seventh ofE G, and may be assumed to be so in quick practice, as the error involved is not greater than would be incurred in the hasty operation of drawing the circle and diagonals.

It may frequently happen that, in consequence of associated[p41]constructions, it may be inconvenient to drawE Gparallel to the sight-line, the square being perhaps first constructed in some oblique direction. In such cases,Q GandE Pmust be determined in perspective ratio by the dividing-point, the lineE Gbeing used as a measuring-line.

[Obs.In drawingFig. 31.the station-point has been taken much nearer the paper than is usually advisable, in order to show the character of the curve in a very distinct form.If the student turns the book so thatE Gmay be vertical,Fig. 31.will represent the construction for drawing a circle in a vertical plane, the sight-line being then of course parallel toG L; and the semicirclesA D B,A C B, on each side of the diameterA B, will represent ordinary semicircular arches seen in perspective. In that case, if the book be held so that the lineE His the top of the square, the upper semicircle will represent a semicircular arch,abovethe eye, drawn in perspective. But if the book be held so that the lineG Fis the top of the square, the upper semicircle will represent a semicircular arch,belowthe eye, drawn in perspective.If the book be turned upside down, the figure will represent a circle drawn on the ceiling, or any other horizontal plane above the eye; and the construction is, of course, accurate in every case.]

[Obs.In drawingFig. 31.the station-point has been taken much nearer the paper than is usually advisable, in order to show the character of the curve in a very distinct form.

If the student turns the book so thatE Gmay be vertical,Fig. 31.will represent the construction for drawing a circle in a vertical plane, the sight-line being then of course parallel toG L; and the semicirclesA D B,A C B, on each side of the diameterA B, will represent ordinary semicircular arches seen in perspective. In that case, if the book be held so that the lineE His the top of the square, the upper semicircle will represent a semicircular arch,abovethe eye, drawn in perspective. But if the book be held so that the lineG Fis the top of the square, the upper semicircle will represent a semicircular arch,belowthe eye, drawn in perspective.

If the book be turned upside down, the figure will represent a circle drawn on the ceiling, or any other horizontal plane above the eye; and the construction is, of course, accurate in every case.]

[Footnote20:Or if the curve is in a vertical plane,Coroll. to Problem IX. As a rectangle may be drawn in any position round any given curve, its position with respect to the curve will in either case be regulated by convenience. See the Exercises on this Problem, in the Appendix,p. 85.]Return to text[Footnote21:Or to its vanishing-point, ifC Dhas one.]Return to text[Footnote22:Of course, by dividing the original rectangle into any number of equal rectangles, and dividing the perspective rectangle similarly, the curve may be approximately drawn without any trouble; but, when accuracy is required, the points should be fixed, as in the problem.]Return to text

[Footnote21:Or to its vanishing-point, ifC Dhas one.]Return to text

[Footnote22:Of course, by dividing the original rectangle into any number of equal rectangles, and dividing the perspective rectangle similarly, the curve may be approximately drawn without any trouble; but, when accuracy is required, the points should be fixed, as in the problem.]Return to text

[p42]PROBLEM XII.TO DIVIDE A CIRCLE DRAWN IN PERSPECTIVE INTO ANY GIVEN NUMBEROF EQUAL PARTS.LetA B,Fig. 32., be the circle drawn in perspective. It is required to divide it into a given number of equal parts; in this case, 20.LetK A Lbe the semicircle used in the construction. Divide the semicircleK A Linto half the number of parts required; in this case, 10.Produce the lineE Glaterally, as far as may be necessary.FromO, the center of the semicircleK A L, draw radii through the points of division of the semicircle,p,q,r, etc., and produce them to cut the lineE GinP,Q,R, etc.From the pointsP Q Rdraw the linesP P′,Q Q′,R R′, etc., through the center of the circleA B, each cutting the circle in two points of its circumference.Then these points divide the perspective circle as required.If from each of the pointsp,q,r, a vertical were raised to the lineE G, as inFig. 31., and from the point where it cutE Ga line were drawn to the vanishing-point, asQ Q′inFig. 31., this line would also determine two of the points of division.[p43][Geometric diagram][Geometric diagram]Fig. 32.If it is required to divide a circle into any number of givenunequal parts (as in the pointsA,B, andC,Fig. 33.), the shortest way is thus to raise vertical lines fromAandBto the side of the perspective squareX Y, and then draw to the vanishing-point, cutting the perspective circle inaandb, the points required. Only notice that if any point, asA, is on the nearer side of the circleA B C, its representative point,a, must be on the nearer side of the circlea b c; and if the pointBis on the farther side of the circleA B C,bmust be[p44]on the farther side ofa b c. If any point, asC, is so much in the lateral arc of the circle as not to be easily determinable by the vertical line, draw the horizontalC P, find the correspondentpin the side of the perspective square, and drawp cparallel toX Y, cutting the perspective circle inc.[Geometric diagram]Fig. 33.COROLLARY.It is obvious that if the pointsP′,Q′,R, etc., by which the circle is divided inFig. 32., be joined by right lines, the resulting figure will be a regular equilateral figure of twenty sides inscribed in the circle. And if the circle be divided into given unequal parts, and the points of division joined by right lines, the resulting figure will be an irregular polygon inscribed in the circle with sides of given length.Thus any polygon, regular or irregular, inscribed in a circle, may be inscribed in position in a perspective circle.

LetA B,Fig. 32., be the circle drawn in perspective. It is required to divide it into a given number of equal parts; in this case, 20.

LetK A Lbe the semicircle used in the construction. Divide the semicircleK A Linto half the number of parts required; in this case, 10.

Produce the lineE Glaterally, as far as may be necessary.

FromO, the center of the semicircleK A L, draw radii through the points of division of the semicircle,p,q,r, etc., and produce them to cut the lineE GinP,Q,R, etc.

From the pointsP Q Rdraw the linesP P′,Q Q′,R R′, etc., through the center of the circleA B, each cutting the circle in two points of its circumference.

Then these points divide the perspective circle as required.

If from each of the pointsp,q,r, a vertical were raised to the lineE G, as inFig. 31., and from the point where it cutE Ga line were drawn to the vanishing-point, asQ Q′inFig. 31., this line would also determine two of the points of division.

[p43][Geometric diagram][Geometric diagram]Fig. 32.

If it is required to divide a circle into any number of givenunequal parts (as in the pointsA,B, andC,Fig. 33.), the shortest way is thus to raise vertical lines fromAandBto the side of the perspective squareX Y, and then draw to the vanishing-point, cutting the perspective circle inaandb, the points required. Only notice that if any point, asA, is on the nearer side of the circleA B C, its representative point,a, must be on the nearer side of the circlea b c; and if the pointBis on the farther side of the circleA B C,bmust be[p44]on the farther side ofa b c. If any point, asC, is so much in the lateral arc of the circle as not to be easily determinable by the vertical line, draw the horizontalC P, find the correspondentpin the side of the perspective square, and drawp cparallel toX Y, cutting the perspective circle inc.

[Geometric diagram]Fig. 33.

It is obvious that if the pointsP′,Q′,R, etc., by which the circle is divided inFig. 32., be joined by right lines, the resulting figure will be a regular equilateral figure of twenty sides inscribed in the circle. And if the circle be divided into given unequal parts, and the points of division joined by right lines, the resulting figure will be an irregular polygon inscribed in the circle with sides of given length.

Thus any polygon, regular or irregular, inscribed in a circle, may be inscribed in position in a perspective circle.

[p45]PROBLEM XIII.TO DRAW A SQUARE, GIVEN IN MAGNITUDE, WITHIN A LARGER SQUARE GIVEN IN POSITION AND MAGNITUDE; THE SIDES OF THE TWO SQUARESBEING PARALLEL.[Geometric diagram]Fig. 34.LetA B,Fig. 34., be the sight-magnitude of the side of the smaller square, andA Cthat of the side of the larger square.Draw the larger square. LetD E F Gbe the square so drawn.JoinE GandD F.On eitherD EorD Gset off, in perspective ratio,D Hequal to one half ofB C. ThroughHdrawH Kto the vanishing-point ofD E, cuttingD FinIandE GinK. ThroughIandKdrawI M,K L, to vanishing-point ofD G, cuttingD FinLandE GinM. JoinL M.ThenI K L Mis the smaller square, inscribed as required.[Footnote23][p46]COROLLARY.[Geometric diagram]Fig. 36.If, instead of one square within another, it be required to draw one circle within another, the dimensions of both being given, inclose each circle in a square. Draw the squares first, and then the circles within, as inFig. 36.[Geometric diagram]Fig. 35.[Footnote23:If either of the sides of the greater square is parallel to the plane of the picture, asD GinFig. 35.,D Gof course must be equal toA C, andD Hequal toB C/2, and the construction is as inFig. 35.]Return to text

[Geometric diagram]Fig. 34.

LetA B,Fig. 34., be the sight-magnitude of the side of the smaller square, andA Cthat of the side of the larger square.

Draw the larger square. LetD E F Gbe the square so drawn.

JoinE GandD F.

On eitherD EorD Gset off, in perspective ratio,D Hequal to one half ofB C. ThroughHdrawH Kto the vanishing-point ofD E, cuttingD FinIandE GinK. ThroughIandKdrawI M,K L, to vanishing-point ofD G, cuttingD FinLandE GinM. JoinL M.

ThenI K L Mis the smaller square, inscribed as required.[Footnote23]

[Geometric diagram]Fig. 36.

If, instead of one square within another, it be required to draw one circle within another, the dimensions of both being given, inclose each circle in a square. Draw the squares first, and then the circles within, as inFig. 36.

[Geometric diagram]Fig. 35.[Footnote23:If either of the sides of the greater square is parallel to the plane of the picture, asD GinFig. 35.,D Gof course must be equal toA C, andD Hequal toB C/2, and the construction is as inFig. 35.]Return to text

[Geometric diagram]Fig. 35.

[Footnote23:If either of the sides of the greater square is parallel to the plane of the picture, asD GinFig. 35.,D Gof course must be equal toA C, andD Hequal toB C/2, and the construction is as inFig. 35.]Return to text

[p47]PROBLEM XIV.TO DRAW A TRUNCATED CIRCULAR CONE, GIVEN IN POSITION AND MAGNITUDE, THE TRUNCATIONS BEING IN HORIZONTAL PLANES, AND THE AXIS OFTHE CONE VERTICAL.LetA B C D,Fig. 37., be the portion of the cone required.[Geometric diagram]Fig. 37.As it is given in magnitude, its diameters must be given at the base and summit,A BandC D; and its vertical height,C E.[Footnote24]And as it is given in position, the center of its base must be given.[Geometric diagram]Fig. 38.Draw in position, about this center,[Footnote25]the square pillar[p48]a f d,Fig. 38., making its height,b g, equal toC E; and its side,a b, equal toA B.In the square of its base,a b c d, inscribe a circle, which therefore is of the diameter of the base of the cone,A B.In the square of its top,e f g h, inscribe concentrically a circle whose diameter shall equalC D. (Coroll. Prob. XIII.)Join the extremities of the circles by the right linesk l,n m. Thenk l n mis the portion of cone required.COROLLARY I.If similar polygons be inscribed in similar positions in the circlesk nandl m(Coroll. Prob. XII.), and the corresponding angles of the polygons joined by right lines, the resulting figure will be a portion of a polygonal pyramid. (The dotted lines inFig. 38., connecting the extremities of two diameters and one diagonal in the respective circles, occupy the position of the three nearest angles of a regular octagonal pyramid, having its angles set on the diagonals and diameters of the squarea d, inclosing its base.)If the cone or polygonal pyramid is not truncated, its apex will be the center of the upper square, as inFig. 26.COROLLARY II.If equal circles, or equal and similar polygons, be inscribed in the upper and lower squares inFig. 38., the resulting figure will be a vertical cylinder, or a vertical polygonal pillar, of given height and diameter, drawn in position.[p49]COROLLARY III.If the circles inFig. 38., instead of being inscribed in the squaresb candf g, be inscribed in the sides of the solid figureb eandd f, those sides being made square, and the lineb dof any given length, the resulting figure will be, according to the constructions employed, a cone, polygonal pyramid, cylinder, or polygonal pillar, drawn in position about a horizontal axis parallel tob d.Similarly, if the circles are drawn in the sidesg dande c, the resulting figures will be described about a horizontal axis parallel toa b.[Footnote24:Or if the length of its side,A C, is given instead, takea e,Fig. 37., equal to half the excess ofA BoverC D; from the pointeraise the perpendicularc e. With centera, and distanceA C, describe a circle cuttingc einc. Thenc eis the vertical height of the portion of cone required, orC E.]Return to text[Footnote25:The direction of the side of the square will of course be regulated by convenience.]Return to text

LetA B C D,Fig. 37., be the portion of the cone required.

[Geometric diagram]Fig. 37.

As it is given in magnitude, its diameters must be given at the base and summit,A BandC D; and its vertical height,C E.[Footnote24]

And as it is given in position, the center of its base must be given.

[Geometric diagram]Fig. 38.

Draw in position, about this center,[Footnote25]the square pillar[p48]a f d,Fig. 38., making its height,b g, equal toC E; and its side,a b, equal toA B.

In the square of its base,a b c d, inscribe a circle, which therefore is of the diameter of the base of the cone,A B.

In the square of its top,e f g h, inscribe concentrically a circle whose diameter shall equalC D. (Coroll. Prob. XIII.)

Join the extremities of the circles by the right linesk l,n m. Thenk l n mis the portion of cone required.

If similar polygons be inscribed in similar positions in the circlesk nandl m(Coroll. Prob. XII.), and the corresponding angles of the polygons joined by right lines, the resulting figure will be a portion of a polygonal pyramid. (The dotted lines inFig. 38., connecting the extremities of two diameters and one diagonal in the respective circles, occupy the position of the three nearest angles of a regular octagonal pyramid, having its angles set on the diagonals and diameters of the squarea d, inclosing its base.)

If the cone or polygonal pyramid is not truncated, its apex will be the center of the upper square, as inFig. 26.

If equal circles, or equal and similar polygons, be inscribed in the upper and lower squares inFig. 38., the resulting figure will be a vertical cylinder, or a vertical polygonal pillar, of given height and diameter, drawn in position.[p49]

If the circles inFig. 38., instead of being inscribed in the squaresb candf g, be inscribed in the sides of the solid figureb eandd f, those sides being made square, and the lineb dof any given length, the resulting figure will be, according to the constructions employed, a cone, polygonal pyramid, cylinder, or polygonal pillar, drawn in position about a horizontal axis parallel tob d.

Similarly, if the circles are drawn in the sidesg dande c, the resulting figures will be described about a horizontal axis parallel toa b.

[Footnote24:Or if the length of its side,A C, is given instead, takea e,Fig. 37., equal to half the excess ofA BoverC D; from the pointeraise the perpendicularc e. With centera, and distanceA C, describe a circle cuttingc einc. Thenc eis the vertical height of the portion of cone required, orC E.]Return to text[Footnote25:The direction of the side of the square will of course be regulated by convenience.]Return to text

[Footnote25:The direction of the side of the square will of course be regulated by convenience.]Return to text

[p50]PROBLEM XV.TO DRAW AN INCLINED LINE, GIVEN IN POSITIONAND MAGNITUDE.Wehave hitherto been examining the conditions of horizontal and vertical lines only, or of curves inclosed in rectangles.[Geometric diagram]Fig. 39.[Geometric diagram]Fig. 40.We must, in conclusion, investigate the perspective of inclined lines, beginning with a single one given in position. For the sake of completeness of system, I give in Appendix II. Article III. the development of this problem from the second. But, in practice, the position of an inclined line may be most conveniently defined by considering it as the diagonal of a rectangle, asA BinFig. 39., and I shall therefore, though at some sacrifice of system, examine it here under that condition.If the sides of the rectangleA CandA Dare given, the slope of the lineA Bis determined; and then its position will depend on that of the rectangle. If, as inFig. 39., the rectangle is parallel to the picture plane, the lineA Bmust be so also. If, as inFig. 40., the rectangle is inclined to the[p51]picture plane, the lineA Bwill be so also. So that, to fix the position ofA B, the lineA Cmust be given in position and magnitude, and the heightA D.[Geometric diagram]Fig. 41.If these are given, and it is only required to draw the single lineA Bin perspective, the construction is entirely simple;thus:—Draw the lineA CbyProblem I.LetA C,Fig. 41., be the line so drawn. Fromaandcraise the vertical linesa d,c b. Makea dequal to the sight-magnitude ofA D. Fromddrawd bto the vanishing-point ofa c, cuttingb cinb.Joina b. Thena bis the inclined line required.[Geometric diagram]Fig. 42.If the line is inclined in the opposite direction, asD CinFig. 42., we have only to joind cinstead ofa bin Fig. 41., andd cwill be the line required.I shall hereafter call the lineA C, when used to define the position of an inclined lineA B(Fig. 40.), the “relative horizontal” of the lineA B.Observation.[Geometric diagram]Fig. 43.In general, inclined lines are most needed for gable roofs, in which, when the conditions are properly stated, the vertical height of the gable,X Y,Fig. 43., is given, and the base line,A C, in position. When these are given, drawA C; raise verticalA D; makeA Dequal to sight-magnitude ofX Y;complete the perspective-rectangleA D B C; joinA BandD C(as by dotted lines in figure); and through the intersection of the dotted lines draw verticalX Y, cuttingD BinY. JoinA Y,C Y; and these lines are the sides of the gable. If[p52]the length of the roofA A′is also given, draw in perspective the complete parallelopipedA′ D′ B C, and fromYdrawY Y′to the vanishing-point ofA A′, cuttingD′ B′inY′. JoinA′ Y, and you have the slope of the farther side of the roof.[Geometric diagram]Fig. 44.The construction above the eye is as inFig. 44.; the roof is reversed in direction merely to familiarize the student with the different aspects of its lines.

Wehave hitherto been examining the conditions of horizontal and vertical lines only, or of curves inclosed in rectangles.

[Geometric diagram]Fig. 39.[Geometric diagram]Fig. 40.

We must, in conclusion, investigate the perspective of inclined lines, beginning with a single one given in position. For the sake of completeness of system, I give in Appendix II. Article III. the development of this problem from the second. But, in practice, the position of an inclined line may be most conveniently defined by considering it as the diagonal of a rectangle, asA BinFig. 39., and I shall therefore, though at some sacrifice of system, examine it here under that condition.

If the sides of the rectangleA CandA Dare given, the slope of the lineA Bis determined; and then its position will depend on that of the rectangle. If, as inFig. 39., the rectangle is parallel to the picture plane, the lineA Bmust be so also. If, as inFig. 40., the rectangle is inclined to the[p51]picture plane, the lineA Bwill be so also. So that, to fix the position ofA B, the lineA Cmust be given in position and magnitude, and the heightA D.

[Geometric diagram]Fig. 41.

If these are given, and it is only required to draw the single lineA Bin perspective, the construction is entirely simple;thus:—

Draw the lineA CbyProblem I.

LetA C,Fig. 41., be the line so drawn. Fromaandcraise the vertical linesa d,c b. Makea dequal to the sight-magnitude ofA D. Fromddrawd bto the vanishing-point ofa c, cuttingb cinb.

Joina b. Thena bis the inclined line required.

[Geometric diagram]Fig. 42.

If the line is inclined in the opposite direction, asD CinFig. 42., we have only to joind cinstead ofa bin Fig. 41., andd cwill be the line required.

I shall hereafter call the lineA C, when used to define the position of an inclined lineA B(Fig. 40.), the “relative horizontal” of the lineA B.

[Geometric diagram]Fig. 43.

In general, inclined lines are most needed for gable roofs, in which, when the conditions are properly stated, the vertical height of the gable,X Y,Fig. 43., is given, and the base line,A C, in position. When these are given, drawA C; raise verticalA D; makeA Dequal to sight-magnitude ofX Y;complete the perspective-rectangleA D B C; joinA BandD C(as by dotted lines in figure); and through the intersection of the dotted lines draw verticalX Y, cuttingD BinY. JoinA Y,C Y; and these lines are the sides of the gable. If[p52]the length of the roofA A′is also given, draw in perspective the complete parallelopipedA′ D′ B C, and fromYdrawY Y′to the vanishing-point ofA A′, cuttingD′ B′inY′. JoinA′ Y, and you have the slope of the farther side of the roof.

[Geometric diagram]Fig. 44.

The construction above the eye is as inFig. 44.; the roof is reversed in direction merely to familiarize the student with the different aspects of its lines.

[p53]PROBLEM XVI.TO FIND THE VANISHING-POINT OF A GIVENINCLINED LINE.If, inFig. 43.orFig. 44., the linesA YandA′ Y′be produced, the student will find that they meet.LetP,Fig. 45., be the point at which they meet.FromPlet fall the verticalP Von the sight-line, cutting the sight-line inV.Then the student will find experimentally thatVis the vanishing-point of the lineA C.[Footnote26]Complete the rectangle of the baseA C′, by drawingA′ C′toV, andC C′to the vanishing-point ofA A′.JoinY′ C′.Now ifY CandY′ C′be produced downwards, the student will find that they meet.Let them be produced, and meet inP′.ProduceP V, and it will be found to pass through the pointP′.Therefore ifA Y(orC Y),Fig. 45., be any inclined line drawn in perspective byProblem XV., andA Cthe relative horizontal (A CinFigs. 39,40.), also drawn in perspective.ThroughV, the vanishing-point ofA V, draw the verticalP P′upwards and downwards.ProduceA Y(orC Y), cuttingP P′inP(orP′).ThenPis the vanishing-point ofA Y(orP′ofC Y).[Geometric diagram]Fig. 45.The student will observe that, in order to find the pointPby this method, it is necessary first to draw a portion of the given inclined line byProblem XV. Practically, it is always necessary to do so, and, therefore, I give the problem in this form.[p54]Theoretically, as will be shown in the analysis of the problem, the pointPshould be found by drawing a line from the station-point parallel to the given inclined line: but there is no practical means of drawing such a line; so that in whatever terms the problem may be given, a portion of the inclined line (A YorC Y) must always be drawn in perspective beforePcan be found.[Footnote26:The demonstration is inAppendix II. Article III.]Return to text

If, inFig. 43.orFig. 44., the linesA YandA′ Y′be produced, the student will find that they meet.

LetP,Fig. 45., be the point at which they meet.

FromPlet fall the verticalP Von the sight-line, cutting the sight-line inV.

Then the student will find experimentally thatVis the vanishing-point of the lineA C.[Footnote26]

Complete the rectangle of the baseA C′, by drawingA′ C′toV, andC C′to the vanishing-point ofA A′.

JoinY′ C′.

Now ifY CandY′ C′be produced downwards, the student will find that they meet.

Let them be produced, and meet inP′.

ProduceP V, and it will be found to pass through the pointP′.

Therefore ifA Y(orC Y),Fig. 45., be any inclined line drawn in perspective byProblem XV., andA Cthe relative horizontal (A CinFigs. 39,40.), also drawn in perspective.

ThroughV, the vanishing-point ofA V, draw the verticalP P′upwards and downwards.

ProduceA Y(orC Y), cuttingP P′inP(orP′).

ThenPis the vanishing-point ofA Y(orP′ofC Y).

[Geometric diagram]Fig. 45.

The student will observe that, in order to find the pointPby this method, it is necessary first to draw a portion of the given inclined line byProblem XV. Practically, it is always necessary to do so, and, therefore, I give the problem in this form.

[p54]Theoretically, as will be shown in the analysis of the problem, the pointPshould be found by drawing a line from the station-point parallel to the given inclined line: but there is no practical means of drawing such a line; so that in whatever terms the problem may be given, a portion of the inclined line (A YorC Y) must always be drawn in perspective beforePcan be found.

[Footnote26:The demonstration is inAppendix II. Article III.]Return to text

[p55]PROBLEM XVII.TO FIND THE DIVIDING-POINTS OF A GIVENINCLINED LINE.[Geometric diagram]Fig. 46.LetP,Fig. 46., be the vanishing-point of the inclined line, andVthe vanishing-point of the relative horizontal.Find the dividing-points of the relative horizontal,DandD′.ThroughPdraw the horizontal lineX Y.With centerPand distanceD Pdescribe the two arcsD XandD′ Y, cutting the lineX YinXandY.ThenXandYare the dividing-points of the inclined line.[Footnote27]Obs.The dividing-points found by the above rule, used with the ordinary measuring-line, will lay off distances on the retiring inclined line, as the ordinary dividing-points lay them off on the retiring horizontal line.Another dividing-point, peculiar in its application, is sometimes useful, and is to be found asfollows:—[p56][Geometric diagram]Fig. 47.LetA B,Fig. 47., be the given inclined line drawn in perspective, andActhe relative horizontal.Find the vanishing-points,VandE, ofAcandA B;D, the dividing-point ofAc; and the sight-magnitude ofAcon the measuring-line, orA C.FromDerect the perpendicularD F.JoinC B, and produce it to cutD EinF. JoinE F.Then, by similar triangles,D Fis equal toE V, andE Fis parallel toD V.Hence it follows that if fromD, the dividing-point ofAc, we raise a perpendicular and makeD Fequal toE V, a lineC F, drawn from any pointCon the measuring-line toF, will mark the distanceA Bon the inclined line,A Bbeing the portion of the given inclined line which forms the diagonal of the vertical rectangle of whichA Cis the base.[Footnote27:The demonstration is in Appendix II.,p. 104.]Return to text

[Geometric diagram]Fig. 46.

LetP,Fig. 46., be the vanishing-point of the inclined line, andVthe vanishing-point of the relative horizontal.

Find the dividing-points of the relative horizontal,DandD′.

ThroughPdraw the horizontal lineX Y.

With centerPand distanceD Pdescribe the two arcsD XandD′ Y, cutting the lineX YinXandY.

ThenXandYare the dividing-points of the inclined line.[Footnote27]

Obs.The dividing-points found by the above rule, used with the ordinary measuring-line, will lay off distances on the retiring inclined line, as the ordinary dividing-points lay them off on the retiring horizontal line.

Another dividing-point, peculiar in its application, is sometimes useful, and is to be found asfollows:—

[p56][Geometric diagram]Fig. 47.

LetA B,Fig. 47., be the given inclined line drawn in perspective, andActhe relative horizontal.

Find the vanishing-points,VandE, ofAcandA B;D, the dividing-point ofAc; and the sight-magnitude ofAcon the measuring-line, orA C.

FromDerect the perpendicularD F.

JoinC B, and produce it to cutD EinF. JoinE F.

Then, by similar triangles,D Fis equal toE V, andE Fis parallel toD V.

Hence it follows that if fromD, the dividing-point ofAc, we raise a perpendicular and makeD Fequal toE V, a lineC F, drawn from any pointCon the measuring-line toF, will mark the distanceA Bon the inclined line,A Bbeing the portion of the given inclined line which forms the diagonal of the vertical rectangle of whichA Cis the base.

[Footnote27:The demonstration is in Appendix II.,p. 104.]Return to text

[p57]PROBLEM XVIII.TO FIND THE SIGHT-LINE OF AN INCLINED PLANE IN WHICH TWO LINES AREGIVEN IN POSITION.[Footnote28]Asin order to fix the position of a line two points in it must be given, so in order to fix the position of a plane, two lines in it must be given.[Geometric diagram]Fig. 48.Let the two lines beA BandC D,Fig. 48.[p58]As they are given in position, the relative horizontalsA EandC Fmust be given.Then byProblem XVI.the vanishing-point ofA BisV, and ofC D,V′.JoinV V′and produce it to cut the sight-line inX.ThenV Xis the sight-line of the inclined plane.Like the horizontal sight-line, it is of indefinite length; and may be produced in either direction as occasion requires, crossing the horizontal line of sight, if the plane continues downward in that direction.Xis the vanishing-point of all horizontal lines in the inclined plane.[Footnote28:Read the Article on this problem in the Appendix,p. 97, before investigating the problem itself.]Return to text

Asin order to fix the position of a line two points in it must be given, so in order to fix the position of a plane, two lines in it must be given.

[Geometric diagram]Fig. 48.

Let the two lines beA BandC D,Fig. 48.

[p58]As they are given in position, the relative horizontalsA EandC Fmust be given.

Then byProblem XVI.the vanishing-point ofA BisV, and ofC D,V′.

JoinV V′and produce it to cut the sight-line inX.

ThenV Xis the sight-line of the inclined plane.

Like the horizontal sight-line, it is of indefinite length; and may be produced in either direction as occasion requires, crossing the horizontal line of sight, if the plane continues downward in that direction.

Xis the vanishing-point of all horizontal lines in the inclined plane.

[Footnote28:Read the Article on this problem in the Appendix,p. 97, before investigating the problem itself.]Return to text

[p59]PROBLEM XIX.TO FIND THE VANISHING-POINT OF STEEPEST LINES IN AN INCLINED PLANE WHOSESIGHT-LINE IS GIVEN.[Geometric diagram]Fig. 49.LetV X,Fig. 49., be the given sight-line.Produce it to cut the horizontal sight-line inX.ThereforeXis the vanishing-point of horizontal lines in the given inclined plane. (Problem XVIII.)JoinT X, and drawT Yat right angles toT X.ThereforeYis the rectangular vanishing-point corresponding toX.[Footnote29]FromYerect the verticalY P, cutting the sight-line of the inclined plane inP.[p60]ThenPis the vanishing-point of steepest lines in the plane.All lines drawn to it, asQ P,R P,N P, etc., are the steepest possible in the plane; and all lines drawn toX, asQ X,O X, etc., are horizontal, and at right angles to the linesP Q,P R, etc.[Footnote29:That is to say, the vanishing-point of horizontal lines drawn at right angles to the lines whose vanishing-point isX.]Return to text

[Geometric diagram]Fig. 49.

LetV X,Fig. 49., be the given sight-line.

Produce it to cut the horizontal sight-line inX.

ThereforeXis the vanishing-point of horizontal lines in the given inclined plane. (Problem XVIII.)

JoinT X, and drawT Yat right angles toT X.

ThereforeYis the rectangular vanishing-point corresponding toX.[Footnote29]

FromYerect the verticalY P, cutting the sight-line of the inclined plane inP.

[p60]ThenPis the vanishing-point of steepest lines in the plane.

All lines drawn to it, asQ P,R P,N P, etc., are the steepest possible in the plane; and all lines drawn toX, asQ X,O X, etc., are horizontal, and at right angles to the linesP Q,P R, etc.

[Footnote29:That is to say, the vanishing-point of horizontal lines drawn at right angles to the lines whose vanishing-point isX.]Return to text

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