[p13]PROBLEM II.

[p13]PROBLEM II.TO DRAW A RIGHT LINE BETWEENTWO GIVEN POINTS.[Geometric diagram]Fig. 6.LetA B,Fig. 6., be the given right line, joining the given pointsAandB.Let the direct, lateral, and vertical distances of the pointAbeT D,D C, andC A.Let the direct, lateral, and vertical distances of the pointBbeT D′,D C′, andC′ B.Then, byProblem I., the position of the pointAon the plane of the picture isa.And similarly, the position of the pointBon the plane of the picture isb.Joina b.Thena bis the line required.[p14]COROLLARY I.If the lineA Bis in a plane parallel to that of the picture, one end of the lineA Bmust be at the same direct distance from the eye of the observer as the other.Therefore, in that case,D Tis equal toD′ T.Then the construction will be as inFig. 7.; and the student will find experimentally thata bis now parallel toA B.[Footnote11][Geometric diagram]Fig. 7.And thata bis toA BasT Sis toT D.Therefore, to draw any line in a plane parallel to that of the picture, we have only to fix the position of one of its extremities,aorb, and then to draw fromaorba line parallel to the given line, bearing the proportion to it thatT Sbears toT D.[p15]COROLLARY II.If the lineA Bis in a horizontal plane, the vertical distance of one of its extremities must be the same as that of the other.Therefore, in that case,A CequalsB C′(Fig. 6.).And the construction is as inFig. 8.[Geometric diagram]Fig. 8.InFig. 8.producea bto the sight-line, cutting the sight-line inV; the pointV, thus determined, is called theVanishing-Pointof the lineA B.JoinT V. Then the student will find experimentally thatT Vis parallel toA B.[Footnote12][p16]COROLLARY III.If the lineA Bproduced would pass through some point beneath or above the station-point,C Dis toD TasC′ D′is toD′ T; in which case the pointccoincides with the pointc′, and the linea bis vertical.Therefore every vertical line in a picture is, or may be, the perspective representation of a horizontal one which, produced, would pass beneath the feet or above the head of the spectator.[Footnote13][Footnote11:For by the constructionA T ∶aT ∷ B T ∶bT;[eqnii]and therefore the two trianglesA B T,a bT, (having a common angleA T B,) are similar.]Return to text[Footnote12:The demonstration is inAppendix II. Article I.]Return to text[Footnote13:The reflection in water of any luminous point or isolated object (such as the sun or moon) is therefore, in perspective, a vertical line; since such reflection, if produced, would pass under the feet of the spectator. Many artists (Claude among the rest) knowing something of optics, but nothing of perspective, have been led occasionally to draw such reflections towards a point at the center of the base of the picture.]Return to text

[Geometric diagram]Fig. 6.

LetA B,Fig. 6., be the given right line, joining the given pointsAandB.

Let the direct, lateral, and vertical distances of the pointAbeT D,D C, andC A.

Let the direct, lateral, and vertical distances of the pointBbeT D′,D C′, andC′ B.

Then, byProblem I., the position of the pointAon the plane of the picture isa.

And similarly, the position of the pointBon the plane of the picture isb.

Joina b.

Thena bis the line required.

If the lineA Bis in a plane parallel to that of the picture, one end of the lineA Bmust be at the same direct distance from the eye of the observer as the other.

Therefore, in that case,D Tis equal toD′ T.

Then the construction will be as inFig. 7.; and the student will find experimentally thata bis now parallel toA B.[Footnote11]

[Geometric diagram]Fig. 7.

And thata bis toA BasT Sis toT D.

Therefore, to draw any line in a plane parallel to that of the picture, we have only to fix the position of one of its extremities,aorb, and then to draw fromaorba line parallel to the given line, bearing the proportion to it thatT Sbears toT D.

If the lineA Bis in a horizontal plane, the vertical distance of one of its extremities must be the same as that of the other.

Therefore, in that case,A CequalsB C′(Fig. 6.).

And the construction is as inFig. 8.

[Geometric diagram]Fig. 8.

InFig. 8.producea bto the sight-line, cutting the sight-line inV; the pointV, thus determined, is called theVanishing-Pointof the lineA B.

JoinT V. Then the student will find experimentally thatT Vis parallel toA B.[Footnote12]

If the lineA Bproduced would pass through some point beneath or above the station-point,C Dis toD TasC′ D′is toD′ T; in which case the pointccoincides with the pointc′, and the linea bis vertical.

Therefore every vertical line in a picture is, or may be, the perspective representation of a horizontal one which, produced, would pass beneath the feet or above the head of the spectator.[Footnote13]

[Footnote11:For by the constructionA T ∶aT ∷ B T ∶bT;[eqnii]and therefore the two trianglesA B T,a bT, (having a common angleA T B,) are similar.]Return to text[Footnote12:The demonstration is inAppendix II. Article I.]Return to text[Footnote13:The reflection in water of any luminous point or isolated object (such as the sun or moon) is therefore, in perspective, a vertical line; since such reflection, if produced, would pass under the feet of the spectator. Many artists (Claude among the rest) knowing something of optics, but nothing of perspective, have been led occasionally to draw such reflections towards a point at the center of the base of the picture.]Return to text

[Footnote11:For by the constructionA T ∶aT ∷ B T ∶bT;[eqnii]and therefore the two trianglesA B T,a bT, (having a common angleA T B,) are similar.]Return to text

[Footnote12:The demonstration is inAppendix II. Article I.]Return to text

[Footnote13:The reflection in water of any luminous point or isolated object (such as the sun or moon) is therefore, in perspective, a vertical line; since such reflection, if produced, would pass under the feet of the spectator. Many artists (Claude among the rest) knowing something of optics, but nothing of perspective, have been led occasionally to draw such reflections towards a point at the center of the base of the picture.]Return to text

[p17]PROBLEM III.TO FIND THE VANISHING-POINT OF A GIVENHORIZONTAL LINE.[Geometric diagram]Fig. 9.LetA B,Fig. 9., be the given line.FromT, the station-point, drawT Vparallel toA B, cutting the sight-line inV.Vis the Vanishing-point required.[Footnote14][p18]COROLLARY I.As, if the pointbis first found,Vmay be determined by it, so, if the pointVis first found,bmay be determined by it. For letA B,Fig. 10., be the given line, constructed upon the paper as inFig. 8.; and let it be required to draw the linea bwithout using the pointC′.[Geometric diagram]Fig. 10.Find the position of the pointAina. (Problem I.)[p19]Find the vanishing-point ofA BinV. (Problem III.)JoinaV.JoinB T, cuttingaVinb.Thena bis the line required.[Footnote15]COROLLARY II.We have hitherto proceeded on the supposition that the given line was small enough, and near enough, to be actually drawn on our paper of its real size; as in the example given in Appendix I. We may, however, now deduce a construction available under all circumstances, whatever may be the distance and length of the line given.[Geometric diagram]Fig. 11.FromFig. 8.remove, for the sake of clearness, the lines[p20]C′ D′,bV, andT V; and, taking the figure as here inFig. 11., draw froma, the lineaRparallel toA B, cuttingB TinR.ThenaRis toA BasaTis toA T.Then—is to— ascTis toC T.Then—is to— asT Sis toT D.That is to say,aRis the sight-magnitude ofA B.[Footnote16][Geometric diagram]Fig. 12.Therefore, when the position of the pointAis fixed ina, as inFig. 12., andaVis drawn to the vanishing-point; if we draw a lineaRfroma, parallel toA B, and makeaRequal to the sight-magnitude ofA B, and then joinR T, the lineR Twill cutaVinb.So that, in order to determine the length ofa b, we need not draw the long and distant lineA B, but onlyaRparallel to it, and of its sight-magnitude; which is a great gain, for the lineA Bmay be two miles long, and the lineaRperhaps only two inches.[p21]COROLLARY III.InFig. 12., altering its proportions a little for the sake of clearness, and putting it as here inFig. 13., draw a horizontal lineaR′and makeaR′equal toaR.Through the pointsRandbdrawR′ M, cutting the sight-line inM. JoinT V. Now the reader will find experimentally thatV Mis equal toV T.[Footnote17][Geometric diagram]Fig. 13.Hence it follows that, if from the vanishing-pointVwe lay off on the sight-line a distance,V M, equal toV T; then draw throughaa horizontal lineaR′, makeaR′equal to the sight-magnitude ofA B, and joinR′ M; the lineR′ Mwill cutaVinb. And this is in practice generally the most convenient way of obtaining the length ofa b.[p22]COROLLARY IV.Removing from the preceding figure the unnecessary lines, and retaining onlyR′ MandaV, as inFig. 14., produce the lineaR′to the other side ofa, and makeaXequal toaR′.JoinXb, and produceXbto cut the line of sight inN.[Geometric diagram]Fig. 14.Then asX R′is parallel toM N, andaR′is equal toaX,V Nmust, by similar triangles, be equal toV M(equal toV TinFig. 13.).Therefore, on whichever side ofVwe measure the distanceV T, so as to obtain either the pointM, or the pointN, if we measure the sight-magnitudeaR′oraXon the opposite side of the lineaV, the line joiningR′ MorX Nwill equally cutaVinb.The pointsMandNare called the “Dividing-Points” of the original lineA B(Fig. 12.), and we resume the results of these corollaries in the following three problems.[Footnote14:The student will observe, in practice, that, his paper lying flat on the table, he has only to draw the lineT Von its horizontal surface, parallel to the given horizontal lineA B. In theory, the paper should be vertical, but the station-lineS Thorizontal (see its definition above,page 5); in which caseT V, being drawn parallel toA B, will be horizontal also, and still cut the sight-line inV.The construction will be seen to be founded on the secondCorollaryof the preceding problem.It is evident that if any other line, asM NinFig. 9., parallel toA B, occurs in the picture, the lineT V, drawn fromT, parallel toM N, to find the vanishing-point ofM N, will coincide with the line drawn fromT, parallel toA B, to find the vanishing-point ofA B.ThereforeA BandM Nwill have the same vanishing-point.Therefore all parallel horizontal lines have the same vanishing-point.It will be shown hereafter that all parallelinclinedlines also have the same vanishing-point; the student may here accept the general conclusion—“All parallel lines have the same vanishing-point.”It is also evident that ifA Bis parallel to the plane of the picture,T Vmust be drawn parallel toG H, and will therefore never cutG H. The lineA Bhas in that case no vanishing-point: it is to be drawn by the construction given inFig. 7.It is also evident that ifA Bis at right angles with the plane of the picture,T Vwill coincide withT S, and the vanishing-point ofA Bwill be the sight-point.]Return to text[Footnote15:I spare the student the formality of thereductio ad absurdum, which would be necessary to prove this.]Return to text[Footnote16:For definition of Sight-Magnitude, seeAppendix I. It ought to have been read before the student comes to this problem; but I refer to it in case it has not.]Return to text[Footnote17:The demonstration is in Appendix II. Article II.p. 101.]Return to text

[Geometric diagram]Fig. 9.

LetA B,Fig. 9., be the given line.

FromT, the station-point, drawT Vparallel toA B, cutting the sight-line inV.

Vis the Vanishing-point required.[Footnote14]

As, if the pointbis first found,Vmay be determined by it, so, if the pointVis first found,bmay be determined by it. For letA B,Fig. 10., be the given line, constructed upon the paper as inFig. 8.; and let it be required to draw the linea bwithout using the pointC′.

[Geometric diagram]Fig. 10.

Find the position of the pointAina. (Problem I.)

[p19]Find the vanishing-point ofA BinV. (Problem III.)

JoinaV.

JoinB T, cuttingaVinb.

Thena bis the line required.[Footnote15]

We have hitherto proceeded on the supposition that the given line was small enough, and near enough, to be actually drawn on our paper of its real size; as in the example given in Appendix I. We may, however, now deduce a construction available under all circumstances, whatever may be the distance and length of the line given.

[Geometric diagram]Fig. 11.

FromFig. 8.remove, for the sake of clearness, the lines[p20]C′ D′,bV, andT V; and, taking the figure as here inFig. 11., draw froma, the lineaRparallel toA B, cuttingB TinR.

ThenaRis toA BasaTis toA T.

Then—is to— ascTis toC T.

Then—is to— asT Sis toT D.

That is to say,aRis the sight-magnitude ofA B.[Footnote16]

[Geometric diagram]Fig. 12.

Therefore, when the position of the pointAis fixed ina, as inFig. 12., andaVis drawn to the vanishing-point; if we draw a lineaRfroma, parallel toA B, and makeaRequal to the sight-magnitude ofA B, and then joinR T, the lineR Twill cutaVinb.

So that, in order to determine the length ofa b, we need not draw the long and distant lineA B, but onlyaRparallel to it, and of its sight-magnitude; which is a great gain, for the lineA Bmay be two miles long, and the lineaRperhaps only two inches.

InFig. 12., altering its proportions a little for the sake of clearness, and putting it as here inFig. 13., draw a horizontal lineaR′and makeaR′equal toaR.

Through the pointsRandbdrawR′ M, cutting the sight-line inM. JoinT V. Now the reader will find experimentally thatV Mis equal toV T.[Footnote17]

[Geometric diagram]Fig. 13.

Hence it follows that, if from the vanishing-pointVwe lay off on the sight-line a distance,V M, equal toV T; then draw throughaa horizontal lineaR′, makeaR′equal to the sight-magnitude ofA B, and joinR′ M; the lineR′ Mwill cutaVinb. And this is in practice generally the most convenient way of obtaining the length ofa b.

Removing from the preceding figure the unnecessary lines, and retaining onlyR′ MandaV, as inFig. 14., produce the lineaR′to the other side ofa, and makeaXequal toaR′.

JoinXb, and produceXbto cut the line of sight inN.

[Geometric diagram]Fig. 14.

Then asX R′is parallel toM N, andaR′is equal toaX,V Nmust, by similar triangles, be equal toV M(equal toV TinFig. 13.).

Therefore, on whichever side ofVwe measure the distanceV T, so as to obtain either the pointM, or the pointN, if we measure the sight-magnitudeaR′oraXon the opposite side of the lineaV, the line joiningR′ MorX Nwill equally cutaVinb.

The pointsMandNare called the “Dividing-Points” of the original lineA B(Fig. 12.), and we resume the results of these corollaries in the following three problems.

[Footnote14:The student will observe, in practice, that, his paper lying flat on the table, he has only to draw the lineT Von its horizontal surface, parallel to the given horizontal lineA B. In theory, the paper should be vertical, but the station-lineS Thorizontal (see its definition above,page 5); in which caseT V, being drawn parallel toA B, will be horizontal also, and still cut the sight-line inV.The construction will be seen to be founded on the secondCorollaryof the preceding problem.It is evident that if any other line, asM NinFig. 9., parallel toA B, occurs in the picture, the lineT V, drawn fromT, parallel toM N, to find the vanishing-point ofM N, will coincide with the line drawn fromT, parallel toA B, to find the vanishing-point ofA B.ThereforeA BandM Nwill have the same vanishing-point.Therefore all parallel horizontal lines have the same vanishing-point.It will be shown hereafter that all parallelinclinedlines also have the same vanishing-point; the student may here accept the general conclusion—“All parallel lines have the same vanishing-point.”It is also evident that ifA Bis parallel to the plane of the picture,T Vmust be drawn parallel toG H, and will therefore never cutG H. The lineA Bhas in that case no vanishing-point: it is to be drawn by the construction given inFig. 7.It is also evident that ifA Bis at right angles with the plane of the picture,T Vwill coincide withT S, and the vanishing-point ofA Bwill be the sight-point.]Return to text[Footnote15:I spare the student the formality of thereductio ad absurdum, which would be necessary to prove this.]Return to text[Footnote16:For definition of Sight-Magnitude, seeAppendix I. It ought to have been read before the student comes to this problem; but I refer to it in case it has not.]Return to text[Footnote17:The demonstration is in Appendix II. Article II.p. 101.]Return to text

[Footnote14:The student will observe, in practice, that, his paper lying flat on the table, he has only to draw the lineT Von its horizontal surface, parallel to the given horizontal lineA B. In theory, the paper should be vertical, but the station-lineS Thorizontal (see its definition above,page 5); in which caseT V, being drawn parallel toA B, will be horizontal also, and still cut the sight-line inV.

The construction will be seen to be founded on the secondCorollaryof the preceding problem.

It is evident that if any other line, asM NinFig. 9., parallel toA B, occurs in the picture, the lineT V, drawn fromT, parallel toM N, to find the vanishing-point ofM N, will coincide with the line drawn fromT, parallel toA B, to find the vanishing-point ofA B.

ThereforeA BandM Nwill have the same vanishing-point.

Therefore all parallel horizontal lines have the same vanishing-point.

It will be shown hereafter that all parallelinclinedlines also have the same vanishing-point; the student may here accept the general conclusion—“All parallel lines have the same vanishing-point.”

It is also evident that ifA Bis parallel to the plane of the picture,T Vmust be drawn parallel toG H, and will therefore never cutG H. The lineA Bhas in that case no vanishing-point: it is to be drawn by the construction given inFig. 7.

It is also evident that ifA Bis at right angles with the plane of the picture,T Vwill coincide withT S, and the vanishing-point ofA Bwill be the sight-point.]Return to text

[Footnote15:I spare the student the formality of thereductio ad absurdum, which would be necessary to prove this.]Return to text

[Footnote16:For definition of Sight-Magnitude, seeAppendix I. It ought to have been read before the student comes to this problem; but I refer to it in case it has not.]Return to text

[Footnote17:The demonstration is in Appendix II. Article II.p. 101.]Return to text

[p23]PROBLEM IV.TO FIND THE DIVIDING-POINTS OF A GIVENHORIZONTAL LINE.[Geometric diagram]Fig. 15.Letthe horizontal lineA B(Fig. 15.) be given in position and magnitude. It is required to find its dividing-points.Find the vanishing-pointVof the lineA B.With centerVand distanceV T, describe circle cutting the sight-line inMandN.ThenMandNare the dividing-points required.In general, only one dividing-point is needed for use with any vanishing-point, namely, the one nearestS(in this case the pointM). But its oppositeN, or both, may be needed under certain circumstances.

[Geometric diagram]Fig. 15.

Letthe horizontal lineA B(Fig. 15.) be given in position and magnitude. It is required to find its dividing-points.

Find the vanishing-pointVof the lineA B.

With centerVand distanceV T, describe circle cutting the sight-line inMandN.

ThenMandNare the dividing-points required.

In general, only one dividing-point is needed for use with any vanishing-point, namely, the one nearestS(in this case the pointM). But its oppositeN, or both, may be needed under certain circumstances.

[p24]PROBLEM V.TO DRAW A HORIZONTAL LINE, GIVEN IN POSITION AND MAGNITUDE, BY MEANS OF ITS SIGHT-MAGNITUDE ANDDIVIDING-POINTS.[Geometric diagram]Fig. 16.LetA B(Fig. 16.) be the given line.Find the position of the pointAina.Find the vanishing-pointV, and most convenient dividing-pointM, of the lineA B.JoinaV.Throughadraw a horizontal linea b′and makea b′equal to the sight-magnitude ofA B. Joinb′M, cuttingaVinb.Thena bis the line required.[p25]COROLLARY I.[Geometric diagram]Fig. 17.Supposing it were now required to draw a lineA C(Fig. 17.) twice as long asA B, it is evident that the sight-magnitudea c′must be twice as long as the sight-magnitudea b′; we have, therefore, merely to continue the horizontal linea b′, makeb′ c′equal toa b′, joincM′, cuttingaVinc, anda cwill be the line required. Similarly, if we have to draw a lineA D, three times the length ofA B,a d′must be three times the length ofa b′, and, joiningd′M,a dwill be the line required.The student will observe that the nearer the portions cut off,b c,c d, etc., approach the pointV, the smaller they become; and, whatever lengths may be added to the lineA D, and successively cut off fromaV, the lineaVwill never be cut off entirely, but the portions cut off will become infinitely small, and apparently “vanish” as they approach the pointV; hence this point is called the “vanishing” point.[p26]COROLLARY II.It is evident that if the lineA Dhad been given originally, and we had been required to draw it, and divide it into three equal parts, we should have had only to divide its sight-magnitude,a d′, into the three equal parts,a b′,b′ c′, andc′ d′, and then, drawing toMfromb′andc′, the linea dwould have been divided as required inbandc. And supposing the original lineA Dbe dividedirregularly into any numberof parts, if the linea d′be divided into a similar number in the same proportions (by the construction given in Appendix I.), and, from these points of division, lines are drawn toM, they will divide the linea din true perspective into a similar number of proportionate parts.The horizontal line drawn througha, on which the sight-magnitudes are measured, is called the “Measuring-line.”And the linea d, when properly divided inbandc, or any other required points, is said to be divided “IN PERSPECTIVE RATIO” to the divisions of the original lineA D.If the lineaVis above the sight-line instead of beneath it, the measuring-line is to be drawn above also: and the linesb′M,c′M, etc., drawndownto the dividing-point. TurnFig. 17.upside down, and it will show the construction.

[Geometric diagram]Fig. 16.

LetA B(Fig. 16.) be the given line.

Find the position of the pointAina.

Find the vanishing-pointV, and most convenient dividing-pointM, of the lineA B.

JoinaV.

Throughadraw a horizontal linea b′and makea b′equal to the sight-magnitude ofA B. Joinb′M, cuttingaVinb.

Thena bis the line required.

[Geometric diagram]Fig. 17.

Supposing it were now required to draw a lineA C(Fig. 17.) twice as long asA B, it is evident that the sight-magnitudea c′must be twice as long as the sight-magnitudea b′; we have, therefore, merely to continue the horizontal linea b′, makeb′ c′equal toa b′, joincM′, cuttingaVinc, anda cwill be the line required. Similarly, if we have to draw a lineA D, three times the length ofA B,a d′must be three times the length ofa b′, and, joiningd′M,a dwill be the line required.

The student will observe that the nearer the portions cut off,b c,c d, etc., approach the pointV, the smaller they become; and, whatever lengths may be added to the lineA D, and successively cut off fromaV, the lineaVwill never be cut off entirely, but the portions cut off will become infinitely small, and apparently “vanish” as they approach the pointV; hence this point is called the “vanishing” point.

It is evident that if the lineA Dhad been given originally, and we had been required to draw it, and divide it into three equal parts, we should have had only to divide its sight-magnitude,a d′, into the three equal parts,a b′,b′ c′, andc′ d′, and then, drawing toMfromb′andc′, the linea dwould have been divided as required inbandc. And supposing the original lineA Dbe dividedirregularly into any numberof parts, if the linea d′be divided into a similar number in the same proportions (by the construction given in Appendix I.), and, from these points of division, lines are drawn toM, they will divide the linea din true perspective into a similar number of proportionate parts.

The horizontal line drawn througha, on which the sight-magnitudes are measured, is called the “Measuring-line.”

And the linea d, when properly divided inbandc, or any other required points, is said to be divided “IN PERSPECTIVE RATIO” to the divisions of the original lineA D.

If the lineaVis above the sight-line instead of beneath it, the measuring-line is to be drawn above also: and the linesb′M,c′M, etc., drawndownto the dividing-point. TurnFig. 17.upside down, and it will show the construction.

[p27]PROBLEM VI.TO DRAW ANY TRIANGLE, GIVEN IN POSITION AND MAGNITUDE, IN AHORIZONTAL PLANE.[Geometric diagram]Fig. 18.LetA B C(Fig. 18.) be the triangle.As it is given in position and magnitude, one of its sides, at least, must be given in position and magnitude, and the directions of the two other sides.LetA Bbe the side given in position and magnitude.ThenA Bis a horizontal line, in a given position, and of a given length.Draw the lineA B. (Problem V.)Leta bbe the line so drawn.FindVandV′, the vanishing-points respectively of the linesA CandB C. (Problem III.)[p28]FromadrawaV, and fromb, drawbV′, cutting each other inc.Thena b cis the triangle required.IfA Cis the line originally given,a cis the line which must be first drawn, and the lineV′bmust be drawn fromV′tocand produced to cuta binb. Similarly, ifB Cis given,Vcmust be drawn tocand produced, anda bfrom its vanishing-point tob, and produced to cuta cina.

[Geometric diagram]Fig. 18.

LetA B C(Fig. 18.) be the triangle.

As it is given in position and magnitude, one of its sides, at least, must be given in position and magnitude, and the directions of the two other sides.

LetA Bbe the side given in position and magnitude.

ThenA Bis a horizontal line, in a given position, and of a given length.

Draw the lineA B. (Problem V.)

Leta bbe the line so drawn.

FindVandV′, the vanishing-points respectively of the linesA CandB C. (Problem III.)

[p28]FromadrawaV, and fromb, drawbV′, cutting each other inc.

Thena b cis the triangle required.

IfA Cis the line originally given,a cis the line which must be first drawn, and the lineV′bmust be drawn fromV′tocand produced to cuta binb. Similarly, ifB Cis given,Vcmust be drawn tocand produced, anda bfrom its vanishing-point tob, and produced to cuta cina.

[p29]PROBLEM VII.TO DRAW ANY RECTILINEAR QUADRILATERAL FIGURE, GIVEN IN POSITION AND MAGNITUDE, IN AHORIZONTAL PLANE.[Geometric diagram]Fig. 19.LetA B C D(Fig. 19.) be the given figure.Join any two of its opposite angles by the lineB C.Draw first the triangleA B C. (Problem VI.)And then, from the baseB C, the two linesB D,C D, to their vanishing-points, which will complete the figure. It is unnecessary to give a diagram of the construction, which is merely that ofFig. 18.duplicated; another triangle being drawn on the lineA CorB C.COROLLARY.It is evident that by this application ofProblem VI.any given rectilinear figure whatever in a horizontal plane may be drawn, since any such figure may be divided into a number of triangles, and the triangles then drawn in succession.More convenient methods may, however, be generally[p30]found, according to the form of the figure required, by the use of succeeding problems; and for the quadrilateral figure which occurs most frequently in practice, namely, the square, the following construction is more convenient than that used in the present problem.

[Geometric diagram]Fig. 19.

LetA B C D(Fig. 19.) be the given figure.

Join any two of its opposite angles by the lineB C.

Draw first the triangleA B C. (Problem VI.)

And then, from the baseB C, the two linesB D,C D, to their vanishing-points, which will complete the figure. It is unnecessary to give a diagram of the construction, which is merely that ofFig. 18.duplicated; another triangle being drawn on the lineA CorB C.

It is evident that by this application ofProblem VI.any given rectilinear figure whatever in a horizontal plane may be drawn, since any such figure may be divided into a number of triangles, and the triangles then drawn in succession.

More convenient methods may, however, be generally[p30]found, according to the form of the figure required, by the use of succeeding problems; and for the quadrilateral figure which occurs most frequently in practice, namely, the square, the following construction is more convenient than that used in the present problem.

[p31]PROBLEM VIII.TO DRAW A SQUARE, GIVEN IN POSITION AND MAGNITUDE, IN AHORIZONTAL PLANE.[Geometric diagram]Fig. 20.LetA B C D,Fig. 20., be the square.As it is given in position and magnitude, the position and magnitude of all its sides are given.Fix the position of the pointAina.FindV, the vanishing-point ofA B; andM, the dividing-point ofA B, nearestS.FindV′, the vanishing-point ofA C; andN, the dividing-point ofA C, nearestS.[p32]Draw the measuring-line througha, and makea b′,a c′, each equal to the sight-magnitude ofA B.(For sinceA B C Dis a square,A Cis equal toA B.)DrawaV′andc′N, cutting each other inc.DrawaV, andb′M, cutting each other inb.Thena c,a b, are the two nearest sides of the square.Now, clearing the figure of superfluous lines, we havea b,a c, drawn in position, as inFig. 21.[Geometric diagram]Fig. 21.And becauseA B C Dis a square,C D(Fig. 20.) is parallel toA B.And all parallel lines have the same vanishing-point. (Note toProblem III.)Therefore,Vis the vanishing-point ofC D.Similarly,V′is the vanishing-point ofB D.Therefore, frombandc(Fig. 22.) drawbV′,cV, cutting each other ind.Thena b c dis the square required.COROLLARY I.It is obvious that any rectangle in a horizontal plane may be drawn by this problem, merely makinga b′, on the measuring-line,Fig. 20., equal to the sight-magnitude of one of its sides, anda c′the sight-magnitude of the other.[p33]COROLLARY II.Leta b c d,Fig. 22., be any square drawn in perspective. Draw the diagonalsa dandb c, cutting each other inC. ThenCis the center of the square. ThroughC, drawe fto the vanishing-point ofa b, andg hto the vanishing-point ofa c, and these lines will bisect the sides of the square, so thata gis the perspective representation of half the sidea b;a eis halfa c;c his halfc d; andb fis halfb d.[Geometric diagram]Fig. 22.COROLLARY III.SinceA B C D,Fig. 20., is a square,B A Cis a right angle; and asT Vis parallel toA B, andT V′toA C,V′ T Vmust be a right angle also.As the ground plan of most buildings is rectangular, it constantly happens in practice that their angles (as the corners of ordinary houses) throw the lines to the vanishing-points thus at right angles; and so that this law is observed, andV T V′is kept a right angle, 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 ofS: but it matters much that the relation of the vanishing-points should be accurate. Their position with respect toSmerely causes the spectator to see a little more or less on one side or other of the house, which may be a matter of chance or choice; but their rectangular relation determines the rectangular shape of the building, which is an essential point.

[Geometric diagram]Fig. 20.

LetA B C D,Fig. 20., be the square.

As it is given in position and magnitude, the position and magnitude of all its sides are given.

Fix the position of the pointAina.

FindV, the vanishing-point ofA B; andM, the dividing-point ofA B, nearestS.

FindV′, the vanishing-point ofA C; andN, the dividing-point ofA C, nearestS.

[p32]Draw the measuring-line througha, and makea b′,a c′, each equal to the sight-magnitude ofA B.

(For sinceA B C Dis a square,A Cis equal toA B.)

DrawaV′andc′N, cutting each other inc.

DrawaV, andb′M, cutting each other inb.

Thena c,a b, are the two nearest sides of the square.

Now, clearing the figure of superfluous lines, we havea b,a c, drawn in position, as inFig. 21.

[Geometric diagram]Fig. 21.

And becauseA B C Dis a square,C D(Fig. 20.) is parallel toA B.

And all parallel lines have the same vanishing-point. (Note toProblem III.)

Therefore,Vis the vanishing-point ofC D.

Similarly,V′is the vanishing-point ofB D.

Therefore, frombandc(Fig. 22.) drawbV′,cV, cutting each other ind.

Thena b c dis the square required.

It is obvious that any rectangle in a horizontal plane may be drawn by this problem, merely makinga b′, on the measuring-line,Fig. 20., equal to the sight-magnitude of one of its sides, anda c′the sight-magnitude of the other.

Leta b c d,Fig. 22., be any square drawn in perspective. Draw the diagonalsa dandb c, cutting each other inC. ThenCis the center of the square. ThroughC, drawe fto the vanishing-point ofa b, andg hto the vanishing-point ofa c, and these lines will bisect the sides of the square, so thata gis the perspective representation of half the sidea b;a eis halfa c;c his halfc d; andb fis halfb d.

[Geometric diagram]Fig. 22.

SinceA B C D,Fig. 20., is a square,B A Cis a right angle; and asT Vis parallel toA B, andT V′toA C,V′ T Vmust be a right angle also.

As the ground plan of most buildings is rectangular, it constantly happens in practice that their angles (as the corners of ordinary houses) throw the lines to the vanishing-points thus at right angles; and so that this law is observed, andV T V′is kept a right angle, 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 ofS: but it matters much that the relation of the vanishing-points should be accurate. Their position with respect toSmerely causes the spectator to see a little more or less on one side or other of the house, which may be a matter of chance or choice; but their rectangular relation determines the rectangular shape of the building, which is an essential point.

[p34]PROBLEM IX.TO DRAW A SQUARE PILLAR, GIVEN IN POSITION AND MAGNITUDE, ITS BASE AND TOP BEING INHORIZONTAL PLANES.LetA H,Fig. 23., be the square pillar.Then, as it is given in position and magnitude, the position and magnitude of the square it stands upon must be given (that is, the lineA BorA Cin position), and the height of its sideA E.[Geometric diagram]Fig. 23.[Geometric diagram]Fig. 24.Find the sight-magnitudes ofA BandA E. Draw the two sidesa b,a c, of the square of the base, byProblem VIII., as inFig. 24.From the pointsa,b, andc, raise vertical linesa e,c f,b g.Makea eequal to the sight-magnitude ofA E.Now because the top and base of the pillar are in horizontal planes, the square of its top,F G, is parallel to the square of its base,B C.Therefore the lineE Fis parallel toA C, andE GtoA B.ThereforeE Fhas the same vanishing-point asA C, andE Gthe same vanishing-point asA B.Fromedrawe fto the vanishing-point ofa c, cuttingc finf.Similarly drawe gto the vanishing-point ofa b, cuttingb ging.Complete the squareg finh, by drawingg hto the vanishing-point ofe f, andf hto the vanishing-point ofe g, cutting each other inh. Thena g h fis the square pillar required.[p35]COROLLARY.It is obvious that ifA Eis equal toA C, the whole figure will be a cube, and each side,a e f canda e g b, will be a square in a given vertical plane. And by makingA BorA Clonger or shorter in any given proportion, any form of rectangle may be given to either of the sides of the pillar. No other rule is therefore needed for drawing squares or rectangles in vertical planes.Also any triangle may be thus drawn in a vertical plane, by inclosing it in a rectangle and determining, in perspective ratio, on the sides of the rectangle, the points of their contact with the angles of the triangle.And if any triangle, then any polygon.A less complicated construction will, however, be given hereafter.[Footnote18][Footnote18:See page 96 (note), after you have readProblem XVI.]Return to text

LetA H,Fig. 23., be the square pillar.

Then, as it is given in position and magnitude, the position and magnitude of the square it stands upon must be given (that is, the lineA BorA Cin position), and the height of its sideA E.

[Geometric diagram]Fig. 23.[Geometric diagram]Fig. 24.

Find the sight-magnitudes ofA BandA E. Draw the two sidesa b,a c, of the square of the base, byProblem VIII., as inFig. 24.From the pointsa,b, andc, raise vertical linesa e,c f,b g.

Makea eequal to the sight-magnitude ofA E.

Now because the top and base of the pillar are in horizontal planes, the square of its top,F G, is parallel to the square of its base,B C.

Therefore the lineE Fis parallel toA C, andE GtoA B.

ThereforeE Fhas the same vanishing-point asA C, andE Gthe same vanishing-point asA B.

Fromedrawe fto the vanishing-point ofa c, cuttingc finf.

Similarly drawe gto the vanishing-point ofa b, cuttingb ging.

Complete the squareg finh, by drawingg hto the vanishing-point ofe f, andf hto the vanishing-point ofe g, cutting each other inh. Thena g h fis the square pillar required.

It is obvious that ifA Eis equal toA C, the whole figure will be a cube, and each side,a e f canda e g b, will be a square in a given vertical plane. And by makingA BorA Clonger or shorter in any given proportion, any form of rectangle may be given to either of the sides of the pillar. No other rule is therefore needed for drawing squares or rectangles in vertical planes.

Also any triangle may be thus drawn in a vertical plane, by inclosing it in a rectangle and determining, in perspective ratio, on the sides of the rectangle, the points of their contact with the angles of the triangle.

And if any triangle, then any polygon.

A less complicated construction will, however, be given hereafter.[Footnote18]

[Footnote18:See page 96 (note), after you have readProblem XVI.]Return to text

[Footnote18:See page 96 (note), after you have readProblem XVI.]Return to text

[p36]PROBLEM X.TO DRAW A PYRAMID, GIVEN IN POSITION AND MAGNITUDE, ON A SQUARE BASE IN AHORIZONTAL PLANE.[Geometric diagram]Fig. 25.LetA B,Fig. 25., be the four-sided pyramid. As it is given in position and magnitude, the square base on which it stands must be given in position and magnitude, and its vertical height,C D.[Footnote19][Geometric diagram]Fig. 26.Draw a square pillar,A B G E,Fig. 26., on the square base of the pyramid, and make the height of the pillarA Fequal[p36]to the vertical height of the pyramidC D(Problem IX.). Draw the diagonalsG F,H I, on the top of the square pillar, cutting each other inC. ThereforeCis the center of the squareF G H I. (Prob. VIII.Cor. II.)[Geometric diagram]Fig. 27.JoinC E,C A,C B.ThenA B C Eis the pyramid required. If the base of the pyramid is above the eye, as when a square spire is seen on the top of a church-tower, the construction will be as inFig. 27.[Footnote19:If, instead of the vertical height, the length ofA Dis given, the vertical must be deduced from it. See the Exercises on this Problem in the Appendix,p. 79.]Return to text

[Geometric diagram]Fig. 25.

LetA B,Fig. 25., be the four-sided pyramid. As it is given in position and magnitude, the square base on which it stands must be given in position and magnitude, and its vertical height,C D.[Footnote19]

[Geometric diagram]Fig. 26.

Draw a square pillar,A B G E,Fig. 26., on the square base of the pyramid, and make the height of the pillarA Fequal[p36]to the vertical height of the pyramidC D(Problem IX.). Draw the diagonalsG F,H I, on the top of the square pillar, cutting each other inC. ThereforeCis the center of the squareF G H I. (Prob. VIII.Cor. II.)

[Geometric diagram]Fig. 27.

JoinC E,C A,C B.

ThenA B C Eis the pyramid required. If the base of the pyramid is above the eye, as when a square spire is seen on the top of a church-tower, the construction will be as inFig. 27.

[Footnote19:If, instead of the vertical height, the length ofA Dis given, the vertical must be deduced from it. See the Exercises on this Problem in the Appendix,p. 79.]Return to text

[Footnote19:If, instead of the vertical height, the length ofA Dis given, the vertical must be deduced from it. See the Exercises on this Problem in the Appendix,p. 79.]Return to text

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