Part 5

Complete the square, and within it inscribe a circle, as in Fig. 62., which is left unlettered that its construction may be clear. At the extremities of this draw vertical lines, which will be the sides of the shaft in its right place. It will be found to be somewhat smaller in diameter than the entire shaft in Fig. 60., because at the center of the square it is more distant than the nearest edge of the square abacus. The curves of the capital may then be drawn approximately by the eye. They are not quite accurate in Fig. 62., there being a subtlety in their junction with the shaft which could not be shown on so small a scale without confusing the student; the curve on the left springing from a point a little way round the circle behind the shaft, and that on the right from a point on this side of the circle a little way within the edge of the shaft. But for their more accurate construction see Notes on Problem XIV.

PROBLEM XI.

It is seldom that any complicated curve, except occasionally a spiral, needs to be drawn in perspective; but the student will do well to practice for some time any fantastic shapes which he can find drawn on flat surfaces, as on wall-papers, carpets, etc., in order to accustom himself to the strange and great changes which perspective causes in them.

The curves most required in architectural drawing, after the circle, are those of pointed arches; in which, however, all that will be generally needed is to fix the apex, and two points in the sides. Thus if we have to draw a range of pointed arches, such as _APB_, Fig. 63., draw the measured arch to its sight-magnitude first neatly in a rectangle, _ABCD_; then draw the diagonals _AD_ and _BC_; where they cut the curve draw a horizontal line (as at the level _E_ in the figure), and carry it along the range to the vanishing-point, fixing the points where the arches cut their diagonals all along. If the arch is cusped, a line should be drawn, at _F_ to mark the height of the cusps, and verticals raised at _G_ and _H_, to determine the interval between them. Any other points may be similarly determined, but these will usually be enough. Figure 63. shows the perspective construction of a square niche of good Veronese Gothic, with an uncusped arch of similar size and curve beyond.

In Fig. 64. the more distant arch only is lettered, as the construction of the nearest explains itself more clearly to the eye without letters. The more distant arch shows the general construction for all arches seen underneath, as of bridges, cathedral aisles, etc. The rectangle _ABCD_ is first drawn to contain the outside arch; then the depth of the arch, _Aa_, is determined by the measuring-line, and the rectangle, _abcd_, drawn for the inner arch.

_Aa_, _Bb_, etc., go to one vanishing-point; _AB_, _ab_, etc., to the opposite one.

In the nearer arch another narrow rectangle is drawn to determine the cusp. The parts which would actually come into sight are slightly shaded.

PROBLEM XIV.

Several exercises will be required on this important problem.

I. It is required to draw a circular flat-bottomed dish narrower at the bottom than the top; the vertical depth being given, and the diameter at the top and bottom.

Let _ab_, Fig. 65., be the diameter of the bottom, _ac_ the diameter of the top, and _ad_ its vertical depth.

Take _AD_ in position equal to _ac_.

On _AD_ draw the square _ABCD_, and inscribe in it a circle.

Therefore, the circle so inscribed has the diameter of the top of the dish.

From _A_ and _D_ let fall verticals, _AE_, _DH_, each equal to _ad_.

Join _EH_, and describe square _EFGH_, which accordingly will be equal to the square _ABCD_, and be at the depth _ad_ beneath it.

Within the square _EFGH_ describe a square _IK_, whose diameter shall be equal to _ab_.

Describe a circle within the square _IK_. Therefore the circle so inscribed has its diameter equal to _ab_; and it is in the center of the square _EFGH_, which is vertically beneath the square _ABCD_.

Therefore the circle in the square _IK_ represents the bottom of the dish.

Now the two circles thus drawn will either intersect one another, or they will not.

If they intersect one another, as in the figure, and they are below the eye, part of the bottom of the dish is seen within it.

To avoid confusion, let us take then two intersecting circles without the inclosing squares, as in Fig. 66.

Draw right lines, _ab_, _cd_, touching both circles externally. Then the parts of these lines which connect the circles are the sides of the dish. They are drawn in Fig. 65. without any prolongations, but the best way to construct them is as in Fig. 66.

If the circles do not intersect each other, the smaller must either be within the larger or not within it.

If within the larger, the whole of the bottom of the dish is seen from above, Fig. 67. _a_.

If the smaller circle is not within the larger, none of the bottom is seen inside the dish, _b_.

If the circles are above instead of beneath the eye, the bottom of the dish is seen beneath it, _c_.

If one circle is above and another beneath the eye, neither the bottom nor top of the dish is seen, _d_. Unless the object be very large, the circles in this case will have little apparent curvature.

II. The preceding problem is simple, because the lines of the profile of the object (_ab_ and _cd_, Fig. 66.) are straight. But if these lines of profile are curved, the problem becomes much more complex: once mastered, however, it leaves no farther difficulty in perspective.

Let it be required to draw a flattish circular cup or vase, with a given curve of profile.

The basis of construction is given in Fig. 68., half of it only being drawn, in order that the eye may seize its lines easily.

Two squares (of the required size) are first drawn, one above the other, with a given vertical interval, _AC_, between them, and each is divided into eight parts by its diameters and diagonals. In these squares two circles are drawn; which are, therefore, of equal size, and one above the other. Two smaller circles, also of equal size, are drawn within these larger circles in the construction of the present problem; more may be necessary in some, none at all in others.

It will be seen that the portions of the diagonals and diameters of squares which are cut off between the circles represent radiating planes, occupying the position of the spokes of a wheel.

Now let the line _AEB_, Fig. 69., be the profile of the vase or cup to be drawn.

Inclose it in the rectangle _CD_, and if any portion of it is not curved, as _AE_, cut off the curved portion by the vertical line _EF_, so as to include it in the smaller rectangle _FD_.

Draw the rectangle _ACBD_ in position, and upon it construct two squares, as they are constructed on the rectangle _ACD_ in Fig. 68.; and complete the construction of Fig. 68., making the radius of its large outer circles equal to _AD_, and of its small inner circles equal to _AE_.

The planes which occupy the position of the wheel spokes will then each represent a rectangle of the size of _FD_. The construction is shown by the dotted lines in Fig. 69.; _c_ being the center of the uppermost circle.

Within each of the smaller rectangles between the circles, draw the curve _EB_ in perspective, as in Fig. 69.

Draw the curve _xy_, touching and inclosing the curves in the rectangles, and meeting the upper circle at _y_.[32]

Then _xy_ is the contour of the surface of the cup, and the upper circle is its lip.

If the line _xy_ is long, it may be necessary to draw other rectangles between the eight principal ones; and, if the curve of profile _AB_ is complex or retorted, there may be several lines corresponding to _XY_, inclosing the successive waves of the profile; and the outer curve will then be an undulating or broken one.

III. All branched ornamentation, forms of flowers, capitals of columns, machicolations of round towers, and other such arrangements of radiating curve, are resolvable by this problem, using more or fewer interior circles according to the conditions of the curves. Fig. 70. is an example of the construction of a circular group of eight trefoils with curved stems. One outer or limiting circle is drawn within the square _EDCF_, and the extremities of the trefoils touch it at the extremities of its diagonals and diameters. A smaller circle is at the vertical distance _BC_ below the larger, and _A_ is the angle of the square within which the smaller circle is drawn; but the square is not given, to avoid confusion. The stems of the trefoils form drooping curves, arranged on the diagonals and diameters of the smaller circle, which are dotted. But no perspective laws will do work of this intricate kind so well as the hand and eye of a painter.

IV. There is one common construction, however, in which, singularly, the hand and eye of the painter almost always fail, and that is the fillet of any ordinary capital or base of a circular pillar (or any similar form). It is rarely necessary in practice to draw such minor details in perspective; yet the perspective laws which regulate them should be understood, else the eye does not see their contours rightly until it is very highly cultivated.

Fig. 71. will show the law with sufficient clearness; it represents the perspective construction of a fillet whose profile is a semicircle, such as _FH_ in Fig. 60., seen above the eye. Only half the pillar with half the fillet is drawn, to avoid confusion.

_Q_ is the center of the shaft.

_PQ_ the thickness of the fillet, sight-magnitude at the shaft’s center.

Round _P_ a horizontal semicircle is drawn on the diameter of the shaft _ab_.

Round _Q_ another horizontal semicircle is drawn on diameter _cd_.

These two semicircles are the upper and lower edges of the fillet.

Then diagonals and diameters are drawn as in Fig. 68., and, at their extremities, semicircles in perspective, as in Fig. 69.

The letters _A_, _B_, _C_, _D_, and _E_, indicate the upper and exterior angles of the rectangles in which these semicircles are to be drawn; but the inner vertical line is not dotted in the rectangle at _C_, as it would have confused itself with other lines.

Then the visible contour of the fillet is the line which incloses and touches[33] all the semicircles. It disappears behind the shaft at the point _H_, but I have drawn it through to the opposite extremity of the diameter at _d_.

Turned upside down the figure shows the construction of a basic fillet.

The capital of a Greek Doric pillar should be drawn frequently for exercise on this fourteenth problem, the curve of its echinus being exquisitely subtle, while the general contour is simple.

[32] This point coincides in the figure with the extremity of the horizontal diameter, but only accidentally.

[33] The engraving is a little inaccurate; the inclosing line should touch the dotted semicircles at _A_ and _B_. The student should draw it on a large scale.

PROBLEM XVI.

It is often possible to shorten other perspective operations considerably, by finding the vanishing-points of the inclined lines of the object. Thus, in drawing the gabled roof in Fig. 43., if the gable _AYC_ be drawn in perspective, and the vanishing-point of _AY_ determined, it is not necessary to draw the two sides of the rectangle, _A′D′_ and _D′B′_, in order to determine the point _Y′_; but merely to draw _YY′_ to the vanishing-point of _AA′_ and _A′Y′_ to the vanishing-point of _AY_, meeting in _Y′_, the point required.

Again, if there be a series of gables, or other figures produced by parallel inclined lines, and retiring to the point _V_, as in Fig. 72.,[34] it is not necessary to draw each separately, but merely to determine their breadths on the line _AV_, and draw the slopes of each to their vanishing-points, as shown in Fig. 72. Or if the gables are equal in height, and a line be drawn from _Y_ to _V_, the construction resolves itself into a zigzag drawn alternately to _P_ and _Q_, between the lines _YV_ and _AV_.

The student must be very cautious, in finding the vanishing-points of inclined lines, to notice their relations to the horizontals beneath them, else he may easily mistake the horizontal to which they belong.

Thus, let _ABCD_, Fig. 73., be a rectangular inclined plane, and let it be required to find the vanishing-point of its diagonal _BD_.

Find _V_, the vanishing-point of _AD_ and _BC_.

Draw _AE_ to the opposite vanishing-point, so that _DAE_ may represent a right angle.

Let fall from _B_ the vertical _BE_, cutting _AE_ in _E_.

Join _ED_, and produce it to cut the sight-line in _V′_.

Then, since the point _E_ is vertically under the point _B_, the horizontal line _ED_ is vertically under the inclined line _BD_.

So that if we now let fall the vertical _V′P_ from _V′_, and produce _BD_ to cut _V′P_ in _P_, the point _P_ will be the vanishing-point of _BD_, and of all lines parallel to it.[35]

[34] The diagram is inaccurately cut. _YV_ should be a right line.

[35] The student may perhaps understand this construction better by completing the rectangle _ADFE_, drawing _DF_ to the vanishing-point of _AE_, and _EF_ to _V_. The whole figure, _BF_, may then be conceived as representing half the gable roof of a house, _AF_ the rectangle of its base, and _AC_ the rectangle of its sloping side.

In nearly all picturesque buildings, especially on the Continent, the slopes of gables are much varied (frequently unequal on the two sides), and the vanishing-points of their inclined lines become very important, if accuracy is required in the intersections of tiling, sides of dormer windows, etc.

Obviously, also, irregular triangles and polygons in vertical planes may be more easily constructed by finding the vanishing-points of their sides, than by the construction given in the corollary to Problem IX.; and if such triangles or polygons have others concentrically inscribed within them, as often in Byzantine mosaics, etc., the use of the vanishing-points will become essential.

PROBLEM XVIII.

Before examining the last three problems it is necessary that you should understand accurately what is meant by the position of an inclined plane.

Cut a piece of strong white pasteboard into any irregular shape, and dip it in a sloped position into water. However you hold it, the edge of the water, of course, will always draw a horizontal line across its surface. The direction of this horizontal line is the direction of the inclined plane. (In beds of rock geologists call it their “strike.”)

Next, draw a semicircle on the piece of pasteboard; draw its diameter, _AB_, Fig. 74., and a vertical line from its center, _CD_; and draw some other lines, _CE_, _CF_, etc., from the center to any points in the circumference.

Now dip the piece of pasteboard again into water, and, holding it at any inclination and in any direction you choose, bring the surface of the water to the line _AB_. Then the line _CD_ will be the most steeply inclined of all the lines drawn to the circumference of the circle; _GC_ and _HC_ will be less steep; and _EC_ and _FC_ less steep still. The nearer the lines to _CD_, the steeper they will be; and the nearer to _AB_, the more nearly horizontal.

When, therefore, the line _AB_ is horizontal (or marks the water surface), its direction is the direction of the inclined plane, and the inclination of the line _DC_ is the inclination of the inclined plane. In beds of rock geologists call the inclination of the line _DC_ their “dip.”

To fix the position of an inclined plane, therefore, is to determine the direction of any two lines in the plane, _AB_ and _CD_, of which one shall be horizontal and the other at right angles to it. Then any lines drawn in the inclined plane, parallel to _AB_, will be horizontal; and lines drawn parallel to _CD_ will be as steep as _CD_, and are spoken of in the text as the “steepest lines” in the plane.

But farther, whatever the direction of a plane may be, if it be extended indefinitely, it will be terminated, to the eye of the observer, by a boundary line, which, in a horizontal plane, is horizontal (coinciding nearly with the visible horizon);—in a vertical plane, is vertical;—and, in an inclined plane, is inclined.

This line is properly, in each case, called the “sight-line” of such plane; but it is only properly called the “horizon” in the case of a horizontal plane: and I have preferred using always the term “sight-line,” not only because more comprehensive, but more accurate; 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 plane, it does not mathematically coincide with it, and the two lines ought not to be considered as theoretically identical, though they are so in practice.

It is evident that all vanishing-points of lines in any plane must be found on its sight-line, and, therefore, that the sight-line of any plane may be found by joining any two of such vanishing-points. Hence the construction of Problem XVIII.

II.

DEMONSTRATIONS WHICH COULD NOT CONVENIENTLY BE INCLUDED IN THE TEXT.

I.

THE SECOND COROLLARY, PROBLEM II.

In Fig. 8. omit the lines _CD_, _C′D′_, and _DS_; and, as here in Fig. 75., from _a_ draw _ad_ parallel to _AB_, cutting _BT_ in _d_; and from _d_ draw _de_ parallel to _BC′_.

Now as _ad_ is parallel to _AB_— _AC_ ∶ _ac_ ∷ _BC′_ ∶ _de_; but _AC_ is equal to _BC′_— ∴ _ac_ = _de_.

Now because the triangles _acV_, _bc′V_, are similar— _ac_ ∶ _bc′_ ∷ _aV_ ∶ _bV_; and because the triangles _deT_, _bc′T_ are similar— _de_ ∶ _bc′_ ∷ _dT_ ∶ _bT_.

But _ac_ is equal to _de_— ∴ _aV_ ∶ _bV_ ∷ _dT_ ∶ _bT_; ∴ the two triangles _abd_, _bTV_, are similar, and their angles are alternate; ∴ _TV_ is parallel to _ad_.

But _ad_ is parallel to _AB_— ∴ _TV_ is parallel to _AB_.

II.

THE THIRD COROLLARY, PROBLEM III.

In Fig. 13., since _aR_ is by construction parallel to _AB_ in Fig. 12., and _TV_ is by construction in Problem III. also parallel to _AB_—

∴ _aR_ is parallel to _TV_, ∴ _abR_ and _TbV_ are alternate triangles, ∴ _aR_ ∶ _TV_ ∷ _ab_ ∶ _bV_.

Again, by the construction of Fig. 13., _aR′_ is parallel to _MV_— ∴ _abR′_ and _MbV_ are alternate triangles, ∴ _aR′_ ∶ _MV_ ∷ _ab_ ∶ _bV_.

And it has just been shown that also _aR_ ∶ _TV_ ∷ _ab_ ∶ _bV_— ∴ _aR′_ ∶ _MV_ ∷ _aR_ ∶ _TV_.

But by construction, _aR′_ = _aR_— ∴ _MV_ = _TV_.

III.

ANALYSIS OF PROBLEM XV.

We proceed to take up the general condition of the second problem, before left unexamined, namely, that in which the vertical distances _BC′_ and _AC_ (Fig. 6. page 13), as well as the direct distances _TD_ and _TD′_ are unequal.

In Fig. 6., here repeated (Fig. 76.), produce _C′B_ downwards, and make _C′E_ equal to _CA_.

Join _AE_.

Then, by the second Corollary of Problem II., _AE_ is a horizontal line.

Draw _TV_ parallel to _AE_, cutting the sight-line in _V_.

∴ _V_ is the vanishing-point of _AE_.

Complete the constructions of Problem II. and its second Corollary.

Then by Problem II. _ab_ is the line _AB_ drawn in perspective; and by its Corollary _ae_ is the line _AE_ drawn in perspective.

From _V_ erect perpendicular _VP_, and produce _ab_ to cut it in _P_.

Join _TP_, and from _e_ draw _ef_ parallel to _AE_, and cutting _AT_ in _f_.

Now in triangles _EBT_ and _AET_, as _eb_ is parallel to _EB_ and _ef_ to _AE_;—_eb_ ∶ _ef_ ∷ _EB_ ∶ _AE_.

But _TV_ is also parallel to _AE_ and _PV_ to _eb_.

Therefore also in the triangles _aPV_ and _aVT_,

_eb_ ∶ _ef_ ∷ _PV_ ∶ _VT_.

Therefore _PV_ ∶ _VT_ ∷ _EB_ ∶ _AE_.

And, by construction, angle _TPV_ = ∠ _AEB_.

Therefore the triangles _TVP_, _AEB_, are similar; and _TP_ is parallel to _AB_.

Now the construction in this problem is entirely general for any inclined line _AB_, and a horizontal line _AE_ in the same vertical plane with it.

So that if we find the vanishing-point of _AE_ in _V_, and from _V_ erect a vertical _VP_, and from _T_ draw _TP_ parallel to _AB_, cutting _VP_ in _P_, _P_ will be the vanishing-point of _AB_, and (by the same proof as that given at page 17) of all lines parallel to it.

Next, to find the dividing-point of the inclined line.

I remove some unnecessary lines from the last figure and repeat it here, Fig. 77., adding the measuring-line _aM_, that the student may observe its position with respect to the other lines before I remove any more of them.

Now if the line _AB_ in this diagram represented the length of the line _AB_ in reality (as _AB_ _does_ in Figs. 10. and 11.), we should only have to proceed to modify Corollary III. of Problem II. to this new construction. We shall see presently that _AB_ does not represent the actual length of the inclined line _AB_ in nature, nevertheless we shall first proceed as if it did, and modify our result afterwards.

In Fig. 77. draw _ad_ parallel to _AB_, cutting _BT_ in _d_.

Therefore _ad_ is the sight-magnitude of _AB_, as _aR_ is of _AB_ in Fig. 11.

Remove again from the figure all lines except _PV_, _VT_, _PT_, _ab_, _ad_, and the measuring-line.

Set off on the measuring-line _am_ equal to _ad_.

Draw _PQ_ parallel to _am_, and through _b_ draw _mQ_, cutting _PQ_ in _Q_.

Then, by the proof already given in page 20, _PQ_ = _PT_.

Therefore if _P_ is the vanishing-point of an inclined line _AB_, and _QP_ is a horizontal line drawn through it, make _PQ_ equal to _PT_, and _am_ on the measuring-line equal to the sight-magnitude of the line _AB_ _in the diagram_, and the line joining _mQ_ will cut _aP_ in _b_.

We have now, therefore, to consider what relation the length of the line _AB_ in this diagram, Fig. 77., has to the length of the line _AB_ in reality.

Now the line _AE_ in Fig. 77. represents the length of _AE_ in reality.

But the angle _AEB_, Fig. 77., and the corresponding angle in all the constructions of the earlier problems, is in reality a right angle, though in the diagram necessarily represented as obtuse.

Therefore, if from _E_ we draw _EC_, as in Fig. 79., at right angles to _AE_, make _EC_ = _EB_, and join _AC_, _AC_ will be the real length of the line _AB_.

Now, therefore, if instead of _am_ in Fig. 78., we take the real length of _AB_, that real length will be to _am_ as _AC_ to _AB_ in Fig. 79.

And then, if the line drawn to the measuring-line _PQ_ is still to cut _aP_ in _b_, it is evident that the line _PQ_ must be shortened in the same ratio that _am_ was shortened; and the true dividing-point will be _Q′_ in Fig. 80., fixed so that _Q′P′_ shall be to _QP_ as _am′_ is to _am_; _am′_ representing the real length of _AB_.

But _am′_is therefore to _am_ as _AC_ is to _AB_ in Fig. 79.

Therefore _PQ′_ must be to _PQ_ as _AC_ is to _AB_.

But _PQ_ equals _PT_ (Fig. 78.); and _PV_ is to _VT_ (in Fig. 78.) as _BE_ is to _AE_ (Fig. 79.).

Hence we have only to substitute _PV_ for _EC_, and _VT_ for _AE_, in Fig. 79., and the resulting diagonal _AC_ will be the required length of _PQ′_.