This section is from the book "Cyclopedia Of Architecture, Carpentry, And Building", by James C. et al. Also available from Amazon: Cyclopedia Of Architecture, Carpentry And Building.

90. Problem XXIII. To construct the shade line of a cylin= der whose axis is perpendicular or parallel to the ground line,

Given the elevation of a cylinder, its axis being AB perpendicular to H. To construct shade lines. Fig. 44.

Let CD be any horizontal line drawn through the cylinder.

Construct the 45° isosceles triangle AGD on the right half of the diameter.

With the radius AG describe the semi-circular arc mGn, cutting the horizontal line CD in the points m and n.

These two points will determine the shade elements mo and np.

91. Problem XXIV. To con= struct the shadow on a plane (parallel to its axis) of a circular cylinder whose axis is either perpendicular, or parallel to the ground line.

Let a=the distance, in the elevation, between the projection of the axis of the cylinder and the projection of the visible shade element. Let b = the distance between the axis of the cylinder and the plane on which the shadow falls, to be obtained from the plan.

Then the distance, between the visible shade element and its shadow on the given plane, will be equal to a+b.

The width of the shadow on the given plane will be equal to 4a.

Given the circular cylinder CDEF (Fig. 45), its axis AB perpendicular to II. To construct its shadow on the V plane which is 1.1/2, inches distant from the axis AB. The shade elements mo and np can be constructed by Problem XXIII. Draw RS the shadow of the shade element np, parallel to np, and distant from it a +1 1/2 inches. The width of the shadow on the given plane will be 4 times the distance An.

FIG.44.

02. Problem XXV. To construct the shadow on a right cylinder of a horizontal line.

a. It will be the arc of a circle of the same radius as that of the cylinder.

b. The center of the circle will be on the axis of the cylinder as far below the given line as that line is in front of the axis.

Given a right circular cylinder CDEF, whose diameter is 1 3/4 inches, and a horizontal line ah, 1 1/4 inches in front of the axis of the cylinder. To construct the shadow. Fig. 46.

F1G.45

F1G-46

FIG.47.

Locate the point o on the axis 1 1/4 inches below avbv. With o as a center, and radius equal to 7/8 inch describe the arc mnp, the required shadow.

93. Problem XXVI. To con= struct the shadow of a verti= cal line on a series of mouldings which are parallel to the ground line.

The Shadow reproduces the actual profile of the mouldings, Given a vertical line avbv which casts a shadow on the moulding M, which is parallel to the ground line, and whose profile is shown in the section ABCD. The line avbv, is 1 inch in front of the fillet AB. To construct its shadow, Fig. 47:

( (instruct the shadow on the fillet AB, of the end of the line av, or any other convenient point in the line, by Problem XVII.

From the point as the shadow of the line reproduces the profile ABCD and we obtain asnsosbs the required shadow.

04. Problem XXVII. To construct the shad= ow on the intrados of a circular arch in section, the plane of the arch being in profile projec= tion.

Let AB (Fig. 48) be the "springing line" of the arch. Let CD be the radius of the curve.

The point F is determined by the construction used in finding the shade element of a cylinder. Problem XXIII. At the point F draw the line GH, with aw inclination to the "horizontal" of 1 in 2. Through the point D draw the 45° line DB. The curve of the line of shadow will be tangent to these two lines at the points F and B. The required shadow is that portion of the curve between the lines DC and IK.

FlG.48

FIG.49.

A similar construction is used in the case of a hollow semi-cylinder when its axis is vertical, except, that the line GH has then an inclination to "the horizontal" of 2 in 1. Fig. 49.

95. Problem XXVIII. To construct the shadow of a spheri= cal hollow with the plane of its face parallel to either of the coordinate planes.

The line of shadow is a semi-ellipse. The projections of the rays of light tangent to the circle determine the major axis. The semi minor axis is equal to 1/3 the radius of the circle.

Given the vertical projection of a spherical hollow, the plane of its face parallel to V. Fig. 50.

Determine the ends of the major axis by drawing the projections of the rays of light tangent to the hollow. The semi-minor axis, oa, equals 1/3 the radius oh. On he and oa construct the semi-ellipse, the required shadow. 96. Problem XXIX. To con= struct the shade line and shadow of a sphere. Fig. 51.

Let the circle whose center is o be the vertical projection of a sphere whose center is at a distance x from the V plane.

The shade line will be an ellipse. The major axis of this ellipse is determined by the projections of the rays of light tangent to the circle. The semi minor axis and two other points can be determined as follows :

Through the points, A, o, and B, draw vertical and horizontal lines, intersecting in the points E and D.

The points E and D are two points in the required shade line.

Through the point E draw the 45° line EF. Through the point F, where this line intersects the circle, and the point B, draw the line FB. The point C, where this line FB intersects the 45° line through the center of the sphere, 0, is the end of the semiminor axis. The shadow of the sphereon the co-ordinate plane will also be an ellipse. The center of this ellipse. os, will be the shadow of the center of the sphere. It will be determined by Problem XV11. The ends of the major axis MX, will be on the projection of the ray of light drawn through the center of the sphere. The minor axis PR will be a line at right angles to this through the point os. lis length will be determined by the projections of the rays of light BR and AP tangent to the circle, and is equal to the diameter of the sphere. The points M and N", which determine the ends of the major axis, are the apexes of equilateral triangles PMR and PNR, constructed on the minor axis as a base.

FIG.50.

97. Problem XXX. To construct the shade line of a torus.

Fig. 52, in elevation: The points 1 and 5 can be determined by drawing the projections of the rays of light tangent to the elevation. Since the shade line is symmetrical on either side of the line MN in plan, the points 3 and 7 can be found from 1 and 5, by drawing horizontal lines to the axis. The points 4 and 8 are determined by the construction used in finding the shade elements of a cylinder. Problem XXIII.

The above points can be determined without the use of plan.

F!G.52

The highest and lowest points in the shade line, 2 and 6, can be found only by use of plan. It is not necessary, as a rule, to determine accurately points 2 and 6. The shade line in plan will be, approximately, an ellipse whose center is O The ends of the major axis R and S, are determined by the projections of the rays of light tangent to the circle. Other points can be determined without the use of the elevation as follows: With center o, construct the plan of a sphere whose diameter equals that of the circle which generated the torus. Determine the shade line by Problem XXIX. Draw any number of radii OE, OF, OG, etc.

On these radii, from the points where they intersect the shade line of the sphere, lay off the distance RT, giving the points c,f and g. These are points on the required shade line.

TOWER CONVERSE MEMORIAL LIBRARY, MALDEN, MASS.

H. H. Richardson, Architect.

Note treatment of shadows in a perspective drawing.

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