distances xr and xv from f to r and f to v in Fig. 165, and the distances y to s and y to u, from g to s and g to u. Take the distance from R to t in Fig. 163 and mark off from h to t in Fig. 165. The curve drawn these

points will give the shape of the opening in the larger pipe.

To make a set of patterns for an oval tea-kettle. When getting out the patterns for an oval tea-kettle to hold 6 quarts, as shown in Fig. 166, first assume

a suitable length and width for the bottom, then a height suitable to the capacity required can be readily ascertained. As 6 quarts contains 3461⁄2 cubic inches, divide this quantity by the number of square inches in the bottom of the kettle and the quotient will be the height required. If the bottom of the kettle be 10 inches long and 7 inches wide, then

the required height for the body of the kettle. Next determine the circumference of the kettle by first finding the circumference of a circle whose diameter is equal to half the sum of the length and width of the oval, thus

10+ 7
2

8.5 X 3.141626.70 inches,

the circumference of the body of the kettle.

The body pattern shown in Fig. 168 is a rectangle 26.7 inches long and 6.3 inches wide, one-half an inch extra depth is allowed for forming cramps to hold the bottom in position while brazing, also one-half inch along the side for the seam. To draw the oval for the bottom, draw the axes AB and CD as in Fig. 167, intersecting at E, and mark the length and width of the oval upon them. Then mark the width CD from A along AB, as at F, and divide the difference between the length and width into three equal parts, as 1, 2, B. From E mark a distance equal to two of these divisions on each side of the center line to the points G and H. Using the distance GH as a radius and G and H as alternate centers, describe the arcs JHK and JGK intersecting at J and K. Draw lines through JG and

JH, and KG and KH. With centers H and G and BH or GA describe the end arcs. With centers J and K, and radius JD or KC, describe the side arcs to join with the end arcs. This completes the oval.

О

*

Fig. 168.

Lay out the pattern for the top shown in Fig. 168 in exactly the same manner, making due allowance for hollowing and edging, as indicated by the dotted lines.

Mark off the opening in the top for the cover, and using the same centers, describe the arcs required to form the cover opening. The pattern for the lid shown in Fig. 168 is also drawn in the same manner. The inner dotted lines are the same dimensions as the opening in the cover. Sufficient allowance should be made for the lap over the rim flange and also for hollowing. To lay out the spout pattern approximately correct, draw a line as in Fig. 169, and set off CD equal in

length to the circumference of the small end of the spout. From A to B mark a length equal to AB in Fig. 166, and through B draw a line at right angles. Make the length EF equal to the circumference EF in Fig. 166, and then draw the quarter circles at E and F, using the same radius as shown in Fig. 166. Draw GC equal to the straight length from the curve to the body on the inside of the spout at G in Fig. 166. From GG mark the lengths to 00 equal to the circumference of the spout at the large end. Join these points to the center line at H, sloping them at the same angle as the base of the spout. Notch the center and cut a cramp at the top and bottom of the seam, as shown. The rim for the lid is a narrow strip of metal equal in length to the circumference of the hole, with a suitable allowance for lap at the seam.

CORNICE WORK.

To describe a pattern for a miter joint at right angles for a semicircular gutter. Let the semicircle ACB, Fig. 170, be the width and depth of the gutter. Draw the line AB, and draw the lines AF and BE at right angles to AB. Join AF and BE by the line FE

at an angle of 45 degrees. Divide the circumference of the semicircle ACB into any number of equal parts, and from these points draw lines parallel to AF, as 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10; then lay off the line AB in Fig. 171, equal in length to the circumference of the semicircle ACB. Erect the lines AF and BE at right angles to AB, and lay off on the line AB the same number of equal spaces as on the circumference of the semi