WINDERS STARTING FROM A CURTAIL STEP.

Fig. 1 shows the plan of a scroll, and wreath to stand over the curtail

step and winders. Divide N N into 9 equal parts. Make N 1 equal 5 of

these parts and draw the largest quadrant. To draw the remainder make

N 1, Fig. 7, equal N 1, Fig. 1; with N as centre and N 1 as radius,

strike the arc. Make 1 2 equal 1 of the 9 parts; from 2 erect a

perpendicular to cut the arc in 7. Join 7 N. Now with N as centre and

N 2 as radius, strike the second arc to cut N 7 in 8; draw 8 3 parallel

to 7 2. Again, with N as centre and N 3 as radius, strike the arc to

cut N 7 in 9; draw 9 4 parallel to 8 3. This process can be continued

as far as required. N 1 will strike the largest quadrant, N 2 the next,

and N 3 the next, and so on until the scroll is completed. The centre

and inside lines of rail will be struck from the same centres. Make the

joint at A, and draw the radius line from 2, which is the centre of

this part of the scroll; draw the tangent A B square to it.

Fig. 2 shows the elevation. Set up the treads and risers as they occur

on the centre line of rail, Fig. 1, and draw the centre falling line so

that it will finish level, about on the line N N, and at a height of

about 1 or 2 inches above the top of the first step. Make 3 7 equal the

stretch-out of the centre line of rail, Fig. 1 from A to C. Draw the

level line through the falling line at 7, and make 3 8 and 8 A equal

C B and B A, Fig. 1, and complete the development of tangents as shown.

Draw the joint at 7 square to the tangent A B. The scroll must be the

distance S S thicker than the thickness of rail, and the joint worked

to the bevel as shown.

Fig. 3 is the face mould for the top side of the plank. Make B F,

Fig. 1, equal 8 F, Fig. 2, and at Fig. 3 make C B F equal C B F and

B A equal B A, Fig. 2, and F A equal F A, Fig. 1. Draw C 11 and A 5

parallel to the horizontal trace F A, and make A 5 and C 11 equal A 5

and C 7, Fig. 1, and draw the major axis through 5 and 11. With 5 as

centre and 5 7, Fig. 1, as radius, strike an arc at 7; draw a line

through 11 and tangent to the arc. Now, if the drawing is correct, 7 11

will equal the height 3 C, Fig. 2. Make 5 4 3 1 6 7 equal corresponding

figures, Fig. 1, and complete the mould as usual. Notice O will be the

centre for sections D E R and C, and S S, Fig. 4, will equal the radius

1 D, Fig. 1. But for the section at the joint A, 2 will be the centre

and S S, Fig. 5, will equal the radius 2 A, Fig. 1. The bevels, &c.,

are got as usual. Make 3 4 5 6 7, Fig. 2, equal C R E D A, Fig. 1. Make

4 4 and 5 5 and 6 6, Fig. 2, equal 1 S and 1 O and 1 W, Fig. 3. Now,

the difference between 4 5 and 6 and the falling line is what each

section is out of the centre of plank.

Fig. 6 is the face mould for the under side of the plank.

[Illustration: PLATE 30.]

Practical Stair Building and Handrailing By the square section and falling line system. · 第 72 / 78 题