Continued.

15 height of the upper part of the wall.

50 length.

750 area of the upper part of the wall, 2 bricks thick. 5 number of half bricks.

3750 area of ditto, a brick thick,

8

4

32 area of the vacuity on the outside, a brick thick.

5 number of windows.

160 area of the vacuities on the outside of five windows, a brick thick.

39 7 area of the vacuity of the inside for 1 window, 2 5 number of windows.

197 11 area of the vacuities for five windows in the inside. half bricks thick.

4

791

8 area of the vacuities on the inside, a brick thick.

4 9

2 6

9 6

2 4 6

11 10 6 area of the vacuity of the recess under each 5 window, 1 brick and a thick.

[brick and a half thick. 59 46 area of the vacuities of the 5 window backs, 1 number of half bricks.

3

178

16 area of the vacuities of the 5 window backs, a

brick thick.

1 i 2

Con

Continued.

11 height of the arch-way, from the pavement to the

add 9 height from the pavement to the springing of the arch.

31

10 width of the arch-way.

3)310

160

103

6

4 area of the vacuity of the arch-way, 3 bricks thick number of bricks thick.

620 area of the vacuity of the arch-way, a brick thick.

791 8

178 1 620

1749 9 areas of all the vacuities, a brick thick.

2666 6

3750

6416 6 area of the whole, a brick thick, as if solid. deduct 1749 9.

3)4666 9 true area of the whole, a brick thick.

-rds. ft.

272)1555 7 (5 195 reduced to the standard thickness. 1360

195

Then 5 195 4500 25726 number of bricks, nearly.

5 195 × 1

= 7 hundred weight of lime, nearly.

5 195 × 2

14 loads of sand, nearly.

I shall, in the following Examples, shew the manner of finding the areas of windows, angle chimnies, and the dif ferent stories of a building.

PROBLEM III.

To measure the vacuity of a window.

Find the area of the outside of the window, and multiply that by the number of half bricks thick, from the face of the sash frame on the outside, to the face of the wall of the same side, to the area so found, at half a brick thick; add the area of the inside vacuity multiplied by the number of half bricks thick, from the face of the sash frame on the outside, to the face of the brick work within the building: also add the area of the vacuity of the recess, the height being taken from the bottom of the sash frame to the floor, and its width the same as the inside vacuity above; multiply this also by the number of half bricks thick, then the sum of these will be the whole vacuity, or void space in the whole window, at half a brick thick; and if required to be reduced to the standard, divide the area so found by 3, and the area of the contents will be reduced to 1 brick and a thick.

EXAMPLE I.

Let Fig. 3, Plate 56, be the plan or horizontal section of a window. Fig. 1, the elevation as would appear within the building. Fig. 2, a vertical section through the middle of the elevation of the window; the height of the outside vacuity is 8ft. and its breadth 4ft. and half a brick thick; the height of the inside vacuity is 8ft. and its breadth 4ft. 9in. and 2 bricks thick, as appears by the plan and section; the recess is 2ft. 9in. high, 4ft. 9in. wide, and half a brick thick, which is also marked upon the plan and section; required the area of the whole va◄ cuity at half a brick thick.

8 height of the outside vacuity.

4 width of the outside vacuity.

32 area of the outside vacuity, a brick thick,

152

area of the inside vacuity, a brick thick,

9 width of the inside vacuity.

29 height of the recess from the floor to the side of the

9 6 369

sash,

13 09 area at 1 brick and a thick.

33

39

2 3 area of the recess, a brick thick.

32 152

223 2 3 area of the whole vacuity, a brick thick.

PROBLEM IV.

To measure any angle chimney, standing equally distant each way from the angle of the room.

Plate 58, Fig. 4. Multiply the breadth AB by the height of the story, and the product by the number of half bricks contained in the half breadth AB, and you will have the solidity at half a brick thick, by deducting the vacuity or opening of the chimney.

To measure an angle chimney, when the plane of its breast intersects the two sides of the room unequally distant from the angle.

Fig. 5. From the points A and B, where the plane of the breast intersects the sides of the room, draw two lines, AE, EB, parallel to the two sides of the room; then multiply either of the lines AE or EB, suppose EB, by the height of the room, and multiply that product by the number of half bricks contained in the other line AE, and deduct the vacuity as before, and the remainder will be the content, at half a brick thick.

PROBLEM VI.

To measure an angle chimney, when the plane of the breast projects out from each wall, and unequally distant from the angle of the

room.

Draw the two lines GF and FH parallel to the two sides of the room, as before; then multiply the breadth FH by the