allowed to deposit took 25 c.c. of the acid solution, then a quantity of baryta water equal to 5 c.c. of the standard baryta water has disappeared, it has been converted to an insoluble form. Since the oxalic acid solution is made of such a strength that each c.c. is equal to 1 c.c. of CO2 it follows that there are 5 c.c. of CO2 in the air under examination. In this as in the other cases it will be convenient to use for the testing or collecting the air large dry bottles with accurately fitting stoppers. After the air has been pumped in, then 50 c.c. of the clear baryta water are added, the stopper inserted and the liquid well distributed in the bottle, by shaking; after a quarter of an hour or longer the contents are transferred to a stoppered cylinder, allowed to stand excluded from the air until the carbonate is completely settled, and then 25 c.c. of the clear liquid removed by a pipette and the alkalinity determined.

Lime water may be used instead of baryta water; a modification of the method also consists in filtering, with special precautions to prevent the farther absorption of CO2, the barium carbonate off, converting it into sulphate by adding a sufficient quantity of sulphuric acid, igniting and weighing; one part of barium sulphate is equivalent to 1891 of CO2.

(63) Estimation of Carbon Monoxide, CO.

It is much to be regretted there is no easy or household test for the presence of this dangerous gas. The best qualitative test is probably that of Vogel: a drop of blood is diluted with water, and the reddish liquid shaken up in a bottle with the air to be examined; the liquid is then examined by the spectroscope, and the position of the bands on the scale noted; a little ammonium sulphide is now added; if carbon monoxide is present, the spectrum will undergo no change. The test is said to detect 03 per

cent.

If a greater quantity of the gas in the air is present than a mere trace, the air, shaken up with a solution of the double chloride of palladium and sodium will produce a black precipitate in the solution, and the quantity present may be approximately estimated by acting on a similar bulk of the solution with air containing known quantities of CO, on similar principles to those detailed at page 98, under the head of "Nephalometric Methods." Before

making the test, sulphuretted hydrogen should be absorbed by passing the air through a solution of lead acetate. Larger quantities of CO are detected and estimated by absorption by cuprous chloride. A solution of cuprous chloride in strong hydrochloric acid absorbs readily CO, forming Berthelot's compound, in which one molecule of CO is united with one molecule of cuprous chloride; on dilution of the cuprous chloride solution, which has absorbed CO with water, and addition of the double salt of palladium and sodium, there is a black precipitate of palladium.

(64) Sulphuretted Hydrogen.

Hydric sulphide is detected in very small quantities in air by either exposing strips of paper moistened by lead acetate to the air or by drawing the air through a solution of the same salt. If there is only a dark colour, but no precipitate, the amount may be estimated on colorimetric principles-that is, the colour may be imitated by acting on a similar bulk of lead acetate solution by water containing known quantities of hydric sulphide,

CHAPTER IX.

HOW TO MEASURE CUBIC SPACE AND TO REPORT ON

VENTILATION.

(65) Cubic space allowed by law or by local regulations.

(66) The Cube of Rectangular and Triangular Spaces.

EVERY sanitary officer should know how to measure cubic

space.

(a) Cubic space of a square or rectangular room.

This is by far the most common case, the rule being simply to

1 To obtain roughly the cubic feet per hour it is usual to multiply by five. Thus the Board Schools will give to the scholars only 500 cubic feet per hour, which is an insufficient quantity.

multiply the three dimensions together-that is, height, length, and breadth.

For example: a room 10 feet high, 15 long, and 12 broad, has a cubic space of 1,800 cubic feet, for 10 x 12 x 15 = 1,800. (b) Cubic space of a room, the upper part of which has a triangular section.

This may either be a lean-to roof, such as Fig. 7, or any other kind of roof, such as Fig. 8.

The first process is to obtain the cube of the rectangular space, abcd, next the area of the triangular space, abe, which, when multiplied by the length, gives the cube of that portion of the room, and the two added together make the total cube.

The area of a triangle, the perpendicular height of which is known (and in rooms the height can always be known), is found by multiplying the base by half the perpendicular height. Thus, in the triangle, abc, Fig. 7, the base, ab, being 10 feet, and cd being 6 feet, the area is 30 feet, for 10 x 3 = 30; or, it may be put in another way: multiply base by perpendicular height and divide the product by two.

Applying this principle to the measurement of a room of which Fig. 8 is a section. Supposing the length is 30 feet, the extreme height, ef, 15 feet, the breadth, cd, 14 feet, and the height of wall, bd, 10 feet, required the cubic space.

First cube, abcd, 10 x 30 x 14 = 4,200; then get area of triangular space, abe, which by our rule is half the perpendicular ex, that is, 5 or 2:5, and multiply this by the base, 14 feet, and the product by the length, that is, 30 feet-2.5 × 14 × 30 = 1,050, which, added to 4,200, 5,250.

=

A shorter method of doing this problem is to consider the sectional area, that of a trapezoid, obtain its area, and then multiply by the length.

=

The rule for obtaining the area of a trapezoid is: multiply the sum of the parallel sides by the perpendicular distance between them, and half the product will be the area. The parallel sides in this instance are ac, ef; the distance between them is half the breadth, 7 feet; then 10+15 x 7 = 175, and 175 × 30 5,250, the same as before. So, again, if a room has either a sectional area, like abcd, Fig. 9, in which the side ab is parallel to the side de, measure off a perpendicular de, add the lengths of de and ab together, multiply by the length of the perpendicular de, halve the product and multiply by the length of the room or the height, according as the trapezoid figure belongs to a section or a ground plan.

a

FIG. 9.

Equilateral Triangular Spaces.-It is useful to remember that, although the area of an equilateral triangle can be found by the general rule, in this special case the area can be calculated without measuring off any perpendicular, if one side only is known; the rule being that the area is equal to half the side squared, and multiplied by the square root of three.

Since the square root of three is 1-732, another way of putting it is halve the side, square it, and multiply it by 1.732. An equilateral triangular space, one side of which was 8 feet, would have an area equal to 27·712 square feet. A cupboard in a room having this area, and 10 feet high, would of course cube out as 277.12 cubic feet.

a

(67) Cubic Space of Rooms, the Ground Plan of which is of an irregular Figure.

In this instance the room may be divided up into triangles, the area of the triangles found and multiplied by the height, or it c may be more convenient to consider the area that of a trapezium. The area of a trapezium is found by multiplying the diagonal by the sum of the perpendiculars let fall upon it from the opposite angles, and halving the product. Supposing, for instance, Fig. 10 is the ground plan of a room 12 feet

FIG. 10.