A. B.'S WATER FRAME, No. 1, TABLED.

In this manner may the tabulation be continued down to the bottom of the Frame.

PROBLEM III.

To Gage a Starch-Maker's Box with a Slider and Cloth.

Take several lengths of the box carefully with the Slider and Cloth in it; and from these deduce a mean length; then having removed the Slider and Cloth, take several lengths again, and find the mean length. The difference between the first and last mean length will be the allowance on the length to be made for the cloth; next, from several breadths, on each side of the Slider with the Cloth on, deduce a mean breadth, which double for the whole breadth of the Frame with the Slider and Cloth in: likewise ascertain the mean breadth

without the Slider and Cloth. The difference of the mean breadths will be the allowance to be made for the Slider and Cloth. This allowance will in general be found to be 7 tenths of an inch; that is, five tenths for the Slider, and two tenths for the Cloth.

The area of the Box is now to be calculated in the usual way, both with and without the Slider, by multiplying the length by the breadth, and dividing the product by 34-8 the number of cubic inches in a pound of Green Starch. This area multiplied by the depth will evidently be the content of the Box in Green Starch pounds. Suppose the length of the Box to be 60.3 inches, and the breadth 11.8 inches with the Slider and Cloth, but 12.3 inches without the Slider, and the entry will be as follows:

A Table of the Box may be constructed by continually adding the area to itself for wet inches, as the mean depth of Starch in the Box is always deduced from the sum of several depths taken with the spit. The most useful Table, however, will be one similar to the Table of the Square Cistern in page 367, having the content for every tenth of an inch, but be

ginning in the Frame at 4 inches deep instead of at 10, as in the Cistern.

OF SPECIFIC GRAVITY.

By SPECIFIC GRAVITY is meant the relation which the weight of a given portion of any kind of body whatever bears to the weight of an equal bulk of any other body whatever.

Now to simplify the doctrine of Specific Gravity it was necessary to assume some body as the standard with which other bodies should be compared; and as soft water is in almost all places of nearly the same weight, it occurred that no other body could be a better or more universal standard than rain or soft water.

Hence, as the gravity or weight of any portion of water is always called 1, the specific gravity of a body twice as heavy as water will be 2, and the specific gravity of a body half as heavy as water will be ⚫5.

A knowledge of specific gravity is of great use in computing the weight of such bodies as by their size are too unwieldy to be weighed, or by their form too irregular to be measured.

METHOD OF FINDING THE SPECIFIC GRAVITIES OF

BODIES.

Of a Solid heavier than Water.

Suspend the body from the arm of a balance by a fine thread, and weigh it accurately in air, then immersing the body in soft water, weigh it again by means of the same balance and thread, noting the weight lost by the immersion of the body in water. The weight in air, divided by the weight lost, will be the Specific Gravity of the Body immersed.

EXAMPLE.

A body weighs 9 drams in air, but when immersed in water it weighs only 8 drams: its specific gravity is required.

SOLUTION.

Weight lost 1 dr.) 9 dr. weight in air

Quotient 9 specific gravity required.

2. Of a Solid lighter than Water.

Then

Affix to the light body something sufficiently ponderous to cause the whole to sink in water. weigh the compound mass and heavier body separately, both in and out of water, carefully noting the weight each loses by immersion. Next, from the greater loss subtract the less, and divide the weight of the light body in air, by the remainder: the quotient will be the Specific Gravity required.

EXAMPLE.

A piece of light substance weighing by itself 3 drams in air, weighs with a portion of metal attached to it, 10 drams in air, and 4.4 drams in water: also the metal by itself weighs 7 drams in air, and 6.2 drams in water. Required the Specific Gravity of the lighter body.

Loss of the compound 5-6 drs. || Loss of the heavy body 8 drs.

But 5.6 •8=

4.8, difference of the losses.

And,

Difference 4.8)3(625 specific gravity of the light body.

3. For the Specific Gravity of a Fluid.

Weigh any solid substance that will sink in the given Fluid, first in air, and then in the Fluid, noting the loss of weight in the body weighed. If this loss be divided by the weight of the heavy body in air, a quotient will arise, which multiplied by the Specific Gravity of the solid body, will be the Specific Gravity of the Fluid.

EXAMPLE.

Required the Specific Gravity of a Fluid in which a solid body (whereof the Specific Gravity is 4) weighs