“AA represents a small scale. It may be taken off at D. Diameter 1½ in., weight 44 grains.
“B a stem of hardened steel wire. Diameter 1⁄100 in.
“E a hollow copper globe. Diameter 28⁄10 in. Weight with stem 369 grains.
“FF a stirrup of wire screwed to the globe at C.
“G a small scale, serving likewise as a counterpoise. Diameter 1½ in. Weight with stirrup 1634 grains.
“The other dimensions may be had from the drawing, which is one-sixth of the linear magnitude of the instrument itself.
“In the construction it is assumed that the upper scale shall constantly carry 1000 grains when the lower scale is empty, and the instrument sunk in distilled water at the temperature of 60° Fahr. to the middle of the wire or stem. The length of the stem is arbitrary, as is likewise the distance of the lower scale from the surface of the globe. But, the length of the stem being settled, the lower scale may be made lighter, and, consequently, the globe less, the greater its distance is taken from the surface of the globe; and the contrary.”
In comparing the densities of different liquids, it is clear that this instrument is precisely equivalent to that of Fahrenheit, and must be employed in the same manner, weights being placed in the top scale only until the hydrometer sinks to the mark on the wire, when the specific gravity of the liquid will be proportional to the weight of the instrument together with the weights in the scale.
In the subsequent portion of the paper above referred to, Nicholson explains how the instrument may be employed as a thermometer, since, fluids generally expanding more than the solids of which the instrument is constructed, the instrument will sink as the temperature rises.
To determine the density of solids heavier than water with this instrument, let the solid be placed in the upper scale pan, and let the weight now required to cause the instrument to sink in distilled water at standard temperature to the mark B be denoted by w, while W denotes the weight required when the solid is not present. Then W − w is the weight of the solid. Now let the solid be placed in the lower pan, care being taken that no bubbles of air remain attached to it, and let w1 be the weight now required in the scale pan. This weight will exceed w in consequence of the water displaced by the solid, and the weight of the water thus displaced will be W1 − w, which is therefore the weight of a volume of water equal to that of the solid. Hence, since the weight of the solid itself is W − w, its density must be (W − w)/(w1 − w).