HYDRAULIC ARCHITECTURE, BOOK III.

stroke of the piston, as the capacity of the receiver is to that of the syringe and the receiver taken together; from which it follows that the expansion The author uses "dilatation" to describe the thinning of the air as it occupies more space. of the air increases after each stroke of the piston, according to the terms of a geometric progression. The ratio of these terms is the same as the capacity of the receiver compared to the combined capacity of the syringe and receiver.

Naming a the capacity of the receiver and b the capacity of both the syringe and receiver together, we have the sequence $a, b^1, \frac{b^2}{a}, \frac{b^3}{a^2}, \frac{b^4}{a^3}$, where the exponents of the numerators of each term represent the number of piston strokes, while the terms themselves express the expansion of the air remaining in the receiver. Now, we know that one can find any desired term of a geometric progression once the first two are known. For example, to find the term corresponding to the fortieth stroke of the piston, I raise the first and second terms to the fortieth power; naming the value we seek as x, we have the proportion $a^{40} : b^{40}

If we assume that the capacity of the receiver is six times that of the syringe, their ratio will be as 6 is to 1; consequently, we have $a = 6$ and $b = 6 + 1 = 7$. To find the value of x (or $\frac{b^{40}}{a^{40}}$), we must use logarithms Logarithms were a vital 18th-century tool for simplifying complex multiplications and divisions into additions and subtractions. to shorten the calculation, which would become very laborious if we had to raise the numbers 6 and 7 to the fortieth power. I assume then that m is the logarithm of $6 = a$, and n is the logarithm of $7 = b$. We then have $40 \times n - 40 \times m = \text{log } x$ instead of $\frac{b^{40}}{a^{40}} = x$. That is to say, one must take from the tables the logarithms of the numbers 7 and 6, which are 0.8450980 and 0.7781512, and multiply their difference (which is 0.0669468) by 40. This gives 2.6778720, which is the logarithm of the number we seek, corresponding to approximately 476. Thus, we have $a^{40} : b^{40} :: 1 : 476$. This shows that after the fortieth stroke of the piston, the air in the receiver will be 476 times more expanded than the air originally enclosed there.

802. When the ratio of the capacity of the receiver to that of the syringe is known, one can also find how many strokes of the piston are required to expand the air in the receiver to a specific determined point. For example, if one asks to expand it 476 times more than it is in its natural state; I name...