viz., 8, 5, and 43-inch shells, .24 of an inch in the same space of time. Now, the logarithms of these two decimal numbers, viz., 9.342423 and 9.380211, are therefore the constant logarithms made use of in the above Rule.

Fuzes are generally marked off, by circular lines, into seconds and fractional parts of a second, so that no time may be lost in measuring and adapting them to the shells for which they are intended.

PROBLEM XLII.

To find the Time that a Red-hot Cannon Ball will take to Cool.

RULE.

If balls of iron be made red-hot, the times of cooling will be as the squares of their diameters. Now, it has been found by experiment, that an iron ball of 2 inches in diameter takes 60 minutes, or 1 hour to cool-hence, as the square of 2 is to 1 hour; so is the square of the diameter of any red-hot ball, to the time, in hours, that it will take to cool; or, as thus, by logarithms:

To twice the diameter of the given red-hot ball, add the constant logarithm 9.397940; the sum, abating 10 in the index, will be the logarithm of the time, in hours, that the ball will take to cool.

Example.

Required the time that a red-hot 24 lbs. ball will take to cool; its diameter being 5.6 inches?

Hence, the time that a red-hot 24 lbs. ball will take to cool is 7.84, or 7 hours, 50 minutes, and 24 seconds.

Note. The constant logarithm is expressed by the arithmetical complement of twice the logarithm of 2.

Remarks.

The velocities communicated to shot of the same weight, with different charges of powder, are nearly as the square roots of the respective weights of the charges,

When shot of different sizes are fired with the same charge of pow der, their velocities are nearly in the inverse ratio of the square roots of their respective weights.

When shot of different sizes are fired with different charges of pow der, the velocities which they acquire are directly as the square roots of the charge of powder, and inversely as the square roots of their respective weights.

When a shot is fired, it is never impelled by the full power of the charge; because a certain portion of the inflamed powder escapes at the vent or touch-hole, and a very considerable portion by windage: besides which, some grains of the powder generally remain unignited, and, consequently, of no expansive force in projecting the shot.

It is almost impossible to find a charge of powder in which the inflammatory matter of each grain is so well proportioned, and so uniformly active, as to make the whole susceptible of the same instantaneous degree of explosion; and hence it is that several grains always pass out of the bore, and escape the fire, uninjured! This assertion, paradoxical as it may appear, is supported by the unquestionable test of experiments, as thus :

When the ground is covered with snow that is a little hardened by frost, let a person discharge a loaded musket, pointed at an object about 40 or 50 yards distant; then, if he walks slowly towards the object, and carefully examines the surface of the snow, he will be certain of finding several grains of powder, each of which will be so sound and perfect as to ignite at the touch of a spark.

Again,-When men are wounded in the face by the accidental firing of a blank cartridge from a musket, or by the unlucky blowing-up of a powder-monkey, several grains of powder will be seen buried, skindeep, in their cheeks, &c. Now, on extracting those with a surgeon's instrument, or the point of a needle, it will be found that each grain is so perfect as to be susceptible of being inflamed on the application of a lighted match.

The expansive force of gunpowder is at the rate of about 6000 feet in one second; but, owing to the quantity lost by windage &c., as mentioned above, it is estimated that not more than three-fourths of that force, or about 4500 feet, are applied to the projection of the ball.

The windage of shot is generally about the of the calibre, or diameter of the bore ; in consequence of this, balls are greatly deflected from the direction in which they are projected:-for, owing to windage, the ball, in its passage through the bore of the gun, seldom escapes without touching some point in its exit at the muzzle :-hence, if the ball strike against the right side, it will be deflected to the left; and, vice versa, should it strike against the left side, it will be deflected to

the right.-Again,-If the ball strike against the upper side, it will be forced downwards, so as to diminish its range; and, conversely, should it strike against the under side, it will be forced upwards, so as to give an increase to its customary range.-Hence, it frequently happens, that in ranges of about half a mile, balls are sometimes deflected to the value of 80 or 100 yards from the objects at which they are aimed.

When a ball passes so freely out of the bore as not to strike any part near the muzzle; then, it will constitute what is termed "a good shot;" but it is manifest that this is entirely owing to chance, and not to the skill of the gunner; for the most experienced artillerist cannot guard against the ordinary deflection of military projectiles.

It may be laid down as a general rule, that the smaller the windage of a gun is, the less will its shot deviate from the line of direction in which it is fired. And, when one has the good fortune to be placed at a gun. that possesses the least comparative degree of windage, he is sure to obtain the credit of being the best marksman in the ship, or in the corps to which he belongs; and thus instances frequently occur in which the commanding officer does not always extend the marks of his approbation to the most deserving gunner.

SOLUTION OF PROBLEMS IN GAUGING.

Gauging is the art of finding the number of gallons, &c., contained in any vessel.

By a Parliamentary statute, it is enacted that there is to be but one general standard gallon throughout Her Majesty's dominions of Great Britain and Ireland; which gallon is to contain 10 lbs. (avoirdupois weight) of distilled water, each pound of which is to weigh 7000 grains (troy weight): hence the standard gallon is to contain 70000 grains (troy weight) of distilled water. Now, since a cubic inch of distilled water weighs 252.458 grains (troy weight), the contents of the standard gallon may be readily reduced to cubic measure, by the following proportion; viz., as 252.458 grains: 1 inch :: 70000 grains: 277.27384357 inches; which, therefore, is the number of cubic inches in the standard gallon. And because the measure of the old standard wine gallon is 231 cubic inches, and that of the old standard ale gallon 282 such inches, we have sufficient data for obtaining proper multipliers for the reduction of the old standard wine and ale measure into the new general standard measure, and conversely. Hence,

277.27384357+2311.20031967) is the general multiplier for reducing

Log.=0.079297

the new standard measure into the old standard wine measure; and,

231-277.27384357=0.83311140) is the general multiplier for reducing

Log.=9.920703

the old standard wine measure into

the new standard measure.

277.27384357-282=0.98324058) is the general multiplier for reducing

Log. 9.992660

the new standard measure into the old standard ale measure; and,

282÷277.27384357=1.01704508) is the general multiplier for reducing

Log. 0.007340

the old standard ale measure into the new standard measure.

Now, the respective multipliers and their corresponding logarithms being thus obtained, the reduction of the old standard wine and ale measure into the new general standard measure, and conversely, may be very readily performed, by means of the following Problems.

PROBLEM I.

To reduce the old standard Wine Measure into the new Imperial Measure.

RULE.

To the lo aithm of the old standard wine gallons add the constant logarithm 9.920703, and the sum will be the logarithm of the new standard g llons.

Example.

Reduce 400 gallons of the old standard wine measure into the new general standard measure.

To reduce the new Imperial Measure into the old standard Wine Measure.

RULE.

To the logarithm of the new standard gallons add the constant logarithm 0.079297, and the sum will be the logarithm of the old standard wine gallons.

Note. From the above Problems it appears that the new general standard gallon is, very nearly, one-fifth greater than the old standard wine gallon.

PROBLEM III.

To reduce the old standard Ale Measure into the new Imperial Measure.

RULE.

To the logarithm of the old standard ale gallons add the constant logarithm 0.007340, and the sum will be the logarithm of the new general standard gallons.

Example.

Reduce 400 gallons of the old standard ale measure into the new general standard measure.

Given number of ale gallons = 400 Log. =2.602060
Constant log.=

0.007340

New standard gallons = 406.82 Log. = 2.609400

PROBLEM IV.

To reduce the new Imperial Measure into the old standard Ale Measure.

RULE.

To the logarithm of the new standard gallons add the constant logarithm 9.992660, and the sum will be the logarithm of the old standard ale gallons.

Example.

Reduce 400 gallons of the new general standard measure into the old standard ale measure.