A.—­An increase of the diameter.  The amount of reacting power of the screw upon the water is hot measured by the number of square feet of surface of the arms, but by the area of the disc or circle in which the screw revolves.  The diameter of the screw of the Rattler being 10 feet, the area of its disc is 78.5 square feet; and with the amount of thrust already mentioned as existing in the first experiment, viz. 8722 lbs., the reacting pressure on each square foot of the screw’s disc will be 108-1/2 lbs.  The immersed midship section being 380 square feet, this is equivalent to 23 lbs. per square foot of immersed midship section at a speed of 9.2 knots per hour.

589. Q.—­In smaller vessels of similar form, will the resistance per square foot of midship section be more than this?

A.—­It will be considerably more.  In the Pelican, a vessel of 109-3/4 square feet of midship section, I estimate the resistance per square foot of midship section at 30 lbs., when the speed of the vessel is 9.7 knots per hour.  In the Minx with an immersed midship section of 82 square feet, the resistance per square foot of immersed midship section was found by the dynamometer to be 41 lbs. at a speed of 8-1/2 knots; and in the Dwarf, a vessel with 60 square feet of midship section, I estimate the resistance per square foot of midship section at 46 lbs. at a speed of 9 knots per hour, which is just double the resistance per square foot of the Rattler.  The diameter of the screw of the Minx is 4-1/2 feet, so that the area of its disc is 15.9 square feet, and the area of immersed midship section is about 5 times greater than that of the screw’s disc.  The diameter of the screw of the Dwarf is 5 feet 8 inches, so that the area of its disc is 25.22 square feet, and the area of immersed midship section is 2.4 times greater than that of the screw’s disc.  The pressure per square foot of the screw’s disc is 214 lbs. in the case of the Minx, and 109-1/2 lbs. in the case of the Dwarf.

590. Q.—­From the greater proportionate resistance of small vessels, will not they require larger proportionate screws than large vessels?

A.—­They will.

591. Q.—­Is there any ready means of predicting what the amount of thrust of a screw will be?

A.—­When we know the amount of pressure on the pistons, and the velocity of their motion relatively with the velocity of advance made by the screw, supposing it to work in a solid nut, it is easy to tell what the thrust of the screw would be if it were cleared of the effects of friction and other irregular sources of disturbance.  The thrust, in fact, would be at once found by the principle of virtual velocities; and if we take this theoretical thrust and diminish it by one fourth to compensate for friction and lateral slip, we shall have a near approximation to the amount of thrust that will be actually exerted.[1]

[1] See Treatise on the Screw Propeller, by J. Bourne, C. E.