Page:Aerial Flight - Volume 1 - Aerodynamics - Frederick Lanchester - 1906.djvu/314

§ 200 and in the first instance let us suppose that this current consists of a quantity of fluid moving en masse with a velocity $$= \mbox{v}_1 .$$ Now the force of propulsion is essentially equal, and of opposite sign, to the resistance experienced by the vessel, action and reaction being equal and opposite; consequently, on the Newtonian basis, the rearward momentum communicated by the propeller will be equal to the forward momentum communicated by the vessel, so that the conditions of propulsion will be satisfied if the propeller impart to the wake current a rearward velocity $$\mbox{v}$$ equal to $$\mbox{v}_1 ;$$ that is to say, the fluid will be brought to rest.

Let us now re-calculate the efficiency as in § 198: we have work done usefully per second $$= \mbox{F}\ V = \mbox{m}\ \mbox{v}\ V ;$$ and energy taken out of the fluid, that is, energy received per second, is $$= \frac{\mbox{m}\ \mbox{v}_1^2}{2} ;$$ or, total energy expended per second $$= \mbox{m}\ \mbox{v}_1\ V - \frac{\mbox{m}\ \mbox{v}_1^2}{2}$$, or,

efficiency $=$

This is greater than unity; the result being, as anticipated, that it is theoretically possible that a vessel should be propelled for a less expenditure of power than that by which it can be towed.

This important result, although not generally known, is not new; it was previously pointed out by Mr. W. Froude in the discussion on a paper by Sir F. C. Knowles (Proc. Inst. C. E. 1871). Froude evidently had also treated the matter quantitatively, since he mentions the theoretical possibility of a negative slip of a screw propeller, from the cause stated, equal to half the positive slip as ordinarily computed.

In the present hypothetical case the influence of the counterwake has not been taken into account, it forming no part of the Newtonian scheme; the conditions are too artificial for the omission to be a matter of any importance, apart from the fact