Page:Aerial Flight - Volume 1 - Aerodynamics - Frederick Lanchester - 1906.djvu/317

Rh helical surface which we regard as the analogue of the horizontal, and $$\mbox{W}$$ (the analogue of $$W$$) the force at right angles thereto. Let $$O\ b$$ be the helical flight path, and $$\gamma$$ the gliding angle; then $$\theta$$, the angle cut off between $$O\ b$$ and the axis of $$x$$, will be the effective pitch angle; that is to say, the line $$O\ b$$ represents the helix along which the blade of the propeller will actually travel, and its pitch will be the effective pitch of the propeller.

Draw the line $$\mbox{F}$$ parallel to the axis of $$y$$ to represent the component of $$\mbox{W}$$ in the direction of motion of the vessel, cut off $$O\ a = \mbox{W} ,$$ and draw $$a\ c$$ perpendicular to $$y ,$$ draw $$a\ b$$ perpendicular to $$O\ a ,$$ and $$b\ d$$ perpendicular to $$O\ x .$$

Then $$O\ a$$ being equal to $$\mbox{W}$$, we have $$a\ c$$ equal to $$\mbox{F} ,$$ the two triangles being equal in every respect. Let us denote $$a\ b = f ,$$ and $$b\ d = h .$$

Now while the blade moves from to h the energy lost will be $$= \mbox{W}\ f ;$$ that is to say, we regard the matter as a case of gliding, to which it is strictly analogous. The energy utilised in propulsion during the same period will be $$= \mbox{F}\ h .$$ (These quantities are indicated by the shaded areas in the figure.)

Now it follows from the construction that—

Rh or, Rh

§ 203. Conditions of Maximum Efficiency.—The conditions of maximum efficiency are attained for the element of the propeller under consideration, when (1) is solved for minimum value.

Firstly, we may note from Equation (1) (and it is otherwise self-evident) that $$\gamma$$ should be made as small as possible; that is