Let _R_ equal the strain per square inch of cross-section of the tube at the point farthest removed from the neutral axis at the instant of rupture. [Illustration: Tube sections. Fig. 1 Fig. 2] Suppose Fig. 1 to represent the half of the tube, and that you are trying to bend it down at the ends. The particles towards the top will be pulled apart, while those at the bottom are crowded together; somewhere between the top and bottom the particles are neither pulled apart nor crowded together. Were the tube solid, the line of these particles would be the neutral axis. In the tube an imaginary line through the centre of the hole does not vary much from said axis. Now _R_π the moment of rupture = ——————(_r_ₑ⁴ − _r_ᵢ⁴), where _r_ₑ and _r_ᵢ (Fig. 4 _r_ₑ _R_π 2) are the exterior and interior radii; ———— is a constant, which 4 we will call _K_, whence we can write moment of rupture = _K_(_r_ₑ² − _r_ᵢ²) (_r_ₑ² + _r_ᵢ²) ÷ _r_ₑ. Here the factor (_r_ₑ² − _r_ᵢ²) is proportional to the area of the annular cross-section and is constant, while the other factor, _r_ᵢ (_r_ₑ² + _r_ᵢ²) ÷ _r_ₑ or, _r_ₑ + ———— _r_ᵢ, though less than 2_r_ₑ, _r_ₑ gets nearer and nearer to 2_r_ₑ as _r_ₑ gets large and _r_ᵢ approaches _r_ₑ. Therefore we have, that in resistance to flexure the tube should be as large in diameter as practicable, which means that it must be as thin as possible. This result is only modified in practice by the necessity of guarding against dinging and also against imperfections in the steel. A surface crack will ruin a very thin tube which otherwise may be harmless in a thicker, but it is safe to say that it is best to use reasonably large thin tubes. Oval tubes are of an advantage only when the direction of the strain is positively known and when it invariably occurs in that direction. Since the tube finds its greatest limit of general resistance in cylindrical form, to alter that form must necessarily weaken it more in one direction than it strengthens it in another. CHAPTER XXII. THE CYCLE IN WAR—STEAM AND ELECTRICITY.