Is it not evident that the ball really rolls on two parallel lines in the groove somewhere between _D_ and _E_, say the lines through _cc_ perpendicular to the plane of the paper? This granted, it follows that points on the ball-surface touching the groove above _c_ are going faster, while those touching below _c_ are going slower than points touching at _c_. Hence, no wonder there is friction. The position of _cc_ is such that the sum of the moments of friction above _cc_ balances the sum of the moments of friction below _cc_. Take axes _OX_, _OY_, as indicated; let the _x_ of _cc_ be _a_, and that of _D D_, _b_; put _d s_ for an element of arc, and let _A_ be the angle between the radius to _d s_ and _OY_. Then the friction on _ds_ is proportional to _ds_ cos _A_ = _dy_, and its moment about _cc_ is proportional to _dy_ (_x − a_), or, _dy_ (_a − x_), according as _ds_ is above or below _cc_. √(1 − _a_²) √(1 − _b_²) Therefore, ∫(_x − a_) _dy_ = ∫(_a − x_) _dy_ 0 √(1 − _a_²) The ball’s radius being unity, the solution of the above equation is,— arc cos _b_ _a_ = ½(——————————— + _b_ √(1 − _b_²)), √(1 − _b_²) which determines _a_ for all values of _b_; that is, determines the points _c_, _c_. It was stated above that _d s_ was proportional to the friction upon itself. Of course, we meant that it was proportional so long as _a_ remained constant. In terms of the unit given at the beginning of this discussion, the friction _ds_ is ———————————————, and the total friction upon the ball is therefore 2_a_ √(1 − _a_²) √(1 − _a_²) 4 ∫(_x − a_) _dy_ 0 arc cos _a_ ————————————————— = ——————————— = 1, 2_a_ √(1 − _a_²) √(1 − _a_²) which is the formula used to calculate our table above. As to the weight balls can safely carry in any bearing, below will be found results of experiments and calculations made by Professor Robinson, of the Ohio State University. This article is the result of careful, exhaustive work, and I am under great obligations for the privilege of introducing it here, as it has never before been in print.