Fig. 103.
It will be found that the points f₁, f₂, &c. are very nearly in a straight line. We assume that, if the apparatus and observations were perfect, the points would lie exactly in a straight line. The object of the construction is to determine the straight line, which on the whole is most close to all the points. If it be true that the friction is proportional to the pressure, this line should pass through the origin a, for then the perpendicular which represents the friction is proportional to the line cut off from a, which represents the load. It will be found that a line at can be drawn through the origin a, so that all the points are in the immediate vicinity of this line, if not actually upon it. A string of fine black silk about 15" long, stretched by a bow of wire or whalebone, is a convenient straight-edge for finding the required line. The circles described about the points f₁, f₂, &c. will facilitate the placing of the silk line as nearly as possible through all the points. It will not be found possible to draw a line through a, which shall intersect all the circles; the best line passes below but very near to the circles round f₁, f₂, f₃, f₄, touches the circle about f₅, intersects the circles about f₆ and f₇, and passes above the circle round f₈. The line should be so placed that its depth below the point which is most above it, is equal to the height at which it passes above the point which is most below it.
From a measure as, a length of 10", and erect the perpendicular st. We find by measurement that st is 2"·7. If, then, we suppose that the friction for any load is really represented by the distance cut off by the line at upon the perpendicular, it follows that
F : R :: 2"·7 : 10".
or F = 0·27 R.
This is the formula from which [Table III]. has been constructed.
[Table IV].
By a careful application of the silk bow-string, x y q can be drawn, which, itself in close proximity to a, passes more nearly through f₁, f₂, &c. than is possible for any line which passes exactly through a. x y q will be found not only to intersect all the small circles, but to cut off a considerable arc from each. Measure off x p a distance of 10", and erect the perpendicular p q; then, if R be the load, and F the corresponding friction, we must have from similar triangles—
| F - | AY | × 1 lb. | |
| 0"·1 | = | PQ | |
| R | PX | ||
By measurement it is found that AY = 0"·14, and PQ = 2"·53.