The formulÆ in the tables on p. 73 and after can be deduced by two methods,—one that of graphical construction, the other that of least squares. The first method is the more simple and requires but little calculation; though neatness and care are necessary in constructing the diagrams. The second method will be described for the benefit of those who possess the requisite mathematical knowledge. The formulÆ used in the preparation of the tables have been generally deduced from the method of least squares, as the results are to a slight, though insignificant, extent more accurate than those of the method of graphical construction. This remark will explain why the figures in some of the formulÆ are carried to a greater number of places of decimals than could be obtained by the other method.

We shall confine the numerical examples to Tables III. and IV., and show how the formulÆ of these tables have been deduced by the two different methods.

Tables V., XIV., XVI., XXI., are to be found in the same manner as Table III.; and Tables VI., IX., X., XI., XV., XVII., XVIII., XIX., XX., XXI., XXII., in the same manner as Table IV.

THE METHOD OF GRAPHICAL CONSTRUCTION.

Table III.

A horizontal line aps, shown on a diminished scale in Fig. 103, is to be neatly drawn upon a piece of cardboard about 14"×6". A scale which reads to the hundredth of an inch is to be used in the construction of the figure. A pocket lens will be found convenient in reading the small divisions. By means of a pair of compasses and the scale, points are to be marked upon the line aps, at distances 1"·4, 2"·8, 4"·2, 5"·6, 7"·0, 8"·4, 9"·8, 11"·2 from the origin a. These distances correspond to the magnitudes of the loads placed upon the slide on the scale of 0"·1 to 1 lb. Perpendiculars to aps are to be erected at the points marked, and distances f1, f2, f3, &c. set off upon these perpendiculars. These distances are to be equal, on the adopted scale, to the frictions for the corresponding loads. For example, we see from Table III., Experiment 3, that when the load upon the slide is 42 lbs., the friction is 12·2 lbs.; hence the point f3 is found by measuring a distance 4"·2 from a, and erecting a perpendicular 1"22. Thus, for each of the loads a point is determined. The positions of these points should be indicated by making each of them the centre of a small circle 0"·1 diameter. These circles, besides neatly defining the points, will be useful in a subsequent part of the process.

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=0·27R.

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—

By measurement it is found that AY=0"·14, and PQ=2"·53.

We have, therefore,

F=1·4+0·253R.

This is practically the same formula as

F=1·44+0·252R,

from which the table has been constructed. In fact, the column of calculated values of the friction might have been computed from the former, without appreciably differing from what is found in the table.

THE METHOD OF LEAST SQUARES.

Table III.

Let k be the coefficient of friction. It is impossible to find any value for k which will satisfy the equation,

F-kR=0,

for all the observed pairs of values of F and R. We have then to find the value for k which, upon the whole, best represents the experiments. F-kR is to be as near zero as possible for each pair of values of F and R.

In accordance with the principle of least squares, it is well known to mathematicians, the best value of k is that which makes

(F1 - k R1)² + (F2 - k R2)² + &c. + (F? - k R?)²

a minimum where f1 and r1, f2 and r2 &c. are the simultaneous values of f and r in the several experiments.

In fact, it is easy to see that, if this quantity be small, each of the essentially positive elements,

(F1-kR1)²,&c.

of which it is composed, must be small also, and that therefore

F-kR

must always be nearly zero.

Differentiating the sum of squares and equating the differential coefficient to zero, we have according to the usual notation,

SR1(F1-kR1)=0;

The calculation of k becomes simplified when (as is generally the case in the tables) the loads R1, R2, &c., R? are of the form,

N, 2N, 3N &c., mN.

In this case,

SR1F1=N(F1+2F2 +3F3+&c.+mF?).

SR1²=N²(1²+2²+&c.+m²)

In the case of Table III.

m=8, N=14,

F1+2F+3F3+mF?=770·9;

whence k=0·27.

Thus the formula F=0·27R is deduced both by the method of least squares, and by the method of graphical construction.

Table IV.

The formula for this table is to be deduced from the following considerations.

No values exist for x and y, so that the equation

F=x+yR shall be satisfied for all pairs of values of F and R, but the best values for x and y are those which make

(F1-x-yR1)²+(F2-x-yR2)² +&c.+(F?-x-yR?)²

a minimum.

Differentiating with respect to x and y, and equating the differential coefficients to zero, we have

S(F1-x-yR1)=0,

SR1(F1-x-yR1)=0.

This gives two equations for the determination of x and y.

Suppose, as is usually the case, the loads be of the form,

N,2N,3N,4N&c.mN,

and making

A=F1+F2+F3+&c.+F?

B=F1+2F2+3F3+&c.+mF?,

we have the equations

Solving these, we find

In the present case,

m=8, N=14, A=138·4, B=770·9;

whencex=1·44

y=0·252,

and we have the formula,

F=1·44+0·252R.

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