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Adjustment of a Balance

I. Suppose the balance is not known to be in adjustment.

Any defect may be due to one of the following causes:

  • The relative position of the beam and pointer and its scale may be wrong. This may arise in three ways: (α) the pointer may be wrongly fixed, (β) the balance may not be level, (γ) the pointer when in equilibrium with the pans unloaded may not point to its zero position. We always weigh by observing the position of the pointer when at rest with the scale pans empty, and then bring its position of equilibrium with the pans loaded back to the same point. It is clear that this comes to the same thing as using a pointer not properly adjusted. In all these cases a will not be equal to α' in equation (1).
  • The arms may not be of equal length, i.e. L not equal to R.
  • The scale pans may not be of equal weight.

We may dispose of the third fault of adjustment first. If the scale pans be of equal weight, there can be no change in the position of equilibrium when they are interchanged; hence the method of testing and correcting suggests itself at once (see p. 101).

The first two faults are intimately connected with each other, and may be considered together. Let the pointer be at its mean position when there is a weight w in P and w'+x in Q, w and w' being weights which are nominally the same, but in which there may be errors of small but unknown amount,

Interchange the weights and suppose now that w in Q balances w'+y, in P, then

And if the pointer stands at zero when the pans are unloaded, we have

Hence equations (3) and (4) become

Multiplying

It will be seen on reference to the figure that L cosα' and R cosα are the projections of the lengths of the arms on a horizontal plane - i.e. the practical lengths of the arms considered with reference to the effect of the forces to turn the beam.

If the balance be properly levelled and the pointer straight α=α', and we obtain the ratio of the lengths of the actual arms. We thus see that, if the pointer is at zero when the balance is unloaded, but the balance not properly levelled, the error of the weighing is the same as if the arms were unequal, provided that the weights are adjusted so as to place the pointer in its zero position. The case in which α = -α' and therefore cosα = cosα' will be an important exception to this; for this happens when the three knife-edges are in one plane, a condition which is very nearly satisfied in all delicate balances. Hence with such balances we may get the true weight, although the middle point of the scale may not be the equilibrium position of the pointer, provided we always make this equilibrium position the same with the balance loaded and unloaded. If we wish to find the excess weight of one pan from a knowledge of the position of the pointer and the sensitiveness of the balance previously determined, it will be a more complicated matter to calculate the effect of not levelling.

We may proceed thus : Referring to equation (1), putting P = Q we get

And since θ=0 when no weights are in the pans, we get

Since α and α' are always very small, we may put cosα' = 1 and sinα'=α', and so on, the angles being measured in circular measure (p. 45).

Neglecting x and the difference between L and R, in the bracket, since these quantities are multiplied by α or α', we have

The error thus introduced is small, unless

is a very large quantity, compared with α, and it well may be so, Since h is small and w+P may be many times K; but α in a well-made balance is generally so small that the effect is practically imperceptible, and if the knife-edges be in a plane, so that α = -α', the correction vanishes.



Last Update: 2011-03-27