Special graduations, marking the position of constant factors which frequently enter into engineering calculations, are found on most slide rules. Usually the values of π = 3·1416 and (π)/(4) = 0·7854—the “gauge points” for calculating the circumference and area of a circle—are marked on the upper scales. The first should be given on the lower scales also. Marks _c_ and _c_^1 are sometimes found on the lower scales at 1·128 = √((4)/(π)) and at 3·568 = √((40)/(π)). These are useful in calculating the contents of cylinders and are thus derived:—Cubic contents of cylinder of diameter _d_ and length _l_ = (π)/(4)_d_^2_l_; substituting for (π)/(4) its reciprocal (4)/(π), the formula becomes (_d_^2)/(1·273 × _l_), and by taking the square root of the fractional part we have (_d_)/(1·128)^2 × _l_. This is now in a very convenient form, since by setting the gauge point _c_ on C to _d_ on D, we can read over _l_ on B the cubic contents on A. This example indicates the principle to be followed in arranging gauge points. Successive multiplication is avoided by substituting the reciprocal of the constant, thus bringing the expression into the form (_a_ × _b_)/(_c_), which, as we know, can be resolved by one setting of the slide. The advantage of dividing _d_ before squaring is also evident. The mark _c_^1 = _c_ × √(10) is used if it is necessary to draw the slide more than one-half its length to the right. A gauge point, M, at 31·83 = (100)/(π) is found on the upper scales of some rules. Setting this point on B to the diameter of a cylinder on A, the circumference is read over 1 or 100 on B or the area of the curved surface over the length on B. As another example of establishing a gauge point, we will take the formula for the theoretical delivery of pumps. If _d_ is the diameter of the plunger in inches, _l_ the length of stroke in feet, and Q the delivery in gallons, we have Q = _d_^2 × (π)/(4) × _l_ × (12)/(277). (N.B.—277 cubic inches = 1 gallon.) Multiplying out the constant quantities and taking its reciprocal, we readily transform the statement into Q = (_d_^2_l_)/(29·4) or ((_d_)/(5·42))^2 × _l_. Hence set gauge point 5·42 on C to _d_ on D and over length of stroke in feet on B, read delivery in gallons per stroke on A; or over piston speed in feet per minute on B, read theoretical delivery in gallons per minute on A. Several examples of gauge points will be found in the section on calculating the weights of metal (see pages 59 and 60). In most cases their derivation will be evident from what has been said above. In the case of the weight of spheres, we have Vol. = 0·5236_d_^3, and this multiplied by the weight of 1 cubic inch of the material will give the weight W in lb. Hence for cast-iron, W = 0·5236 × _d_^3 × 0·26, which is conveniently transformed into W = (_d_ × _d_^2)/(7·35) as in the example on page 60.