This section is from the book "Practical Sheet And Plate Metal Work", by Evan A. Atkins. Also available from Amazon: Practical Sheet And Plate Metal Work.

The setting out of the patterns for a copper (Fig. 230) conveniently comes in with our consideration of the sphere.

A diagram of the vessel is shown in Fig. 231, from which it will be seen how to obtain the sizes of bottom and sides. The radius for the bottom disc can be measured from the diagram, or can be calculated as follows: -

Disc radius = AC =

The height of the plate for the body is determined by the property of the sphere and cylinder mentioned in connection with Fig. 226. The body will be made up in either one, two, or more pieces, according to the diameter of the copper. The patterns are not shown set out, as their shapes being so simple there is no need for it.

Fig. 231.

After the body is roughly riveted together the bottom part can be worked over or razed in on a bench bar, as shown in Fig. 232. (It will, perhaps, not be here out of place to state that the bench bar in a general shop is probably the most useful tool to have, as it can be used for a score or more different purposes. If it has a square-tapered hole in the flat end, it can be made to carry various heads and do similar work to the horse in a floor-block.)

The stretching or throwing off of the brim is explained by Fig. 233. This can either be done on a head stake as shown, on an anvil, or on the edge of the flat end of bench bar. In stretching the brim the depth of flange should first be marked around the inside. When throwing the brim off on the stake, care should be taken that the more intense part of the hammer-blow falls near the edge of metal, as the greatest amount of stretch must, of course, take place on the outside of the brim.

The position of the wrinkling circle for the bottom, and thus its diameter (Fig. 231) can be determined by the following rule: "To find the distance of the neutral circle from the bottom, deduct the chord from the arc, and multiply this difference by the radius and divide the product by the arc." That is -

Fig. 232.

Fig 233.

CN = radius O D x (arc A C D - chord A D) / arc ACD

Before proceeding to completely rivet up, the bottom should be attached with a few tacking rivets. In finishing the joints care should be taken that the seams are properly countersunk, and also that the tails of rivets are drawn up, so that the inside of the copper shall be completely smooth.

Sketches of hammers are shown in Fig. 234, the two on the left being types of hollowing or blocking hammers, the centre one a stretching or razing hammer, and the right-hand pair planishing hammers. It is, perhaps, hardly necessary to point out that there are hundreds of different shapes and sizes of the above, the form of hammer used depending upon the strength of material and kind of job in hand.

Fig 234.

Capacity of a Copper.

Before proceeding to show how to calculate the number of gallons that a copper of the shape shown in Fig. 230 will hold, it will be necessary to explain how to find the volume of a sphere.

One of the simplest aids to the remembering of how to find the volume of a sphere is in the peculiar relation that exists between the volumes of a cone, sphere, and cylinder when their diameters and heights are equal. Imagine that Fig. 235 represents these three solids, then the relative volumes of cone, sphere, and cylinder will be as 1 is to 2 is to 3. So that, having found the volume of the cylinder, the sphere will be two-thirds, and the cone one-third of it. It will also be seen that the sphere is just twice the volume of the cone. The above relation facilitates the work in calculating the cubic contents of a vessel with hemispherical or conical ends.

If a vessel has hemispherical ends, all that is necessary is to add two-thirds of the diameter on to the cylindrical portion, and calculate its volume.

If a vessel has a conical bottom or top pointing outwards, then its volume can be found by adding one-third the height of the cone on to the cylindrical portion, and calculating as before. If a vessel has a conical bottom pointing inwards, then, of course, one-third the height of cone would be deducted from the length.

It is sometimes convenient to use the ordinary mensuration rules, such as -

Volume of sphere -

= π d3 / 6 = π 4 r3 / 3 = r3 x 4.1888

Fig. 235.

Volume of cone-

= π d2 h /12 = π r2 h / 3 = r2h X 1.0472

Volume of cylinder-

= π d2 h /4 = πr2 h = r2 h X 3.1416

Where d = diameter, r = radius, and h = height.

In practice, it is handier to use some such rules as the following, which have been calculated on the basis of 277.274 c. in. to the gallon. Taking dimensions in feet -

Gallons in cylinder = r2 h x 19.6 " sphere = r3 x 26.11

" cone = r2 h x 6.53

If the dimensions are in inches, then the following multipliers must be used-

Gallons in cylinder = r2 h X .01133 " sphere = r3 X .01511

" cone = r2h X .00378

Taking one example to illustrate their use. Suppose we require to find the number of gallons in a hemispherical bowl of 20 in. diameter. Then -

Gallons = 10 X 10 X 10 X .01511 / 2 = 7.55 = 7½ (nearly).

After the above explanation, let us come back to the copper. Suppose it is 3 ft. diameter and 3 ft. 6 in. deep. Then deducting half the diameter from the depth, this will give us 2 ft. for the length of the cylindrical part. Adding two-thirds the depth of the hemispherical bottom on to the cylindrical portion, this will give us an equivalent cylinder of 3 ft. length. The cubical contents will, therefore, be-

3/2 X 3/2 X 3 X 19.6 = 132.3 = 132¼ gallons.

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