This section is from the book "The Tinman's Manual And Builder's And Mechanic's Handbook", by Isaac Ridler Butt. Also available from Amazon: The Tinman's Manual And Builder's And Mechanic's Handbook.

Problem X.

To describe an Ellipse by Means of a Carpenter's Square, or a piece of notched Lath.

Having drawn two lines to represent the diameters of the ellipse required, fasten the square so that the internal angle or meeting of the blade and stock shall be at the centre of the ellipse. Then take a piece of wood or a lath, and cut it to the length of half the longest diameter, and from one end cut out a piece equal to half the shortest diameter, and there will then be a piece remaining at one end equal to the difference of the half of the two diameters. Place this projecting piece of the lath in such a manner that it may rest against the square, on the edge which corresponds to the two diameters; then, turning it round horizontally, the two ends of the projection will slide along the two internal edges of the square, and if a pencil be fixed at the other end of the lath, it will describe one quarter of an ellipse. The square must then be moved for the successive quarters of the ellipse, and the whole figure will thus be easily formed.

This method of forming an ellipse is a good substitute for the usual plan, and the figure thus produced is more accurate than that made by passing a pencil round a string moving upon two pins or nails fixed in the foci, for the string is apt to stretch, and the pencil cannot be guided with the accuracy required.

There are many other methods of drawing ellipses, or more properly ovals, but we can only notice two of those in common use.

1. By ordinates, or lines drawn perpendicular to the axis. Having formed the two diameters, divide the axis, or larger diameter, into any number of equal parts, and erect lines perpendicular to the several points. Next draw a semicircle, and divide its diameter into the like number of equal parts; that is, if the larger diameter or axis of the intended ellipse be divided into twenty equal parts, then the semicircle must be divided into the like number. As the diameter of the semicircle is equal to the shorter diameter of the ellipse, or conjugate axis, perpendiculars may be raised from these divisions of the diameter, or the semicircle, till they meet the circumference; and the different perpendiculars, which are called ordinates, may be erected like perpendiculars, on the axis of ellipse. Joining the several points together, the ellipse is described; and the more accurately the perpendiculars are formed the more exact will be the ellipse.

2. By intersecting arches. Take any point in the axis, and with a radius equal to the distance of that point from one extremity of the axis, and with one of the foci as a centre, describe an arc; then with the distance of the assumed point in the axis from the other end of it, and with the other focus as a centre, describe another arc intersecting The former, and the point of intersection will be a point in the ellipse. By assuming any number of points in the axis, any number of points on the curve may be found, and these united will give the ellipse. This process is founded on the property of the ellipse; that if any two lines are drawn from the foci to any point in the curve, the length of these lines added together will be a constant quantity, that is, always the same in the same ellipse.

Problem XI.

To find the Centre and the two Axes of an Ellipse. Let a b c d be an ellipse, it is required to find its centre. Draw any two lines, as e f and G H, parallel and equal to each other. Bisect these lines as in the points I and k. and bisect I k as in l,. From l, as a centre, draw a circle cutting the ellipse in four points, 1, 2, 8, 4. Now L is the centre of the ellipse. But join the points 1, 3, and 2, 4; and bisect these lines as in M and n. Draw the line m n, and produce it to a and b, and it will be the transverse axis, Draw c D through l, and perpendicular to ab, and it will be the conjugate or shorter axis.

Fig. 9.

Problem XII.

To draw a flat Arch by the intersection of Lines, having the Opening and Spring or Rise given.

Let a d b be the opening, and c d its spring or rise. In the middie of A b, at d, erect a perpendicular d e, equal to twice c d, its rise; and from e draw e a and e b, and divide a e and b e into any number or equal parts, as a, b, c, and 1, 2, 3. Join Ba, 3c, 2b, and 1 a, and it will form the arch required.

Fig. 10.

The more parts a e and b e are divided into, the greater will be the accuracy of the curve.

Many curves may be made in the same manner, according to the position of the lines a e and e b; and if instead of two lines drawn from a and b, meeting in e, a perpendicular be erected at the same points, and two lines be then drawn from the ends of these perpendiculars meeting in an angle, and these lines be divided into any number of equal parts, the points of the adjacent lines may be joined, and a curve will be formed resembling a gothic arch. The demonstration already given is therefore very useful to the workman, as he may vary the form of the curve by altering the position of the lines, either with respect to the angles which they make with each other, or their proportional lengths.

Problem XIII.

To find the Form or Curvature of a raking Moulding that shall unite correctly with a level one.

Let A b c d be part of the level moulding, which we will here suppose to be an ovolo, or quarter round; a and c, the points where the raking moulding takes its rise on the angle;F C G, the angle the raking; moulding makes with the horizontal one. Draw C f at the given angle, and from a draw a e parallel to it; continue b a to h, and from c make c h perpendicular to a h. Divide c h into any number of equal parts, as 1,2, 3, and draw lines parallel to h a, as 1 a, 2b, 3c; and then in any part of the raking moulding, as I, draw I k perpendicular to e a, and divide ik into the same number of equal parts h c is divided into; and draw 1 a, 2 b, 3 c, parallel to e a. Then transfer the distances la, 2b , 3 c, and a curve drawn through these points will be the form of the curve required for the raking moulding

Fig. 11.

We have here shown the method to be employed for an ovolo; but it is just the same for any other formed moulding, as a cavetto, semi-recta, etc. It may be worthy remark, that, after the moulding is worked, and the mitre is cut in the mitre-box for the level moulding, the raking moulding must be cut, either by the means of a wedge formed to the required angle of the rake, or a box made to correspond to that angle: and if this be accurately done, the mitre will be true, and the moulding in all its members correspond to the level moulding. The plane in which the raking moulding is situated is square to that of the level one. This is always the case in a pediment, the mouldings of which correspond with the return.

Problem XIV.

To find the Form or Curvature of the Return in an open or broken nice, and let the form and size of the moulding be as in the last problem, and as shown at d A B H. From d drop a perpendicular on c B, and draw D E perpendicular to D C, or parallel to C B; and let d e be equal to e i (Fig. 11). Then from e draw E f, parallel to d a, and divide E F into the same Dumber of parts as I k (Fig 11), at 1 a, 2B, 3 c, and transfer the distances 1 a, 2 b, 3 c, as in Fig. l1 . Then a curve line drawn through the points a, b, c, will be the form of the return for the moulding of the open pediment.

Pediment.

Let ab c be the angle which the pediment makes with the cor-

Fig. 12.

The mitre for the return is cut in the usual manner, but that of tHE pediment is cut to the proper angle of its inclination, as in the last problem. Infixing the mitre, the portion E D G of the return must be cut away, to make it come flush with the top of the pediment moulding.

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