Problem VI.

To divide a given Line into any Number of Parts, which Parts shall de in the same Proportion to each other as the Parts of some other given Line, whether those Parts are equal or unequal.

Let A b be the given line which it is required to divide in the same manner and proportion as the line c D, whether the parts are equal or unequal. On the base line C D, form an equilateral triangle in the manner already described in a former problem. Then take the distance A b, and with e as a centre, describe the arc f g, and join the points f and g, and f g shall be equal to A b. Now, if from the points h I k, which are the divisions of the line c, we draw lines to e, as h e, I e, and k e, these lines will cut f g in the points a b c, which will divide the line F g into parts proportionate to the divisions of the line c d.

Fig. 6.

Problem VII.

On a given Line to draw a Polygon of any Number of Sides, so that that Line shall be one Side of a Polygon; or, in other words, to find the Centre of a Circle which shall circumscribe any Polygon, the Length of the Side of the Polygon being given.

We shall here show, in a tabular form, the length of the radius of a circle, which shall contain the given line, as a side of the required polygon; and here we will suppose the line to be divided into one thousand equal parts, and the radius into a certain Dumber of like parts. The radius of the circle for different figures will be as follows: -

 For an inscribed Triangle 577 701 Pentagon 850 Ilexagon 1000 Heptagon 1152 Octagon ...... 1306 ½ Enneagon 1462
 Decagib ................... 1618 Endecagon ............. 1775 Dodecagon ............... 1932

By this table, the workman may, with a simple proportion, find the radius of a circle which shall contain a polygon, one side being given: thus, if it be required to draw a pentagon, the side given being fifteen inches, we may say as 1000 is to 15, so is 850, the tabular number for a pentagon, to 12 inches and seventy-five hundredth parts of an inch, or seven-tenths and a half o fa tenth of an inch.

We may here give another table for the construction of polygons, one in which the radius of the circumscribing circle is given. If it be required to find the side of the inscribed polygon, the radius being one thousand parts, the sides of the different polygons will be according to the following scale: -

 The Triangle ............. 1732 Square .. ............... 1414 Pentagon ............... 1175 Hexagon .............. 1000 Heptagon .................... 867½ Octagon ............ 765 Enneagon ............ 684 Decagon .............. 618 Endecagon ............. 563½ Dodecagon ................ 517½

Here, as in the case already mentioned, the law of proportion applies, and the statement may be thus made: as one thousand is to the number of inches contained in the radius of the given circle, so is the tabular number for the required polygon to the length of one of its sides in inches. Thus, let it be supposed that we have a circle whose radius in inches is 30, and that we wish to inscribe an octagon within it; then say as 1000 is to 30 inches, so is 765 to 22 inches and 95-100 parts of an inch, the length of the side of the required octagon.

Method of Drawing Curved Lines.

We will now introduce a few remarks upon the method of drawing curved lines, and also give some rules for finding the forms of mouldings when they are to mitre together, that is to say, of raking mouldings, and of bevel work in general. It will also be necessary to make a few remarks upon the form of ribs for domes and groins, a knowledge of which is so necessary to the builder, that without it the workman cannot correctly execute his task. It is hardly necessary to state, that all these mechanical operations are founded upon geometrical principles; and, unless he is acquainted with these, the workman cannot hope to succeed in his attempt to excel in his art,- one which is necessary for the comfort and convenience of all communities.

Problem VIII.

To draw on Ellipse with the Rule and Compasses, the transverse and conjugate Diameters being given; that is, the Length and Width.

Let a b be the transverse or longest diameter; c d the conjugate or shortest diameter; and o the point of their intersection, that is, the centre of the ellipse. Take the distance oc or oD; and, taking A as one point, mark that distance A e upon the line a O. Divide

Fig. 7.

0 e into three equal parts, and take from a f, a distance e f, equal to one of those parts. Make O g equal to O f. With the radius f g, and f and g as centres, strike arcs which shall intersect each other in the points I and h. Then draw the lines h f k, h g m, and I f l, i g n. With f as a centre, and the radius a f, describe the arc

L a k; and, from g as a centre, with the same radius, describe the arc m b n. With the radius h c, and h as a centre, describe the arc k c m; and, from the point I, with the radius I D, describe the arc l d m. The figure a c b d is an ellipse, formed of four arcs of circles.

Problem IX.

To draw an Ellipse by means of two Concentric Circles.

Fig. 8.

Let A b be the transverse, and e f the conjugate diameter, and o the centre of an ellipse to be drawn. From O with the radius O A, describe the circle a c b d, and from the same centre describe another circle g e h f. Divide the outer circle into any number of equal parts; the greater the number, the more exact will be the ellipse: and they should not be less than twelve. From each of these divisions draw lines to the centre o, as a o, b o, c o. Then, from a, b, c, etc, draw lines perpendicular to a b, and from the corresponding points in the inner circle, that is, from the points marked 1, 2, 3, etc, draw lines parallel to a b. Draw a curve through the points where these lines intersect each other, and it will be an ellipse.

In the diagram to which this demonstration refers, only one quarter of the ellipse is lettered, but the process described in relation to that must be carried round the circles, as is shown in the dotted and other lines.