Home >> Encyclopedia-of-architecture-1852 >> Portic1 to Sir Christopher Wren >> Q I E P

Q I E P Q

equal, angle, rafters, draw, roof and weight

E P Q, Q I, are to one another as the numbers 1,3, 5, 7, and E 0, E P, E Q, E I, are as the square numbers 1, 4, 8, 16, but the lines o D, P C, Q B, are to one another as 1, 2, 3, 4; therefore the abscissas E E P. E Q, E I, are as the squares of the ordinates o D, P C, Q B, and the points A, B, C, D, E, are in the curve of a parabola.

In the same manner it may be shown that this is the case, whatever be the number of ordinates.

a roof of this construction may be described to any given height and vertical angle, or to a given width and height with any number of rafters on each side.

describe a roof with any given number of rafters on each side, of a given width and height, so that all the weights suspended from the angular points of the rafters in vertical equidistant lines, may angular the rafters in equilibrio.

Figure 3.-Let there be four rafters on each side, let I N be half the width, and I E, the height. Draw N T and E T parallel to I E and I x; divide N T into four equal parts, isr f f e, e d, d and draw d g E, e h n,/ i E: likewise divider x into four equal parts, t c, c b, It a, a x, and draw r g, b b, a i, parallel to I E. Join E 9, g h, h i and these lines will be the rafters of half the roof required. For the demonstration, see the article CONIC SECTIONS.

Proposition IV. Figure 4.-Suppose it were required to construct a curb roof to have the bottom rafter to the upper rafter as 2 to 3, to a given vertical angle at the top, and a given width A a.

Now the weight on the upper angle is to the weight on le, 2 It I+I Athe lower angle, is to - that is 3+3 =3 is 2 2 2to this is in the proportion of 6 to 5, or the half weight at n is to the bottom weight at 1, as 3 to 5.

Bisect A B by the perpendicular c D, and make the angle A E e equal to half the vertical angle, or the angle E A C, equal to its complement. ,Nlake E n to E e as 5 to c; join D A and D ; in A take any point, F; draw F G parallel to A E, making A F to F C. as '2 to 3; draw A G cutting C D at ii,

and it t parallel to F G or E A, cutting A D at 1; K equal to and join u II, then A T II K B iS the contour of the roof required. This is so evident from its construction, that its does not require demonstration.

Proposition V. Figure 5.—To describe a roof with four equal rafters, that shall be in equilibrio by the weight of the of a given width A E, and height F c.

Join c and bisect it in II by a perpendicular, D HG, meeting A E in G ; on o, as a centre, with the distance G E or G c, describe the circle CEO. Draw K a 1 parallel to F E, to meet the vertical line o C in K, and the circle in r. Draw I D C, and join D E, than make the side C B A similar to C D E, and A B C, C D E, will be the rafters of the roof required.

For, in Figure 6, complete the parallelogram c D Q B, and join B D, I F, and draw C L perpendicular to C F, and equal to F G ; on L, with the distance G E describe the circle N I F, meeting the vertical line at s and ; produce E D to meet it also in st, and B C to P.

Then because K F is equal to K c, and R C equal R Q, the triangles a I F and c D Q, are similar ; therefore I F is parallel to D Q, and because the two segments N t F and C E o are equal to one another, the angle N I F is equal to the angle G E 0, equal to twice the angle C E F, or twice the alternate angle E C L, equal to ECD± D C L; but E C D is equal to halt' the external angle M D C, and D C L is half the angle D C P; equal to D C Q. Therefore the angle N I F is equal to the angles ITDC±CDQ, equal to the angle 111DQ. conse quently, c s;:co; C M, but c F and C N are equal, there fore c Q and c m are equal ; but c Q is to C M as the weight on c is to the weight on B or D, therefore the weights on c and B are equal, and the rafters A B, D C, C D. D E, are in equilibrio.