But by the similar triangles TAD, TPM, TCE, and TBF, the sides AT, PT, CT, and BT are proportional to the four perpendiculars AD, PM, CE, BF.

Therefore AD: PM::CE: BF.

1092. THEOREM IX. The two radius vectors meeting the curve in the same point will make equal angles with a tangent passing through that point. (Fig. 439.)

For, by Theor. VII.,

By Cor. 2. Theor. V.,
By equality,

CA: CP:: CT: CA;

CA: CP::CF: CA+ FM;
CT: CF:: CA: CA+ FM;

By division and composition, CF-CT: CF+ CT:: FM: 2CA

That is,

+ FM;
FT:ƒT:: FM:ƒR;

And by the similar triangles TFM, TƒR, FT: ƒT:: FM: ƒR. Therefore fR is equal to fM; consequently the angle fRM is equal to the angle fMR: and because ƒR is parallel to ƒM, the angle FMT is equal to the angle fRM; therefore the angle FMT is equal to the angle ƒ RM.

1093. PROBLEM I. To describe an hyperbola by means of the end of a ruler moveable on a pin F (fig 440.) fixed in a plane, with one end of a string fixed to a point E in the same plane, and the other extremity of the string fastened to the other end C of the ruler, the point C of the ruler being moved towards G in that plane.

While the ruler is moving, a point D being made to slide

Fig. 439.

along the edge of the ruler, kept close to the string so as to keep each of the parts CD, DE of the string stretched, the point D will describe

the curve of an hyperbola.

If the end of the ruler at F (fig. 441.) be made moveable about the point E, and the string be fixed in F and to the end C of the ruler, as before, another curve may be described in the same manner, which is called the opposite hyperbola: the points E and F, about which the ruler is made to revolve, are the foci.

There are many occasions in which the use of this conic section occurs in architectural details. For instance, the profiles of many of the Grecian mouldings are hyperbolic; and in conical roofs the forms are by intersections such that the student should be well acquainted with the methods of describing it.

1094. PROE. II. Given the diameter AB, the abscissa BC, and the double ordinate DE in position and magnitude, to describe the hyperbola. (Fig. 442.)

Through B draw FG parallel to DE, and draw DF and EG parallel to AB. Divide DF and DC each into the same number of equal parts, H and from the points of division in BF draw lines to B, also from the points of division in DC draw straight lines to A; then through the points of intersection found by the lines drawn through the corresponding points draw the curve DB. In like manner the curve EB may be drawn so that DBE will form the curve on each side of the diameter AB. If the point A be considered as the vertex, the opposite hyperbola HAI may be described in the same manner, and thus the two curves formed by cutting the opposite cones by the same plane will be found. By the theorists, the hyperbola has been considered a proper figure of equilibrium for an arch whose office is to support a load which is greatest at the middle of the arch, and diminishes towards the abutments. This, however, is matter of consideration for another part of this work.

F

D123 C

Fig. 442.

OF THE PARABOLA.

1095. DEFINITIONS.-1. The parameter of the axis of a parabola is a third proportional to the abscissa and its ordinate.

2. The focus is that point in the axis where the ordinate is equal to the semi-parameter. 3. The diameter is a line within the curve terminated thereby, and is parallel to the axis.

4. An ordinate to any diameter is a line contained by the curve and that diameter parallel to a tangent at the extremity of the diameter.

1096. THEOREM I. ordinates.

In the parabola, the abscissas are proportional to the squares of the

M

Let QVN (fig. 443.) be a section of the cone passing along the axis, and let the dire trix RX pass through the point Q perpendicular to QN, and let the parabolic section be ADI meeting the base QIND of the cone in the line DI, and the diameter QN in the point H; also let KML be a section of the cone parallel to the base QIN intersecting the plane VQN in the line KL, and the section ADI in PM. Let P be the point of concourse of the three planes QVN, KML, AHI, and let H be the point of concourse of the three planes QVN, KML, AHI; then, because the planes VRX and ADI are parallel, and the plane VQN is perpendicular to the plane VRX, the plane ADI is also perpendicular to the plane VQN. Again, because the plane R QIN is perpendicular to the plane QVN, and the plane KML is parallel to the plane QIN, the plane KML is perpendicular to the plane QVN; therefore the common sections PM and HI are perpendicular to the plane VQN; and because the plane KML is parallel to the plane QIN; and these two planes are intersected by the plane QVN, their common sections KL and QN are parallel. Also, since PM and H are each perpendicular to the plane QVN, and since KL is the common section of th planes QVN, KML, and QN in the common section of the planes QVN, QIN; therefor PM and HI are perpendicular respectively to KL and QN.

Consequently

AP: AH:: PM2: HI2,

For, by the similar triangles APL, AHN, AP: AH:: PL: HN,

Or

But, by the circle

And, by the circle

Therefore

Therefore, by substitution,

Q

D

Η

Fig. 443.

Coroll. By the definition of the parameter, which we shall call P,

1097. THEOREM II. As the parameter of the axis is to the sum of any two ordinates, so is the difference of these ordinates to the difference of their abscissas.

That is, P HI+ PM:: HI-PM: AH-AP.

Multiplying the first of these equations by AP and the second by AH,

Subtract the corresponding numbers of the first equation, and P(AH-AP) = HI?— PM2. But the difference of two squares is equal to a rectangle under the sum and difference of their sides.

Therefore P(AH-AP) (HI+ PM) (HI-PM).
Consequently P HI+PM ::HI-PM: AH-AP;

Or, by drawing KM parallel to AH, we have GK= PM + HI, and KI=HI-PM; and since PH AH-AP; P: GK::KI: PH, or KM.

Coroll. Hence, because

And since
Therefore, by multiplication, KM × HIo= GK × KI × AH, or
AH: KM::HI2: GK × KI.

So that any diameter MK is as the rectangle of the segments GK,
KI of the double ordinate GI. From this a simple method has been
used of finding points in the curve, so as to describe it.

1098. THEOREM III. The distance between the vertex of the curve and

Fig. 445.

the focus is equal to one fourth of the parameter.

Let LG (fig. 445.) be a double ordinate passing through the focus, then LG is the parameter. For by the definition of parameter AF: FG::FG ; P=2FG

Therefore 2AF=FG=LG;

Consequently AF = 4LG.

il

1099. THEOREM IV. The radius vector is equal to the sum of the distances between the focus and the vertex, and between the ordinate and the vertex. (Fig. 446.)

Coroll. 1. If through the point G (fig. 447.) the line GQ be drawn perpendicular to the axis, it is called the directrix of the parabola.

By the property shown in this theorem, it appears that if any line QM be drawn parallel to the axis, and if FM be joined, the straight line FM is equal to QM; for QM is equal to GP.

T

Coroll. 2. Hence, also, the curve is easily described by points. Take AG equal to AF (fig. 447.), and draw a number of lines M, M perpendicular to the axis AP; then with the distances GP, GP, &c. as radii, and from F as a centre, describe arcs on each side of AP, cutting the lines MM, MM, &c. at MM, &c.; then through all the points M, M, M, &c. draw a curve, which will be a parabola.

1100. THEOREM V. If a tangent be drawn from the vertex of an ordinate to meet the axis produced, the subtangent PT (fig. 448.) will be equal to twice the distance of the ordinate from the vertex.

QQQ Q G Q Q QQ

Mk

M

M

If MT be a tangent at M, the extremity of the ordinate PM; then the sub-tangent PT is equal to twice AP. For draw MK parallel to

Then, by Theor. II.,

And as MKI, TPM are similar,

AH,

KM: KI:: GK::P;
KM: KI::PT : PM.

Therefore, by equality,

And by Cor. Theor. I.,

Therefore, by equality,

P: PM::GK: PT;

P: PM:: PM : AP.
AP PT:: PM: GK.

But when the ordinates HI and PM coincide, MT will become a tangent, and GK will become equal to twice PM.

Therefore AP: PT:: PM: 2PM, or
PT=2AP.

T

T

From this property is obtained an easy and accurate method of drawing a tangent to any point of the curve of a parabola. Thus, let it be required to draw a tangent to any point M in the curve. Produce PA to T (fig. 449.), and draw MP perpendicular to PT, meeting AP in the point P. Make AT equal to AP, and join MT, which will be the tangent required.

1101. THEOREM VI. The radius vector is equal to the distance between the focus and the intersection of a tangent at the vertex of an ordinate and the axis produced.

Produce PA to T (fig. 450.), and let MT be a tangent at M; then will FT=FM.

Coroll. 1. If MN be drawn perpendicular to MT to meet the axis in N, then will FN=FM=FT. For draw FH perpendicular to MT, and it also bisects MT, because FM = FT; and since HF and MN are parallel, and MT is bisected in H, the line TN will also be bisected in F. It therefore follows that FN=FM = FT.

Coroll. 2. The subnormal PN is a constant quantity, and it is equal to half the para meter, or to 2AF. For since TMN is a right angle,

Therefore

2AP or TP: PM:: PM: PN.

But, by the definition of parameter, AP: PM:: PM; P;
Therefore

PN=P

Coroll. 3. The tangent of the vertex AH is a mean proportional between AF and AP For since FHT is a right angle, therefore AH is a mean proportional between AF and AT and since AT=AP, AH is a mean proportional between AF and AP. Also FH is : mean proportional between FA and FT, or between FA and FM.

Coroll. 4. The tangent makes equal angles with FM and the axis AP, as well as with FC and CI.

1102. THEOREM VII. A line parallel to the axis, intercepted by a double ordinate and c tangent at the vertex of that ordinate, will be divided by the curve in the same ratio as the lin itself divides the double ordinate.

Let QM (fig. 451.) be the double ordinate, MT the tangent, AP the axis, GK the intercepted line divided by the curve in the point I;

then will GI IK::MK: KQ

For by similar triangles MKG, MPT; MK: KG:: PM: PT,

By the definition of parameter,

Therefore, by equality,

And again, by equality,

And by division,

or 2AP;

P: PM:: PM: 2AP;

P: MK:: PM: KG;
PM: MK::2AP: KG;
MK: KQ:: GI: IK.

1109. PROBLEM I. To describe a parabola.

If a thread, equal in length to the leg BC (fig. 452.) of a right angle or square, be fixed to the end C, and the other end of the thread be fixed to a point F in a plane, then if the square be moved in that plane so that the leg AB may slide along the straight line GH, and the point D be always kept close to the edge BC of the square, and the two parts FD and DC of the string kept stretched, the point D will describe a curve on the plane, which will be a parabola.

B

T

Fig. 452.

1104. PROB. II. Given the double ordinate DE and the abscissa BC in position and magnitude, to describe a parabola.

Through B (figs. 453, 454.) draw FG parallel to DE,and DF and EG parallel to CD

Divide DC and

DF each into

the points of

C

Fig. 454.

division in DF draw lines to B. Through the points of division in DC draw lines parallel to BC, and through the points of intersection of the corresponding lines draw a curve, and complete the other half in the same manner; then will DBE be the complete curve of the parabola. The less BC is in proportion to CD, the nearer the curve will approach to the arc of a circle, as in fig. 422.; and hence we may describe the shaft of a column, or draw a flat segment of a circle. 1105. PROB. III. The same parts being given, to describe the parabola by the intersection of straight lines.

2 3 C
Fig. 453.

the curve for diminishing

Produce CB to F (fig. 455.), and make BF equal to BC. Join FD and FE. DF and FE in the same proportion, or into the same number of equal parts. Let the divisions be numbered from D to F, and from F to E, and join every two corresponding points by a straight line; then the intersection of all the straight lines will form the parabola required.

1106. PROB. IV. To draw a straight line from a given point in the curve of a parabola, which shall be a tangent to the curve at that point.

BC in f: produce cB to g, and make Bg equal to Bf, and join ge, then will ge be the tangent required. In the same manner DH will be found to be a tangent at D. be drawn perpendicular to the tangent ge, then will eK be also perpendicular to the curve, and in the proper direction for a joint in the masonry of a parabolic arch.

1107. The uses of the parabolic curve in architecture are many. The theorists say that it is the curve of equilibrium for an arch which has to sustain a load uniformly diffused over its length, and that therefore it should be included in the depth of lintels and flat arches; and that it is nearly the best form for suspension and other bridges, and for roofs. It is also considered the best form for beams of equal strength. It may be here also remarked, that is is the curve described by a projectile, and that it is the form in which a jet of water is delivered from an orifice made in the side of a reservoir. So is it the best curve for the reflection of light to be thrown to a distance. In construction it occurs in the intersection of conic surfaces by planes parallel to the side of the cone, and is a form of great beauty for the profiles of mouldings, in which manner it was much used in Grecian buildings.

GENERAL METHOD OF DETERMINING AND DESCRIBING THE SPECIES OF CONIC SECTIONS.

1108. In a conic section, let there be given the abscissa AB (fig. 457.), an ordinate BC. and a tangent CD to the curve at the extremity of the ordinate to determine the species of the conic section, and to describe the figure.

Draw AD parallel to BC, and join AC (Nos. 1. and 2.). Bisect AC in E, and produce DE and AB, so as to meet in F when DE is not parallel to AB; then in the case where DE will meet AB or AB produced in F, the point F will be the centre of an ellipsis or hyperbola. In this case produce AF to G, and make FG equal to FA; then if the ordinate BC and the centre be upon the same side of the apex A, the curve to which the given parts belong is an ellipsis; but if they be on different sides of it, the curve is an hyperbola. When the line DE (No. 3.) is parallel to AB, the figure is a parabola.

C

No.1

G

D

No-1 A
D

No. 2

D

No 3

Fig. 458.

D

1109. In a conic section, the abscissa AB (fig. 458.), an ordinate BC, and a point D in the curve being given, to determine the species of the curve, and thence to describe it. Draw CG parallel to AB (Nos. 1. and 2.), and AG parallel to BC. Join AD, and produce it to meet CG in e. Divide the ordinate CB in f in the same proportion as CG is divided, then will Cf: fB:: Ce: eG. Join Df, and produce it or fD to meet AB or BA in h; then if the points D and h fall upon opposite sides of the ordinate BC, the curve is an ellipsis; but if D and h fall upon the same side of the ordinate BC, the curve will be an hyperbola. If Df (No. 3.) be parallel to AB, the curve will be a parabola. In the case of the ellipsis and hyperbola, Ah is a diameter; and therefore we have a diameter and ordinate to describe the curve.

SECT. V.

DESCRIPTIVE GEOMETRY.

1110. The term Descriptive Geometry, first used by Monge and other French geometers to express that part of the science of geometry which consists in the application of geometrical rules to the representation of the figures and the various relations of the forms of bodies, according to certain conventional methods, differs from common perspective by the design or representation being so made that the exact distance between the different points of the body represented can always be found; and thus the mathematical relations arising from its form and position may be deduced from the representation. Among the English writers on practical architecture, it has usually received the name of projection, from the circumstance of the different points and lines of the body being projected on the plane of representation; for, in descriptive geometry, points in space are represented by their orthographical projection on two planes at right angles to each other, called the planes of projec tion, one of which planes is usually supposed to be horizontal, in which case the other is vertical, the projections being called horizontal or vertical, according as they are on one or other of these planes.

1111. In this system, a point in space is represented by drawing a perpendicular from it to each of the planes of projection; the point whereon the perpendicular falls is the