NOTE.

THE two following series of propositions concerning the Ovals belong to this period, and afford an illustration of Maxwell’s mode of working in those early days. See above, p. 87. The only alterations made, with the exception of corrections of obvious slips of the pen, consist in a sparing insertion of stops. The MS., as will be seen by the facsimile at p. 104, has almost no punctuation.

I. Oval.

Definition 1.—If a point move in such a manner that m times its distance from one point, together with n times its distance from another point, may be equal to a constant quantity, it will describe a curve called an Oval.

A geometric diagram labeled 3.2 showing an oval shape with points A, B, and C connected by lines forming a triangle inside it, alongside a separate straight vertical line marked F at its base.

Definition 2.—The two points are called the foci, and the numbers signified by m and n are called the powers of the foci.

Definition 3.—The line joining the foci is called the axis.

Proposition 1—Problem.

To describe an oval with given foci given multiples, and given constant quantity

Let normal upper A and normal upper B be the given foci, 3 and 2 the multiples, and upper E upper F the constant quantity, it is required to describe an oval. At normal upper A and normal upper B erect two infinitely small cylinders. Take a perfectly flexible and inextensible thread, without breadth or thickness, equal to upper E upper F; wind it round the focal cylinders and another movable cylinder normal upper C, so that the number of plies between normal upper A and normal upper C may be equal to m, that is 3, and the number between normal upper B and normal upper C equal to n or 2. Now move normal upper C in such a manner that the thread may be quite tight, and an oval will be described by the point.

For take any point normal upper C. There are m plies of thread between normal upper A and normal upper C, and n plies between normal upper B and normal upper C, which taken together make up the thread or the constant quantity, therefore m upper A upper C plus n upper B upper C equals upper E upper F.

The focus normal upper A, which has the greatest number of plies is called the greater focus, and normal upper B is called the less focus.

Proposition 2—Theorem.

The greater focus is always within the oval, but the less is within, on, or without the curve, according as the distance between the foci, multiplied by the power of the greater focus, is less equal or greater than, the constant quantity.

A series of three egg/oval shapes, each bisected by a vertical line from a top point A to a bottom point, with a small circular diagram labeled x and r to the left and a horizontal baseline connecting all the shapes at bottom.

The greater focus is always within the Oval, for suppose it to be at normal upper A, Fig. 1, then m upper A upper C plus n upper B upper C equals constant quantity equals m upper A upper D plus m upper D upper C plus n upper B upper C, and m upper A upper D plus n upper D upper B equals constant quantity equals m upper A upper D plus n upper D upper C plus n upper B upper C equals m upper A upper D plus m upper D upper C plus n upper B upper C, and n upper D upper C equals m upper D upper C, and m equals n, but m greater than n.

The less focus normal upper B is within the oval when m upper A upper B less than upper E upper F the constant quantity.

For it is evident that upper B upper C left parenthesis Fig period 2 right parenthesis equals StartFraction upper E upper F minus m upper A upper B Over m plus n EndFraction

It is in the curve when m upper A upper B equals upper E upper F, this is evident.

It is without the curve when m upper A upper B greater than upper E upper F, for upper A upper C equals StartFraction upper E upper F minus n upper A upper B Over m minus n period EndFraction

Proposition 3—Theorem.

If a circle be described with a focus for a center, and the constant quantity divided by the power of that focus for a radius, the distance of any point of the oval from the other focus is to the distance from the circle as the power of the central focus is to the power of the other.

The circle upper E upper H upper P is described with the centre normal upper B and radius = constant quantity, divided by the power of normal upper B. At any point normal upper C comma upper A upper C colon upper C upper H colon colon power of normal upper B: power of normal upper A.

A two-part geometric diagram showing two large circles, each containing a smaller oval-shaped curve with labeled points and intersecting lines radiating from interior points.

Let m equals power of normal upper A, and n equals power of normal upper B comma n upper B upper H equals constant quantity equals n upper B upper C plus n upper C upper H, and n upper B upper C plus m upper A upper C equals constant quantity equals n upper B upper C plus n upper C upper H, take away n upper B upper C, and n upper H upper C equals m upper A upper C, therefore upper H upper C colon upper C upper A colon colon m colon n. QED.

A geometric diagram showing two concentric circles labeled P (outer) and a smaller inner circle, with points A, B, C, and N connected by lines forming a small triangle near the top and a line extending to the circle's edge.

Cor. 1.—When the powers of the foci are equal the curve is an ellipse.

Cor. 2.—When the less focus is at an infinite distance the curve is an ellipse, for the circle becomes a straight line; and

Cor. 3.—When the greater focus is at an infinite distance, the curve is an hyperbola for the same reason.

Proposition 4—Theorem.

A geometric diagram showing two concentric circles with a smaller star-like or triangular shape formed by points C, B, D, L, N, and A connected by intersecting lines near the center.

When the less focus is in the curve, angle will be formed equal to the vertical angle of an isosceles triangle, of which the side is to the perpendicular on the base, the power of the greater focus to that of the less.

For let a circle be described, as in Prop. 3, it is evident that it will pass through normal upper B. Take indefinitely small arcs upper C upper B equals upper B upper D, join upper C upper A and upper D upper A, join upper E upper H. upper E upper C colon upper E upper B equals power of normal upper B colon power of normal upper A, and upper E upper C equals upper B upper O, therefore upper E upper B colon upper B upper O, power of normal upper A colon power of normal upper B.

Proposition 5—Problem.

A point normal upper A, and a point normal upper B in the line upper B upper C being given, to find a point in the line as normal upper D, so that m upper A upper D plus n upper B upper D may be a minimum.

Take a line upper H upper P, raise upper H upper X perpendicular, and from normal upper X as a centre describe a circle with a radius equals StartFraction m upper X upper H Over n EndFraction so that m colon n colon colon upper X upper P colon upper X upper H.

A geometric diagram showing a horizontal line with multiple labeled points (B, S, R, D, E, C) converging via straight lines to an apex point A, alongside a right-angle construction with points b, B, and P forming a small rectangle on the right.

Draw upper A upper C perpendicular to upper B upper C, and make an angle upper C upper A upper D = upper H upper P upper X. normal upper D is the required point. If not, take any point normal upper E on the opposite side of normal upper D from normal upper B, make upper A upper T equals upper A upper E. Join upper T upper E, and make upper D upper T upper L = right angle, then upper D upper T upper E greater than right angle upper D upper T upper L, because upper A upper T upper E is less than a right angle (1.16), therefore normal upper L is within normal upper E. And in the triangle upper D upper T upper L comma upper D upper T upper L equals upper P upper H upper X and upper T upper D upper L equals upper H upper X upper P, therefore the triangles are similar, and upper T upper D: upper D upper L equals upper X upper H colon upper X upper P right parenthesis equals n: m semicolon and n upper D upper L equals m upper D upper T, then n upper D upper E greater than m upper D upper T, add n upper B upper D and m upper T upper A or m upper E upper A, then n upper B upper D plus n upper D upper E plus m upper E upper A greater than n upper B upper D plus m upper D upper T plus m upper T upper A and n upper B upper D plus m upper D upper A less than n upper B upper E plus m upper A upper E.

Now take a point normal upper R on the other side. Join upper A upper R, and cut off upper A upper O equals upper A upper D, make upper S upper R upper D equals upper H upper X upper P equals upper A upper D upper C. normal upper S will be beyond upper A upper R. Draw upper D upper S perpendicular to upper A upper D, then it will be perpendicular to upper R upper S, and below the line upper O upper D, then upper R upper S upper D similar to upper X upper H upper P and upper R upper S: upper R upper D equals left parenthesis upper X upper H colon upper X upper P equals right parenthesis n: m, and m upper R upper S equals n upper R upper D, but upper S upper R less than upper R upper V less than upper R upper O and n upper R upper D left parenthesis equals m upper R upper S right parenthesis less than m upper R upper O, add n upper B upper R and m upper O upper A or m upper D upper A, and n upper B upper R plus n upper R upper D plus m upper D upper A less than n upper B upper R plus m upper R upper O plus m upper O upper A and n upper B upper D plus m upper D upper A less than n upper B upper R plus m upper R upper A. QED.

Proposition 6—Problem.

A geometric diagram showing an elongated oval or vesica shape formed by overlapping curved arcs, with a vertical axis connecting points A at top and B at bottom, and a diagonal line passing through a point labeled C or O on the left side.

To draw a tangent to an oval from a focus without:—

Take m for the power of the greater focus, and n for that of the less, and find the angle upper A upper D upper B (Prop. 5). Upon upper A upper B describe a segment of a circle containing an equal angle. Join normal upper B and normal upper C, the point where it cuts the oval, normal upper B normal upper C is a tangent. For take any point normal upper O, join upper A upper O, m upper A upper O plus n upper O upper B greater than m upper A upper C plus n upper C upper B, therefore normal upper O is without the oval.

A complex geometric diagram with two large overlapping circles, each containing smaller intersecting circles and ovals, with numerous labeled points (D, E, C, A, F, B, K, T, X) connected by an intricate web of lines.

Proposition 7—Problem.

To draw a tangent to a given Oval, the foci and the ratio being given:—

It is required to draw a tangent at normal upper C to the oval upper C upper N upper T, ratio m colon n. From normal upper B the less focus describe a circle as in (Prop. 3). Join upper C upper A, join upper C upper B, and produce to normal upper D, then upper D upper C colon upper C upper A colon colon m colon n. Join upper A upper D and produce it. Bisect the angle upper D upper C upper A by the line upper C upper O. upper D upper O colon upper O upper A colon colon left parenthesis upper D upper C colon upper C upper A colon colon right parenthesis m colon n. Draw upper C upper K perpendicular to upper C upper O, and describe the circle upper O upper C upper L upper K. Because upper O upper C upper K is a right angle, upper O upper C upper K equals upper D upper C upper O plus upper K upper C upper B equals upper O upper C upper A plus upper A upper C upper K. But upper O upper C upper A equals upper D upper C upper O therefore upper K upper C upper B equals upper A upper C upper K and (6.3) upper K upper A colon upper K upper D colon colon upper C upper A colon upper C upper D colon colon upper A upper O colon upper O upper D therefore upper K upper A colon upper K upper D colon colon upper A upper O colon upper O upper D, and at any point normal upper L in the circle, upper A upper L colon upper D upper L colon colon upper A upper O colon upper O upper D left parenthesis 6 normal upper F right parenthesis, and upper A upper L colon upper D upper L colon colon n colon m; and the circle is wholly without the oval, and can only touch it in the two points normal upper C and normal upper P, where upper D upper B cuts the oval; for suppose the circle coincided with the oval at normal upper L, StartLayout 1st Row Join upper B upper L and produce to normal upper X comma then upper A upper L colon upper D upper L colon colon n colon m 2nd Row And by Prop period 3 upper A upper L colon upper X upper L colon colon n colon m EndLayout right brace therefore upper D upper L equals upper X upper L semicolon but upper D upper L greater than upper X upper L left parenthesis 3.7 right parenthesis, therefore the oval is within the circle. Therefore draw upper E upper C upper F a tangent to the circle, and as it is without the circle it is without the oval.

A simple geometric diagram showing an oval/teardrop shape with a point inside connected by multiple lines to two points labeled with letters, forming a fan-like arrangement of triangles.

Scholium.—Join upper A upper C, join upper B upper C, and produce till upper A upper C colon upper C upper D colon colon n colon m. Join upper A upper D, bisect upper A upper C upper D by upper C upper O. Draw upper O upper E perpendicular to upper A upper O. Make upper O upper C upper E equals upper E upper O upper C. It is evident from the proposition that upper E upper C upper F is the tangent.

A geometric diagram showing two large circles side by side, each containing smaller intersecting curves and labeled points connected by a web of lines, with a handwritten label Oval and S Curves near the center.

If a line upper A upper E be cut in normal upper C and normal upper B, so that upper A upper B colon upper B upper C colon colon upper A upper E colon upper C upper E, and two semicircles upper B upper O upper E, upper B upper X upper E be described on upper B upper E, and upper A upper O, upper C upper O be drawn to normal upper O in the circumference, and perpendiculars upper D upper H, upper D upper L, be drawn from the centre, upper D upper H colon upper D upper L colon colon upper A upper B colon upper B upper C.

For upper A upper B colon upper B upper C colon colon upper A upper E colon upper C upper E therefore upper A upper B colon upper A upper E colon colon upper B upper C colon upper C upper E therefore upper A upper E plus upper A upper B right parenthesis equals left parenthesis 2 upper A upper B plus 2 upper B upper D right parenthesis equals 2 upper A upper D colon upper A upper B colon colon left parenthesis upper C upper E plus upper B upper C equals right parenthesis 2 upper B upper D colon upper B upper C and upper A upper D colon upper A upper B colon colon upper B upper D colon upper B upper C therefore upper A upper D minus upper A upper B equals right parenthesis upper B upper D colon upper A upper B colon colon left parenthesis upper B upper D minus upper B upper C equals right parenthesis upper C upper D colon upper B upper C therefore upper B upper D colon upper C upper D colon colon upper A upper B colon upper B upper C.

Draw upper D upper T perpendicular to upper O upper B, then as it bisects the base it bisects the angle upper O upper D upper B and upper O upper D upper T equals upper B upper D upper T, then in the triangles upper O upper T upper P, upper D upper P upper L comma upper O upper T upper P equals upper D upper L upper P and upper O upper P upper T equals upper D upper P upper L therefore upper P upper D upper L equals upper P upper O upper T. But upper P upper O upper T equals upper B upper O upper A (6 F Cor.) and upper B upper O upper A equals upper S upper O upper H equals upper P upper D upper L, and in the triangles upper S upper O upper H, upper S upper T upper D comma upper S upper H upper O equals upper S upper T upper D, and upper O upper S upper H equals upper T upper S upper D therefore upper S upper O upper H equals upper S upper D upper T and upper S upper D upper T equals upper P upper D upper L. But upper O upper D upper T equals upper B upper D upper T therefore upper C upper D upper L equals upper O upper D upper H and upper D upper H upper O equals upper D upper L upper C therefore upper H upper O upper D comma upper C upper D upper L are equiangular, and upper H upper D colon upper D upper L colon colon left parenthesis upper D upper O equals upper B upper D colon upper C upper D, but upper B upper D colon upper C upper D colon colon upper A upper B colon upper B upper C therefore upper H upper D colon upper D upper L colon colon upper A upper B colon upper B upper C.

Cor. Sine upper H upper O upper D colon sine upper D upper O upper L colon colon upper A upper B colon upper B upper C.

A geometric diagram showing a large circle containing a smaller oval curve with multiple labeled points connected by intersecting lines radiating from a central point.

Proposition 9—Theorem.

If lines be drawn from the foci to any point in the oval, the sines of the angles which they make with the perpendicular to the tangent are to one another as the powers of the foci.

sine upper D upper C upper E colon sine upper A upper C upper E colon colon power of normal upper A colon power of normal upper B.

For describe the circle upper C upper X upper T as in Prop. 7, so that upper D upper T colon upper T upper A colon colon upper D upper X colon upper A upper X, then upper S upper C the tangent to the circle is a tangent to the oval, and upper C upper E the radius is perpendicular to it; then by Prop. 8 Corollary, sine upper D upper C upper E colon sine upper A upper C upper E colon colon left parenthesis upper D upper T colon upper T upper A colon colon right parenthesis power of normal upper A colon power of normal upper B.

A series of four labeled geometric diagrams showing curved lens-like or oval shapes formed by converging lines from points to a central curve, with one figure including horizontal parallel lines suggesting light rays.

As the ratio of the sines of incidence and refraction is invariable for the same medium, and as this ratio is as 2 to 3 in glass; and as in the oval at Fig. 1 the powers of the foci are as 2 to 3; if the oval were made of glass, rays of light from normal upper B would be refracted to normal upper A. If now a circle be described from normal upper A, the case will not be altered, for the rays are perpendicular to the circle as in fig. 2.

The ellipse and hyperbola, figs. 3 and 3 [bis], cause parallel rays to converge, for (Prop. 3) they are ovals with one of the foci at an infinite distance.

Fig. 4 is a combination of 2 hyperbolas, and rays from normal upper A are refracted to normal upper B.

A geometric diagram showing two concentric circles with a point inside connected by lines to a labeled point A outside the circles, accompanied by the inequality notation m > n.

II. Meloid and Apioid.

If a point normal upper C move so that m upper C upper A tilde n upper C upper B equals normal a constant quantity, it will describe a curve called a meloid when m upper C upper A greater than n upper C upper B, and an apioid when n upper C upper B greater than m upper C upper A. The points normal upper A and normal upper B are called the foci.

Proposition 1—Problem.

To describe a Meloid, Fig. 1, or Apioid, Fig. 2.

Take a rigid straight rod upper A upper D, centered at normal upper A. At normal upper D erect a small cylinder upon the rod, and also another at normal upper B on the paper. Then take a flexible and inextensible thread of such a length that m upper A upper D tilde n times the thread = constant quantity; then wrap the thread round normal upper B and normal upper D and a movable cylinder normal upper C, so that there may be n plies between normal upper B and normal upper C, and m plies between normal upper C and normal upper D; if now upper A upper D be turned round and the thread kept tight, the point normal upper C will describe a meloid or apioid.