Differential geometry can be used in the design of springs
Differential geometry is the investigation and characterization of the local properties of curves and surfaces in space; that is, in the neighbourhood of a point on the curve or surface. It can be extended to the properties of space itself, but we shall not go that far. We'll study curves in Euclidean three-dimensional space, which are characterized by curvature and torsion, and illustrate the less familiar property, torsion, by appeal to helical springs. Then, we shall investigate the curvature of surfaces. All these things will be done by relating these properties to the derivatives of the parametric equations specifying the curves or surfaces.
Another article treats curves in a plane, which can be described by introducing a rectangular coordinate system and using the parametric equations x = x(t) and y = y(t). First, let's review the properties of a plane curve. A tangent vector to the curve is (dx,dy) = (x'dt,y'dt). Here, we represent a vector by its components within parentheses, separated by commas. Later, we shall also use the same notation for the scalar product of the two vectors between the parentheses. The element of arc length ds = √(x'2 + y'2)dt. This can be integrated, at least in principle, to give s as a function of t. Then, we can express the curve with arc length as parameter: x = x(s), y = y(s). This so greatly simplifies the algebra that we shall assume the curve is specified in this way in what follows. The vector a = dr/ds = (x',y') is a unit vector, since x'2 + y'2 = (dx/ds)2 + (dy/ds)2 = (ds/ds)2 = 1, and is called the tangent unit vector. The change in a with a movement ds along the curve is (x"ds,y"ds). Since the change in a unit vector is perpendicular to the vector and corresponds to a rotation of dθ, dθ/ds = √(x"2 + y"2). The unit vector b = r"/|r"| is perpendicular to a, and is called the normal vector. Indeed, (x',y')·(x",y") = x'x" + y'y" = (1/2)(d/ds)(x'2 + y'2) = (1/2)(d/ds)(1) = 0. We can write the derivative dθ/ds = |b| = dθ/Rdθ) = 1/R. The length R defined in this equation is the radius of curvature, and its reciprocal, 1/R is the curvature itself. The normal vector b points toward the centre of curvature. For a straight line, x" = y" = 0, so that R = ∞ and the curvature is zero, as expected.
As a concrete example, consider a circle of radius b and centre at the origin. This circle can be described by x = b cos 2πt, y = b sin 2πt, where b is the radius and the parameter t is the number of revolutions around the circle. Then, dx/dt = -2πb sin 2πt, dy/dt = 2πb cos 2πt, so (dx/dt)2 + (dy/dt)2 = (2πb)2 and ds = 2πb dt, or s = 2πbt, which is no surprise. Now we can describe the circle by x = b cos(s/b), y = b sin(s/b). Clearly, x'2 + y'2 = 1, and the tangent unit vector is [-sin(s/b),cos(s/b)], which is clearly a unit vector and perpendicular to the radius vector. The second derivatives are x" = (-1/b)cos(s/b), y" = (-1/b)sin(s/b). The length of this vector is 1/b, and b = [-cos(s/b),-sin(s/b)], which is directed towards the origin. 1/b = 1/R, so the radius of curvature is b, again no surprise.
Let us now go into curves in three dimensions. We now have three parametric equations, x = x(t), y = y(t), z = z(t). The arc length ds = √[(dx/dt)2 + (dy/dt)2 + (dz/dt)2]dt. If this can be integrated, we can express the curve with arc length as parameter, x = x(s), y = y(s), z = z(s). Then, a = dr/ds = (x',y',z') is the tangent vector. The plane perpendicular to this vector is called the normal plane. The curve is represented by a point in this plane, and passes through it perpendicularly. The vector (x",y",z") = r" is perpendicular to the tangential unit vector, which can easily be seen by dotting it into r'. A tangent plane is any plane containing the tangent. The tangent plane that contains r" is locally equivalent to the plane of a plane curve, and is called the osculating plane. In the neighborhood of the point under consideration, the curve lies approximately in this plane. The unit vector b = r"/|r"| is called the normal unit vector, and lies in the osculating plane. The unit vector c = a x b is perpendicular to the osculating plane, and lies in the normal plane. It is called the binormal. The plane perpendicular to the normal unit vector is called the rectifying plane, and c lies in this plane.
We now have an orthonormal triad of unit vectors, the tangent, normal and binormal, and three mutually perpendicular planes, the osculating plane, the rectifying plane and the normal plane, at every point of the curve. The curvature 1/R can be defined exactly as for plane curves, 1/R = |r"|, so that it reduces to the desired form for plane curves. For space curves, we have the possibility that the osculating plane varies in orientation. We define the torsion as the change in angle of the osculating plane with movement along the curve, so that 1/T = dθ/ds, where dθ is the change in direction of c in a distance ds along the curve, or 1/T = |c'|. The torsion is the reciprocal of the length T, which does not have a vivid geometrical interpretation. It is, however, dimensionally correct. At any point, the curve is penetrating the osculating plane, and the sign of T gives the direction of this penetration, whether in the direction of c or opposite to it.
It's very instructive to consider a concrete example. Probably the most useful and commmon space curve is the helix. This curve is found in screw threads, helical springs, and corkscrews. In fact, when I looked for a physical example the first helix I hit upon was a corkscrew. To describe a helix, all we need is to add the z-equation z = at to the parametric equations of a circle. The constant a is the pitch of the helix, the amount it advances per turn. With our preceding equations, this gives a right-handed helix that rotates clockwise as t increases, as seen from the starting point. If you look at the helix from the other direction, the rotation is also clockwise, so right-handedness is an intrinsic property of this helix. By a slight change in the y equation, we can obtain a left-handed helix, which cannot be superimposed on the right-handed helix. A screw with right-handed helical threads advances into a piece threaded with a right-handed helix when it is rotated clockwise. A left-handed screw will not fit into a right-handed nut whichever way it is rotated, but advances into a left-handed nut when rotated anticlockwise. The turnbuckle of a screen-door tension brace has a right-handed screw on one end, and a left-handed screw on the other.
Examine your helix, and picture how the osculating plane changes as you go along the helix. This can be done in imagination or with a sketch, but is easier to see with an actual helix. Its inclination to the z-axis remains constant, but the binormal rotates about the z-axis, making one turn for each revolution. Imagine one osculating plane, and look at how the curve passes through it. A helix can be unwound to make a right triangle, with legs a and 2πb. From this triangle, the inclination of the osculating planes is θ = tan-1(a/2πb). The arc length of one turn of the helix is √(a2 + 4π2b2).
In terms of arc length, the helix is x = b cos(2πs/p), y = b sin(2πs/p), z = as/p, where p = √(a2 + 4π2b2) is the length of one turn of the helix. Then, x' = -(2πb/p) sin(2πs/p), y' = (2πb/p) cos(2πs/p), z' = a/p. It is easy to verify that x'2 + y'2 + z'2 = 1, so that the unit tangential vector is a = (x',y',z'). Then, x" = -(2π/p)2b cos(2πs/p), y" = -(2π/p)2b sin(2πs/p), z" = 0. Therefore, 1/R = (2π/p)2b. Inserting the value of p, we have 1/R = 4π2b/[a2 + (2πb)2]. If a → 0, 1/R → 1/b, the expected result.
The torsion can be obtained from the definition. In fact, 1/T = 2πsin(θ)/p, where θ is the inclination of the osculating plane and p the arc length of one turn. This gives 1/T = 2πa/[a2 + (2πb)2]. As a → 0, 1/T → = 0, as expected. T is taken as positive for a right-handed helix, and negative for a left-handed helix. A helix is characterized by a constant torsion.
The changes in the reference unit vectors with displacement along the curve can be expressed in terms of R and T. The definitions of these quantities give da/ds = b/R and dc/ds = -b/T. Since c = a x b, we can also find the change in b in terms of these two derivatives. In fact, db/ds = -a/R + c/T. These three formulas are the Frenet-Serret formulas. They show that the parameters R and T are sufficient to describe the local behavior of a curve in space, which is an important result. Note that if 1/T is zero, then c is constant, and the curve is a plane curve. Torsion is an essential accompaniment of a space curve. If 1/R is zero, the curve is a straight line, and 1/T has no significance.
In general the torsion is given by 1/T = r'·r" x r'''/|r"|. Note that the third derivative enters in this expression, and the sign of T has been specified. If the result is worked out for a helix, it will be found to agree with our previous choice. We note that the torsion of a helix is a constant, which makes working with it very easy. Derivations of these formulas will be found in Widder.
This definition of torsion agrees with the usual definition in strength of materials as twist per unit length, as in the design of helical springs. A helical spring is in the form of a helix with a certain torsion. When the spring is extended or compressed, the torsion changes, and the change in torsion times the total length of the spring wire is the torsional deflection. Consider a long circular rod of diameter d and length L. If a torque Q is applied at one end of the rod and the other is held fixed, the end of the rod at which Q is applied will rotate by an amount θ = QL/GJ, where G is the modulus of rigidity of the material (in psi, for example) and J is the polar moment of inertia of area, πd4/32 for a cylindrical rod. For steel, G = 11.5 x 106 psi. The same rotation occurs in a spring where the rod is wound into a helix of radius b, and the axial force F is applied at the centre of the spring, so that Q = Fb. The helical form is much easier to handle than the straight rod, and the deflection can be larger without exceeding the elastic limit.
The torsion of the helix is 1/T = 2πa/p, where p is the length of rod for one turn of the helix. If there are N turns, and the pitch a changes by da, then the total change in length of the spring will be δ = Nda. The change in the torsion will be d(1/T) = (2π/p2)da, so that θ = d(1/T)(Np) = (2π/p)δ. We are assuming that the material of the spring is fixed to the osculating plane. When we equate the two expressions for θ, we find (2π/p)δ = FbpN/GJ, or δ = Fbp2N/GJ. If the spring is relatively "flat" so that p = 2πb, we get δ = 8F(2b)2N/Gd4, which can be compared with the result obtained in machine design texts. The spring constant k = F/δ = Gd4/8(2b)3N. The ratio of the helix diameter to the rod diameter, 2b/d, is called the spring index. The solid height of a spring is the height when compressed so that all turns touch. The free length is the length when under no load. The ends of springs are prepared in different ways to suit the application.
Now let's talk about surfaces in space. We are interested in finding the normal vector and tangent plane to the curve at a point on the surface, the length of curves drawn on the surface, and the curvature of the surface. All this can be done fairly easily with differential geometry. First, it is necessary to specify the surface. Two useful ways are, first, to define the surface implicitly by an equation f(x,y,z) = 0, and, second, to specify the surface parametrically by three equations r = r(u,v). The two parameters u,v are a two-dimensional coordinate system on the surface. For example, they may be the latitude and longitude of a point r on the surface of a sphere. An equation z = f(x,y) is easily expressed in the first form, f(x,y) - z = 0, and is a special case of the parametric representation r = [x, y, f(x,y)]. We shall assume that all functions have continuous second derivatives, and are single-valued.
Using the parametric representation, we can find two tangent vectors by differentiation. These are eu = ∂r/∂u and ev = ∂r/∂v. For notational ease, we shall denote these by xu and xv, and the radius vector simply by x. These are the directions on the surface corresponding to changes du and dv in the parameters. We assume that these two directions are not the same, as they certainly would not be in any useful parametrization. Then, the local normal to the surface is in the direction of xu x uv. The unit normal vector is ζ. It satisfies (ζ,uu) = 0 and (ζ,uv) = 0. The tangent plane to the surface is (x - x0)·ζ, and a normal line is x = x0 +tζ. These expressions all contain vectors, of course.
If we use f(x,y,z) = 0 and do not single out any coordinate for special treatment, we can find the normal by differentiating this expression, to find fxdx + fydy + fzdz = 0. This restricts the vector (dx,dy,dz) to lie in the tangent plane. Therefore, the vector grad f = (fx,fy,fz) is normal to the surface. In fact, ζ = grad f/|grad f|.
For further investigation, we shall use x = x(u,v) [a vector!]. Then, as we have seen, ζ = xu x xv. There is an arbitrary sign, whose effect we shall ignore. A differential vector in the tangent plane is then dx = xudu + xvdv. The square of distance in the tangent plane, ds2 is the length of this vector, or ds2 = (xu,xu)du2 + 2(xu,xv)dudv + (xv,xv)dv2. We have used (xu,xv) = (xv,xu). The expressions in parentheses are, of course, scalar products. This may be expressed more concisely as ds2 = Edu2 + 2Fdudv + Gdv2. This equation, called the metric form for the sphere, expresses length on the surface as a function of du and dv. The angle between two curves on the surface can be expressed in terms of these quantities in a simlar way, by expanding (dx1,dx2). If one curve is v = h(u) and the other is v = k(u), the result is cos θ = (E + Fh' + Fk' + Gh'k')/[√(E + 2Fh' + Gh'2)√(E + 2Fk' + gk'2)].
To study the curvature of the surface, we consider the curvature of normal sections of the surface. Normal sections are the intersection of the surface with planes that contain the surface normal. We must prepare ourselves to accept that the curvature will depend on the direction of the normal plane, and that the curvature of the surface cannot be expressed in terms of a single parameter. To find the curvature, we must consider changes in the direction of the normal, dζ = ζudu + ζvdv. Let us form -(dx,dζ) = -(xu,ζ u)du2 + (xu,ζv)dudv + (xv,xu)dvdu + (xv,xv)dv2. The motivation for considering this scalar product will be seen in the sequel. The scalar products can be expressed in terms of ζ by differentiating the conditions that ζ is normal to xu and xv. For example, (ζu,xu) + (ζ,xuu) = 0. This introduces the second derivatives that are essential for expressing curvature. We find that (dx,dζ) = edu2 + 2fdudv + gdv2, where e = (ζ,xuu), f = (ζ,xuv), g = (ζ,xuu). These are just the components of the second derivative vectors in the direction of the normal.
Suppose the tangent vector for a certain curve on the surface is α = dx/ds, where ds is arc length along the curve. Since (α,ζ) = 0, we find by differentiation that (dα/ds,ζ) + (α,dζ/ds) = 0. The derivative of α is given by the Frenet-Serret formulas as dα/ds = β/r = ±ζ/r. Since ζ is a unit vector, we then find that 1/r = ±(dx/ds,dζ/ds) = ±[(dx,dζ)/ds2. We have already found expressions for the scalar products in this expression in terms of du and dv. In fact, 1/r = [edu2 + 2fdudv + gdv2] / [Edu2 + 2Fdudv + Gdv2]. We can also specify a certain direction with the ratio h' = dv/du. The maximum and minimum values of 1/r are called the principal curvatures. Their product is the Gaussian curvature, and their average is the mean curvature.
As a concrete example, consider a sphere of radius a, where x = a sin u cos v, y = a sin u sin v, z = a cos u. Here, u and v are the polar angles. In this case, it is easy to see that the normal vector is (sin u cos v, sin u sin v, cos u). The first derivatives are xu = (a cos u cos v, a cos u sin v, -a sin u) and xv = (-a sin u sin v, a sin u cos v, 0). If we had not realized that the normal could be obtained easily, it could be found from the cross product of these tangent vectors. This gives E = a2, F = 0, G = a2 sin2u, so ds2 = a2du2 + a2sin2u dv2. We can see by inspection that this result is valid, considering displacements along a meridian and along a circle of latitude: ds2 = (adu)2 + (a sin u dv)2. If u and v are orthogonal coordinates, then F = 0.
The second derivatives are xuu = (-a sin u cos v, -a sin u sin v, -a cos u), xuv = (-a cos u sin v, a cos u cos v, 0), xvv = (-a sin u cos v, -a sin u sin v, 0). Now we find e = a, f = 0, g = a sin2 u. The curvature is 1/r = (1/a)[(1 + sin2u h'2)/(1 + sin2u h'2)] = 1/a, independent of direction. This is, of course, expected, and verifies our equations.
The reader may wish to consider a cylinder of radius a. Take u as the polar angle, and v as the vertical coordinate, so that x = a cos u, y = a sin u, z = v. Then, E = a2, F = 0, G = 1, e = a, f = g = 0. The curvature is 1/r = a/(a2 + h'2). The maximum curvature occurs for h' = 0, 1/r = 1/a, and the minimum for h' = ∞, where 1/r = 0. The Gaussian curvature is zero, while the mean curvature is 1/2a.
Since the metric form is ds2 = a2dφ2 + a2cos2φdθ2 for a sphere where we have now taken φ = latitude and θ = longitude, the ratio of the north-south distance dy to an east-west distance dx in a small displacement is dy/dx = dφ/cos φ dθ. This is the slope of the displacement, or the tangent of its azimuth from north. Whenever F = 0, the coordinates are rectangular and we can form this ratio simply by looking at the metric form. Let's now consider a cylinder wrapped around the sphere, where z replaces the latitude φ, and ds2 = a2dθ2 + dz2. Here, dy/dx = dz/adθ.
Now suppose we map the sphere on the cylinder, so that the point φ,θ on the sphere is represented by the point z,θ on the cylinder. This can be done by a arbitrary z = z(φ), but a very useful mapping results when angles are preserved. That is, two displacements that make an angle w on the sphere are to make the same angle when mapped onto the cylinder. This will be true when the angle between the meridian and a displacement is preserved, since we only have to refer the displacements to the meridian. This means that the slopes of the displacements are to be the same. We know the slopes, so we have the requirement that dφ/cos φ dθ = dz/adθ, or dz/dφ = a/cos φ. This integrates to z = a ln tan(φ/2 + π/4) + C. If z = 0 when φ = 0, then C = 0. This is the desired relation between z and φ, which gives a Mercator map. The scale is determined by the factor a. Distance representing longitude is given by x = a θ.
A straight line on a Mercator map is called a loxodrome, Greek for "slanting course." It is a course of constant compass heading, very convenient for navigation. It represents the shortest distance only when approximately north-south or east-west, since the scale of a Mercator map is not constant, but changes with latitude. Any long course is divided into a series of loxodromes. A Mercator map is not good for comparing regions at different distances from the equator, but always gives an accurate appreciation of the shape of limited areas. It is not applicable to polar regions. It is an interesting exercise to compare loxodromes on a Mercator map with courses on a globe with the same end points. No good way to represent the whole globe on a plane map has yet been found. Look at the various attempts displayed in atlases; the best tear the surface of the globe into disconnected regions.
D. V. Widder, Advanced Calculus (New York: Dover, 1989). Chapter 3.
V. M. Faires, Design of Machine Elements, 4th ed. (New York: Macmillan, 1965). Chapter 6, pp. 184-187. Helical springs.
Composed by J. B. Calvert
Created 16 January 2005
Last revised 17 January 2005