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Speed vs. velocity

For some background, see here and here.

Velocity is defined as: “The time rate of change of position of a body; it is a vector quantity having direction as well as magnitude.” Speed is defined as: “The time rate of change of position of a body without regard to direction; in other words, the magnitude of the velocity vector.” (McGraw-Hill Dictionary of Physics, 3rd ed.)

However, it’s not that simple. A common example shows the problem:

When something moves in a circular path (at a constant speed …) and returns to its starting point, its average velocity is zero but its average speed is found by dividing the circumference of the circle by the time taken to move around the circle. This is because the average velocity is calculated by only considering the displacement between the starting and the end points while the average speed considers only the total distance traveled. Wikipedia

So the average speed is not the magnitude of the velocity (which is zero in this case) but something else – the travel distance divided by the travel time.

The question is whether the speed over a finite interval should be the magnitude of the displacement divided by the time interval or the arc length divided by the time interval (i.e., the integral of the norm of the velocity function over the time interval). The answer should be the latter, although the former is implied by the common definition of speed.

It is better to define speed as the ratio of the arc length (travel distance) divided by the arc time (travel time). In short, speed is that which is measured by a speedometer.

Dimensions and units

A dimension is informally regarded as the number of coordinates needed to specify the location of a body or point. That may suffice for a mathematical dimension, but a physical dimension is a dimension of something, that is, some unit. In that sense, the dimensions of force are different from the dimensions of velocity.

However, units such as force and velocity use the same dimensions of space and time. They have a common notion of direction as their basis. A northward force and a northward velocity are in the same direction. In that sense, it is common to speak of space and time as the only physical dimensions.

It is possible to use the speed of light to translate spatial units into temporal units and vice versa. This is done in relativity: the invariant interval may be expressed in length units by multiplying time by the speed of light or in time units by dividing lengths by the speed of light. So space and time are integrated as spacetime dimensions.

Outside the technical usage of relativity, length and time units are distinguished because they are physically different. The method of measuring them differs. The common measures of motion are related to time, not space. The philosophy of time is different from the philosophy of space.

For vectors of rates, such as velocity, different component rates lead to different vectors. Consider a displacement/distimement north 90 km in 3 hrs and east 30 km in 2 hours. The rates for each component dimension are 90/3 = 30 km/hr north and 60/2 = 30 km/hr east. However, the time rates use the overall time of 5 hrs and get 90/5 = 18 km/hr north and 60/5 = 12 km/hr east. The space rates use the total distance of 150 km to get 150/3 = 50 km/hr north and 150/2 = 75 km/hr east (which would normally be expressed as hr/km).

That is, for time rates, the dimensions apply to space, and for space rates the dimensions apply to time. For the rates by component, the dimensions apply to velocity itself rather than space or time.

Addendum: If we consider the progress in the direction of the end point in time, the velocity denominator would be the square root of (3²+2²) = √13. The corresponding numerator would be the square root of 90²+30² = √95. The resulting velocity magnitude would be √(95/13) = 7.2 km/hr.

Algebraic relativity

Relativity may be derived as an algebraic relation among differentials. Consider motion in the x spatial dimension, with a differential displacement, dx, differential velocity displacement, dv, and arc (elapsed) time t:

dx² = (dx/dt)²dt² = dv²dt² =  d(vt)².

Let there be a constant, c:

dx² = d(vt)² = d(cvt)²/c² = d(ct)² (v/c)² = d(ct)² (1 – (1 – v²/c²)).

Let γ² = 1/(1 – v²/c²). Then

dx² = d(vt)² = d(ct)² (1 – 1/γ²) = d(ct)² – d(ct)²/γ² = d(ct)² – d(ct/γ)² = d(ct)² – d()²,

where τ = t/γ. This may be rearranged as

d()² = d(ct)² – dx²,

which equals the Lorentz invariant, d()² = ds².


Alternately, consider motion in the t temporal dimension, with a differential displacement, dt, differential velocity displacement, dv, and arc length x:

dt² = (dt/dx)²dx² = d(1/v)²dx² =  d(x/v)².

Let there be a constant, c:

dt² = d(x/v)² = d(x/cvc² = d(x/c)² (c/v)² = d(x/c)² (1 – (1 – c²/v²)).

Let λ² = 1/(1 – c²/v²). Then

dt² = d(x/v)² = d(x/c)² (1 – 1/λ²) = d(x/c)² – dx²/(λc)² = d(x/c)² – d(x/λc)² = d(x/c)² – d(σ/c)²,

where σ = x/λ. This may be rearranged as

d(σ/c)² = d(x/c)² – dt²,

which equals the Lorentz invariant, d(σ/c)² = dw².


Equivalently, consider motion in the ξ temporal dimension, with a differential displacement, , differential celerity displacement, du, and elapsed space x:

² = (/dx)²dx² = du²dx² =  d(ux)².

Let there be a constant, k = 1/c:

² = d(ux)² = d(kux)²/k² = d(kx)² (u/k)² = d(kx)² (1 – (1 – u²/k²)).

Let λ² = 1/(1 – u²/k²). Then

² = d(ux)² = d(kx)² (1 – 1/λ²) = d(kx)² – d(kx)²/λ² = d(kx)² – d(kx/λ)² = d(kx)² – d()²,

where σ = x/λ. This may be rearranged as

d()² = d(kx)² – ²,

which equals the Lorentz invariant, d()² = dw².

Measures of motion

This post follows others such as the one here and here. A background document is here.

One-dimensional kinematics is like traveling in a vehicle, and on the dashboard are three instruments: (1) a clock, (2) an odometer, and (3) a speedometer. In principle the speedometer reading can be determined from the other instruments, so let’s focus on a clock and an odometer. The clock measures time, which will be used to measure travel time. The odometer measures distance, or length, which will be used to measure travel distance.

From this information one can establish a reference frame to describe space and time. The reference frame includes a point of reference, called the origin (or destination depending on the context), and a unit vector for each dimension, from which one can determine a position vector in space or time. A position vector is the sum of position coordinates times unit vectors. For example, r(t) = x(t) î, with position vector r, position coordinate x, and unit vector î.

The path or route that the vehicle takes is mathematically a curve or arc. The travel distance or length of a trip is its arc length along the path, which is measured by an odometer. The travel time of a trip is its arc time along the path, which is measured by a clock or stopwatch. If the path is a mathematical curve, it can be integrated to find the arc length or arc time:

s = s(t) = ∫ || r′(τ) || , from 0 to t, with arc length s(t) and position vector r(t); or

t = t(s) = ∫ || w′(σ) || , from 0 to s, with arc time t(s) and time position vector w(s).

An interval of time is the arc time from time t1 to time t2. Similarly, an interval of space is the arc length from position s1 to position s2. If time or space are compressed to 1D, then an interval equals the difference between the endpoints: Δt = t2t1 or Δs = s2s1.

The speed of a body is the ratio of the arc length to the arc time: Δst. The pace of a body is the ratio of the arc time to the arc length: Δts.

The displacement of a body during a time interval is defined as the vector change in the position (x) of the body, which for 1D is:

Δrr(t2) − r(t1) = (x(t2) − x(t1)) î ≡ Δx(t) î.

Similarly, the distimement of a body during a space interval is defined as the vector change in the time position (ξ) of the body, which for 1D is:

Δww(s2) − w(s1) = (ξ(s2) − ξ(s1)) î ≡ Δξ(s) î.

The x-component of the average velocity for a time interval Δt is defined as the displacement Δx divided by the time interval Δt:

vavg ≡ Δxt.

Similarly, the ξ-component of the average celerity for a space interval Δs is defined as the distimement Δξ divided by the space interval Δs:

uavg ≡ Δξs.

The x-component of instantaneous velocity at time t is given by the slope of the tangent line to the curve of position vs. time at time t:

vx(t) = dx/dt.

The instantaneous velocity vector is then: v(t) = vx(t) î. Or more generally: v(t) = Σj vj(t) îj.

Similarly, the ξ-component of instantaneous celerity at position s is given by the slope of the tangent line to the curve of time vs. position at position s:

uξ(s) = ds/.

The instantaneous velocity vector is then: u(s) = uξ(s) î. Or more generally: u(t) = Σj uj(s) îj.

The x-component of the instantaneous acceleration at time t is the slope of the tangent line at time t of the graph of the x-component of the velocity as a function of the time: a(t) ≡ dv/dt. The instantaneous acceleration vector at time t is then a(t) ≡ a(t) î.

Similarly, the ξ-component of the instantaneous prestination at position s is the slope of the tangent line at position s of the graph of the ξ-component of the celerity as a function of the position: b(s) ≡ du/ds. The instantaneous prestination vector at position s is then b(s) ≡ b(s) î.

Space, time, arc length, and arc time

Let there be a displacement vector r that is a parametric function of arc time t so that r = r(t). Then define s as the arc length of r so that

s = s(t) = ∫ || r′(τ) || ,

where the integral is from 0 to t. Let us further assume that s is bijective so that the inverse function is t.

Let there also be a distimement vector w that is a parametric function of arc length s so that w = w(s). Then t is the arc time of w if

t = t(s) = ∫ || w′(σ) || ,

where the integral is from 0 to s. Now the arc time derivative of r, that is, the derivative of s with respect to t is

s′ = ds/dt = || r′(t) ||.

And the arc length derivative of w, that is, the derivative of t with respect to s is

t′ = dt/ds = || w′(s) ||.

From the inverse function theorem we have that

t′ = dt/ds = 1/|| r′(t) || = 1/(ds/dt).

And also that

s′ = ds/dt = 1/|| w′(s) || = 1/(dt/ds).

Putting these together we find

s′ = ds/dt = || r′(t) || = 1/|| w′(s) || = 1/(dt/ds).

And also that

t′ = dt/ds = || w′(s) || = 1/|| r′(t) || = 1/(ds/dt).

We then have

s = s(t) = ∫ || r′(τ) || = ∫ 1/|| w′(σ) || dσ,

where the first integral is from 0 to t and the second integral is from 0 to s. And also that

t = t(s) = || w′(σ) || = 1/|| r′(τ) || ,

where the first integral is from 0 to s and the second integral is from 0 to t.

Because of the difficulty of inverting s(t), this shows a bypass is available. That is, from r(t) we find

t(s) = ∫ 1/|| r′(τ) || ,

and from w(s) we find

s(t) = ∫ 1/|| w′(σ) || .

The interpretation is that the space vector, r, is a position vector function of the arc time, t, and the time vector, w, is a time vector function of the arc length, s.

Polar coordinates for time-space

This post follows the material on polar coordinates from MIT Open Courseware, here. Instead of the space position vector r, we’ll use the time position vector w, and replace (arc) time with arc length, s.

In polar coordinates, the time position of an eventicle A is determined by the value of the radial duration to the origin, w, and the angle that the radial line makes with an arbitrary fixed line, such as the tx axis (x axis with time metric). Thus the trajectory of an eventicle will be determined if we know w and θ as a function of s, i.e., w(s) and θ(s). The directions of increasing w and θ are defined by the orthogonal unit vectors ew and eθ.

The time position vector of an eventicle has a magnitude equal to the radial duration, and a direction determined by ew. Thus

w = w ew.

Since the vectors ew and eθ are clearly different from point to point, their variation will have to be considered when calculating the celerity and prestination. Over an infinitesimal interval of arc length ds, the coordinates of time point A will change from (w, θ) to (w + dw, θ + ).

We note that the vectors ew and eθ do not change when the coordinate w changes. Thus dew/dw and eθ/dw = 0. On the other hand, when θ changes to (θ + ), the vectors ew and eθ are rotated by an angle .

A mathematical approach to obtaining the derivatives of the unit vectors is to express ew and eθ in terms of their Cartesian components along i and j. We have that

ew =  cos θ i + sin θ j

eθ = – sin θ i + cos θ j.

Therefore, when we differentiate we obtain

dew/dw = 0,   dew/ = – sin θ i + cos θ j = eθ

deθ/dw = 0,   deθ/ = – cos θ i – sin θ j = –ew.

Celerity vector

We can now differentiate w = w ew with respect to arc length and write

u = w′ = wew + w ew′,

where the prime indicates differentiation with respect to arc length. Or, using the above expression for eθ, we have

u = w′ = wew + eθ.

Here, uw = w′ is the radial celerity component, and uθ = ′ is the circumferential celerity component. We also have that  The radial component is the rate at which w changes magnitude, or stretches, and the circumferential component is the rate at which w changes direction, or swings.

Differentiating again with respect to arc length, we obtain the prestination

b = u′ = wew + wew′ + w′θ′eθ + wθ″eθ + wθ′eθ′.

Using the previous relations, we obtain

b = u′ = (w″wθ′²) ew + (wθ″ + 2w′θ′) eθ,

where bw = (w″wθ′²) is the radial prestination component, and bθ = (wθ″ + 2w′θ′) is the circumferential prestination component. Also, we have that

Change of basis

In many practical situations, it will be necessary to transform the vectors expressed in polar coordinates to Cartesian coordinates and vice versa. Since we are dealing with free vectors, we can translate the polar reference frame for a given point (w, θ) to the origin, and apply a standard change of basis procedure. This will give, for a generic vector A,

Equations of motion

In two dimensional polar coordinates, the surge and prestination vectors are Γw = Γwew + Γθeθ and b = bwew + bθeθ. Thus, in component form, with vass n, we have

Γw = nbw = n (w″wθ′²)

Γθ = nbθ = n (wθ″ + 2w′θ′).

Intrinsic coordinates for time-space

This post follows the introduction to intrinsic coordinates given here, and changes it for time-space (1D space + 3D time). So rather than the 3D space position vector r, we’ll use the 3D time position vector, w, and we’ll switch the arc time, t, with the arc length, s.

We follow the motion of a point using a time position vector w(s) whose position along a known trajectory in time is given by the scalar function t(s) where t(s) is the arc time along the curve. We obtain the celerity, u, from the space rate of change of the vector w(s) following the particle:

We identify the scalar as the magnitude of the celerity u and dw/dt as the unit vector tangent to the curve at the point t(s). Therefore we have

u = u et,

where w(s) is the time position vector, is the pace, et is the unit tangent vector to the trajectory, and t is the arc time coordinate along the trajectory.

The unit tangent vector can be written as et = dw/dt. The prestination vector, b, is the derivative of the celerity vector with respect to arc length. Since dw/dt depends only on t, using the chain rule we can write

The second derivative d²w/dt² is another property of the arc time. We shall see that it is related to the radius of curvature.

Taking the space derivative of u, an alternate expression can be written in terms of the unit vector et as

The vector et is the local unit tangent vector to the curve which changes from point to point in time. Consequently, the space derivative of et will, in general, be nonzero. The space derivative of et can be written as

In order to calculate the derivative of et, we note that, since the magnitude of et is constant and equal to one, the only changes that et can have are due to rotation, or swinging. When we move from t to t + dt, the tangent vector changes from et to et + det. The change in direction can be related to the angle (the differential angle).

The direction of det, which is perpendicular to et, is called the normal direction. On the other hand, the magnitude of det will be equal to the length of et (which is one), times . Thus, if en is a unit normal vector in the direction of det, we can write

det = en.

Dividing by dt yields

Here κ = /dt is a local property of the curve, called the time curvature, and ρ = 1/κ is called the radius of time curvature.

From the above we have that

Finally, we have that the prestination can be written as

Here is the tangential component of the prestination, and bn = u²/ρ is the normal component of the prestination. Since bn is the component of the prestination pointing towards the center of curvature, it is sometimes referred to as the centripetal prestination. When bt is nonzero, the celerity vector changes magnitude, or stretches. When bn is nonzero, the celerity vector changes direction, or swings. The modulus of the total prestination can be calculated as

Relationship between t, u, and b

The quantities t, u, and bt are related in the same manner as the quantities t, u, and b for rectilinear motion. In particular we have that This means that if we have a way of knowing bt, we may be able to integrate the tangential component of the motion independently.

The vectors et and en, and their respective coordinates t (tangent) and n (normal), define two orthogonal directions. The plane defined by these two directions is called the osculating plane. This plane changes from point to point and can be thought of as the plane that locally contains the trajectory (Note that the tangent is the current direction of the celerity, and the normal is the direction into which the celerity is changing).

In order to define a right-handed set of axes we need to introduce an additional unit vector which is orthogonal to et and en. This vector is called the binormal, and is defined as eb = et × en.

At any point in the trajectory, the time position vector, the celerity, and prestination can be referred to these axes. In particular, the celerity and prestination take very simple forms:

u = u et

The difficulty of working with this reference frame stems from the fact that the orientation of the axis depends on the trajectory itself. The time position vector, w, needs to be found by integrating the relation dw/ds = u as follows:

where w0 = w(0) is given by the initial condition.

We note that, by construction, the component of the prestination along the binormal is always zero. When the trajectory is planar, the binormal stays constant (orthogonal to the plane of motion). However, when the trajectory is a time curve, the binormal changes with t. It will be shown below that the derivative of the binormal is always along the direction of the normal. The rate of change of the binormal with t is called the time torsion, σ. Thus

We see that whenever the torsion is zero, the trajectory is planar, and whenever the curvature is zero, the trajectory is linear.

Radius of curvature and torsion for a trajectory

In some situations the trajectory will be known as a curve of the form y = f(x). The radius of curvature in this case can be computed according to the expression,

Since y = f(x) defines a planar curve, the torsion σ is zero. On the other hand, if the trajectory is known in parametric form as a curve of the form w(s), where s can be arc length or any other parameter, then the radius of curvature ρ and the torsion σ can be computed as

Equations of motion in intrinsic coordinates

Newton’s second law is a vector equation, Γ = nb (with surge Γ and vass n; cf. F = ma), which can now be written in intrinsic coordinates. In tangent, normal, and binormal components, tnb, we write Γ = Γt et + Γn en and b = bt et + bn en. We observe that the positive direction of the normal coordinate is that pointing toward the center of curvature. Thus, in component form, we have

Note that, by definition, the component of the prestination along the binormal direction, eb, is always zero, and consequently the binormal component of the surge must also be zero.

Inverting motion curves

The mathematical problem is this: given a curve with distance coordinates that are parametric functions of time (duration), find the reparametrization of the curve with duration (time) coordinates that are parametric functions of distance. Symbolically, given the regular curve α(t) = (a1(t), …, an(t)), find β(s) = (b1(s), …, bn(s)) such that bi(s) = t(ai) for i = 1, …, n, where t is time (duration) and s is distance. (Greek letters are used for vectors, Roman letters for scalars.)

The solution is to invert each coordinate function and express them in terms of a common parameter. That is, set each ai(t) = s and solve for t to get t = ai-1(s) = bi(s) for the inverse coordinates in parametric form.

For example, consider a projectile fired from height h with velocity v at angle θ. The path of the projectile is represented by a parametric equation

α(t) = (a1(t), a2(t)) = (vt cos(θ), h + vt sin(θ) – ½gt²),

where g is the acceleration of gravity. Setting s = vt cos(θ) and s = h + vt sin(θ) – ½gt²), then solving for t results in the inverse coordinates, which are in two parts:

β(s) =(s/(v cos(θ)), (v sin(θ) + sqrt(2gh – 2gs + v² sin²(θ)))/g) going up, and

β(s) =(s/(v cos(θ)), (v sin(θ) – sqrt(2gh – 2gs + v² sin²(θ)))/g) coming down.

The spatial position vector α(t) corresponds to a temporal position vector β(s). As there are multiple dimensions of space, so there are multiple dimensions of time. But the time in multidimensional space is a scalar, and the space in multidimensional time is a scalar.

Curves for space and time, continued

The following is a continuation and revision of the previous post, here.

Based on the differential geometry part of the book Shape Interrogation for Computer Aided Design and Manufacturing by Nicholas M. Patrikalakis and Takashi Maekawa of MIT. A pdf version in parallel is here.

Let a three-dimensional curve be expressed in parametric form as x = x(t); y = y(t); z = z(t); where the coordinates of the point (x, y, z) of the curve are expressed as functions of a parameter t (time) within a closed interval t1tt2. The functions x(t), y(t), and z(t) are assumed to be continuous with a sufficient number of continuous derivatives.

In vector notation the parametric curve can be specified by a vector-valued function r = r(t), where r represents the position vector (i.e., r(t) = (x(t), y(t), z(t)).

curve1

Displacement Δr connecting points A and B on parametric curve r(t).

Consider a segment (displacement) of a parametric curve r = r(t) between two points P(r(t)) and Q(r(tt)) as shown in the figure above. As point B approaches A or in other words Δt → 0, the length s becomes the differential arc length of the curve:

ds = |dr/dt| dt = | rt | dt = √(rtrt) dt,

where the superscript t denotes differentiation with respect to the arc time parameter t. The vector rt = dr/dt is called the tangent vector at point A.

Then the arc length, s, of a segment of the curve between points r(t0) and r(t) can be obtained as follows:

s(t) = ∫ ds = ∫ √(rtrt) dt = ∫ √((dx/dt)2 + (dy/dt)2 + (dz/dt)2) dt.

The magnitude of the tangent vector is

| rt | = ds/dt = v.

Hence the unit tangent vector is

Ts = rt / | rt | = (dr/dt) / (ds/dt) = dr/dsrs,

where the superscript s denotes differentiation with respect to the arc length parameter, s.

If r(s) is an arc length parametrized curve, then rs(s) is a unit vector, and hence rsrs = 1. Differentiating this relation, we obtain rsrss = 0, which states that rss is orthogonal to the tangent vector, provided it is not a null vector. The unit vector

Ns = rss(s) / |rss(s)| = Tss(s)/|Tss(s)|,

which has the direction and sense of rss(s) is called the unit principal normal vector at s. The plane determined by the unit tangent and normal vectors Ts(s) and Ns(s) is called the osculating plane at s. The curvature is

κs ≡ 1/ρ = |rss(s)|,

and its reciprocal ρ is called the radius of curvature at s. It follows that

rss = Tss = κs Ns.

The vector ks = rss = Tss is called the curvature vector, and measures the rate of change of the tangent along the curve. By definition κs is nonnegative, thus the sense of the normal vector is the same as that of rss(s). For a three-dimensional curve, the curvature is

κs = |rt × rtt| / | rt |³.


Let a three-dimensional curve be expressed in parametric form as X = X(s); Y = Y(s); Z = Z(s); where the coordinates of the point (X, Y, Z) of the curve are expressed as functions of a parameter s (length) within a closed interval s1ss2. The functions X(s), Y(s), and Z(s) are assumed to be continuous with a sufficient number of continuous derivatives.

In vector notation the parametric curve can be specified by a vector-valued function w = w(s), where w represents the position vector (i.e., w(s) = (X(s), Y(s), Z(s)).

curve2

Distimement Δw connecting points C and D on parametric curve w(s).

Consider a segment (distimement) of a parametric curve w = w(s) between two points C(w(s)) and D(w(ss)) as shown in the figure above. As point D approaches C or in other words Δs → 0, the length t becomes the differential arc time of the curve:

dt = |dw/ds| ds = | ws | ds = √(wsws) ds,

where ws = dw/ds, which is called the tangent vector at point C. Then the arc time, t, of a segment of the curve between points w(s0) and w(s) can be obtained as follows:

t(s) = ∫ dt = ∫ √(wsws) ds = ∫ √((dX/ds)2 + (dY/ds)2 + (dZ/ds)2) ds.

The vector ws = dw/ds is called the tangent vector at point C. The magnitude of the tangent vector is

| ws | = dt/ds = u.

Hence the unit tangent vector is

Ttws / | ws | = (dw/ds) / (dt/ds) = dw/dtwt.

If w(t) is an arc length parametrized curve, then wt(t) is a unit vector, and hence wtwt = 1. Differentiating this relation, we obtain wtwtt = TtTtt = 0, which states that wtt is orthogonal to the tangent vector, provided it is not a null vector. The unit vector

Nt = wtt(t) / |wtt(t)| = Ttt(t)/|Ttt(t)|,

which has the direction and sense of wtt(t) is called the unit principal normal vector at t. The plane determined by the unit tangent and normal vectors Tt(t) and Nt(t) is called the osculating plane at t. The curvature is

κt ≡ 1/ρ = |wtt(t)| = |Ttt(t)|,

and its reciprocal ρ is called the radius of curvature at t. It follows that

wttTtt = κt Nt.

The vector kt = wttTtt is called the curvature vector, and measures the rate of change of the tangent along the curve. By definition κ is nonnegative, thus the sense of the normal vector is the same as that of wtt(t). For a three-dimensional curve, the curvature is

κt = |ws × wss| / | ws |³.


Here are some useful formulae of the derivatives of arc length, s, and the arc time, t:

v = st = ds/dt = | rt | = (rtrt)1/2 = 1/| ws | = 1/(wsws)1/2,

a = stt = dst/dt = (rtrtt) / (rtrt)1/2 = – (wswss) / (wsws)4/2,

sttt = dstt/dt = [(rtrt)(rtrttt + rttrtt) – (rtrtt)²] / (rtrt)3/2

= – [(wsws)(wswsss + wsswss) – 4(wswss)²] / (wsws)7/2,

u = ts = dt/ds = 1/| rt | = 1/(rtrt)1/2 = | ws | = (wsws)1/2,

b = tss = dts/ds = – (rtrtt) / (rtrt)4/2 = (wswss) / (wsws)1/2,

tsss = dtss/ds = – [(rtrt)(rtrttt + rttrtt) – 4(rtrtt)²] / (rtrt)7/2

= [(wsws)(wswsss + wsswss) – (wswss)²] / (wsws)3/2.

Curves for space and time

The following is slightly modified from the differential geometry part of the book Shape Interrogation for Computer Aided Design and Manufacturing by Nicholas M. Patrikalakis and Takashi Maekawa of MIT.

A plane curve can be expressed in parametric form as x = x(t); y = y(t); where the coordinates of the point (x, y) of the curve are expressed as functions of a parameter t (time) within a closed interval t1tt2. The functions x(t) and y(t) are assumed to be continuous with a sufficient number of continuous derivatives. The parametric representation of space curves is: x = x(t); y = y(t); z = z(t); t1tt2.

In vector notation the parametric curve can be specified by a vector-valued function r = r(t), where r represents the position vector (i.e., r(t) = (x(t), y(t), z(t)).

curve

Displacement Δr connecting points P and Q on parametric curve r(t).

Let us consider a segment (displacement) of a parametric curve r = r(t) between two points P(r(t)) and Q(r(tt)) as shown in the figure above. As point Q approaches P or in other words Δt → 0, the length s becomes the differential arc length of the curve:

ds = |dr/dt| dt = | r | dt = √(rr) dt.

Here the dot denotes differentiation with respect to the parameter t. Therefore the arc length of a segment of the curve between points r(t0) and r(t) can be obtained as follows:

s(t) = ∫ ds = ∫ √(rr) dt = ∫ √(x2(t) + y2(t) + z2(t)) dt.

The vector dr/dt is called the tangent vector at point P. The magnitude of the tangent vector is

| r | = ds/dt = v.

Hence the unit tangent vector becomes

T = r / | r | = (dr/dt) / (ds/dt) = dr/ds.

Here the prime ¹ denotes differentiation with respect to the arc length, s. We list some useful formulae of the derivatives of arc length s with respect to parameter t and vice versa:

v = s = ds/dt = | r | = (rr)1/2,

a = s•• = ds/dt = (rr) / (rr)1/2,

s = ds/dt = [(rr)(rr + rr) – (rr)²] / (rr)3/2,

u = t¹ = dt/ds = 1/| r | = 1/(r • r)1/2,

b = t¹¹ = d/ds = – (rr) / (rr)4/2,

t¹¹¹ = dt¹¹/ds = – [(rr)(rr + rr) – 4(rr)²] / (rr)7/2.

If r(s) is an arc length parametrized curve, then (s) is a unit vector, and hence = 1. Differentiating this relation, we obtain r¹¹ = 0, which states that r¹¹ is orthogonal to the tangent vector, provided it is not a null vector.

The unit vector

N = r¹¹(s)/|r¹¹(s)| = (s)/|(s)|,

which has the direction and sense of (s) is called the unit principal normal vector at s. The plane determined by the unit tangent and normal vectors T(s) and N(s) is called the osculating plane at s. The curvature is

κ ≡ 1/ρ = |r¹¹(s)|,

and its reciprocal ρ is called the radius of curvature at s. It follows that

r¹¹ = = κN.

The vector k = r¹¹ = is called the curvature vector, and measures the rate of change of the tangent along the curve. By definition κ is nonnegative, thus the sense of the normal vector is the same as that of r¹¹(s).

For a space curve, the curvature is

κ = |r × r| / |r|³.