# Envelope

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 Revision as of 08:33, 24 May 2012 (edit)← Previous diff Revision as of 11:20, 24 May 2012 (edit) (undo)Next diff → Line 8: Line 8: |ImageDescElem= |ImageDescElem=
- In geometry, an '''envelope''' of a {{EasyBalloon|Link=family of curves|Balloon=A family of curves is a set of curves having the same function, except for one or more variable parameter. For example, $y = -x + a$ with variable a. See image: [[image:family.png|350px]]}} is the boundary of the area "swept" by the curve when we change the variable parameter in the curve function. +
- + In geometry, an '''envelope''' of a {{EasyBalloon|Link=family of curves|Balloon=A family of curves is a set of curves having the same function, except for one or more variable parameter. For example, $y = -x + a$ with variable a. See image: [[image:family.png|350px]]}} is a curve that is '''tangent''' to each member of the family at some point. Visually, it's the boundary of these curves' "sweeping area". - For example, suppose there is a ladder leaning on a wall. The ladder starts to slide down because someone steps on it. What will be the shape of the area swept by the ladder before he hits the ground? + - {{{!}} border=0 cellpadding=10 cellspacing=10 + + {{{!}} border=0 cellpadding=0 cellspacing=0 + {{!}}{{Anchor|Reference=Figure1-1|Link=[[Image:Ladderman.jpg|left|thumb|200px|Figure 1-1:
Ladder and man]]}} + {{!}}{{!}}For example, suppose there is a ladder leaning on a wall. The ladder starts to slide down because someone steps on it (see [[#Figure1-1|Figure 1-1]] ). What will be the shape of the area "swept" by the ladder before it hits the ground?

A simulation of this process is shown below. {{!}}- {{!}}- - {{!}}{{Anchor|Reference=Figure1-1|Link=[[Image:EnvelopeAnim.gif|center|thumb|250px|Figure 1-1:
The complete astroid]]}} + {{!}}{{!}}{{Anchor|Reference=Figure1-3|Link=[[Image:Astroid.png|left|thumb|280px|Figure 1-3:
The complete astroid]]}} {{!}}} {{!}}} - Surprisingly, the area swept by a moving straight line does not have a straight boundary (see [[#Figure1-1|Figure 1-1]]). The envelope is the first-quadrant portion of an {{EasyBalloon|Link=astroid|Balloon=Astroid is the image of function $x^{2/3} + y^{2/3} = a^{2/3}$}}. If we complete the envelope by letting the ladder sweep across the other 3 quadrants, we can see that it shapes like a star, which is why we call it an astroid (See [[#Figure1-2|Figure 1-2]]).
+ Surprisingly, the area swept by a moving straight line does not have a straight boundary (see [[#Figure1-2|Figure 1-2]]). The envelope is the first-quadrant portion of an {{EasyBalloon|Link=astroid|Balloon=Astroid is the image of function $x^{2/3} + y^{2/3} = a^{2/3}$}}. One may notice that, this Astroid envelope is always tangent to ladder during the sliding process, as stated in the definition. -
+ + + If the ladder is allowed to slide in other three quadrants, we will get a complete star-shaped envelope(See [[#Figure1-3|Figure 1-3]]). In fact, the name '''astroid''' comes from the Greek word for "star". + + =A gallery of beautiful envelopes= =A gallery of beautiful envelopes= + ==Envelope of lines== ==Envelope of lines== - We can get more interesting envelopes if we play with the geometry: + + A moving straight line can have curves as its envelope. {{{!}}border="0" cellpadding=5 cellspacing=5 {{{!}}border="0" cellpadding=5 cellspacing=5 Line 30: Line 38: {{!}}} {{!}}} - In [[#Figure2-1|Figure 2-1]], point A lies on line l, M is midpoint of OA, and line m is perpendicular to OA. + In [[#Figure2-1|Figure 2-1]], line m, our '''sweeping line''', is perpendicular to segment OA at its midpoint M. Point O is fixed in space. If we slide point A along line l, line m will sweep out a '''parabola''', with point O as its focus and line l as its directrix. If we slide point A along line l, line m will sweep out a '''parabola''', with point O as its focus and line l as its directrix. Line 62: Line 70: ==Envelope of circles== ==Envelope of circles== - +

---- ---- Line 82: Line 90: {{!}}} {{!}}} - In [[#Figure3-3|Figure 3-3]], l is a vertical line that passes the center of circle O. Circle A has its center on Circle O and is '''tangent''' to l at B + Repeat the process in [[#Figure3-1|Figure 3-1]], except our sweeping circle is now '''tangent''' to vertical line l. - If we slide point A along the bold green Circle O, we will get a '''nephroid''' as the envelope of the sweeping circle. + The result is a '''Nephroid''', which is the Greek word for "kidney-shaped". - Nephroid is one of the [[Roulette|Roulettes]]. The word "nephroid" means ''kidney-shaped''. + Astroid, Cardioid, and Nephroid all belong to '''Roulettes''', which means they can also be constructed by rolling a circle around another. For more information about Roulettes, please go to [[Roulette|this page]].

---- ---- Line 95: Line 103: {{!}}} {{!}}} - In [[#Figure3-3|Figure 3-3]], F1 and F2 are foci of the blue hyperbola. A is midpoint of segment F1F2. The sweeping circle O has its center on the hyperbola, and passes A. + In [[#Figure3-3|Figure 3-3]], our sweeping circle O has its center on the '''hyperbola''', and passes a fixed point A, which is the midpoint of foci segment F1F2. If we slide O along the hyperbola, we will get a '''Lemniscate''' as the envelope of the sweeping circle. If we slide O along the hyperbola, we will get a '''Lemniscate''' as the envelope of the sweeping circle. - Lemniscate is an eight-shaped curve discovered by Jacob Bernoulli[http://en.wikipedia.org/wiki/Jakob_Bernoulli]. It has the polar equation of the form $r^2 = 2a^2cos(2\theta)$ + Lemniscate is an eight-shaped curve discovered by Jacob Bernoulli For an introduction to Lemniscate and how it was discovered, please go to [http://en.wikipedia.org/wiki/Lemniscate_of_Bernoulli this page].. It has the polar equation of the form $r^2 = 2a^2cos(2\theta)$

---- ---- Line 108: Line 116: {{!}}} {{!}}} - Note that if in [[#Figure3-5|Figure 3-5]], instead of having A as midpoint of foci segment, we move it to an arbitrary position inside the hyperbola, then we will get a variation of "lemniscate", which has a funny shape like a bunny's ears. + Note that if in [[#Figure3-5|Figure 3-5]], instead of having A as midpoint of foci segment, we move it to an arbitrary position between the two halves of the hyperbola, then we will get a variation of "lemniscate", which has a funny shape like a bunny's ears. Line 114: Line 122:

- The following envelopes have more complicated mechanism than previous ones. But they are also more interesting. + The following envelopes have more complicated mechanism than previous ones. But in return they are also more interesting. Line 365: Line 373: As pointed out in the [[#Beginning|main image]], the envelope of all particles' trajectories in an exploding firework is a '''parabola'''. Here comes the explanation: As pointed out in the [[#Beginning|main image]], the envelope of all particles' trajectories in an exploding firework is a '''parabola'''. Here comes the explanation: + {{Anchor|Reference=Figure8-1|Link=[[Image:FireworkEnvelope.png|center|thumb|700px|Figure 8-1
Simulation of the blue-aerial-shell firework]]}} {{Anchor|Reference=Figure8-1|Link=[[Image:FireworkEnvelope.png|center|thumb|700px|Figure 8-1
Simulation of the blue-aerial-shell firework]]}} Line 371: Line 380: For the envelope to be parabolic, we have to make several assumptions: For the envelope to be parabolic, we have to make several assumptions: + :*The firework is consisted of many shinning particles, each projected from the origin at the same time, with same velocity '''v'''. :*The firework is consisted of many shinning particles, each projected from the origin at the same time, with same velocity '''v'''. Line 376: Line 386: :*Particles are subject to constant gravity, with gravitational acceleration g. :*Particles are subject to constant gravity, with gravitational acceleration g. - :*Air friction can be neglected. In fact, this turns out to be an arguable assumption. Most firework particles are relatively small and light, so they could be significantly deflected by air friction. However, the case with air friction is way too complicated for this page. Besides, air friction can be neglected, at least on some big and heavy firework particles. So we can still accept this assumption and see what happens. + :*Air friction can be neglected. In fact, this turns out to be an arguable assumption. Most firework particles are relatively small and light, so they could be significantly deflected by air friction. However, the case with air friction is way too complicated for this page. Besides, air friction can be neglected, at least for some fireworks with big and heavy particles such as blue-aerial-shell. So we can still accept this assumption and see what happens. -

+ + + With the assumptions above, we can write out the trajectory of one particular particle: With the assumptions above, we can write out the trajectory of one particular particle:

## Revision as of 11:20, 24 May 2012

Blue-aerial-shell
Field: Geometry
Image Created By: skylighter.com
Website: skylighter.com

Blue-aerial-shell

This is a beautiful blue-aerial-shell firework filling the sky. Each particle of the firework follows a parabolic trajectory, and together they sweep a parabolic area.

# Basic Description

In geometry, an envelope of a family of curves is a curve that is tangent to each member of the family at some point. Visually, it's the boundary of these curves' "sweeping area".

 Figure 1-1: Ladder and man For example, suppose there is a ladder leaning on a wall. The ladder starts to slide down because someone steps on it (see Figure 1-1 ). What will be the shape of the area "swept" by the ladder before it hits the ground?A simulation of this process is shown below. Figure 1-2: Demonstration of a sliding ladder Figure 1-3: The complete astroid

Surprisingly, the area swept by a moving straight line does not have a straight boundary (see Figure 1-2). The envelope is the first-quadrant portion of an astroid. One may notice that, this Astroid envelope is always tangent to ladder during the sliding process, as stated in the definition.

If the ladder is allowed to slide in other three quadrants, we will get a complete star-shaped envelope(See Figure 1-3). In fact, the name astroid comes from the Greek word for "star".

# A gallery of beautiful envelopes

## Envelope of lines

A moving straight line can have curves as its envelope.

 Figure 2-1 Figure 2-2 Parabola gif coming later

In Figure 2-1, line m, our sweeping line, is perpendicular to segment OA at its midpoint M. Point O is fixed in space.

If we slide point A along line l, line m will sweep out a parabola, with point O as its focus and line l as its directrix.

 Figure 2-3 Figure 2-4Ellipse gif coming later

Repeat the process in Figure 2-1,except point A now slides on a circle.

The result is an ellipse, with O and B as its focuses.

 Figure 2-5 Figure 2-6 Hyperbola gif coming later

Repeat the process in Figure 2-3,except point B is outside the circle.

The result is a hyperbola, with O and B as its focuses.

So far we have got all of the three Conic Section Curves as envelopes of straight lines. However, the envelopes are in no ways restricted to straight lines. Circle, Ellipse, and other curves can make fantastic envelopes as well.

## Envelope of circles

We can make more interesting envelopes if we use circles, rather than straight lines, as our sweeping curve.

 Figure 3-1 Figure 3-2 Cardioid gif coming later

In Figure 3-1, Bold green circle O is fixed in space; A and B are two points on the fixed circle. Our sweeping circle is centered at A, and passes B.

If we fix point B and slide point A along the fixed circle, we will get a Cardioid as the envelope of the sweeping circle.

 Figure 3-3 Figure 3-4 Nephroid gif coming later

Repeat the process in Figure 3-1, except our sweeping circle is now tangent to vertical line l.

The result is a Nephroid, which is the Greek word for "kidney-shaped".

 Figure 3-5 Figure 3-6 Lemniscate gif coming later

In Figure 3-3, our sweeping circle O has its center on the hyperbola, and passes a fixed point A, which is the midpoint of foci segment F1F2.

If we slide O along the hyperbola, we will get a Lemniscate as the envelope of the sweeping circle.

Lemniscate is an eight-shaped curve discovered by Jacob Bernoulli [1]. It has the polar equation of the form $r^2 = 2a^2cos(2\theta)$

 Figure 3-7 a variation of "lemniscate" Figure 3-8 Aha! I have ears like a lemniscate!

Note that if in Figure 3-5, instead of having A as midpoint of foci segment, we move it to an arbitrary position between the two halves of the hyperbola, then we will get a variation of "lemniscate", which has a funny shape like a bunny's ears.

## More complicated envelopes

The following envelopes have more complicated mechanism than previous ones. But in return they are also more interesting.

### 1.Astroid again, but this time using ellipses

Recall that in Figure 1-1, we showed how to construct an astroid using a line segment sliding on coordinate axes. Actually there is another way to generate the same astroid: using a family of ellipses.

First, let's look at the family of ellipse ${x^2 \over c^2} + {y^2 \over (1-c)^2 } = 1$:

 Figure 4-1 Figure 4-2

Here, because there is a variable parameter c in the equation of the ellipse, this function refers to a family of curves. For every different c value, we will get a different ellipse.

For example, in Figure 4-1, since $c < {1\over2}$, then $(1-c)^2 > c^2$, so here the major axis is y-axis, and the minor axis is x-axis.

However, if we choose another $c = 0.89$, which is larger than $1\over2$, then we will have $c^2 > (1-c)^2$, which makes y-axis the major axis.

What will happen if we let c vary from 0 to 1 continuously and trace the area swept by these ellipses? Well, turns out that it's also an astroid, exactly like the one we constructed before.

Figure 4-3
Astroid as envelope of ellipses
gif coming later

For the math behind this, please go to the More Mathematical Explanation section.

### 2.Deltoid as envelope of Wallace-Simson lines

The Wallace-Simson line is related to a simple theorem in geometry proposed by William Wallace [1] in 1796. The theorem itself has nothing special, but with a little manipulation we can get a gorgeous envelope out of it.

Here is a brief introduction of the Wallace-Simson line, which will be our sweeping line:

 Figure 5-1Perpendicular Projections Figure 5-2The Wallace-Simson Line (orange)

In Figure 5-1, first we draw an arbitrary triangle, then draw its circumscribed circle O. For an arbitrary point M on the circumscribed circle, we make perpendicular projections of M onto 3 sides of the triangle (extent line segment if not inside triangle), intersecting at P, Q, and R.

Wallace claimed that the three projections are on the same straight line (see the orange line in Figure 5-2). This line is called Wallace-Simson Line. A proof of this theorem can be found here [2].

Since M is an arbitrary point on circle O, we can certain move it along the circle, and points P, Q, and R are also going to move, since they are perpendicular projections of point M. So we will have a sweeping Wallace-Simson Line, and its envelope is a deltoid, as shown in the following image:

Figure 5-3
Deltoid as envelope of Wallace-Simson Line
gif coming later

Notice that deltoid, like cardioid and nephroid, belongs to the Roulette Family.

This envelope was firstly discovered and proved by Swiss mathematician Jakob Steiner. For the math behind this please go here [3].

# A More Mathematical Explanation

Note: understanding of this explanation requires: *Calculus

An envelope of [...]

An envelope of a family of curves is the boundary of the area swept by these curves when we change the variable parameter t.

However, if we want a more mathematical explanation of envelope, we have to redefine it in a more mathematical way, because two serious problems arise with the old definition when we dive into more math:

First, not all 2-D curves can be represented as a single function y = f(x) with a variable parameter t. For example, a circle $x^2 + y^2 = a^2$ can not be expressed unless we use two functions, $y = \sqrt{a^2 - x^2}$ and $y = - \sqrt{a^2 - x^2}$. So we have to introduce new ways to describe curves.

Second, "boundary of the area swept by these curves" is a rough description in everyday language. It's not something that we can use to derive mathematical formula and equations. So we have to be clear about we mean by "sweep", "boundary", and so on.

In the rest of this section, we are going to deal with these two problems one by one, and show how can we get a good mathematical explanation of envelopes using the new definition.

## Resolve the first problem: the power of level sets

The first problem is easily resolved if we describe 2-D curves using level sets, rather than images of one-variable functions y = f(x).

In Multivariable Calculus, the level set of a two-variable function F(x,y) at height C is defined as the set of points (x,y) that satisfy the condition F(x,y) = C. For example, instead of writing y = 2x - 1 for a line, we could write 2x - y = 1, in which F(x,y) = 2x - y and C = 1.

In general, level sets are more powerful than images of single functions when we need to describe 2-D curves, since all single variable functions $y = f(x)$ can be written in the level set form $F(x,y) = f(x) - y = 0$, but the converse is not true. Some of the level sets can not be rewritten as y = f(x) unless we use multiple functions (see the circle in Figure 6-1), others can't be reduced at all (see the level set curve in Figure 6-2).

 Figure 6-1How to represent circles and lines Figure 6-2How to represent a complicated 2-D curve

Because of these advantages of level sets, in the rest of this section we will use $F(x,y) = C$, rather than $y = f(x)$, to describe a family of curves. At least for the purpose of envelopes, level set is sufficient to describe all 2-D curves that we care about.

## Resolve the second problem: the boundary condition

The next question is, given a family of level set curves F(x,y,t) = C with variable parameter t, how can we find its boundary and express it in mathematical language?

The answer is given by the boundary condition, which states that:

For a family of level set curves F(x,y,t) = C with variable parameter t, it's envelope, or boundary of sweeping area, must satisfy the condition ${\partial F(x,y,t) \over \partial t} = 0$

To see why this is true, let's look more carefully at the ladder problem. The following two images are different phases of a ladder captured in the sliding process.

 Figure 6-3Sliding process captured Figure 6-4Intersections lying on envelope

In Figures 6-3 and 6-4, different phases of a sliding ladder are indicated using different colors. One may immediately notice that if we connect the intersections of different phases, we will get a curve close to the astroid envelope. It is not exactly the envelope because the time intervals between different captured phases are finite. However, if we make the time intervals shorter, the curve connecting them will get closer and closer to the envelope.

This is not mere coincidence. Same thing happens in the elliptic envelope of astroid (see Figure 6-5) . Actually it can be shown that if we choose two phases with infinitely small time interval, then their intersection must lie on the envelope. The argument goes as following:

Figure 6-5
Elliptic envelope revisited

In Figure 6-5, it's obvious that in one phase, the only segment that contributes to the envelope is between its intersections with the previous and next phase. For example, in the green ellipse, which has variable parameter t = 0.40, the segment above point A is covered by the orange envelope, thus not contributing to the envelope. Similarly the segment below point B does not contribute since it's covered by the blue ellipse. So the envelope is actually consisted of many "chunks" of segments that's not covered by other phases.

Now consider what happens if we increase the number of captures, and shorten each time interval:

• Two neighboring phases will approximate each other.
• Two intersections A and B will be closer.
• Segment AB will be shorter and smoother.

Eventually, when the number of captures goes to infinity, points A and B will meet together, and segment AB will be "squeezed" into one point on the envelope (call this point P). Recall that P is the intersection of two neighboring phases:

$F(x,y,t) = C$ and $F(x,y,t + dt) = C$

which gives us:

$F(x,y,t + dt) - F(x,y,t) = 0$

${\partial F(x,y,t) \over \partial t} dt = 0$

${\partial F(x,y,t) \over \partial t} = 0$

This is the boundary condition we are trying to prove. The same argument goes for every other envelope.

One may also notice that, since the envelope only touches each curve at this single "squeezed" point, it is tangent to every curve in the family. In fact, envelope can also be defined as a curve that is tangent to every one of a family of curves.

## Conclusion and Application

Now we have the family of level set curves:

• $F(x,y,t) = C$

and the boundary condition:

• ${\partial F(x,y,t) \over \partial t} = 0$

Since every point on the envelope must satisfy both equations, we can combine them to solve for a 2-D envelope curve. However, the calculation involved is rather long and complicated, so here I will only prove the envelope of a sliding ladder is an Astroid.

### Proof for the Astroid envelope

Figure 6-6

See Figure 6-6, the length of the ladder is a. For simplicity we will only consider the envelope in first quadrant.

Choose x-coordinate of point A as the variable parameter t. So y-coordinate of point B is $\sqrt{a^2 - t^2}$

Thus the equation of line AB, our sweeping curve, is:

Eq. 1        $F(x,y,t) = {x \over t} + {y \over \sqrt {a^2 - t^2}} = 1$

Differentiate Eq.1 to get the boundary condition:

${\partial F(x,y,t) \over \partial t} = -{x \over t^2} + {yt \over (\sqrt {a^2 - t^2})^3} = 0$

From which we can get:

Eq. 2        ${x \over t^3} = {y \over (\sqrt {a^2 - t^2})^3}$

Substitute Eq.1 into Eq.2 twice, first substitute x, and then y, we get:

${t \over x^{1/3}} = {\sqrt {a^2 - t^2} \over y^{1/3}} = a^{2/3}$

Substitute back into Eq.1, we get:

${x^{2/3} \over a^{2/3}} + {y^{2/3} \over a^{2/3}} = 1$

$x^{2/3} + y^{2/3} = a^{2/3}$ , the equation of an Astroid.

Other proofs are similar.

# Why It's Interesting

Although envelope looks like a pure math concept, it does have some interesting applications in various areas, such as Microeconomics, Applied Physics, and String Arts:

## Application in Microeconomics: the Envelope Theorem

Figure 7-1
The Envelope Theorem

Economists often deal with maximization or minimization problems: to maximize benefit, minimize cost, maximize social revenue, and so on. However, the problem is that there are so many variable parameters in economics. How many men should I hire? How much land should I buy or rent? Should I invest more money to buy new machines, or should I just do with old ones? Because of all these variable parameters, economists often end up doing maximization or minimization of a family of curves, rather than a single curve (see Figure 7-1)

So here comes the Envelope Theorem. It allows economists to find the envelope of a family of curves first, and then determine the maximum or minimum value on the envelope. Since no points go beyond the envelope, this point must be the absolute extrumum among the whole family of curves

## Application in Physics: Envelope of Waves

 Figure 7-2A typical beating wave with high frequency and slowly changing amplitude Figure 7-3An audio signal may be carried by an AM or FM radio wave

In physics, if we combine two wave of almost the same wavelength and frequency, we will get a beating wave (see Figure 7-2). For such a wave, physicists usually care more about its envelope, rather than the wave itself, since the envelope is what people will actually hear, or see. For example, the two branches of a tuning fork are almost, but not exactly identical. So if a tuning fork starts to vibrate, its two branches will produce two slightly different sound waves. The superposition of these two waves is a beating sound wave with varying amplitude. This is why people can hear "beats" when they strike a tuning fork.

A similar mechanism is used in Amplitude Modulation broadcasting. Different waves are superposed with each other to form a sinusoidal carrier wave with changing amplitude, which can be used to carry audio signals (see Figure 7-3). For more about broadcasting, please go here.

## Application in String Arts

 Figure 7-4String arts use straight lines to reprensent curves Figure 7-5A 3-D String Art Product

String Arts is a material representation of envelope, in which people arrange colored straight strings to form complicated geometric figures.

# How the Main Image Relates

As pointed out in the main image, the envelope of all particles' trajectories in an exploding firework is a parabola. Here comes the explanation:

Figure 8-1
Simulation of the blue-aerial-shell firework

Figure 8-1 shows a simulation of the exploding process. Blue parabolas are trajectories of particles, and the red parabola is their envelope.

For the envelope to be parabolic, we have to make several assumptions:

• The firework is consisted of many shinning particles, each projected from the origin at the same time, with same velocity v.
• Particles are subject to constant gravity, with gravitational acceleration g.
• Air friction can be neglected. In fact, this turns out to be an arguable assumption. Most firework particles are relatively small and light, so they could be significantly deflected by air friction. However, the case with air friction is way too complicated for this page. Besides, air friction can be neglected, at least for some fireworks with big and heavy particles such as blue-aerial-shell. So we can still accept this assumption and see what happens.

With the assumptions above, we can write out the trajectory of one particular particle:

$y = x{\tan \theta} - x^2{{g \over 2v^2}(1 + \tan^2 \theta)}$,

in which $\theta$ is the angle of projection. This is a result from simple mechanics. For more about projectile trajectory, please go here.

For now, let's leave Physics behind and focus on the curves themselves. In this trajectory, $\theta$ is the variable parameter. If we denote $\tan \theta$ by $t$, we can write out a family of curves (in level set form):

Eq 1        $F(x,y,t) = y - tx + {{g \over 2v^2}(1 + t^2)x^2} = 0$

Differentiate to get the boundary condition (see the More Mathematical Explanation section):

Eq 2        $F(x,y,t) = {gx^2 \over v^2}t - x = 0$

Substitute Eq 2 into Eq 1, after doing some algebra we can get:

$y = { v^2 \over 2g } - {gx^2 \over 2v^2}$,

which gives us a parabolic envelope of the particles' trajectories.

As we have discussed before, this is not true for all fireworks. Because of air friction, most fireworks have an envelope more like a sphere. Nonetheless, this parabolic pattern can be seen somewhere else, such as fountains or explosions. This analysis is also useful in the study of "safe domains" in projectile motion.

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