# Involute

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Involute
Field: Geometry
Image Created By: Chengying Wang

Involute

An involute of a circle can be obtained by rolling a line around the circle in a special way.

# Basic Description

Imagine you have a string attached to a point on a fixed curve. Then, tautly wind the string onto the curve. The trace of the end point on the string gives an involute of the original curve, and the original curve is called the evolute of its involute.

The animation on the right gives an example of an involute of a circle. We can see that a straight line is rotating around the center circle. This is like unwinding a string from a pole and keeping it taut the whole time. The trace of the end point of the line is the involute of the circle. In the image, it is the red spiral.

The involute is also the of a selected point on a line that rolls (as a tangent) along a given curve. Notice that while the rolling object can be anything (point, line, curve) for roulettes, it has to be a line to roll out involutes. For more information about roulettes, you can refer to the roulette image page.

## History of Involute

The involute was first introduced by Huygens in 1673 in his Horologium Oscillatorium sive de motu pendulorum, in which he focused on theories about pendulum motion.

Christiaan Huygens was a Dutch mathematician, astronomer and physicist. In 1656, he invented and built the world's first pendulum clock, which was the basic design for more accurate clocks in the following 300 years. Huygens had long realized that the period of a simple pendulum's oscillations was not constant, but rather depended on the magnitude of the movements. In other words, if the releasing height of the pendulum changes, the time for one oscillation changes as well. In search of improvements to the pendulum clocks, Huygens showed that the cycloid was the solution to this issue in 1659. He found that for a particle (subject only to gravity) to slide down a cycloidal curve without friction, the time it took to reach the lowest point of the curve was independent of its starting point. But how can we make sure the pendulum oscillates along a cycloid, instead of a circular arc? This was the point where the involute came in.

In Horologium Oscillatorium sive de motu pendulorum, Huygens proved that the involute of a cycloid is another cycloid (you will see an example in the More Mathematical Explanation section). To force the pendulum bob to swing along a cycloid, the string needed to "unwrap" from the evolute of the cycloid. As shown in the image on the right, he suspended the pendulum from the cusp point between two cycloidal semi-arcs. As a result, the pendulum bob traveled along a cycloidal path that was exactly the same as the cycloid to which the semi-arcs belonged. Thus, the time needed for the pendulum to complete swings was the same regardless of the swing amplitude. 

## To Draw an Involute

The involute of a given curve can be approximately drawn following the instructions below (there will be an example afterwards):

• Draw a number of tangent lines to the given curve.
• Pick a pair of neighboring tangent lines and set their intersection as the center. Then, with an endpoint at the point where one of those tangent lines meets the curve, draw an arc bounded by the two tangent lines. We call the line whose tangent point to the curve is also on the arc L1, the other tangent line L2, and the newly constructed arc Arc1.
• Pick the neighboring tangent of L2 and call it L3. Set the intersection of L2 and L3 as the center. Then draw an arc bounded by these two tangents, using a radius that will make this arc join Arc1. In other words, the radius would be the distance between the point where L2 meets L3 and the point where L2 meets Arc1.
• Repeat the step above for the rest of the tangents: pick the neighboring tangent, draw the arc; pick the next neighboring tangent, draw another arc...

Here is an illustration of the construction procedure above:  1. The black curve that points A, B, C, D, E, F lie on is the original curve. Draw the tangent line at each of the six points respectively. 2. We pick the tangents at points A and B (lines L1 and L2) as the first pair of neighbors. They intersect at point X. Then, setting point X as the center, draw an arc that passes point A and is bounded by the two tangents. We now have $\overset{\frown}{AA_1}$ (called Arc1 in the description above).  3. The next pair of neighboring tangents would be the tangent lines at points B and C (lines L2 and L3). Their intersection is point Y. Setting Y as the center, draw $\overset{\frown}{A_1A_2}$ so that A2 lies on the tangent of point C. 4. Repeat the steps above for the remaining tangents. We get the red curve, which is an approximate involute of the black curve.

This method does not produce the accurate involute curve because, for each of the line segments between the original curve and the involute (e.g. BA1, CA2, DA3...), instead of using the length of arc of the original curve as its length, we use the sum of the segments of the tangents. To be more concrete, the length of segment BA1 should equal the length of arc AB; the length of segment CA2 should equal the length of arc BC. This is because the line segment represents the part of the string that has just been unwound. It was originally wrapped along the fixed curve. As a result, its length should equal the length of the part of the original curve that it "covered" before. However, with the construction process described above, the lengths of these line segments are in fact sums of tangent segments. For example, the length of BA1 is found by adding XA and BX, which is shorter than it is supposed to be. Therefore, the distance between the involute and the original curve is not perfectly accurate. Fortunately, this error gets smaller as we construct more tangent lines that are closer to each other.

# A More Mathematical Explanation

In this section, we will derive the general formula for involutes, [...]

In this section, we will derive the general formula for involutes, and provide various examples of involutes and their equations. We will also introduce some properties of involute curves.

## General Formula for Involutes

This section derives the general equation for involutes of curves: ${\color{Gray}\vec r_{inv} = \vec r_{cur} - s\vec T}$
As an example, the equation of the involute of a circle is going to be derived immediately in the examples of involutes section.

Recall that we can think of the involute as the path of the end point of a string that is unwound from a fixed curve. So the tangent line segment we see between a point on the original curve and its corresponding point on the involute can be considered the part of the string that has just been unwound from the fixed curve. Therefore, the length of this line segment equals the distance traveled by the contact point between the unwound string and the fixed curve (the tangent point).

If we are given a curve, we construct its involute by unwinding a string tautly from the curve. For the point on the original curve that has Cartesian coordinates (f(t ), g(t )), its corresponding point on the involute (the endpoint of the string) can be found with this formula: $\vec r_{inv} = \vec r_{cur} - s\vec T$

where $\vec r_{cur}$ is the for a point on the original curve (where the unwound part of the string touches the curve), $\vec T$ is the to the original curve at this point (the current direction of the string), s is the distance traveled by this point so far (how much of the string has been unwound), and $\vec r_{inv}$ is the position vector for the corresponding point on the involute curve (the position of the end of the string). The image on the right takes a circle as an example and explains these variables visually. How each part is calculated is going to be explained below.

Note: For the following lines, we will use f to represent f(t ) so that the formulas look neater. Similarly, g(t ), f′(t ), g′(t ) are shortened to be g, f′ and g′ respectively.

Generally, the tangent vector for a curve with a position vector $\vec {x}(t)$ is defined as $\frac{d\vec{x}}{dt}$. If you are not very familiar with calculus, you can check these webpages to learn more about tangent vectors, derivatives, and velocity vector.
In the case we have, the tangent vector is $\frac{d\vec r_{cur}}{dt} = (f', g')$ Thus, the unit tangent vector is: $\vec T = \frac{\frac{d\vec r_{cur}}{dt}}{\left\vert\frac{d\vec r_{cur}}{dt}\right\vert}=\frac{(f',g')}{\sqrt{f'^2+g'^2}}$

The distance traveled by the contact point can be calculated by taking the integral of its speed. In the horizontal direction, its velocity is the derivative f′; in the vertical direction, its velocity is g′. Therefore, the velocity of the point written as a vector is (f′, g′) (We can see that its velocity is exactly the tangent vector). Its speed is: $\text{speed} = \sqrt{f'^2+g'^2}$

Thus, the point has traveled a distance of $s = \int_a^t\,ds = \int_a^t \frac{ds}{dt}\,dt = \int_a^t\sqrt{f'^2+g'^2}\,dt$

where f(a ) is the point where the involute and the curve intersect (the starting point of the unwinding process).

Therefore, if we write the vectors using Cartesian coordinates and plug in the results we get above for each term in the equation, we have $(x,y) = (f,g) - \int_a^t\sqrt{f'^2+g'^2}\,dt \cdot \frac{(f',g')}{\sqrt{f'^2+g'^2}} = (f,g)-\frac{s\cdot(f',g')}{\sqrt{f'^2+g'^2}}$

Hence, the parametric equation for the involute is $x = f - \frac{sf'}{\sqrt{f'^2+g'^2}}$ $y = g - \frac{sg'}{\sqrt{f'^2+g'^2}}$

In the examples of involutes section, you will see the derivation of the equation for the involute of a circle as an example.

## Examples of Involutes The image on the left shows the Involute of a Circle. It resembles an Archimedean spiral.
In the animation, we can see there is a point moving on the circle. The length of the rotating line segment should equal the distance this point has traveled.
This is easier to understand if we imagine again we are unwinding a string. The line segment we see is the part of the string that has just been unwound from the circle, and we can call the point where the string just stopped touching the original curve the contact point. Since the part of the string used to be wrapped on the curve, it should equal the length of the part it used to cover, that is, the distance the contact point has traveled as the string unwinds. This property holds true for all other types of curves as well, as we shall see.
The parametric equation for the involute of a circle with radius a is: $x = a(\cos t + t \cdot \sin t)$ $y = a(\sin t - t \cdot \cos t)$

We know that the parametric equation for a circle with radius a is: $x = a \cos t$ $y = a \sin t$

For any point on the circle, its position vector is $\vec r_{cur} = (a \cos t, a \sin t)$

Then its tangent vector (also the velocity of the point on the circle) is $\frac{d \vec r_{cur}}{dt} = (-a \sin t, a \cos t)$

The unit tangent vector is $\vec T = \frac{(-a \sin t, a \cos t)}{\sqrt{(-a \sin t)^2 + (a \cos t)^2)}} = (- \sin t, cos t)$

The speed of the point on the circle is $\sqrt{(-a \sin t)^2 + (a \cos t)^2)} = a$

The distance traveled by the point on the circle from its starting point is calculated by taking the integral of its speed: $s = \int _0^t \text{speed}\,dt = \int _0^t a \,dt = at$

Plugging these numbers into the general formula, which is $\vec r_{inv} = \vec r_{cur} - s\vec T$, we get the position vector of the corresponding point on the involute: $\vec r_{inv}$ $= (a \cos t, a \sin t) - at(- \sin t, cos t)$ $= a(\cos t + t \sin t, \sin t - t \cos t)$

Therefore, the parametric equation for the involute of a circle with radius a is: $x = a(\cos t + t \cdot \sin t)$ $y = a(\sin t - t \cdot \cos t)$

We can go through the same procedure to get the equations for all the involute curves below, but we are not going to derive all these equations in this page.

The involute of a parabola looks like the images on the left.

For example, if the parabola is $x = t$ $y = t^2$

its involute curve is $x = \frac{1}{4} (2t - \frac{\operatorname{arcsinh}\ 2t}{\sqrt{1 + 4t^2}})$ $y = -\frac{t \cdot \operatorname{arcsinh}\ 2t}{2\sqrt{1+4t^2}}$ On the other hand, if the parabola is $x = t^2$ $y = t$

its involute curve is $x = -\frac{t \cdot \operatorname{arcsinh}\ 2t}{2\sqrt{1+4t^2}}$ $y = \frac{1}{4} (2t - \frac{\operatorname{arcsinh}\ 2t}{\sqrt{1 + 4t^2}})$ The involute of an astroid is another astroid that is half of its original size and rotated 1/8 of a turn.

For example, if the parametric equation for the astroid is $x = \cos ^3 t$ $y = \sin ^3 t$

The involute of the astroid is $x = -\frac{1}{4}\cos t (-2 + \cos 2t )$ $y = \frac{1}{4} (2 + \cos 2t ) \sin t$ The involute of a cycloid is a shifted copy of the original cycloid.

If the cycloid is $x = t - \sin t$ $y = 1 - \cos t$

its involute curve is $x = t + \sin t$ $y = 3 + \cos t$ The involute of a cardioid is a mirrored, but bigger cardioid.

For example, the cardioid is given as $x = (1 + \cos t) \cos t$ $y = (1 + \cos t) \sin t$

Its involute is $x = \frac{1}{2} (1 + 6\cos t - 3\cos 2t)$ $y = 6 \sin \left (\frac{t}{2} \right )^2 \sin t$

For more examples of involutes, you can visit WolframMathWorld -- Involutes and Evolutes

## Properties of Involutes

 The normal of the involute is tangent to the original curve. From the procedure of drawing an involute, we know that the involute curve is joined by arcs that are bounded by pairs of tangent lines of the original curve and centered at their intersections. Therefore, the tangent lines are perpendicular to the arcs they bound (the radii of the arcs lie on these tangent lines). The image on the right shows the approximate involute curve that we drew. The line segments XA, BA1, CA2, DA3, EA4, FA5, GA6 are tangent to the original curve (the black curve). We know that XA is perpendicular to the red curve at point A because it is the radius of the circle that arc AA1 belongs to, and each radius is perpendicular to the arc at its endpoint. For similar reasons, BA1 is perpendicular to the red curve at point A1, and so on. Therefore, all the tangent line segments we constructed are perpendicular to the approximate involute curve. If we draw the involute curve accurately, we have an infinite number of tangent line segments and arcs. Using the same reasoning as above, we know that these tangent segments are all perpendicular to the circular arcs they meet. In other words, at each point on the involute, its normal is tangent to the original curve. Another way to understand this property is to imagine we are unwinding a string from a fixed curve. Because the string is held taut, the string should always be tangent to the fixed curve at their contact point and is perpendicular to the path of its end point. Therefore, we can conclude that the normal of the involute is tangent to the original curve. We can also think of involute as a curve orthogonal to all the tangents to a given curve. All involutes of a curve are parallel to each other. Every curve has many involutes because the initial point, where the involute intersects the original curve, can be chosen arbitrarily. The various involute curves obtained by choosing different initial points are parallel curves. In other words, they are a constant distance apart (this distance is measured along their common normal). For example, the image on the right shows various involutes of the same circle. The red curve represents one involute and the green one represents another. You would notice that instead of a spiral, each involute in the image has a cusp and this cusp separates the involute into two symmetrical parts. One part is what we have seen in this page (the main image and illustration of how to draw involutes). The other part is made by unwinding the string in the opposite direction. If you pick any two of the various involutes of the circle and find their common normal, the distance between the the points where the normal meets the two involutes would be a constant.
Why are the various involutes of a given circle parallel to each other?

We can approach this question through string unwinding again.

Since each involute of the circle is symmetrical, we can look at one side of the cusp first. That is, we unwind two strings from a circle following the same direction. The endpoints of the strings are initially at points A and B (see the image below on the left). After some unwinding, the line that starts at point A will pass point B. Then, it is easy to see that the involute curves are parallel as desired. Starting at point B, we can also think of the case as a long string (the one drawing out the red involute) and a short string (the one drawing out the blue involute) being unwound simultaneously from the circle, and the starting point for both of them is point B (see the image below on the right). Therefore, they are always a constant distance apart. The unwinding process starts at points A and B for the two strings. A long string and a short string are unwound simultaneously.
For similar reasons, curves on the other side of the cusps (formed by unwinding the string in an opposite direction) are a constant distance apart as well. If we put the two parts of the involute curves together, we can see that all the involutes of a circle are parallel to each other.

# Why It's Interesting

 One of the most commonly used gearing systems today is the involute gear. The image on the right is an example of such a gear. In an involute gear system, it is desired that the two wheels should revolve as if the two pitch-circles were rolling against each other. This effect can be achieved if the teeth profiles are drawn as involutes of the base-circles and the tops of the teeth are arcs of circles that are concentric with, and bigger than the pitch circles. To see why involute gears can revolve as if the pitch-circles are rolling against each other, consider two points, Q′ and R′, which will rotate to points Q and R in the same time interval. Point P is the point where the two teeth are in contact (it's called the pitch-point and will be explained later). If Q′Y and R′Z are the tangent lines to the base-circles at Q′ and R′, respectively, then $Q'Y + ZR' = QP + PR$ Due to how the involute is constructed (unwinding the string), we know that $Q'Y = \overset{\frown}{Q'Q} + QP$ and $ZR' = PR - \overset{\frown}{R'R}$ Therefore, $\overset{\frown}{Q'Q} = \overset{\frown}{R'R}$ Because Q′ and R′ move the same distance in the same time interval, we can conclude that Q and R move with equal velocities. Therefore, points on the pitch-circles will also move with same velocities. In the animation to the right, we can see that each pair of gear teeth has an instant contact point, called the pitch-point (point P in the hidden section above), and it moves along one single line as the gears rotate. This line is called the line of action and it is the common internal tangent of the two circles. In other words, the involutes of the two circles are always tangent to each other at a point on their common internal tangent. Because of this design, a constant velocity ratio is transmitted and the fundamental law of gearing is satisfied. Also, having all the contact points on a single straight line results in a constant force and pressure, while for gear teeth of other shapes, the relative speeds and forces change as teeth engage, resulting in vibration, noise, and excessive wear. Lastly, involute gears have the advantage that they are easy to manufacture since all the teeth are uniform. # Teaching Materials

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