# Parabolic Reflector

(Difference between revisions)
 Revision as of 12:37, 17 June 2009 (edit)← Previous diff Current revision (09:42, 26 June 2012) (edit) (undo) (28 intermediate revisions not shown.) Line 1: Line 1: - {{Image Description + {{Image Description Ready |ImageName=Parabolic Reflector Dish |ImageName=Parabolic Reflector Dish + |Image=Solardish.jpg |Image=Solardish.jpg - |ImageIntro=Solar Dishes such as the one shown use a paraboloid shape to focus the incoming light into a single collector. + + |ImageIntro=Solar Dishes such as the one shown use a parabolic shape to focus the incoming light into a single collector. + |ImageDescElem= |ImageDescElem= - [[Image:Parfocdir3.JPG|thumb|300px|left|Incoming beams of light perpendicular to the directrix bounce off the dish directly towards the focus.]] + The geometry of a [[Parabola|parabola]] makes this shape useful for solar dishes. If the dish is facing the sun, beams of light coming from the sun are essentially parallel to each other when they hit the dish. Upon hitting the surface of the dish, the beams are reflected directly towards the focus of the parabola, where a device to absorb the sun's energy would be located. - [[Image:Solarparab4.JPG|thumb|300px|left|Note that incoming beams reflect 'over' the line perpendicular to the parabola at the point of contact.]] + [[Image:Parfocdir3.JPG|thumb|300px|left|Figure 1: Incoming beams of light perpendicular to the directrix bounce off the dish directly towards the focus.]] - The geometry of a parabola makes this shape useful for solar dishes. If the dish is facing the sun, beams of light coming from the sun are essentially parallel to each other when they hit the dish. Upon hitting the surface of the dish, the beams are reflected directly towards the focus of the parabola, where a device to absorb the sun's energy would be located. + [[Image:Solarparab4.JPG|thumb|300px|left|Figure 2: Note that incoming beams reflect 'over' the line perpendicular to the parabola at the point of contact.]] - We can see why beams of light hitting the parabola-shaped dish will reflect towards the same point. A beam of light reflects 'over' the line perpendicular to the parabola at the point of contact. In other words, the angle the light beam makes with the perpendicular when it hits the parabola is equal to the angle it makes with same perpendicular after being reflected. + We can see why beams of light hitting the parabola-shaped dish will reflect towards the same point. A beam of light reflects 'over' the line perpendicular to the parabola at the point of contact. In other words, the angle the light beam makes with the perpendicular when it hits the parabola is equal to the angle it makes with same perpendicular after being reflected, as shown in Figure 2. - Near the bottom of the parabola the perpendicular line is nearly vertical, meaning an incoming beam barely changes its angle after being reflected, allowing it to reach the focus above the bottom part of the parabola. Further up the parabola the perpendicular becomes more horizontal, allowing a light beam to undergo the greater change in angle needed to reach the focus. + Near the bottom of the parabola the perpendicular line is nearly vertical, meaning an incoming beam is still nearly vertical after being reflected. Being nearly vertical allows it to reach the focus above the bottom part of the parabola. Further up the parabola the perpendicular becomes more horizontal, allowing a light beam to take on the more horizontal angles needed to reach the focus. - |ImageDesc= + - [[Image:Parabdiagram3.JPG|thumb|400px|right|Diagram for the proof]] + - [[Image:Parabdiagram4.JPG|thumb|400px|right|'''A''' represents equal angles: the line normal to the parabola has the same slope relative to the y-axis as the line tangent to the parabola has relative to the x-axis, as shown in the diagram.]] + - The fact that a parabolic reflector can collect light in this way can be proven. A proof follows: + - ::Begin with the equation of a parabola with focus at (0,p): + Parabolic reflectors are really parabolas rotated about their axis of symmetry (the y axis on this page's diagrams) to form a bowl-like shape known as a paraboloid. This shape is used in many modern application because of its ability to collect incoming information efficiently. For example, TV dishes reflect incoming television signals towards a receiver centered at the focus of the dish. - ::*$x^2=4py$ + - ::Taking the derivative with respect to x gives the slope of the tangent at any point on the parabola: + Parabolic reflectors can also work in reverse: if a light emitter is placed at the focus and shined inward towards the parabola, the light will be reflected straight out of the parabola, with the beams of light traveling parallel to each other. Headlights on cars often use this effect to shine light directly forward. - ::*$\frac{x}{2p} = \frac{dy}{dx}$ + |ImageDesc= - ::This slope of the tangent line is relative to the x-axis. When the slope is zero, the tangent line is parallel to the x-axis. The line normal to the parabola has the same slope relative to the y-axis as the line tangent to the parabola has relative to the x-axis. + The fact that a parabolic reflector can collect light in this way can be proven. We can show that any beam of light coming straight down into a parabola will reflect at exactly the angle needed to hit the focus, as follows: - ::We can use this slope to find the angle between the normal and the y-axis, which is the same as the angle between the normal and an incoming beam of light. The desired angle $\omega$ can be expressed as: + [[Image:Parabdiagram3.JPG|thumb|400px|right|Figure 3]] - ::*$\tan\omega = \frac{x}{2p}$ (Equation 1) + [[Image:Parabdiagram4.JPG|thumb|400px|right|Figure 4: '''A''' represents equal angles: the line normal to the parabola makes the same angle with the y-axis as the line tangent to the parabola has relative to the x-axis.]] - ::As mentioned previously, a beam of light is reflected 'over' the normal. This means that the angle the beam of light takes relative to a vertical line is equal to two times the angle the normal makes with the same vertical line. + - ::We now must show that the direction the light takes after being reflected is exactly the angle needed to hit the focus. Notice that geometrically, this angle $2\omega$ satisfies the relationship + - ::* $\tan2\omega = \frac{x}{p-x^2/4p}$ + - ::Which, by a trigonometric identity, also implies + - ::* $\tan2\omega = \frac{x}{p-x^2/4p} =\frac{2\tan\omega}{1-\tan\omega^2}$ (Equation 2) + - ::We now manipulate our Equation 1 for $\tan\omega$ to show its equivalence to Equation 2: + - ::* $2tan\omega = 2\frac{x}{2p} = \frac{x}{p}$ , and + - ::* $1 - \tan\omega^2 = 1 - (\frac{x}{2p})^2 = 1 - \frac{x^2}{4p^2}$ . Combining these two expressions gives + - ::* $\frac{\frac{x}{p}}{1 - \frac{x^2}{4p^2}} = \frac{x}{p-x^2/4p}$ + - ::Which is the same as the former expression for + + :'''Step 1''' + ::We begin with the equation of a parabola with focus at (0,p). + ::$x^2=4py$ + :'''Step 2''' + ::We take the derivative with respect to x, giving the slope of the tangent at any point on the parabola: + ::$\frac{x}{2p} = \frac{dy}{dx}$ + ::The slope of this tangent line is relative to the x-axis: when the slope is zero, the tangent line is parallel to the x-axis. The line normal to the parabola makes the same angle with the y-axis as the line tangent to the parabola has relative to the x-axis, as shown in Figure 4. + :'''Step 3''' + ::We use this slope to find the angle between the normal and the y-axis, which is the same as the angle between the normal and an incoming beam of light. + ::The desired angle $\theta$ can be expressed as: + ::$\tan\theta= \frac{x}{2p}$ (Equation 1) + ::As mentioned previously, a beam of light is reflected 'over' the normal, as shown in Figure 2. This means that the angle the beam of light takes relative to a vertical line is equal to two times the angle the normal makes with the same vertical line. + :'''Step 4''' + ::We now must show that the direction the light takes after being reflected is exactly the angle needed to hit the focus. + ::Notice from Figure 3 that geometrically, the angle needed to hit the focus is equal to $2\theta$, and satisfies the relationship + ::$\tan2\theta= \frac{x}{p-x^2/4p}$ + :'''Step 5''' + ::We use a trigonometric identity to rewrite the equation in Step 4: + ::$\tan2\theta= \frac{x}{p-x^2/4p} =\frac{2\tan\theta}{1-\tan^2\theta }$ (Equation 2) + :'''Step 6''' + ::We now manipulate Equation 1's expression for $\tan\theta$ to show its equivalence to Equation 2 (that is, to show the angle $\theta$ in Equation 1 is the same as the angle $\theta$ in Equation 2). + ::$2tan\theta= 2\frac{x}{2p} = \frac{x}{p}$ , and + ::$1 - \tan^2\theta = 1 - (\frac{x}{2p})^2 = 1 - \frac{x^2}{4p^2}$. + :'''Step 7''' + ::We combine the two expressions in Step 6. + ::$\frac{2\tan\theta}{1-\tan^2\theta} = \frac{\frac{x}{p}}{1 - \frac{x^2}{4p^2}} = \frac{x}{p-x^2/4p}$ + ::Which is the same as the expression in Equation 2. + :Therefore, a beam of light will hit the parabola's focus after being reflected. + |other=Elementary Trigonometry and Calculus. |AuthorName=Energy Information Administration |AuthorName=Energy Information Administration |SiteURL=http://www.eia.doe.gov/cneaf/solar.renewables/page/solarthermal/solarthermal.html |SiteURL=http://www.eia.doe.gov/cneaf/solar.renewables/page/solarthermal/solarthermal.html |Field=Geometry |Field=Geometry - |InProgress=Yes + |InProgress=No }} }}

## Current revision

Parabolic Reflector Dish
Field: Geometry
Image Created By: Energy Information Administration
Website: [1]

Parabolic Reflector Dish

Solar Dishes such as the one shown use a parabolic shape to focus the incoming light into a single collector.

# Basic Description

The geometry of a parabola makes this shape useful for solar dishes. If the dish is facing the sun, beams of light coming from the sun are essentially parallel to each other when they hit the dish. Upon hitting the surface of the dish, the beams are reflected directly towards the focus of the parabola, where a device to absorb the sun's energy would be located.
Figure 1: Incoming beams of light perpendicular to the directrix bounce off the dish directly towards the focus.
Figure 2: Note that incoming beams reflect 'over' the line perpendicular to the parabola at the point of contact.

We can see why beams of light hitting the parabola-shaped dish will reflect towards the same point. A beam of light reflects 'over' the line perpendicular to the parabola at the point of contact. In other words, the angle the light beam makes with the perpendicular when it hits the parabola is equal to the angle it makes with same perpendicular after being reflected, as shown in Figure 2.

Near the bottom of the parabola the perpendicular line is nearly vertical, meaning an incoming beam is still nearly vertical after being reflected. Being nearly vertical allows it to reach the focus above the bottom part of the parabola. Further up the parabola the perpendicular becomes more horizontal, allowing a light beam to take on the more horizontal angles needed to reach the focus.

Parabolic reflectors are really parabolas rotated about their axis of symmetry (the y axis on this page's diagrams) to form a bowl-like shape known as a paraboloid. This shape is used in many modern application because of its ability to collect incoming information efficiently. For example, TV dishes reflect incoming television signals towards a receiver centered at the focus of the dish.

Parabolic reflectors can also work in reverse: if a light emitter is placed at the focus and shined inward towards the parabola, the light will be reflected straight out of the parabola, with the beams of light traveling parallel to each other. Headlights on cars often use this effect to shine light directly forward.

# A More Mathematical Explanation

Note: understanding of this explanation requires: *Elementary Trigonometry and Calculus.

The fact that a parabolic reflector can collect light in this way can be proven. We can show that an [...]

The fact that a parabolic reflector can collect light in this way can be proven. We can show that any beam of light coming straight down into a parabola will reflect at exactly the angle needed to hit the focus, as follows:

Figure 3
Figure 4: A represents equal angles: the line normal to the parabola makes the same angle with the y-axis as the line tangent to the parabola has relative to the x-axis.
Step 1
We begin with the equation of a parabola with focus at (0,p).
$x^2=4py$
Step 2
We take the derivative with respect to x, giving the slope of the tangent at any point on the parabola:
$\frac{x}{2p} = \frac{dy}{dx}$
The slope of this tangent line is relative to the x-axis: when the slope is zero, the tangent line is parallel to the x-axis. The line normal to the parabola makes the same angle with the y-axis as the line tangent to the parabola has relative to the x-axis, as shown in Figure 4.
Step 3
We use this slope to find the angle between the normal and the y-axis, which is the same as the angle between the normal and an incoming beam of light.
The desired angle $\theta$ can be expressed as:
$\tan\theta= \frac{x}{2p}$ (Equation 1)
As mentioned previously, a beam of light is reflected 'over' the normal, as shown in Figure 2. This means that the angle the beam of light takes relative to a vertical line is equal to two times the angle the normal makes with the same vertical line.
Step 4
We now must show that the direction the light takes after being reflected is exactly the angle needed to hit the focus.
Notice from Figure 3 that geometrically, the angle needed to hit the focus is equal to $2\theta$, and satisfies the relationship
$\tan2\theta= \frac{x}{p-x^2/4p}$
Step 5
We use a trigonometric identity to rewrite the equation in Step 4:
$\tan2\theta= \frac{x}{p-x^2/4p} =\frac{2\tan\theta}{1-\tan^2\theta }$ (Equation 2)
Step 6
We now manipulate Equation 1's expression for $\tan\theta$ to show its equivalence to Equation 2 (that is, to show the angle $\theta$ in Equation 1 is the same as the angle $\theta$ in Equation 2).
$2tan\theta= 2\frac{x}{2p} = \frac{x}{p}$ , and
$1 - \tan^2\theta = 1 - (\frac{x}{2p})^2 = 1 - \frac{x^2}{4p^2}$.
Step 7
We combine the two expressions in Step 6.
$\frac{2\tan\theta}{1-\tan^2\theta} = \frac{\frac{x}{p}}{1 - \frac{x^2}{4p^2}} = \frac{x}{p-x^2/4p}$
Which is the same as the expression in Equation 2.
Therefore, a beam of light will hit the parabola's focus after being reflected.