# Parabolic Reflector

(Difference between revisions)
 Revision as of 13:14, 16 June 2009 (edit)← Previous diff Revision as of 13:22, 16 June 2009 (edit) (undo)Next diff → Line 13: Line 13: |ImageDesc=The fact that a parabolic reflector can collect light in this way can be proven. A rough proof follows: |ImageDesc=The fact that a parabolic reflector can collect light in this way can be proven. A rough proof follows: - ::Begin with the equation of a parabola in terms of the location of the focus: + ::Begin with the equation of a parabola in terms of the location of the focus at (0,p): ::*$x^2=4py$ ::*$x^2=4py$ ::Taking the derivative with respect to x gives the slope of the tangent at any point on the parabola: ::Taking the derivative with respect to x gives the slope of the tangent at any point on the parabola: Line 19: Line 19: ::The line normal to the parabola at any point is perpendicular to the tangent line, having slope ::The line normal to the parabola at any point is perpendicular to the tangent line, having slope ::*$-\frac{2p}{x}$ ::*$-\frac{2p}{x}$ - ::The angle that this perpendicular line makes with the horizontal (made positive for simplicity) is: + ::The angle that this perpendicular line makes with the x-axis(made positive for simplicity) is: ::*$\theta = \arctan \frac{2p}{x}$ ::*$\theta = \arctan \frac{2p}{x}$ - ::The angle that this perpendicular line makes with the vertical is then + ::The angle that this perpendicular line makes with the y-axis is then ::*$\frac{\pi}{2} - \arctan \frac{2p}{x} = \omega$ ::*$\frac{\pi}{2} - \arctan \frac{2p}{x} = \omega$ ::This angle $\omega$ is the angle that the incoming beam of light would make with the perpendicular line. Two times $\omega$ gives the angle that the incoming beam of light makes with the beam after it is reflected. This is because the beam makes the angle $\omega$ with the perpendicular and after being reflected makes the same angle on the opposite side of the perpendicular. ::This angle $\omega$ is the angle that the incoming beam of light would make with the perpendicular line. Two times $\omega$ gives the angle that the incoming beam of light makes with the beam after it is reflected. This is because the beam makes the angle $\omega$ with the perpendicular and after being reflected makes the same angle on the opposite side of the perpendicular. - ::The angle the reflected beam makes with the horizontal, an angle we may compare to the angle needed to reach the focus, is then + ::The angle the reflected beam makes with x-axis, an angle we may compare to the angle needed to reach the focus, is then ::*$\frac{\pi}{2} - 2\omega = \frac{\pi}{2} -2(\frac{\pi}{2}-\arctan\frac{2p}{x}) = -\frac{\pi}{2}+2\arctan\frac{2p}{x} =\alpha$ ::*$\frac{\pi}{2} - 2\omega = \frac{\pi}{2} -2(\frac{\pi}{2}-\arctan\frac{2p}{x}) = -\frac{\pi}{2}+2\arctan\frac{2p}{x} =\alpha$ ::We compare this angle to the angle needed to reach the focus from the point of contact with the parabola. The focus is located at point (0,p) and the point of contact is at $(x,x^2/4p)$ The angle that a line through these two points makes with the horizontal, which should be the same as $\alpha$ above, is ::We compare this angle to the angle needed to reach the focus from the point of contact with the parabola. The focus is located at point (0,p) and the point of contact is at $(x,x^2/4p)$ The angle that a line through these two points makes with the horizontal, which should be the same as $\alpha$ above, is ::*$\arctan\frac{p-x^2/4p}{x}$ ::*$\arctan\frac{p-x^2/4p}{x}$ + When graphed, these two expressions, -\frac{\pi}{2}+2\arctan\frac{2p}{x} [/itex] and $\arctan\frac{p-x^2/4p}{x}$ are identical, providing a numerical verification of light reflecting towards the focus of a parabolic dish. |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

## Revision as of 13:22, 16 June 2009

Parabolic Reflector Dish
Solar Dishes such as the one shown use a paraboloid 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.

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.

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.

# A More Mathematical Explanation

The fact that a parabolic reflector can collect light in this way can be proven. A rough proof follo [...]

The fact that a parabolic reflector can collect light in this way can be proven. A rough proof follows:

Begin with the equation of a parabola in terms of the location of the focus at (0,p):
• $x^2=4py$
Taking the derivative with respect to x gives the slope of the tangent at any point on the parabola:
• $\frac{x}{2p} = \frac{dy}{dx}$
The line normal to the parabola at any point is perpendicular to the tangent line, having slope
• $-\frac{2p}{x}$
The angle that this perpendicular line makes with the x-axis(made positive for simplicity) is:
• $\theta = \arctan \frac{2p}{x}$
The angle that this perpendicular line makes with the y-axis is then
• $\frac{\pi}{2} - \arctan \frac{2p}{x} = \omega$
This angle $\omega$ is the angle that the incoming beam of light would make with the perpendicular line. Two times $\omega$ gives the angle that the incoming beam of light makes with the beam after it is reflected. This is because the beam makes the angle $\omega$ with the perpendicular and after being reflected makes the same angle on the opposite side of the perpendicular.
The angle the reflected beam makes with x-axis, an angle we may compare to the angle needed to reach the focus, is then
• $\frac{\pi}{2} - 2\omega = \frac{\pi}{2} -2(\frac{\pi}{2}-\arctan\frac{2p}{x}) = -\frac{\pi}{2}+2\arctan\frac{2p}{x} =\alpha$
We compare this angle to the angle needed to reach the focus from the point of contact with the parabola. The focus is located at point (0,p) and the point of contact is at $(x,x^2/4p)$ The angle that a line through these two points makes with the horizontal, which should be the same as $\alpha$ above, is
• $\arctan\frac{p-x^2/4p}{x}$

When graphed, these two expressions, -\frac{\pi}{2}+2\arctan\frac{2p}{x} [/itex] and $\arctan\frac{p-x^2/4p}{x}$ are identical, providing a numerical verification of light reflecting towards the focus of a parabolic dish.