Help with parabolic reflector

J

jwpaine

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Mar 10, 2007
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Hi.

For a non-school related project, I want to make a parabolic reflector that reflects light from like 25 mirrors to a single point.

Now: lets say I want to build a single parabola as defined by x^2

I want my depth to be 5, so: 5 = x^2 thus my diameter would be 2(sqrt(5))

D = 2(sqrt(5))
d = 5

Now according to my research, my focus would be close to: F= (D^2)/16(5) =
(0, 1/4)

Now here is what I need a tad bit of help on. For every point on my parabola, I need to find the angle that my little pieces of mirror should be glued at. (I believe it would be called finding the instantaneous rate of change at a given point) Or the line tangent to the parabola at that point... (tell me if I'm wrong)

I don't have any calculus experience.. I'm in Algebra III and having fun exploring beyond our in-class instruction, after school.

How would I go about finding the angle that each mirror should glued to to reflect the light to my focus of (0, 1/4).

I plan on having 4 parabolas arranged so that al parabolas' vertex meet at (0,0) and on each of the 4 parabolas I will build stands to mount the mirror at a specific angle, at that specific point.


Thanks for the help. I don't need construction ideas... only help mathematically, finding the angle of each mirror.


EDIT: This is what I was thinking.
Find the angle of the reflection from a point on the parabola to the point of the focus. And then glue my mirror at an angle, perpendicular to the angle of calculated reflection. ???
That way....the center of my mirror (the part glued) would be at the same angle as the parabola at that exact point on the curve.


Thanks, John.
 
Hello, John!

For a non-school related project, I want to make a parabolic reflector
that reflects light from like 25 mirrors to a single point. .Good for you!

Now: lets say I want to build a single parabola as defined by \(\displaystyle y \:=\:x^2\)

I want my depth to be 5, so: 5 = x^2 thus my diameter would be 2(sqrt(5))

Diameter = 2(sqrt(5)), depth = 5 .Right!

Now according to my research, my focus would be close to:
. . F= (D^2)/16(5) = (0, 1/8)
No, the focus is at (0, 1/4)


Now here is what I need a tad bit of help on. For every point on my parabola,
I need to find the angle that my little pieces of mirror should be glued at.

I believe it would be called finding the instantaneous rate of change at a given point,
or the slope of the line tangent to the parabola at that point... .Yes!

I don't have any calculus experience.. I'm in Algebra III and having fun
exploring beyond our in-class instruction, after school. .Great!

How would I go about finding the angle that each mirror should glued
to reflect the light to my focus of (0, 1/4)?
Click to expand...

For the parabola \(\displaystyle \,y\:=\:x^2\), the slope at any point \(\displaystyle x\) is given by: \(\displaystyle \:y' \:=\:2x\)


Examples:

At \(\displaystyle (1,\,1)\), the slope is: \(\displaystyle \,y'\,=\,2(1)\,=\,2\)

At \(\displaystyle (-1,\,1)\), the slope is \(\displaystyle \,y'\,=\,2(-1)\,=\,-2\)

At \(\displaystyle (\pm1.5,\,2.25)\), the slope is \(\displaystyle \,y'\,=\,2(\pm1.5)\,=\,\pm3\)


Good luck with your project!

 
Thanks a lot skeeter!
Thanks a lot soroban!


For the parabola \(\displaystyle \,y\:=\:x^2\), the slope at any point x is given by: \(\displaystyle \:y' \:=\:2x\)
Click to expand...

Is that called differentiating? I've seen that done on similar problems. Multiply the exponent by its coefficient and drop the degree by 1?

Thanks again, John.
 
HI.

I have decided on a depth of 4" and a diameter of 12" this gives me three points: (0,0),(-12,4),(12,4)

I found the equation for the parabola that goes through the three points by creating a system of equations for each x and y value in the ordered pair, and then solving for the three unknown coefficients, a,b,c.

I now have my quadratic (1/36)x^2

Now question: How do I find the distance between (12,4) and (0,0) along my quadratic curve: (1/36)x^2

Thanks, John.
 
Yeah, I know about the distance formula for a straight line....I knew I would need some type of calculus to find the distance along the curve.

Thanks for the links, Stapel.

Way beyond me... I guess I'll just guesstimate and live with a small percent error.

John.
 
Calc comes in handy here;

The arc length formula.

\(\displaystyle \L\\\int_{a}^{b}\sqrt{1+[\frac{d}{dx}]^{2}}dx\)

Find the derivative of x^2/36. It is x/18.

Square it and get x^2/324.

Integrate:

\(\displaystyle \L\\\int_{0}^{12}\sqrt{1+\frac{x^{2}}{324}}dx\)

I skipped the calculations required to arrive at the answer. Since you haven't had calc, you may not follow it.

\(\displaystyle \L\\\frac{1}{18}\int_{0}^{12}\sqrt{x^{2}+324}dx\approx{12.84}\)
 
Thanks a lot, Galactus!

That's beyond me, but it gave me a good indication of how far off my estimation was.

I can't wait to start calculus in college.

John.
 
I can't wait to start calculus in college.
Click to expand...


That sure isn't a phrase one hears much. That is great. We'll be here when you do.

All I hear is, "what am I ever going to use this for?". "Math sucks". "I hate math".

No one ever dare say they like math. Then they get looks as if they just said they enjoy being kicked in the groin.
 
Hey JW. You don't have to wait until college to begin learning calc. I self-taught myself Calc I in my spare time, then tested out of it. You could begin learning about limits and derivatives anyway.
 
Thanks guys!

I'll check out that book.

I currently have Schaum's Outlines precalculus and Schaum's Outlines calculus. I got a discount by buying them together.... the precalculus book is a great "outline" book, and I use it regularly for supplemental problem sets and explanations for the work we are doing in-class.

I have been accepted to the University of Maine, but I am planing on taking calculus this summer at a local community college...something to keep me busy between summer work and get me prepared for this fall.

I want to dedicate my life to mathematics. I may not be an elite mather, but I sure do enjoy learning new material, and looking back on how far I have come as a student.

Cheers,
John.
 
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