# Math help please - parabola

• posted on July 17, 2005, 3:28 pm

I'm trying to build a circular parabolic dish solar concentrator. So far I've designed a simple hub to which I'll attach radial ribs. These ribs will support reflective "petals" and an outer support ring.
I know that the general formula for a parabola is:
(x - h) ^ 2 = 4 * a * (y - k), where
(h,k) are the coordinates of the vertex and a is the distance from the vertex to the focus and the distance from the vertex to the directrix (and one-fourth the length of the latus rectum).
What I need is an algebraic expression representing the distance along the parabola from the vertex to any point (x,y) on the parabola so I can lay out the shape of a petal for cutting from flat stock.
-- Morris Dovey DeSoto Solar DeSoto, Iowa USA http://www.iedu.com/DeSoto/solar.html
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• posted on July 17, 2005, 4:49 pm

<snip>
Hi Morris,
Answer is kinda mathy so I posted a jpg snapshot of the mathcad screen in abpw
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• posted on July 17, 2005, 6:02 pm
snipped-for-privacy@yahoo.com (in snipped-for-privacy@giganews.com) said:
| || I'm trying to build a circular parabolic dish solar || concentrator. So || far I've designed a simple hub to which I'll attach radial || ribs. These || ribs will support reflective "petals" and an outer support || ring. || || I know that the general formula for a parabola is: || || (x - h) ^ 2 = 4 * a * (y - k), where | | <snip> | | Answer is kinda mathy so I posted a jpg snapshot of the | mathcad screen in abpw
Mark...
Thank you - that's *exactly* what I need!
It's been more than three decades since I last needed to work with definite integrals; and a bit of review is in order - but your help will allow me to complete a prototype in just a day or two.
"Mathy" is ok. Trial and error cutting with a 16-petal assembly would have been really nasty. Having the general formula will allow me to make reflectors with /any/ number of "petals" and even build accurate reflectors with concentric rings of petals.
-- Morris Dovey DeSoto Solar DeSoto, Iowa USA http://www.iedu.com/DeSoto/solar.html
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• posted on July 17, 2005, 6:20 pm

glad to be of help. MathCad does all that symbolically, and I've got some ancient version of it. Can't imagine what it does now.
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• posted on July 17, 2005, 5:08 pm
David Merrill

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• posted on July 17, 2005, 6:14 pm
David Merrill (in ogwCe.182504\$xm3.89741@attbi_s21) said:
| || snip... || || What I need is an algebraic expression representing the distance || along the parabola from the vertex to any point (x,y) on the || parabola so I can lay out the shape of a petal for cutting from || flat stock. | | http://www.google.com/search?hl=en&q=parabola+%22arc+length%22
Dave...
Thanks - I'd done a google search and hadn't found an article that I could regognize as a solution to my problem (may be a vocabulary problem on my part) and guessed (correctly) that this would be a good forum in which to ask.
It's interesting to note that a query to rec.woodworking produced a usable solution immediately, while the same query to sci.math (where it's probably more topical) hasn't produced any response at all.
-- Morris Dovey DeSoto Solar DeSoto, Iowa USA http://www.iedu.com/DeSoto/solar.html
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• posted on July 17, 2005, 7:59 pm
You have probably already noted that there are several forms for expressing the equation of a parabola, some perhaps more convenient than others for deriving the associated arc length expression in a compact form for programming your Shopbot.
You might find these sites helpful for optical effects visualization: http://www.geocities.com/thesciencefiles/parabola/focus.html http://www.cut-the-knot.org/ctk/Parabola.shtml Amateur telescope makers (ATM) are often very interested in parabolic mirrors as in Newtonian telescopes.
David Merrill

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• posted on July 17, 2005, 5:19 pm

I'm not checking my work, but I'll give it a shot. First, let's simplify a bit by locating your vertex at the origin. Your parabola is then y x^2/4a. The formula for arc length can be derived to be
L = int( x1, x2 ) sqrt( 1 + (dy/dx)^2) dx, where int( x1, x2 ) is the definite integral from x1 to x2 on your parabola.
dy/dx = x/2a (dy/dx)^2 = x^2/4a^2
so L = int( x1, x2 ) sqrt( 1 + x^2/4a^2 ) dx = 1/2a int( x1, x2 ) sqrt( 4a^2 + x^2 ) dx
I looked up the indefinite integral in my CRC math tables to be
int() sqrt( x^2 + c^2) dx = 1/2[ x sqrt( x^2 + c^2 ) + c^2 ln( x + sqrt( x^2 + c^2 ) ]
In our case c = 4a^2. Since you want your arc length to be from the vertex, we can let x1 = 0. So our final formula becomes:
L = 1/4a[ x sqrt( x^2 + (2a)^4 ) + (2a)^4 ln( x + sqrt( x^2 + (2a)^4 ) ]
Check the algebra. I haven't had lunch yet so maybe I'm not thinking clearly.
- Owen -
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• posted on July 17, 2005, 6:36 pm
Owen Lawrence (in snipped-for-privacy@quag.dido.ca) said:
| I'm not checking my work, but I'll give it a shot. First, let's | simplify a bit by locating your vertex at the origin. Your | parabola is then y = x^2/4a. The formula for arc length can be | derived to be | | L = int( x1, x2 ) sqrt( 1 + (dy/dx)^2) dx, where int( x1, x2 ) is | the definite integral from x1 to x2 on your parabola. | | dy/dx = x/2a | (dy/dx)^2 = x^2/4a^2 | | so L = int( x1, x2 ) sqrt( 1 + x^2/4a^2 ) dx | = 1/2a int( x1, x2 ) sqrt( 4a^2 + x^2 ) dx | | I looked up the indefinite integral in my CRC math tables to be | | int() sqrt( x^2 + c^2) dx = 1/2[ x sqrt( x^2 + c^2 ) + c^2 ln( x + | sqrt( x^2 + c^2 ) ] | | In our case c = 4a^2. Since you want your arc length to be from | the vertex, we can let x1 = 0. So our final formula becomes: | | L = 1/4a[ x sqrt( x^2 + (2a)^4 ) + (2a)^4 ln( x + sqrt( x^2 + | (2a)^4 ) ] | | Check the algebra. I haven't had lunch yet so maybe I'm not | thinking clearly.
Owen...
Thank you. Even without lunch you did better than I managed. :-)
-- Morris Dovey DeSoto Solar DeSoto, Iowa USA http://www.iedu.com/DeSoto/solar.html