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Math help please - parabola
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 |
On 17-Jul-2005, "Morris Dovey" wrote: 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 Hi Morris, Answer is kinda mathy so I posted a jpg snapshot of the mathcad screen in abpw ml |
http://www.google.com/search?hl=en&q...2arc+length%22
David Merrill "Morris Dovey" wrote in message ... 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. -- Morris Dovey DeSoto Solar DeSoto, Iowa USA http://www.iedu.com/DeSoto/solar.html |
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. 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 - |
David Merrill (in ogwCe.182504$xm3.89741@attbi_s21) said:
| "Morris Dovey" wrote in message | ... || 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...2arc+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 |
On 17-Jul-2005, "Morris Dovey" wrote: Mark... Thank you - that's *exactly* what I need! 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. ml |
Owen Lawrence (in ) 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 |
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/thesciencef...ola/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. And you probably already know about this site that I stumbled across (DAGS "solar concentrators"): http://www.redrok.com/main.htm David Merrill "Morris Dovey" wrote in message ... 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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