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
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload

<snip>
Hi Morris,
Answer is kinda mathy so I posted a jpg snapshot of the mathcad screen in abpw
ml
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
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.
You made my day!
-- Morris Dovey DeSoto Solar DeSoto, Iowa USA http://www.iedu.com/DeSoto/solar.html
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload

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
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
http://www.google.com/search?hl=en&q=parabola+%22arc+length%22
David Merrill

Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
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
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
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.
And you probably already know about this site that I stumbled across (DAGS "solar concentrators"): http://www.redrok.com/main.htm
David Merrill

Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload

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 -
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
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
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload

Site Timeline

HomeOwnersHub.com is a website for homeowners and building and maintenance pros. It is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.