MD

"Morris Dovey"

17/07/2005 10:28 AM

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


This topic has 8 replies

OL

"Owen Lawrence"

in reply to "Morris Dovey" on 17/07/2005 10:28 AM

17/07/2005 1:19 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.

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 -

MD

"Morris Dovey"

in reply to "Morris Dovey" on 17/07/2005 10:28 AM

17/07/2005 1:14 PM

David Merrill (in ogwCe.182504$xm3.89741@attbi_s21) said:

| "Morris Dovey" <[email protected]> wrote in message
| news:[email protected]...
|| 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

DM

"David Merrill"

in reply to "Morris Dovey" on 17/07/2005 10:28 AM

17/07/2005 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.

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" <[email protected]> wrote in message
news:[email protected]...
>
> 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
>
>

MD

"Morris Dovey"

in reply to "Morris Dovey" on 17/07/2005 10:28 AM

17/07/2005 1:02 PM

[email protected] (in [email protected]) said:

| On 17-Jul-2005, "Morris Dovey" <[email protected]> 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>
|
| 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

k

in reply to "Morris Dovey" on 17/07/2005 10:28 AM

17/07/2005 6:20 PM


On 17-Jul-2005, "Morris Dovey" <[email protected]> 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

k

in reply to "Morris Dovey" on 17/07/2005 10:28 AM

17/07/2005 4:49 PM


On 17-Jul-2005, "Morris Dovey" <[email protected]> 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

MD

"Morris Dovey"

in reply to "Morris Dovey" on 17/07/2005 10:28 AM

17/07/2005 1:36 PM

Owen Lawrence (in [email protected]) 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

DM

"David Merrill"

in reply to "Morris Dovey" on 17/07/2005 10:28 AM

17/07/2005 5:08 PM

http://www.google.com/search?hl=en&q=parabola+%22arc+length%22

David Merrill

"Morris Dovey" <[email protected]> wrote in message
news:[email protected]...
> 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
>
>


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