you didnt say how close it needs to be, but if it needs to be close,
everyone is assuming you have an ellipse and the formula that has been given
is based on that assumption. an ellipse has definition involving foci and
such. it is possible to have an oval shape that is not an ellipse.
for truly irregular shapes, its probably best to use graph paper and just
start counting squares. the closer it needs to be, the smaller the squares
on the graph paper. OR you could get tricky and start dunking the thing in
water and start calculating area based on displacement.
anyway, fair warning about using that formula. or maybe you meant to say
ellipse.
randy
"kklaric" <[email protected]> wrote in message
news:[email protected]...
> Dear All,
>
> I need some help calculating the area of an oval.
>
> If I have an oval that is 5" x 7" what is the surface area?
>
> Is it greater than or less than the surface area of a 6.5" circle?
> (pi*(6.5/2)^2=33.18")
>
> Thanks,
>
> David.
>
>
AutoCAD calls it 27.4889 . A 6.5 circle has 33.183 sq. in.
"kklaric" <[email protected]> wrote in message
news:[email protected]...
> Dear All,
>
> I need some help calculating the area of an oval.
>
> If I have an oval that is 5" x 7" what is the surface area?
>
> Is it greater than or less than the surface area of a 6.5" circle?
> (pi*(6.5/2)^2=33.18")
>
> Thanks,
>
> David.
>
>
In article
<[email protected]>,
[email protected] says...
> Dear All,
>
> I need some help calculating the area of an oval.
>
> If I have an oval that is 5" x 7" what is the surface area?
>
> Is it greater than or less than the surface area of a 6.5" circle?
> (pi*(6.5/2)^2=33.18")
>
> Thanks,
>
> David.
I couldn't remember, but curiosity got the better of me so- did a quick
search and found this:
http://www.pool-bright.com/pool-area.php?poolshape=oval
Yeah, it's for pools, but it seems to work if you plug in 0 for depth.
Vic
btw- it works out to be 29 for the oval, and 34 for a 6.5 round.
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"kklaric"
snip
> I need some help calculating the area of an oval.
>
> If I have an oval that is 5" x 7" what is the surface area?
If these are the width and height:
A=Pi x 5/2 x 7/2
=27.457
: "The only person sure of himself is the man who wishes to
: leave things as they are, and he dreams of an impossibility"
: -- George M. Wrong.
:
: Angelo Castellano emails - statsone@sympatico dot gov
: gov to be replaced with ca
: www.reliable-quality.com
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A = Pi * a * b where a & b are the semimajor(half the long) and semiminor(half the
short) axes.
Art
"kklaric" <[email protected]> wrote in message
news:[email protected]...
> Dear All,
>
> I need some help calculating the area of an oval.
>
> If I have an oval that is 5" x 7" what is the surface area?
>
> Is it greater than or less than the surface area of a 6.5" circle?
> (pi*(6.5/2)^2=33.18")
>
> Thanks,
>
> David.
>
>
In article <[email protected]>, "kklaric" <[email protected]> wrote:
>Dear All,
>
>I need some help calculating the area of an oval.
>
>If I have an oval that is 5" x 7" what is the surface area?
Do you mean an ellipse (a "stretched" circle, for lack of a better
description), or an oval (a rectangle with the corners rounded off)?
>
>Is it greater than or less than the surface area of a 6.5" circle?
>(pi*(6.5/2)^2=33.18")
>
If you mean an ellipse, the area is pi * (a/2) * (b/2) = pi * 5/2 * 7/2 = 27.5
square inches.
If you mean an oval, the area is between 29.6 and 35 square inches, depending
on the radius of curvature of the corners.
--
Regards,
Doug Miller (alphageek-at-milmac-dot-com)
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