What is the formula for finding the square inch area of an oval?

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  • #1
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I am trying to find a simple formula for finding the surface area in square inches of an oval.

L x W x .80?

Thanks for any help!
 

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  • #3
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? I get something neat:

[tex] A = \frac{\pi}{2} W(W + 2L) [/tex]

With W and L being measured in inches.

That's supposing that an oval consists of a half circle and a half ellipse.
 
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  • #4
CRGreathouse
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The area of an ellipse (two-dimensional "flattened" circle) is [tex]\scriptstyle{lw}\pi/4[/tex].

The surface area of an ellipsoid (three dimensional "flattened" sphere) is more complex; see the formula linked above.
 
  • #5
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Thanks for the links! I am not a math whiz...lol.

I am trying to find the surface area of a "oval" throttle body blade on a EFI vehicle?

What measurements will I need?
 
  • #6
uart
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The difficulty here stang is that an "oval" is not really a precisely defined mathematical entity. See http://mathworld.wolfram.com/Oval.html

Various well defined shapes such as ellipses and cycloids are "oval" like, but other mathematical forms could also be considered as oval, In other words the term "oval" is a little too broad to give an actual formula for surface area.

BTW. If the throttle body blade happens to fit snugly, at an oblique angle, inside a cylinder then it is in fact exactly an ellipse. As GRG has already pointed out the area of an ellipse is length times breadth times Pi divide 4. This is actually fairly close to what you originally guessed (L x W x 0.8), but it's actually closer to (L x W x 0.785).
 
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  • #7
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Okay, guys, resurrecting an old thread.

How can one find the area of a D shaped port?

Multiply the long side of the D, and the width, and then multiply by .75?
 
  • #8
CRGreathouse
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How can one find the area of a D shaped port?

Multiply the long side of the D, and the width, and then multiply by .75?
If it's a semicircular D, then the constant will be pi/4 ~= 0.7854. If there is a semicircle, then a (short) straight segment along the top and bottom, then the constant will be somewhat larger -- perhaps 0.8?
 

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