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March 17th, 2011, 02:33 PM   #1
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Area for this shape

Hey I'm wondering if there is a equation or formula for finding the area to this certain shape. I've searched everywhere, and cannot find one. I've also tried to work it, but can't seem to figure it out.

Thanks in advance!

[attachment=0:2a7l3flc]odd cresent oval.png[/attachment:2a7l3flc]
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March 17th, 2011, 03:28 PM   #2
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Re: Area for this shape

Do you happen to know if the arcs are circular, elliptical, parabolic, hyperbolic, etc.?
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March 17th, 2011, 06:09 PM   #3
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Re: Area for this shape

Thank you so much for replying. What I'm trying to do is calculate the area of a half open open ball valve like this one. Can you see the shape I am trying to calculate the area of? Does that help?

[attachment=0:vfg4ap75]150px-Ball.PNG[/attachment:vfg4ap75]
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March 18th, 2011, 12:27 AM   #4
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Re: Area for this shape

Quote:
Originally Posted by NASAorbust
Thank you so much for replying. What I'm trying to do is calculate the area of a half open open ball valve like this one. Can you see the shape I am trying to calculate the area of? Does that help?

[attachment=0:vlmuxvqi]150px-Ball.PNG[/attachment:vlmuxvqi]
If the ball valve is really a circle, the surface you're trying to calculate the area is an ellipse. What's the radius of the valve? Is 0.5"? if it's half open the angle between the original normal vector and the "actual" is 45 and we can calculate the area knowing only this(or, equivalently, the diameter of the valve).
 
March 18th, 2011, 02:56 PM   #5
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Re: Area for this shape

Yes, the radius for the fully open valve is 0.5 inches. Could you explain how to calculate the area using that info? Could I use the equation for the area of an ellipse? Area = Pi*A*B
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March 18th, 2011, 03:38 PM   #6
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Re: Area for this shape

Quote:
Originally Posted by NASAorbust
Yes, the radius for the fully open valve is 0.5 inches. Could you explain how to calculate the area using that info? Could I use the equation for the area of an ellipse? Area = Pi*A*B
Yeah, you can use this. Use A as 0.5 inch (semimajor axis) and B as, I think, ( semiminor axis ). You can see that using elementary geometry. I can post the explanation if you want to.
 
March 18th, 2011, 03:50 PM   #7
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Re: Area for this shape

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Originally Posted by akpc
. . . the surface you're trying to calculate the area is an ellipse.
I disagree. The area in question is the area shared by two overlapping circles, both of radius 0.5 inches. What you want to do is calculate the area of the segment of one circle and multiply that by two.

http://en.wikipedia.org/wiki/Circular_segment#Area

R = 1/2, ? = ? - 2arctan(1/2), (radian measure used)

Area = 2 * 1/8 * (? - 2arctan(1/2) - 4/5) ? 0.3535743589.

You may need to account for the curvature of the ball.
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March 18th, 2011, 04:58 PM   #8
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Re: Area for this shape

Quote:
Originally Posted by greg1313
Quote:
Originally Posted by akpc
. . . the surface you're trying to calculate the area is an ellipse.
I disagree. The area in question is the area shared by two overlapping circles, both of radius 0.5 inches. What you want to do is calculate the area of the segment of one circle and multiply that by two.

http://en.wikipedia.org/wiki/Circular_segment#Area

R = 1/2, ? = ? - 2arctan(1/2), (radian measure used)

Area = 2 * 1/8 * (? - 2arctan(1/2) - 4/5) ? 0.3535743589.

You may need to account for the curvature of the ball.
Why the area is this?

We can prove that the area is an ellipse, and it's quite simple, if you want to I can do this. Try to think in that problem in another view: You're rotating a circle about a diameter and you don't change your look. The diameter(now major axis) will be always constant, the the minor axis will decrease. Rotating the sphere you'll get a family of ellipses and when you completely open the valve you'll get a straight vertical segment(a degenerated ellipse), right?
 
March 18th, 2011, 05:11 PM   #9
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Re: Area for this shape

A simple diagram should suffice to show the area is not an ellipse.
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March 18th, 2011, 05:21 PM   #10
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Re: Area for this shape

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Originally Posted by greg1313
A simple diagram should suffice to show the area is not an ellipse.
actually you're right. I took a calm look to the figure he posted and it was really not what I was thinking. I was thinking in a butterfly valve, not a ball one.

Like this.

 
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