Area of a segment of a circle

[gixerhowie]

It should be obvious to me but it isn't. My problem is this. I have a 130m long cylindrical water vessel with sealed ends laid flat. The vessel is 1.4m in diameter. The vessel has 400mm of water sitting in the bottom at its deepest point. I need to calculate the percentage of the overal volume that the water is occupying in the tank. Ignoring the length, i'm struggling to calculate the area of the segment. Drawing on AutoCad, i know the answer is 0.3783sq.m. which equates to 21% of the circles area. But how do i do the maths?

 

[Deleted member 4993]

It should be obvious to me but it isn't. My problem is this. I have a 130m long cylindrical water vessel with sealed ends laid flat. The vessel is 1.4m in diameter. The vessel has 400mm of water sitting in the bottom at its deepest point. I need to calculate the percentage of the overal volume that the water is occupying in the tank. Ignoring the length, i'm struggling to calculate the area of the segment. Drawing on AutoCad, i know the answer is 0.3783sq.m. which equates to 21% of the circles area. But how do i do the maths?

Draw the circle with center at O.

Draw a cord AB - to indicate the water level.

Join OA and OB.

Drop a perpendicular to AB from O - and extend it to the edge of the circle. Let it intersect AB at D.

OD = 300

OA = 700

Angle AOD = cos[sup:236tjd9r]-1[/sup:236tjd9r](300/700)

Now you know measure of angle AOB.

Find the area of the sector AOB (Piece of a pie). ................................(1)

Find the area of the triangle AOB...................................................(2)

(2) - (1) will give you the area of the water level.

 

[Denis]

I have a 130m long cylindrical water vessel with sealed ends laid flat. The vessel is 1.4m in diameter.
The vessel has 400mm of water sitting in the bottom at its deepest point.

I must be missing something...
Volume = pi r^2 h = pi(.7^2)(130) = ~200.12 ; how can you have 400 in there ?

 

[soroban]

Hello, gixerhowie!

I have a 130m long cylindrical water vessel with sealed ends laid flat.
The vessel is 1.4m in diameter.
The vessel has 400mm of water sitting in the bottom at its deepest point.
I need to calculate the percentage of the overal volume that the water is occupying in the tank.
Ignoring the length, i'm struggling to calculate the area of the segment.

              * * * 
          *           *
        *               *
       *                 * 

      *         O         *
      *         *         *
      *    r  * | *  r    *
            *   |   *  
       *  *     |     *  *
      A * - - - + - - - * B
          *     |d    * 
              * * *

The center of the circle is \(\displaystyle O.\)
The radius is: .\(\displaystyle OA = OB = r\)
The depth of the water is \(\displaystyle d.\)


Look at that triangle more closely . . .

                  O
                  *
                * | *
           r  * @ | @ *  r
            *     |     *
          *       |r-d    *
        *         |         *
    A * - - - - - * - - - - - * B
                  C

The height of the triangle is: \(\displaystyle OC \,=\,r-d\)

Let \(\displaystyle \theta \,=\,\angle AOC \,=\,\angle BOC\)

\(\displaystyle \text{Then: }\:\cos\theta \:=\:\frac{r-d}{r} \quad\Rightarrow\quad \theta \:=\:\cos^{-1}\!\left(\frac{r-d}{r}\right) \quad (\theta\text{ is in }radians.)\)


\(\displaystyle \text{The area of the sector is: }\:A_{\text{sector}} \:=\:\tfrac{1}{2}r^2(2\theta) \:=\: r^2\theta\)

\(\displaystyle \text{The area of the triangle is: }\:A_{\Delta} \:=\:\tfrac{1}{2}r^2\sin2\theta\)


\(\displaystyle \text{Therefore: }\;A \;=\;r^2\theta - \tfrac{1}{2}r^2\sin2\theta\)

. . \(\displaystyle \text{where: }\;\theta \:=\:\cos^{-1}\left\!\!(\tfrac{r-d}{r}\right)\)

 

[gixerhowie]

Thanks chaps. I found this way aswell:

.7sq/2*(pie/180*124-sin 124)

but this way i need to obviously calculate the angle from the circle centre to the insersection of the base line.

It works which is good. Learn something new everyday. I knew i should have paid more attention at school!