Ellipse [Resolved]

Hi,

I need to be able to calculate the amount of liquid remaining in an oval tank at various levels, so that I can graduate a dipstick.

I have obtained some help as follows: (I have difficulty in copying some of the symbols and have explained them where relevent)


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let's say your ellipse has it's centre on the origin, and goes through the points

x=0,y=b; x=0,y=-b; x=a,y=0; x=-a,y=0

i.e. it has width 2a and height 2b. now the equation of the ellipse is given by

(x/a)^2 + (y/b)^2 = 1

i can prove this from something else if you want, but usually it is taken as the defenition of an ellipse.

now you want to find the area between the line y=k and the ellipse, k<b. to do this rearrange the equation:

x=a@(1 - (y/b)^2 ) ("@" should be a square root sign which I do not have on my keyboard)

and concentrate on y and x both greater than 0. now the area between the "top" of the ellipse and the line y=k is given by

y=b y=k a@(1 - (y/b)^2 ) dy. ("y=b" & "y=k" are interlined and preceded by a squiggle which looks like a vertically extended "S")

you should be able to do this integral with a simple substitution - let y = bsinq ("q" is the greek letter Theta which stands for an angle)"
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My problem is that my knowledge of Integration is virtually non-existent. Could someone please help me to interpret what I actually have to work out so that I can write a programme to do it.

For an example, assume an elipse 2.5 metres wide and 1.5 metres in height.

Obviously all I need to know is how to calculate the relative area of the ellipse and the finding the volume will be easy.

VB Code:

Private Function Test(ByVal r1 As Decimal, _
                            ByVal r2 As Decimal, ByVal h As Decimal) As Decimal
        Dim decTemp As Decimal = (Math.PI * r1 * r2)
 
        decTemp -= (r1 * r2) * Math.Asin(1 - (h / r1))
 
        decTemp -= (r1 - h) * Math.Sqrt(h * (2 * r1 - h))
 
        Return decTemp
 
    End Function

how does this work? I adapted it from code that i used for a circle and i dont got very good math skillz.

VB Code:

'Tested it like this
 
Msgbox (2.5*1.5*Math.PI)
Msgbox Test(2.5,1.5,2.5)

try the BEST integral resource:

http://integrals.wolfram.com/

Otherwise, it's a bit hard to read that, would you be able to draw it in a picture, so it makes a bit more sense? Thanks

Hi ABX.


Many thanks. Please confirm what it is that your formula produces. I have tried several calculations and I think that it calculates the area between the line representing the level of the liquid and the horizontal axis - and assumes that the tank would be over half full. Am I correct?

Hi sql_lall,

"try the BEST integral resource:

http://integrals.wolfram.com/

Otherwise, it's a bit hard to read that, would you be able to draw it in a picture, so it makes a bit more sense? Thanks"

Imagine an ellipse x=0 y=4; x=0 y=-4; x=7 y=0; x=-7 y=0.

A line K lies at y=3

I need to know one of the following: the area lying between:

1. Under K and within the Ellipse; or

2. Above K and within the ellipse or

3. Between K and the horizontal (x) axis of the ellipse.

The site you gave is certainly impressive, but I just cannot see how to use it at the moment.

Many thanks.

I was not sure as I am not a math person :blush: so i googled and found this: http://mathforum.org/library/drmath/view/55296.html

A = a*b*(Pi/2 + Arcsin[d/a]) + b*d*sqrt(a^2-d^2)/a

A=Area of ellipes

a=Radius 1
b=Radius 2



so to apply it...

VB Code:

Private Function TestIt(ByVal r1 As Decimal, ByVal r2 As Decimal, ByVal d As Decimal) As Decimal
        'A = a*b*(Pi/2 + Arcsin[d/a]) + b*d*sqrt(a^2-d^2)/a
 
            Return r1 * r2 * (Math.PI / 2 + Math.Asin(d / r1)) + _
            (r2 * d * Math.Sqrt(r1 ^ 2 - d ^ 2) / r1)
 
    End Function

Hi ABX,


Brilliant site:) :) :)

Thanks a million.