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# Area of part of ellipse watch

1. Hi,

I have a project to graduate the dipstick for an eliptical oil tank.

Can anyone tell me if there is a formula for calculating the area of the portion of an ellipse contained in the space between a line parrellel to the horizontal axis, and the circumference?

Many thanks.
2. do you mean this area?
Attached Images

3. (Original post by elpaw)
do you mean this area?
Yes, I think that is what he meant.
4. Well there is a problem in that your dip stick will have readings closer together near the middle, since the increase in depth does not increase with the increase in volume linearly, due to the biconvex shape of the tank. To work out a specific formula, you'll need to know the exact shape of the tank.
5. hI,

Many thanks. I do know the exact measurements. Start off with 7' * 4' ellipse, 15 ' long.
6. This is your project. And who uses inches?
7. A = 1/16*[(l²+h²)(π-arcsin[4d/h.√(h/d-1)-8d²/h².√(h/d-1)])-(l²-h²)(4d/h.√(h/d-1)-8d²/h².√(h/d-1))]-4d/h.√(h/d-1).(h/2-d)

where h is the height of the ellipse, l is the length of the ellipse, and d is the distance from the top of the ellipse to your parallel line. remember angles are in radians.
8. (Original post by elpaw)
A = 1/16*[(l²+h²)(π-arcsin[4d/h.√(h/d-1)-8d²/h².√(h/d-1)])-(l²-h²)(4d/h.√(h/d-1)-8d²/h².√(h/d-1))]-4d/h.√(h/d-1).(h/2-d)

where h is the height of the ellipse, l is the length of the ellipse, and d is the distance from the top of the ellipse to your parallel line. remember angles are in radians.
crikey
9. (Original post by mik1a)
crikey
i know, there's probably a way to simplify it, im just too tired to do it, and ive got a lot of optics questions to do for tomorrow 9am
10. Hi,

"A = 1/16*[(l²+h²)(π-arcsin[4d/h.√(h/d-1)-8d²/h².√(h/d-1)])-(l²-h²)(4d/h.√(h/d-1)-8d²/h².√(h/d-1))]-4d/h.√(h/d-1).(h/2-d)

where h is the height of the ellipse, l is the length of the ellipse, and d is the distance from the top of the ellipse to your parallel line. remember angles are in radians."

Many thanks elpaw. I reckon I would have worked that out by the year 3000!!!
11. (Original post by taxes)
Hi,

"A = 1/16*[(l²+h²)(π-arcsin[4d/h.√(h/d-1)-8d²/h².√(h/d-1)])-(l²-h²)(4d/h.√(h/d-1)-8d²/h².√(h/d-1))]-4d/h.√(h/d-1).(h/2-d)

where h is the height of the ellipse, l is the length of the ellipse, and d is the distance from the top of the ellipse to your parallel line. remember angles are in radians."

Many thanks elpaw. I reckon I would have worked that out by the year 3000!!!
but dont use it yet, it could be wrong, i havent checked it.
12. (Original post by elpaw)
A = 1/16*[(l²+h²)(π-arcsin[4d/h.√(h/d-1)-8d²/h².√(h/d-1)])-(l²-h²)(4d/h.√(h/d-1)-8d²/h².√(h/d-1))]-4d/h.√(h/d-1).(h/2-d)

where h is the height of the ellipse, l is the length of the ellipse, and d is the distance from the top of the ellipse to your parallel line. remember angles are in radians.
WOW!
13. (Original post by elpaw)
A = 1/16*[(l²+h²)(π-arcsin[4d/h.√(h/d-1)-8d²/h².√(h/d-1)])-(l²-h²)(4d/h.√(h/d-1)-8d²/h².√(h/d-1))]-4d/h.√(h/d-1).(h/2-d)

where h is the height of the ellipse, l is the length of the ellipse, and d is the distance from the top of the ellipse to your parallel line. remember angles are in radians.

ok, now i feel insignificant
14. (Original post by bono)
WOW!
its just P4 polar integration
15. (Original post by elpaw)
its just P4 polar integration
Thought so - but I didn't want to steal your limelight!

Ben
16. Hi mik1a

(Original post by mik1a)
This is your project. And who uses inches?
99% of true Englishmen!!! Seriously, most householders with oil tanks prefer Imperial. By the way, you have obviously forgotten that ' is feet and " is inches!!!
17. Not forgotten, just thankfully I never learnt the needlessly complicated imperial system.
18. "A = 1/16*[(l²+h²)(π-arcsin[4d/h.√(h/d-1)-8d²/h².√(h/d-1)])-(l²-h²)(4d/h.√(h/d-1)-8d²/h².√(h/d-1))]-4d/h.√(h/d-1).(h/2-d)

where h is the height of the ellipse, l is the length of the ellipse, and d is the distance from the top of the ellipse to your parallel line. remember angles are in radians"

Hi elpaw,

Could you give me some explanation of the formula? It may be the way the forum prints certain characters.

What is the meaning of the "." in "[4d/h.√(h/d-1)" etc? I would not think you mean the 4d/h root of "h/d-1)".

Should there be brackets around "-8d²/h².√(h/d-1)])-(l²-h²)(4d/h.√(h/d-1)-8d²/h².√(h/d-1))" or did you intend the subtraction of
"-8d²/h²" to be carried out in isolation (as it appears in the formula)?

Why do you make a reference to "Radians"? The only angle reference is "[4d/h.√(h/d-1)-8d²/h².√(h/d-1)]". Do you mean I should convert that to radians?

19. (Original post by mik1a)
Not forgotten, just thankfully I never learnt the needlessly complicated imperial system.

Hi Mik1a,

But you have learned it, whether you know it or not. Virtually all formula derived from practical calculations, eg Pi are found from imperial type measurements rather than decimal. The Imperial is by far a more natural system, whilst the Metric is entirely unnatural. (e.g. the way the metric system decides the standard length of 1 metre, Originally a fraction (an Imperial term!!) of the Earth's circumference, now a physical specimin kept in Paris!!)
20. Hi All,

Hi all,

Some one kindly found the following for me

________________________________ _________________
"Subject: Re: Partial area of an ellipse

Thanks for writing to Ask Dr. Math, Phil.

The formula you seek is the following. Suppose the major and minor
axes of the ellipse have length 2*a and 2*b, and you seek the area
between the tangent at one vertex and a parallel line a + d units from
it, perpendicular to the major axis. Here -a <= d <= a. Then:

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

Notice that when d = a, and the parallel line is the tangent at the
other vertex, you get A = a*b*Pi, the well-known formula for the area
of the entire ellipse, and when d = 0, and the parallel line is the
minor axis, you get A = a*b*Pi/2, half the area of the whole ellipse.

To work with a tangent and a line both perpendicular to the minor
axis, swap a and b in the above discussion.

- Doctor Rob, The Math Forum
http://mathforum.org/dr.math/ "

________________________________ ____________________________

This seems to be totally outside of integration and produces a finite value - unless anyone knows otherwise.

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