Depth of liquid in hemisphere

Hi,

I have a hemisphere which is filled with a known volume of water. i.e. it's known to be a % of the volume of the whole hemisphere.

I need to work out the depth from the bottom of the tank.

I know it involves using the usual 2/3 * pi() * r^2 formula, but somewhere along the line the angle from the centre to the top of the liquid where it touches the tank is needed. i.e. the ACOS() or other trig. function is involved.

Can someone come up with the general formula.

i.e. depth = Volume * function

See attached.

Regards

The volume of a spherical cap is

=PI() * h^2 * (3*radius - h) / 3

... for h=0 to 2*radius

That's not the whole answer -- I'll be back.

Assuming you know the volume of the sphere.

The formula for partial volume of a sphere is

v = (pi/3)(d^2)(3r-d)

(same equation as shg's)

where d is the depth of the liquid

The full volume of a sphere is

V = 4pi(r^2)/3

Where I get stuck is solving for d when you have a known fraction of the full volume (p).

(pi/3)(d^2)(3r-d) = pV

Hi,

Thanks to all so far for their input but I believe the solution is a non linear one and will involve the use of a trig function, most likely the ACOS() function, to work out the angle between the perpendicular from the centre of the top plane of the hemisphere to the bottom of the hemisphere, and the line from the centre of the top plane to the point on the edge where the top of the water meets the edge.

It's non linear since the volume of liquid when the level rises from say empty to 1 foot, is less than the increase in volume when the liquid rises for the second foot. Similarly when the level rises for the third foot the volume increase is more than the volume increase from 1-2 feet (assuming of course that the hemisphere is not yet full).

I know that the volume of the hemisphere is 2/3*pi()*r^3 but a level half way between the bottom and the top contains different volumes above and below halfway. I'm trying to establish the formula for the depth of the liquid when I know the the volume of liquid as a % of the maximum capacity. e.g. what's the height when the henisphere is say half full in terms of volume?

Regards

Hi,

Yes thanks for pointing out the typo. It should of course have been r^3 not r^2

Regards

http://keisan.casio.com/has10/SpecExec.cgi shows how to calculate it the other way ie with the round bit cut off but you should be able to subtract that from the whole maybe?

Thanks Martin,

Was there a typo in that URL? It returns a page not found error message.

Regards

try this one
http://tiny.cc/08qgm

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Col B is an input, the ratio of the liquid depth to the sphere radius.

Col C computes the volume relative to the volume of a full hemisphere. In B3 and down,

=k^2 * (3 - k) / 2

Col D recovers the value in col B. In D3 and down,

=2 * COS(ACOS(1 - V) / 3 - 2*PI() / 3) + 1

Derivation attached.

EDIT: It occurs to me that you could get to the same answer a lot more directly by computing the area of the chord as a function of the height, integrating from 0 to pi to get the volume, and then solving for the height from the volume.

Hello Shg,

Thank you very much for that most detailed response. I'm really grateful.
Funnily enough whilst taking a bath last night (shades of Archimedes ), I'd been pondering the subject and was coming round to the pragmatic solution of creating the table you have so kindly worked out and using it as a lookup table. I was going to see how much the maximum error might be which I may be able to live with.

As you say the ultimate solution is some calculus which was going to be my final resort.

Regards

Not sure if I understand your follow-up Richard.

My point about integrating to get the volume was just that you could probably get the the same formula I ended up with more quickly if you computed the volume swept by the area, and then solved for the height.

The final formula to get normalized height from normalized volume is just simple trig -- no lookup table needed.

Thanks shg,

Yes ignore my previous remark. To avoid what I saw as a complicated transposition I'd originally thought of building a small granularity table of known volume proportions and the resultant liquid depth proportions, and using that as a lookup. I realise as you say that that's not necessary.

As an addendum how would the formula

=2 * COS(ACOS(1 - V) / 3 - 2*PI() / 3) + 1

vary if instead of a hemisphere, i.e. where depth = radius, the bottom shape of this tank is an ellipsoid where the depth is a smaller proportion of the radius. It's still a circular tank but the smaller radius (or depth) of the ellipsoid is say 10 and the radius 20, i.e. a squashed hemisphere?

I've tried playing around with the formula but can't seem to get the right results.

the bottom shape of this tank is an ellipsoid where the depth is a smaller proportion of the radius.

Do you mean an oblate spheroid (like the earth) or prolate ellipsoid (like a rugby ball),

1. ... with the axis vertical? That sounds doable, the volume is swept by the segment of an ellipse.

2. ... with the axis horizontal? That sounds hard.

And while we're at it, what's this for?

Hi,

Like a horizontal rugby ball, or to be precise the lower half of it.

Viewed from above looking straight down the plan view is a perfect circle. Looking at it from the side the height is less than the width.

Would you believe it's a pub argument? - Yes I know I know, with the euro collapsing around us we should have better things to do, but it was getting late (and heated after a few pints) and I happened to bet someone that there would be a fairly easy trig function to calculate it but and importantly it would involve an arc sin or cos function.

That will teach me......

Like a horizontal rugby ball, or to be precise the lower half of it.

Viewed from above looking straight down the plan view is a perfect circle.

A rugby ball would only look circular from above if it were standing on a pointy end, not lying on its side.

Sorry, I mixed up the descriptions. I should have said that the profile only looks like half a rugby ball from the side. The use of a rugby ball in hindsight was a bad one.

This is essentially an ellipsoid (I think that's the name) constructed from a 2 dimensional circle, and a smaller third radii perpendicular to the circle
I'm attaching a picture which hopefully helps. Looked at from above the plan view is a circle. We're looking from the side and slightly above the plane of the top.

Regards

OK, definitely a volume of revolution.

It may take me a few days to get back to this. Going to see Notre Dame vs Navy this weekend in South Bend.

Does this site shed any help?

http://en.wikipedia.org/wiki/Spherical_cap

Hi Marvin,

Thanks for the input, however I don't think the Wiki item quite addresses the point. The volume of a hemisphere is the well known 2/3 * pi() * r^3 (and a squashed sphere is similar).

The requirement however is to determine, for a known volume percentage of the whole volume, the proportional split of the radius. A hemisphere filled with liquid to 50% of it's volume does not mean that the level of the liquid is 50% of the radius. For this reason the arc COS function comes into play (or so my pub bet maintains!)

Regards

You could always cheat and use SOLVER

See this workbook (I haven't thoroughly tested it yet)
Change the values in A2:B2

@Richard:

The result for an prolate or oblate ellipsoid, oriented so as to be a circle in plan view, is the same as for the sphere. I can't decide if that seems obvious or curious.

I get half of your winnings.

shg.

Thanks for that - and your other posts. I suppose in hindsight it perhaps is obvious and I should have seen it.

You're very welcome to a pint or two and if I'm ever in your Texas location I'll be sure to let you know.

pip, pip.

Richard