I'm loaded with codeine and my math and typing skills might not be up
to their usual standards.

https://www.dropbox.com/s/kvt3n3wmit0ideg/Sphere.jpg?dl=0

Suppose we have a spherical tank of r=1, and it is filling with
liquid. I can estimate angle a, where the fill intersects the wall.

Given a, what is the fluid height H, and what per centage of the
volume is filled?

fredag den 16. september 2022 kl. 16.51.24 UTC+2 skrev John Larkin:
> I'm loaded with codeine and my math and typing skills might not be up
> to their usual standards.
>
> https://www.dropbox.com/s/kvt3n3wmit0ideg/Sphere.jpg?dl=0
>
> Suppose we have a spherical tank of r=1, and it is filling with
> liquid. I can estimate angle a, where the fill intersects the wall.
>
> Given a, what is the fluid height H, and what per centage of the
> volume is filled?

h must be 1-cos(a)

area is a bit more involved
https://en.wikipedia.org/wiki/Circular_segment

On Fri, 16 Sep 2022 08:05:26 -0700 (PDT), Lasse Langwadt Christensen
<langwadt@fonz.dk> wrote:

>fredag den 16. september 2022 kl. 16.51.24 UTC+2 skrev John Larkin:
>> I'm loaded with codeine and my math and typing skills might not be up
>> to their usual standards.
>>
>> https://www.dropbox.com/s/kvt3n3wmit0ideg/Sphere.jpg?dl=0
>>
>> Suppose we have a spherical tank of r=1, and it is filling with
>> liquid. I can estimate angle a, where the fill intersects the wall.
>>
>> Given a, what is the fluid height H, and what per centage of the
>> volume is filled?
>
>h must be 1-cos(a)

Is that affected by the sphere shape?
>
>area is a bit more involved
>https://en.wikipedia.org/wiki/Circular_segment

And volume even worse!

(the fluid is of course vitreus humor, and the gas is S2F6)

On Fri, 16 Sep 2022 08:29:47 -0700, John Larkin
<jlarkin@highlandSNIPMEtechnology.com> wrote:

>On Fri, 16 Sep 2022 08:05:26 -0700 (PDT), Lasse Langwadt Christensen
><langwadt@fonz.dk> wrote:
>
>>fredag den 16. september 2022 kl. 16.51.24 UTC+2 skrev John Larkin:
>>> I'm loaded with codeine and my math and typing skills might not be up
>>> to their usual standards.
>>>
>>> https://www.dropbox.com/s/kvt3n3wmit0ideg/Sphere.jpg?dl=0
>>>
>>> Suppose we have a spherical tank of r=1, and it is filling with
>>> liquid. I can estimate angle a, where the fill intersects the wall.
>>>
>>> Given a, what is the fluid height H, and what per centage of the
>>> volume is filled?
>>
>>h must be 1-cos(a)
>
>Is that affected by the sphere shape?

Of course not; must be the drugs making me stupid.

>
>>
>>area is a bit more involved
>>https://en.wikipedia.org/wiki/Circular_segment
>
>And volume even worse!

Still true.

On Fri, 16 Sep 2022 10:22:46 -0700, John Larkin
<jlarkin@highlandSNIPMEtechnology.com> wrote:

>On Fri, 16 Sep 2022 08:29:47 -0700, John Larkin
><jlarkin@highlandSNIPMEtechnology.com> wrote:
>
>>On Fri, 16 Sep 2022 08:05:26 -0700 (PDT), Lasse Langwadt Christensen
>><langwadt@fonz.dk> wrote:
>>
>>>fredag den 16. september 2022 kl. 16.51.24 UTC+2 skrev John Larkin:
>>>> I'm loaded with codeine and my math and typing skills might not be up
>>>> to their usual standards.
>>>>
>>>> https://www.dropbox.com/s/kvt3n3wmit0ideg/Sphere.jpg?dl=0
>>>>
>>>> Suppose we have a spherical tank of r=1, and it is filling with
>>>> liquid. I can estimate angle a, where the fill intersects the wall.
>>>>
>>>> Given a, what is the fluid height H, and what per centage of the
>>>> volume is filled?
>>>
>>>h must be 1-cos(a)
>>
>>Is that affected by the sphere shape?
>
>Of course not; must be the drugs making me stupid.
>
>>
>>>
>>>area is a bit more involved
>>>https://en.wikipedia.org/wiki/Circular_segment
>>
>>And volume even worse!
>
>Still true.

Great problem !

boB

On Friday, September 16, 2022 at 1:22:54 PM UTC-4, John Larkin wrote:
> On Fri, 16 Sep 2022 08:29:47 -0700, John Larkin
> <jla...@highlandSNIPMEtechnology.com> wrote:
>
> >On Fri, 16 Sep 2022 08:05:26 -0700 (PDT), Lasse Langwadt Christensen
> ><lang...@fonz.dk> wrote:
> >
> >>fredag den 16. september 2022 kl. 16.51.24 UTC+2 skrev John Larkin:
> >>> I'm loaded with codeine and my math and typing skills might not be up
> >>> to their usual standards.
> >>>
> >>> https://www.dropbox.com/s/kvt3n3wmit0ideg/Sphere.jpg?dl=0
> >>>
> >>> Suppose we have a spherical tank of r=1, and it is filling with
> >>> liquid. I can estimate angle a, where the fill intersects the wall.
> >>>
> >>> Given a, what is the fluid height H, and what per centage of the
> >>> volume is filled?
> >>
> >>h must be 1-cos(a)
> >
> >Is that affected by the sphere shape?
> Of course not; must be the drugs making me stupid.
> >
> >>
> >>area is a bit more involved
> >>https://en.wikipedia.org/wiki/Circular_segment
> >
> >And volume even worse!
> Still true.

True- mathematical abilities suffer.

Getting back to that vertical cross-section thru the center. The radius to the fluid level makes an angle a with the vertical. As pointed out, the height is 1-cos(a). Also, the line drawn from the vertical through the center to the point of intersection is sin(a). Classically you then look at little horizontal cylinders of revolution that have radii sin(a) and thickness d(ifferential)h. Then the volume of cyclinder is pi*sin^2(a)*dh. Since h=1-cos(a), dh=sin(a)da, making the volume, as a function a, pi*sin^2(a)*sin(a)da. You volume is the integration (summation) of this function from a=0 to a=ao. the function in full being pi*sin^3(a) integrated with respect to a from a=0 to a=ao.
Looking up an answer that is:

1/3 cos^3(a)-cos(a)+constant , evaluated at ao and subtract of the evaluation at a=0 ( which is -2/3). So the volumes is then:
Volume=v=v(a)= pi*(1/3cos^3(a)-cos(a)+2/3)
Crude check:
at a=0 volume computes to 0 so that's right.
at a=90o volume computes to pi*2/3
Knowing the volume of sphere is 4/3*pi(R^3=1), half the sphere is pi*2/3
so the rough checks check out.

So to answer question about percentage of the volume filled, assuming by volume you mean the whole sphere, that would be:

pi*(1/3cos^3(a)-cos(a)+2/3)/(4/3*pi)= 1/4cos^3(a)-3/4cos(a)+1/2

Not going to try to simplify that any further...