Quick Math question

[Ekliptix]

I have a sphere.
It's radius is 0.6m
It's total volume is 0.90478m^3
It's floating in a liquid and 0.61037m^3 of the sphere is submersed.

I need to know the distance the sphere is exposed above the liquid.

How?

[Ekliptix]

66 views. Is this impossible?

I think the Volumetric formula has to me manipulated somehow.
V = [4(pi)r^3]3
I think the r value needs to be turned into diameter and diameter needs to be broken into (r+distance sphere below water+ distance sphere above water).

Gah, fluid dynamics is owning me with this one.

[Audio_Rookie]

so you have 2 distance, above and below water right... Im having grade 12 flashbacks.

d = 2r, or r=1/2d

rxrxr = .5dx.5dx.5d


[4(pi)r^3]/3

Rbelow + Rabove = 1.20M

Rbelow = .60m - Rabove

V = [4(pi)Rabove^3]/3 + [4(pi)(.60m - Rabove)^3]/3

V = 4/3(pi)Rabove^3 + 4/3(pi)(.36m - .60mRabove - .60mRabove + Rabove^2)(.60m - Rabove)

V = 4/3(pi)Rabove^3 + 4/3(pi)(.36m - 1.20mRabove + Rabove^2)(.60m - Rabove)

V = 4/3(pi)Rabove^3 + 4/3(pi)( .216m - .36Rabove - .72mRabove + 1.2Rabove^2 + .60Rabove^2 - Rabove^3)

V = 4/3(pi)Rabove^3 + 4/3(pi)( .216m - 1.08Rabove + 1.8Rabove^2 - Rabove^3)

The problem I see is that you cant calculate the volume above or below the fluid using this formula because your not calculating the volum of a sphere, your calculating the volume of a semi-sphere...

perhaps it starts out something like that :dunno: ..I have never taken fluid dynamics..but this question sounds familiar....perhaps I will have more info after I go to school and fool with my calculator a little and actually think about it.

Im thinking all those calculations up there are a waste of time as i calculating 2 differant spheres, not 2 semispheres or parts of 1 sphere.

lets try this....
We know .5Dunder + .5Dover = .60M
.5Dunder = .60m - .5Dover where Dunder>Dover
V = 4/3(pi)((.5Dunder + .5Dover)^3)
V = 4/3(pi)((.25Dunder^2 + .25DunderDover + .25DoverDunder + .25Dover^2)
V = 4/3(pi)((.25Dunder^2 + .50DunderDover + .25Dover^2)(.5Dunder + .5Dover))
V = 4/3(pi)((.125Dunder^3 + .125Dunder^2Dover + .25Dunder^2Dover + .25Dover^2Dunder +.125Dover^2Dunder + .125Dover^3))
V= 4/3(pi)(.125Dunder^3 + .125Dover^3 + .375Dunder^2Dover + .375Dover^2Dunder)

We still know .5Dunder + .5Dover = .60M where Dunder is distance below fluid and D over is distance above...It also looks like perhaps some derivatives are gonna have to be taken...but im pretty screwed without any of my school stuff to do that.....besides...I gotta go to advanced algrebra class now...

[G]

I'll take a stab at this

Sphere1 has total volume of 0.90477 m^3

Now 0.61037m^3 is submerged therefor 0.90477 - 0.61037 = 0.29441 m^3 above water

If you draw it out you can visualize that the part of the sphere that is above the water can be seen as half of another sphere. We will call this Sphere2, now you know half the volume of Sphere2 is 0.29441 m^3 so all you need to do is find the radius of Sphere2 then you will know the height above the water.

The volume of Sphere2 is 2(0.29441) = 0.58882 m^3, now using the formula v=(pi/6)d^3 we can solve for d

d^3 = 6v/pi
d = the cube root of 6v/pi
d = (6x0.58882/3.14159) ^ 1/3
d = 1.03991 m
r = d/2
r = 0.51995 m

Now if you drop a perpendicular line from the top of Sphere2 to the centre (which is the water line) of Sphere2 you will find the radius which is 0.51995 m which is the height above water.

I am not 100% how right this is but it made a lot of sense to me.

[Ekliptix]

Audio_Rookie, I'm still getting my head around your calculations.
Thanks.


G, the problem with that is that sphere2 would not be shaped like a perfect sphere, it'd be more like a discus. The value you got is based on a real sphere shape which is taller/volume then that of the real disk shaped sphere2.

I appreciate the idea though.

[Audio_Rookie]

ya, I thought about that method to...but then I thought, if more than 1/2 the volume of the sphere is under water than you couldnt count the part of the sphere above water to be 1/2 a sphere because the sphere wouldnt be uniform....it would be cone shaped.

It would be nice if that worked...but in reality I would think that the answer will be slightly less than that, as your assuming a cone shaped object to be a sphere, and giving it a greater radius than it actually has.

back to doing guassian elimination with 12 variables. Having a wireless laptop is awsome...

[Ekliptix]

Someone suggested this formula but I'm trying to see how it makes sense:


r = (h^2+(s)^2)/(2h)

V = (Pi/6)(3(s)^2+h^2)h

h is what you want to solve for, is the height above the water. r is the radius (0.6m) and (s) is the radius of the circle made where the sphere contacts the water (the waterline on the sphere). ]

So: V=0.90478-0.61037=0.29441m^3=(pi/6)(3(s)^2+h^2)h

and

.6=(h^2+(s)^2)/(2h)

Solve the second equation for (r1) in terms of h, and plug that into the first equation, slove for h and you'll have your answer.

[Audio_Rookie]

lets try this....
We know .5Dunder + .5Dover = .60M
.5Dunder = .60m - .5Dover where Dunder>Dover
V = 4/3(pi)((.5Dunder + .5Dover)^3)
V = 4/3(pi)((.25Dunder^2 + .25DunderDover + .25DoverDunder + .25Dover^2)
V = 4/3(pi)((.25Dunder^2 + .50DunderDover + .25Dover^2)(.5Dunder + .5Dover))
V = 4/3(pi)((.125Dunder^3 + .125Dunder^2Dover + .25Dunder^2Dover + .25Dover^2Dunder +.125Dover^2Dunder + .125Dover^3))
V= 4/3(pi)(.125Dunder^3 + .125Dover^3 + .375Dunder^2Dover + .375Dover^2Dunder)

We still know .5Dunder + .5Dover = .60M
.5Dunder = .60m - .5Dover
so lets substitute...and see where it gets us.
V = 4/3(pi)(.125((.60m - .5Dover)(.60m - .5Dover)(.60m - .5Dover))) + .125Dover^3 + (.375((.60m - .5Dover)(.60m - .5Dover)(Dover)) + (.375)((Dover^2)(.60m - .5Dover))

now after you foil everything out and waste 10 minutes...luckily I have 10 minutes to waste because im in class....math is to easy once you know what your doing....to bad im not totally sure what im doing here... you get...

V = 4/3(pi)((.125(.36m - .3Dover - .3Dover + .25Dover^2)(.60m - .5Dover)) + .125Dover^3 + (.375((.36m -.3Dover - .3Dover + .25Dover^2)(Dover)) + (.375((.60Dover^2 - .5Dover^3)))

V = 4/3(pi)((.125(.36m - .6Dover + .25Dover^2)(.60m - .5Dover)) + .125Dover^3 + (.375((.36m - .6Dover + .25Dover^2)(Dover)) + .225Dover^2 - .1875Dover^3

V = 4/3(pi)((.125( .216m - .18Dover - .36Dover + .3Dover^2 + .15Dover^2 - .125Dover^3))) + .125Dover^3 + (.375((.36Dover - .6Dover^2 + .25Dover^3))) + .225Dover^2 - .1875Dover^3

V = 4/3(pi)((.125(.216m - .54Dover + .45Dover^2 - .125Dover^3))) + .125Dover^3 + .135Dover - .225Dover^2 + .09375Dover^3 + .225Dover^2 - .1875Dover^3

V = 4/3(pi)(.027m - 0.0675Dover + .05625Dover^2 - .015625Dover^3 + .125Dover^3 + .135Dover - .225Dover^2 + 0.09375Dover^3 + .225Dover^2 - .1875Dover^3)

collect like terms and add/subtract......hope it helps some....I showed way more work than I ever do on any of my assignemtns, but hopefully it makes it easier to udnerstand. There are alot of places for error, so I might have messed up a multiplication....

[69cougar]

:banghead: :banghead: :banghead:

Why did I eveb look?

Now I feel like :barf:

:guns: :guns: Stupid Math

[Audio_Rookie]

doh, made a big mistake way up there, substituted same formula in twice...what was I thinking.. why monday...

r = (h^2+(s)^2)/(2h)

V = (Pi/6)(3(s)^2+h^2)h

dont have any idea where he got those from, but they look like they could work.

looks like your basicalls gonna have to solve for S, then plugg in the big ass formula of S into the first formula/(or seconds if you solved for first), do all the great foiling and stuff, and h will be your answer.

assuming r1 is h with respect to s.....sounds like a guaranteed derivative question..

[Alterac]

http://mathworld.wolfram.com/images/eps-gif/SphericalCap_1001.gif

Then use

http://mathworld.wolfram.com/images/equations/SphericalCap/equation6.gif

and solve for h

radius, r = 0.6m
sphere is submersed, Vcap = 0.61037m^3


I think. :)

[nambis]

^^^^ exactly.

and just to clarify, this is NOT a fluid dynamics problem but rather a simple and straightforward geometry problem :D

[Ekliptix]

Originally posted by nambis
^^^^ exactly.

and just to clarify, this is NOT a fluid dynamics problem but rather a simple and straightforward geometry problem :D

heck, I've never seen that before. :dunno:

Thanks everyone, I'm going to bed now.;)