Anyone good at calculus?

Dondrei

Diabloii.Net Member
Re: Anyone good at calculus?

I don't get it, if it's a cylinder lying on its side then it's just the area of a circular segment times the length. Integrating would be trivial.
 

jakotaco

Diabloii.Net Member
Re: Anyone good at calculus?

I don't get it, if it's a cylinder lying on its side then it's just the area of a circular segment times the length. Integrating would be trivial.
Assuming the tank is full, yes. But as he asked I guess it is not. Although area times length is probably just as good as throwing integrals into the mix...



 

ffejrxx

Diabloii.Net Member
Re: Anyone good at calculus?

yes, but he doesnt know (or care about) the angle
he wants to figure the partial area of the circle from known radius, depth

then to get volume, multiply by the length of the tank
 

AeroJonesy

Diabloii.Net Member
Re: Anyone good at calculus?

This question would be much more interesting if the tank sat on an incline.
 

AeroJonesy

Diabloii.Net Member
Re: Anyone good at calculus?

Well it would be easy if we were given *any* fractional volume of the tank as being full.
 

jakotaco

Diabloii.Net Member
Re: Anyone good at calculus?

Oh, so that is a circular segment. Well, hmm I guess it should be possible to express the angle as a function of depth and radius.

I'll give it a try, but don't hold me responsible for any miscalculations (or typoes), I don't even have a paper at hand to go through my calculation...

So from Dondrei's link we get that:

Area = 1/2 * radius^2 * (angle - sin(angle))

Where the angle is the angle between the centrum of the circle and the ends of the water in such a way that for an empty tank angle = 0 and for a full tank angle = 2pi.

If we knew how wide the water was at the top (width) it would be "trivial." (4*arctan(1/2*width/depth) = angle)

And the width can probably be calculated through the intersecting chords theorem somehow. (depth*(2*radius-depth) = width^2*1/4)
=> width = 4*sqrt(depth*(2*radius-depth)
=> angle = (4*arctan(1/2*(4*sqrt(depth*(2*radius-depth)/depth)
=>Area = 1/2 * radius^2 * ((4*arctan(1/2*(4*sqrt(depth*(2*radius-depth)/depth) - sin((4*arctan(1/2*(4*sqrt(depth*(2*radius-depth)/depth)))

If this, (which I doubt somehow) is correct then we have a formula for the area that can be multiplied with the length of the cylinder. And yes, I guess it can be made a lot neater. Didn't use any papers for notes and calculations so i probably made some silly mistake aswell.

Well well, only took two days after my exams and then I was working on maths problems again... thought this was supposed to be my "free time".
 

AeroJonesy

Diabloii.Net Member
Re: Anyone good at calculus?

width is easy though. From a practical standpoint, if you can measure the height, you can just measure how wide the tank is at that point. No math required!
 

ffejrxx

Diabloii.Net Member
Re: Anyone good at calculus?

or if the tank is partially burried, use a dipstick to measure height, pull it out and you get the liquid level on the stick
 

Dondrei

Diabloii.Net Member
Re: Anyone good at calculus?

yes, but he doesnt know (or care about) the angle
he wants to figure the partial area of the circle from known radius, depth

then to get volume, multiply by the length of the tank
Depth? You mean from the centreline to the bottom of the tank? Use that to determine the angle. Think about it. Geometry is fun. :crazyeyes:

This question would be much more interesting if the tank sat on an incline.
That's what I thought. Still wouldn't require integrals though (in fact that would probably make it harder).

Now, if there were ripples in the water...

or if the tank is partially burried, use a dipstick to measure height, pull it out and you get the liquid level on the stick
Bah, applied science, get the hell out of here.



 
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