[RESOLVED] Circular segment height?

So i have an assignment at school. the first part of the assignment is finding out how many litres of liquid are in a cylinder. I have the following variables:

d = 50 cm (diameter of circle)
L = 120 cm (length of cylinder)
h = 10 cm (height of circular segment)

The Figure to the left is sideview of the cylinder, and the figure to the right is front view.
Attachment 120177

first part is easy.
Find out how much liquid is in the cylinder

Code:

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A=r^2/2 * ((pi*v)/180 - sin(v))
A=312,5*(1,854587-0,96)=279,5581 cm^2

V_(cylinder) = 279,5581 cm^2 * 120 cm = 33546,97 cm^3
V_(cylinder) = (33546,97 cm^3)/(1000 cm^3/L) = 33,54697 L

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now the second part is where i'm stuck at.
i need to find the new height of the circular segment, if i add 50 L liquid. so our new V_(cylinder) is 50 + 33,54697 = 83,54697 L.

how would i go about doing this? i can't exactly isolate V from the previous formula?

Not sure why you wrote your formula in terms of v (which you didn't quite define) rather than h when you want to solve for h. Either way you have a transcendental equation, which will at best have an inverse in terms of special functions. The upshot is, you should have a graphing calculator or numerical solver find the solution numerically. In terms of v (using radians, which is more natural than degrees), your equation is V = 312.5 * 120 * (v - sin(v))/1000, and you want this to be 83.54697, so Wolfram Alpha says the answer is v = 2.67646 (radians); figure out the equation for h in terms of v to find h.

sorry, i didn't quite think it through while i posted. v is angle.

this is the formula:
http://upload.wikimedia.org/math/a/7...7540d5faa5.png


in the first part, we know that the height of the circular segment is 10, and then we need to find what the area A of that is, so we can find out how much liquid is in the cylinder (there were 33 liters), therefore I use the formula that I did. then we add 50 L of liquid to the cylinder so that the liquid is now 83 Liters, however, we don't know the new height of the circular segment. the reason I want to find the new angle (v) is so I can use that to find the height of the new circular segment.

I know that the new height of the circular segment is 19,2 cm (that's what the checklist says) and the new angle, i found using a what-if analysis in excel, is 153,3499 degrees.

when i know the new angle, i can use this formula to find the height:

Code:

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h=r*(1-COS(v/2))
=25*(1-cos(153,3499/2))
=25*(1-cos(76,67495))
=25*(1-0,230475)
=25*0,769525
=19,238125 cm

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again v is angle. and it's in degrees, not radians.

So, basically what i want to know, is there a way to find the angle or the height of the circular segment, without using maple, excel or other software like that?

when i isolate the formula, this is what i get, and it's useless:

Code:

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A=r^2/2 * ((pi*v)/180 - sin(v))
A/(r^2/2)=((pi*v)/180 - sin(v))

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(the excel file i used for what-if analysis, see attachment)

Yes, I understood your original post. I'm saying to find v or h requires solving a transcendental equation, which in general will have to be done numerically (or using a special function, though you probably won't have access to the correct one without using a non-standard library). It's not like you can just throw the quadratic formula at it or take an inverse cosine. I did forget to divide by 1000 for the cm^3 to L conversion in my original post; I've updated my Wolfram Alpha link to reflect the correct version.

Anywho, the equation you want to solve is V = 312.5 * 120 * (v - sin(v))/1000 = 37.5 * (v - sin(v)). I'll continue to use radians since programs almost universally take input to trig functions in radians; feel free to convert to degrees. In any case, the right-hand side has weakly positive derivative, so it's monotonically increasing, so finding the solution numerically is actually very easy, and there is precisely 1 solution. At v=0, the right-hand side is 0. At v = V/37.5 + 1, the right-hand side is V + 37.5 * (1 - sin(v)), which is >= V. Hence the solution is between 0 and V/37.5 + 1; call these points A and B. If at v=(A+B)/2 the right-hand side is < V, you know the solution is between (A+B)/2 and B, so replace A with (A+B)/2 and repeat. On the other hand, if at v=(A+B)/2 the right-hand side is >= V, you know the solution is between A and (A+B)/2, so replace B with (A+B)/2 and repeat. This gives you one digit of binary precision per iteration; stop whenever you wish. This general method is called "quadrature".

Of course you can use cos/arccos to translate between v and h, so this suffices. Let me know if you need more details.