Isolating Theta in Area of a Circle Segment

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Nasai

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Basically what i have to do is find out the equation for what the water level will be when a given volume of water is poured into a cylinder on its side (imagine a tanker truck). i dont need to worry about three dimensions, because the area of water coverage on the circles on either end of the cylinder is directly proportional to the volume of water in the cylinder (since the cylinder is on it's side, the length portion of the cylinder can be eliminated... the volume increase is represented by the area increase of any circular cutsection of the cylinder).

I figured the best way to find this out would be to use the equation for the area of a circle segment (a section of a circle is a portion of a circle cut by a chord or secant) to find the angle that is produced by two lines intersecting the points where the water level meets the wall of the circle, then taking the isosceles triangle formed by those two lines and the chord, splitting it down the middle and simply using cosine with my angle divided by 2 to find the length of the triangle's height. this would effectively give me the height of the water level for a given area of a circle (assuming the area starts at the bottom of the circle and follows the rules of gravity). at this point if you feel i've missed a simpler way to find the water level of water in a cylinder on it's side for a given volume of water, go ahead and skip t he rest of this post and feel free to fill me in xD.

anyway, this is all good, but the equation for the area of a circle segment gives you the area if you know the radius and the angle. obviously, i need to change the equation to give the angle by inputting area (i already know the radius anyway). i ran into a brick wall when i couldnt find a way to isolate the angle in the equation (this is where the angle is used to find the height of the triangle i mentioned earlier, which is subtracted from the radius to find the water level):
(assume A = area, r = radius of the circle, u = theta)
The equation of the area of a circle segment is: A = r[sup:3iw28yxw]2[/sup:3iw28yxw]u/2 - (r[sup:3iw28yxw]2[/sup:3iw28yxw]sin u)/2 (basically the area of the sector at the given angle (rad) subtracted by the triangular portion of that sector).
this can be expressed as: A = r[sup:3iw28yxw]2[/sup:3iw28yxw](u - sin u)/2
and then: A/.5r[sup:3iw28yxw]2[/sup:3iw28yxw] = u - sin u
and that's where i get stuck. i need u by itself on one side. that way, i can plug that function into u in the following equation to find the water level:
r - r[sup:3iw28yxw]2[/sup:3iw28yxw]cos(u/2) = h

again if there's an easier way to solve my original problem let me know. otherwise my whole problem comes down to isolating the angle in the equation for the area of a circle segment. thanks in advance
 
Nasai said:
Basically what i have to do is find out the equation for what the water level will be when a given volume of water is poured into a cylinder on its side (imagine a tanker truck). i dont need to worry about three dimensions, because the area of water coverage on the circles on either end of the cylinder is directly proportional to the volume of water in the cylinder (since the cylinder is on it's side, the length portion of the cylinder can be eliminated... the volume increase is represented by the area increase of any circular cutsection of the cylinder).

I figured the best way to find this out would be to use the equation for the area of a circle segment (a section of a circle is a portion of a circle cut by a chord or secant) to find the angle that is produced by two lines intersecting the points where the water level meets the wall of the circle, then taking the isosceles triangle formed by those two lines and the chord, splitting it down the middle and simply using cosine with my angle divided by 2 to find the length of the triangle's height. this would effectively give me the height of the water level for a given area of a circle (assuming the area starts at the bottom of the circle and follows the rules of gravity). at this point if you feel i've missed a simpler way to find the water level of water in a cylinder on it's side for a given volume of water, go ahead and skip t he rest of this post and feel free to fill me in xD.

anyway, this is all good, but the equation for the area of a circle segment gives you the area if you know the radius and the angle. obviously, i need to change the equation to give the angle by inputting area (i already know the radius anyway). i ran into a brick wall when i couldnt find a way to isolate the angle in the equation (this is where the angle is used to find the height of the triangle i mentioned earlier, which is subtracted from the radius to find the water level):
(assume A = area, r = radius of the circle, u = theta)
The equation of the area of a circle segment is: A = r[sup:2xn280l1]2[/sup:2xn280l1]u/2 - (r[sup:2xn280l1]2[/sup:2xn280l1]sin u)/2 (basically the area of the sector at the given angle (rad) subtracted by the triangular portion of that sector).
this can be expressed as: A = r[sup:2xn280l1]2[/sup:2xn280l1](u - sin u)/2
and then: A/.5r[sup:2xn280l1]2[/sup:2xn280l1] = u - sin u<<< That is a non-linear equation - did you learn how to solve those?
and that's where i get stuck. i need u by itself on one side. that way, i can plug that function into u in the following equation to find the water level:
r - r[sup:2xn280l1]2[/sup:2xn280l1]cos(u/2) = h

again if there's an easier way to solve my original problem let me know. otherwise my whole problem comes down to isolating the angle in the equation for the area of a circle segment. thanks in advance
Click to expand...
 
<<< That is a non-linear equation - did you learn how to solve those?
Click to expand...
yes. it's the trig that gets me. probably has something to do with a trig identity. i dont know though, im the one that needs help with it xD
 
There are two formulas you can use.

The area of a circle segment is given by \(\displaystyle A=\frac{1}{2}r^{2}({\theta}-sin{\theta})\)

The depth is known as a middle ordinate. It's formula is given by \(\displaystyle d=r(1-cos\frac{\theta}{2})\)

Now, if you know the radius and area of the segment, you can solve the top equation for theta and sub into the middle ordinate formula.

Then you have your depth.

Let's use a simple example. Let's let the radius = 1 and the area is 1.

\(\displaystyle \frac{1}{2}(1)^{2}({\theta}-sin{\theta})=1\)

\(\displaystyle {\theta}=2.55419595284\)

Sub this into the formula for depth:

\(\displaystyle (1)(1-cos(2.55419595284/2)=0.710505816972\)

The depth is about 8-1/2 inches. See?.

Of course for volume you would have to implement the length of the tank.
 
galactus said:
There are two formulas you can use.

The area of a circle segment is given by \(\displaystyle A=\frac{1}{2}r^{2}({\theta}-sin{\theta})\)

The depth is known as a middle ordinate. It's formula is given by \(\displaystyle d=r(1-cos\frac{\theta}{2})\)

Now, if you know the radius and area of the segment, you can solve the top equation for theta and sub into the middle ordinate formula.

Then you have your depth.

Let's use a simple example. Let's let the radius = 1 and the area is 1.

\(\displaystyle \frac{1}{2}(1)^{2}({\theta}-sin{\theta})=1\)

\(\displaystyle {\theta}=2.55419595284\)

Sub this into the formula for depth:

\(\displaystyle (1)(1-cos(2.55419595284/2)=0.710505816972\)

The depth is about 8-1/2 inches. See?.

Of course for volume you would have to implement the length of the tank.
Click to expand...

well yes, i suppose this is an alternate way of finding the depth using the angle. however, i dont understand how you solved for theta in the area of a circle segment equation. if i knew how to do that, my problem would already be solved :). could you explain how you did that without having isolated theta?
 
I know, it's not easy algebraically. I used Newton's method. Well, actually, I used my calculator to solve for theta, but to do it by manually you can use Newton's method. That works well. Use it with an initial guess of 2.5 and it converges very quickly. Do you know Newton's method?

I will use the example I used before with a radius of 1 and an area of 1, for simplicity.

\(\displaystyle \frac{1}{2}(1)(x-sin(x))=1, \;\ x-sin(x)=2, \;\ x-sin(x)-2=0\)

\(\displaystyle f(x)=x-sin(x)-2\). I will use x instead of theta for ease of typing.

\(\displaystyle f'(x)=1-cos(x)\)

Newton's method is \(\displaystyle x_{n+1}=x_{n}-\frac{f(x_{n})}{f'(x_{n})}\)

\(\displaystyle 2.5-\frac{(2.5)-sin(2.5)-2}{1-cos(2.5)}=2.5546720113\)

\(\displaystyle 2.5546720113-\frac{2.5546720113-sin(2.5546720113)-2}{1-cos(2.5546720113)}=2.55419598709\)

That ought to be good enough. Of course, you could use a guess of 1.5 or whatever and it would still converge.

\(\displaystyle 2.55419598709 \;\ radians \;\ = \;\ 146.344 \;\ degrees\)
 
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