Volume of a Cylindrical TankDate: 3 Feb 1995 07:34:24 -0500 From: Blockwood Subject: Volume of partially filled cylinder on its side? Okay, here's a stumper. I have to keep an inventory of how much is kept in a farm of tanks outside my school. The tanks are cylindrical, which would be no problem if they were standing on end, but they're not, they're lying on their sides. xx x x x------------x Liquid level which is variable, x x can go up or down x x x x x x xx Imagine the above is a circle. I know the diameter of the circle, and I can measure from the top of the circle down to the liquid level. How can I use this information to derive the amount of liquid in the circle? I hope to make a formula I can use in a spreadsheet. Thanks so much. Date: 3 Feb 1995 19:51:37 GMT From: Dr. Math Subject: Re: Volume of partially filled cylinder on its side? Hello there! Thanks for writing to Dr. Math! You asked a good question. Before we start, let's go over the "law of sines" (we'll be using it shortly). Given a triangle ABC where A, B, and C are vertices, the following relation holds: sin A sin B sin C -------- = ------- = ------- A B C where sin A means sin of the angle at vertex A, and A (when it appears in the denominator) is the length of the side of the triangle opposite A. Okay, now on to the problem. You did a nice job with your drawing on the computer, so I'm not going to draw another picture. Instead, you draw a picture on paper and label the following: Call the distance from the top of the circle to the liquid level h. Call the radius of the circle r. (these are given quantities) Now, draw in the center of the circle. Draw a line connecting the center of the circle to the top of the circle. Then draw two more lines from the center of the circle to the points where the liquid level lines intersect the circle. Note that these lines are just radii, so they have length r. Now you have 3 triangles drawn, 2 of which are identical and make up the third. Look at one of the small triangles. Call the angle at the center angle z. One of the other angles is a 90 degree angle. Call the third angle y. Now, note that the distance between the center of the circle and the top of the circle is r (that's the definition of radius). We already said that the distance between the top of the circle and the liquid level was h, so that means the distance between the center of the circle and the liquid level must be r-h, right? So, our small triangle has a side of length r-h opposite the angle y, a side of length r opposite the 90 degree angle, and a side of length x opposite angle z. x is an unknown quantity we will need later on in the problem. By the law of sines, we have the following relations: r r-h x ----- = -------- = ----- * sin 90 sin y sin z sin 90 = 1, so we have: r-h ----- = r sin y Solve this equation for y to get: y = arcsin [(r-h)/r] Note also that z = 90 - y. And, sin z = sin (90 - y) = cos y = cos [arcsin [(r-h)/r]] = (1/r)Sqrt(2rh - h^2) (this is by trig identities and stuff...if you have questions, write back). So, then equation * becomes: rx r = -------------- Sqrt (2rh - h^2) So, x = Sqrt (2rh - h^2) Now the problem is a lot easier. To find the area of the whole circle, let's first consider the area contributed by the segment we've been talking about that is of angle 2z from the center. The area of this segment occupied by liquid is just the area of the 2 small triangles. We know the base of each triangle is x (see above) and the height is r-h. So, the total area contributed by this segment is x(r-h). Now consider the area outside of the segment. That is, we want the area of a portion of the circle. If we measure z in degrees we want the portion of the circle that takes up 360 - 2z degrees of the circle. So, the fraction we are dealing with here is: 360 - 2z -------- 360 The area of the portion we want then is: 360 - 2z -------- (Pi)r^2 360 Now add these two areas together and you are done. Thus, the total area is going to be: 360 - 2(90 - arcsin [(r-h)/r]) ------------------------------- (Pi)r^2 + (r-h)Sqrt (2rh - h^2) 360 You can simplify this to get something that looks nicer, but I'll leave that to you. Please do not hesitate to write back if anything is confusing or doesn't seem right, okay? Hope this helps! - Sydney, "Dr. Math" |
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