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Height of a tank liquid by percent fill
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Our forum has been available for many years. In September 2014 we decided to switch it to read-only mode. Month after month we saw less posts with questions and answers from real people and more spam posts. We were spending more and more resources cleaning the spam until there were less them 1 legitimate message per 100 spam posts. Then we decided it's time to stop.
All the posts in the forum will be available and searchable. We understand there are a lot of useful information and we aren't going to remove anything. As for the new questions, you can always ask them on convert-me.com FaceBook page
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Height of a tank liquid by percent fill
by Daddio » Thu Jun 19, 2008 9:01 pm
I have a cylindrical tank but on it's side level and flat with a known diameter and volume. I need to solve for the actual physical height of a particular percent of the volume of the tank. For example, half full is half the length of the circumference but it's not linear beyond that point in either direction. If I need to know how high the liquid is if the tank is 28% full, what would be the formula? Thanks in advance.
- Daddio
Re: Height of a tank liquid by percent fill
by Guest » Fri Jun 20, 2008 2:13 am
Daddio wrote:I have a cylindrical tank but on it's side level and flat with a known diameter and volume. I need to solve for the actual physical height of a particular percent of the volume of the tank. For example, half full is half the length of the circumference but it's not linear beyond that point in either direction. If I need to know how high the liquid is if the tank is 28% full, what would be the formula? Thanks in advance.
It is not a pretty formula, and you may have to solve graphically, as it is easier to go from depth to percent full; hard to solve the other way.
If d is depth, and R is radius (half the diameter), introduce computational variable s = 1 - d/R.
(s = 1 when tank is empty, 0 when half full, and -1 when full.)
then V/Vmax = 1/(pi) * [arccos(s) -s*SQRT(1 - s²)] (Your calculator MUST be in radian mode when taking the arccos)
The arccos term varies from 0 to pi radians, the second term is a correction, but zero at s = -1, 0, 1.
I would just at intervals of s = 0.1 or 0.05 and estimate graphically. But I suppose you could differentiate the above equation (messy) and use Newton's method to solve iteratively.
- Guest
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