Liquid height for a given volume.

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oilytin

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Apr 21, 2020
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Hi, First of all apologies if the is posted in the wrong category (not sure if it should have been in geometry)
I use a spreadsheet which contains the following formula to calculate the volume of a liquid for a given level in a horizontal cylindrical tank.
what i would like to do is make D the subject to allow me to do the opposite and calculate the level for a given volume.
\(\displaystyle V=L(R^{2} cos^{-1}\bigg(\frac{R-D}{R}\bigg)-(R-D)\sqrt{2RD-D^{2}})\)
I jumped into this thinking i should be at least be able to make some progress but i just cant begin to understand how to untangle this.
Could someone give me some pointers if its possible (Mathway was unable to help)
Thanks
Anthony
 
oilytin said:
Hi, First of all apologies if the is posted in the wrong category (not sure if it should have been in geometry)
I use a spreadsheet which contains the following formula to calculate the volume of a liquid for a given level in a horizontal cylindrical tank.
what i would like to do is make D the subject to allow me to do the opposite and calculate the level for a given volume.
\(\displaystyle V=L(R^{2} cos^{-1}\bigg(\frac{R-D}{R}\bigg)-(R-D)\sqrt{2RD-D^{2}})\)
I jumped into this thinking i should be at least be able to make some progress but i just cant begin to understand how to untangle this.
Could someone give me some pointers if its possible (Mathway was unable to help)
Thanks
Anthony
Click to expand...
Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

https://www.freemathhelp.com/forum/threads/read-before-posting.109846/#post-486520

Please share your work/thoughts about this assignment.
 
This is as much as I have tried at the moment.

\(\displaystyle \frac{V}{L}=R^{2} cos^{-1}\bigg(\frac{R-D}{R}\bigg)-(R-D)\sqrt{2RD-D^{2}}\)

\(\displaystyle \frac{V}{LR}=R cos^{-1}\bigg(\frac{R-D}{R}\bigg)-(R-D)\sqrt{2RD-D^{2}}\)

\(\displaystyle \bigg(\frac{V}{LR}\bigg)^2=R^2 cos^{-1}\bigg(\frac{R-D}{R}\bigg)^2-(R-D)(2RD-D^{2})\)

I really dont have a lot of confidence that i have got these first steps correct.
 
oilytin said:
This is as much as I have tried at the moment.

\(\displaystyle \frac{V}{L}=R^{2} cos^{-1}\bigg(\frac{R-D}{R}\bigg)-(R-D)\sqrt{2RD-D^{2}}\)

\(\displaystyle \frac{V}{LR}=R cos^{-1}\bigg(\frac{R-D}{R}\bigg)-(R-D)\sqrt{2RD-D^{2}}\)

\(\displaystyle \bigg(\frac{V}{LR}\bigg)^2=R^2 cos^{-1}\bigg(\frac{R-D}{R}\bigg)^2-(R-D)(2RD-D^{2})\)

I really dont have a lot of confidence that i have got these first steps correct.
Click to expand...
Unfortunately, those steps are not correct.

You said you are using a spreadsheet. If it is MS-Excel, there is a function there called "goal-seek". You should be able to use that and get an estimation of the "inverted number" - I assume D in this case.

Please try that and let us know what you find.
 
Thanks for confirming my suspicions, quickly realised i had bitten off more than i could chew.
The goal seek function you mentioned does indeed find an acceptable value (accurate enough for my purposes!) will look further into how to automate this in the spread sheet. i did have a backup plan of just using a lookup table but thought a inverted equation would be neater.
Thanks for the pointer, out of interest does it look possible to make D the subject (with a few more maths skills than i have!)
Anthony
 
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