Sump volume - any geometry majors?

[packman9111]

I'm using a 72g bow front for sump and would like to know volume.It's 48x18-12x22 (18 at widest and 12 at ends) with skimmer section 15" high and 17" long on one end and fuge 15 high and 14 long on the other, return in center taking the remaining 48" length (~15") with about 7" deep water. I'm thinking about 40 or 50 gal, but would be nice to be more exact in knowing the system capacity.

Thanks - Rick

 

[sediener]

http://home.comcast.net/~jdieck1/volcalc.html

that might help...

- Steve

 

[Anonymous]

Well I'm not going to do ALL the little bits, but its a trapazoid shape? 18 at one end 12 at the other? so 48* (18+12)/2) * 22 = 68gallons, then just subtract out volumes.

 

[Anonymous]

I would treat it as a rectangle using the 12" width. If you want to be more exact threat the arc sections in each compartment as triangles. Do the composit area of a triangle and a rectangle then multiply by the height of the water column in that section. This should give you an accurate approximation. Just for reference the 72 bow has 3.27 gal per inch of height.

 

[RobertoVespucci]

Yeah, I would just add a known quantity of water large enough to come up the side a measurable amount. Like 5 gallons. Measure the height of the water. Add 5 more and measure again. Now you should be able to closely estimate the amount of water in the tank on a per inch basis.

 

[Anonymous]

It hard to calculate the precise area without know whether or not the front is a circular arch and radius if it is circular. To under estimate the area and therefore volumn you might first take the rectangle at the back and then add the base and height of the front arch as a triangle. Add those two areas together times the height to get an underestimate of the volumn. Should put you in the ball park anyway.



Edit: well I could at least run the numbers

assuming a rectangular tank with a bow front. ends 12" center 18" deap, 48" wide and 7" water height.


48x12=576 in2
48*(18-12)=288 in2
576+288=864 in2 *7in=6048in3

6048*.00433gallons/in2=26 gallons. Actual will be slightly higher.

Of course the assumes an arched front. don't know where the 12x22 came from so probably is all wrong. LOL

 

[Anonymous]

Sheesh, just do like Roberto said and fill it with 5 gallon buckets Easier on the brain.