I know this is not the usual tech-related question but it does involve some formula making in Excel. I have a series of tanks (aproximately 500 barrels each) with a somewhat strange shape and I need to be able to calculate the amount that is contained in each one based on the level of the liquid contained in each one. Today I'm using a simple volume formula that I use on normal tanks but these ones have a ladder so it's kinda complex.
I'm attaching a simple image of the side and top views of the tanks so you have an idea of what I'm talking about. What I would like to do is divide/split the tanks in 3 areas so I can re-measure each tank (they'r not exactly the same size) and use the same formula for every tank. If I do it like this, all I have to do is calculate the volume of 2 rectangles and 1 3D triangle.
The problem is that I don't know how to measure or how to make a formula that progressively calculates the volume in that triangle part (below the stairs).
I'd say the easiest way to calculate this is to view the tank as one very large cuboid, and then to SUBTRACT the triangular prism volume occupied by the stairs.
D = fluid depth
Lx = total internal length of tank
Wx = total internal width of tank
Lt = length of triangular area below stairs (to fluid depth)
Ws = width of stairs
On that basis,
total tank volume is obviously:
Lx * Wx * D
and volume of area lost due to the stairs is:
1/2 x (Lx * D) * Ws
The only tricky bit is to calculate Lx
and you could do that with either trig:
Tan (a) = Opp/Adj [where a is the stair angle]
Opp would equal the fluid depth D, and Adj would = Lx
rearranged would give Lx = D/Tan(a)
Or you could use pythagoras
- all depending how easy it is to measure your tank.
Danny Child
by the way... I can't see all your comment starting with "The 3 areas where I have to calculate the ...."
It might also help if you have any sample dimensions for us to work with?
Danny Child
minor correction:
and volume of area lost due to the stairs is:
1/2 x (Lx * D) * Ws
should be
and volume of area lost due to the stairs is:
1/2 * (Lx * D) * Ws
Thanks, Caracena - I'll work on this a bit later, no time now, but will come back to you. I'm a metric boy, but can happily convert all your info above. What's your desired final unit of capacity - US gallons, etc?
Cesar Aracena
ASKER
Hi Dan. We use metric too hehe. Here they are in metrics:
As for the final result, we use m3 (cubic meters).
I'm trying to apply the formulas you stated above but I'm missing where you use Lt. Actually I'm a little confused about what Lt means. Is it the new "adjacent" side created by the fluid's level?
Ok, here is a 3D drawing of one of the tanks. There are like 5 different models of tanks but this is the more complex so I grabbed this one to try and understand how to measure the volume correctly under those stairs.
There's an image of the whole tank, the part of the stairs, the same part with measures and another one with a triangle I would have to draw in my mind in order to "close" the triangle.
The stairs themselves are not there... they just go on top of the ramp.
D = fluid depth
Lx = total internal length of tank
Wx = total internal width of tank
Lt = length of triangular area below stairs (to fluid depth)
Ws = width of stairs
On that basis,
total tank volume is obviously:
Lx * Wx * D
and volume of area lost due to the stairs is:
1/2 x (Lx * D) * Ws
The only tricky bit is to calculate Lx
and you could do that with either trig:
Tan (a) = Opp/Adj [where a is the stair angle]
Opp would equal the fluid depth D, and Adj would = Lx
rearranged would give Lx = D/Tan(a)
Or you could use pythagoras
- all depending how easy it is to measure your tank.