Is this a 50g Tank?

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parrotfish100

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Feb 22, 2004
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I just got a new fishtank from a freind, and she claims that it is a 50 gallon tank. I haven't really been around them much, so i don't know how big they are supposed to be; looks like a 50g to me...

Dimensions:

Height- 18 3/4 inches

Width- 13 inches

Length- 48 inches

Can you help?
 

RioXingu

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Mar 30, 2004
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If your not on-line at the time:
For rectangular tanks--
(L x W x H)/231 = capacity in gallons.
L, W, & H in inches

Hex, bow-fronts and other shaped tanks are left as an exercise for the student.

Bill in WI
 

RioXingu

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DEmigh--
You are correct, sir!
In my spare time, I worked out this formula for hex tanks:

Volume (gallons) = (2.6 x L x L x H)/231

where L=length of any side in inches, H=height of tank in inches.

OR:
V=( A x P x H)/462

where A=distance from center to the middle of one side, P=perimeter (sum of the sides), H=height of tank.
The mathematical term for A is the apothem.
All measurements in inches.

I'm still working on bow-fronts.

Bill in WI
 

OxyBomb

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hey bill where in WI are you from? I am in La Crosse. and thats way cool you did those math things. I was goign to go searching for them soon. im trying to come up with ideas for tanks I want to have built.
 

DEmigh

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Originally posted by RioXingu
[...]I'm still working on bow-fronts.[...]
Awesome! Even though you left it as an exercise for the student you couldn't resist the urge to be prepared to check their work. Have you considered teaching? :laugh: I wish some of my Geometry students would take up the aquarium hobby, and be stuck buying hex tanks with no listed capacity, maybe it would help them to find some relevance for their studies. :rolleyes:

Since you're still working on bow-fronts, may I propose a couple of approaches to the problem?

I think we can safely assume that the bow shape is an elliptical curve, so we can use the area of an ellipse (1/2 * minor axis * 1/2 * major axis * Pi), the wrinkle on this approach is extrapolating the major and minor axes of the ellipse in question from the small part that actually forms the front of the aquarium.

Another possibility, less elegant but much easier, is to "simplify" the bow front to a triangle. Determining the altitude of said "triangle" requires only a sufficiently long straightedge. Having found the altitude, we now insert the time honored "fudge factor" into our formula. Instead of (1/2 * base * altitude) we nudge the 1/2 upward. I propose 7/12 as it is the median value between two limits that can be established on our fudge factor.

{For the upper limit, I am using an alternate form for the volume of a sphere, namely (2/3 * the volume of the right-circular cylinder that would exactly contain the sphere in question)}

Anything you have to say on this topic will be of interest to me :D
 

OrionGirl

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When did this turn into a math forum? What the heck am I doing here? Auntie Em, is that you??? :confused: :confused:



Kidding--great info! Get everything posted pretty and I'll make you sticky. :laugh: Or at least, I'll make the thread a sticky. :D
 

RioXingu

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Actually, I was kind of hoping that the bow-front curve of the aquarium could be assumed to be that special case of the ellipse, you know, a circle?
It would make the math all that much simpler.
All you would need to know is the radius of that circle which forms the bow-front.
Then the area of the part of a circle known as a segment can be calculated.
A segment is the part of a circle bounded by a chord (line between any two points on a circle), and the circle itself.
I am pressed for time now; I will post again on this subject later.

Bill (going off to work) in WI
 
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