Surface area?

[Skittyfish]

I need to know how to find out the surface area of a 72 gallon bowfront (math challenged ).
Please somebody help me.

TIA:D

[NJ Devils Fan]

I have that tank, just got everything for it a few days ago. As far as finding surface area, i'm stumped.

[JamisonBWolsh]

Dont you take the width and times it by the length and that would give the total surface area in inches?

[Skittyfish]

OK here are the specs
48" long
24" high
13" deep at ends
18" deep at middle

BEAUTIFUL Tank huh?

I know somebody out there is a math genius....:D

[NJ Devils Fan]

No because its a bow front, so the length of one side is different from the length of the other side.

[NJ Devils Fan]

Actually, the exact measurments are 48"x22"x18(12 on the ends)". I would guess you would take one of those flexible tape measurers and get the difference of the front and the back(48") and then divide the difference in two and add it to the 48", then do the LxW.

I just did it and got 588" of surface area. I did 49" x 12".

[Gumby7]

785 - by CAD.

Gumby7

[Skittyfish]

THANKS THANKS THANKS!

[NJ Devils Fan]

Gumby, what's CAD?

[slipknottin]

just wondering... but what do you need to find surface area for?

[Skittyfish]

just information to clutter my brain with...
I saw a fomula for deciding how much wattage you need for plants...
you take the surface area x distance of light to gravel x (type of plant) divided by 11


Low light plants .08
Low to moderate .12
moderate to bright .18
birght .27

I just wanted to see what it came out as

so 785 x 19 x .12 /11= 162.7 watts of light for low to moderate light plants.

That for 72gallons is 2.25 watts per gallon

Knowledge can be deadly .....


Thanks for you help :D

[NJ Devils Fan]

Dude, it's 588, not 785. You have to remember that the tank isn't 18" wide all the way, only at the widest part of the tank. Thats what I took into affect when I got the average of the front and the back because they are 2 different lengths, then I times the average length by the width. I am sure that what I did is right. Also, it shouldn't be 19", it's 22" high.

So, for moderate plants, the formula should be:

588 x 22 x .12 / 11= 141.12 watts.

[wetmanNY]

...unless you're using the new efficient fluorescents that put out much more light for the watts of electricity they consume...

then the old watts-per-X rules-of-thumb aren't going to be working any more..

[quick061]

yes 785 sounds better than 588.

hell if you just take the 48"x13" (for the shallowest part) then that's 624"^2. then you'd have to measure the circumfrence of the bow but it makes sense that its about another 150"^2.

[Gumby7]

Originally posted by NJ Devils Fan
Gumby, what's CAD?

Computer Aided Drafting, specifically AutoCAD.

Trust me I've been earning a good living as a geometrics designer (highways, railways etc.) using AutoCAD for the past 12 years.

(I did assume the front glass is bent into a circular curve and not something more elaborate like an elipse or parabola).

Gumby7

[slipknottin]

Originally posted by NJ Devils Fan
Dude, it's 588, not 785.


How do you get that?

a 55 gallon tanks surface area is 576. A 75's (48x18) surface area is 864.

785 sounds right to me.

[NJ Devils Fan]

Quick was on the right track, I don't know why I didn't think of this before, it's basic geometry. You get the area of the tank without the bow front part(12"x48") which is 576, then get the area of the bow front, which is a semicircle((pi)R squared divided by 2) which is 56.55, and add them and get 632.55.

Ergo, meaning therefore, the surface of the 72 gallon bow front tank is 632.55 inches squared. (could I sound any more like a nerd? ;))

[Bantam]

NJDevils fan; the bowfront isn't a semi circle, its smaller than that. i've no idea how to work it out but 785 sounds about right to me

[quick061]

Originally posted by NJ Devils Fan
Quick was on the right track,
...

i didn't take all those years of calc for nothing ;)

[Fishiebusiness]

Originally posted by NJ Devils Fan
Quick was on the right track, I don't know why I didn't think of this before, it's basic geometry. You get the area of the tank without the bow front part(12"x48") which is 576, then get the area of the bow front, which is a semicircle((pi)R squared divided by 2) which is 56.55, and add them and get 632.55.

Ergo, meaning therefore, the surface of the 72 gallon bow front tank is 632.55 inches squared. (could I sound any more like a nerd? ;))

As stated, the bow is clearly not a semi circle. Its a small slice of one side of a circle that has a radius much larger and harder to determine than from the length of the tank.

Just trust gumby and his AutoCAD, it works!

[NJ Devils Fan]

(|
That is what the tank basically looks like. The bent part is a semi sircle because a circle doesn't have to be perfectly round like this-o. It can also look like this- 0, or this- ().

[carpguy]

The bow is an arc, not a semicircle (halfcircle). If the bow was a semicircle the tank would be 36" wide at its widest and the center of the circle would be in the plane of the front corners. The circle the bow is part of is about 8.5 feet in diameter and its center is about 2.75 feet behind the tank. You're figuring for a rectangular tank with a small bump out in the middle.

[Fishiebusiness]

Originally posted by NJ Devils Fan
(|
That is what the tank basically looks like. The bent part is a semi sircle because a circle doesn't have to be perfectly round like this-o. It can also look like this- 0, or this- ().

The formula you state: ((pi)R squared divided by 2) applies only to perfect half circles, not to half ovals, half ellipses, or half zero's.

[Gumby7]

Originally posted by Fishiebusiness


As stated, the bow is clearly not a semi circle. Its a small slice of one side of a circle that has a radius much larger and harder to determine than from the length of the tank.

Just trust gumby and his AutoCAD, it works!

Thanks for the vote of confidence!

PS the radius is 51" and the center point is 33" behind the back of the tank. If you're still interested you can verify this by placing the tank on the floor mark out the center point 33" behind the tank. Then hold the zero end of a tape measure on that point and try to sweep the front bow radius as 51".

Check out http://www.constructionworknews.com/field-manual/1-2.html "Segment of a circle" to get a feel for why is not as simple as pi*r^2.

PSS I be sceptical of that lighting equation. I see no good reason why the "divide by 11" is not already factored into the 0.12 etc factor. That is, why not use 0.0109 instead of 0.12/11? Are you sure something isn't missing like a pair of brackets? Often formulas which require two lines (like a denominator and a numerator) get screwed up on websites due to font issues etc.

Gumby7.14159265...

[Skittyfish]

The formula came from www.thekrib.com
eiither in the plant area or the lighting, I can't remember. and it could have had something else.

Sorry to create such a ruckus..


Thanks for your replies:D