How to calculate an Octagon shape tank? height is 18" and I guess L & W is 6 ". it's about 2 " per side of the octagon
Volume
- Thread starter MrArapaima
- Start date
Assuming that you have an even sided octagon:
Area of Octagon (A) = 4.828(a*a), where a = length of each side of the octagon
Volume of Octagon = Area (A) * Height (H)
So: Volume = 4.828(a*a)*H or 4.828(2"*2")*18" = 347.6 cu in
Since this is an aquarium site, I'll further assume you are looking for how much water the Oct will hold:
1 gallon = 231 cu in
So: Gallons = 347.6 cu in * 1g/231 cu in = 1.5 gallons (5.7 liters)
Area of Octagon (A) = 4.828(a*a), where a = length of each side of the octagon
Volume of Octagon = Area (A) * Height (H)
So: Volume = 4.828(a*a)*H or 4.828(2"*2")*18" = 347.6 cu in
Since this is an aquarium site, I'll further assume you are looking for how much water the Oct will hold:
1 gallon = 231 cu in
So: Gallons = 347.6 cu in * 1g/231 cu in = 1.5 gallons (5.7 liters)
Except that the original poster is in London, therefore need the volume in Imperial gallons. The conversion factor for imperial gallons is 277.42 cubic inches per Imperial gallon.
Good eye there RTR! I notice the poster was from the UK, which is why I threw in the litre at the end. Shoulda' went back & posted the other version of gallons
Someday we'll all use litres, but despite all those years in the lab, I still think tanks in US gallons.