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volume of the fish tank
Posted: 03 Nov 2005, 19:16
by mitmop
does anyone know how to work out the volume of my fish tank. I need the formula to work out a triangle shape.
Thanks
Cant find it anywhere on the net,
Dawn
Posted: 03 Nov 2005, 19:45
by Silurus
If it is shaped like a prism, then it's just area of the base x height.
area of prism
Posted: 03 Nov 2005, 19:58
by mitmop
thanks for the reply sorry to sound realy thick but how do i work out the area for instance..
on of the triangle is 83cm by 83cm by 118 cm and depth is 60 cm so how do i work that out?
thanks
Dawn
Posted: 03 Nov 2005, 20:07
by Silurus
That works out to 206658.4 cubic centimeters, I think.
This would be about 206.7 liters.
Posted: 03 Nov 2005, 20:08
by zenyfish
1/2 * b* h = area of triangle
b=118
h=((82)**2 + (118/2)**2))**1/2 = 101.01
=> 357540 cm**3
Posted: 03 Nov 2005, 20:16
by mitmop
Sorry still confused, I have to be right as we treating the tank for disease (this is only part of the tank as due to its shape we had to brak it down into triangles and rectangles, we are ok with the latter)
D
Posted: 03 Nov 2005, 20:23
by Silurus
Actually, the correct h should be
((82)**2 - (118/2)**2))**1/2 = 58.38 (using Pythagoras' theorem to calculate the height of the triangle)
Volume should then be (58.38 x 118)/2 x 60.
Posted: 03 Nov 2005, 20:34
by mitmop
Thanks for your help so far, please could you tell me the volume in litre.
plus the same for the next triangle which is 22 x 22 x 28cm.
Thanks
D
Posted: 03 Nov 2005, 20:37
by Silurus
As I mentioned earlier, that works out to 206.7 liters.
The area of the second triangle is about 237.6 square centimeters. Multiply that by the height of the tank (in cm) and divide the total you get by 1000 to get the volume in liters.
Posted: 03 Nov 2005, 20:40
by zenyfish
Silurus wrote:
Volume should then be (58.38 x 118)/2 x 60.
I would check the 58.38?
Posted: 03 Nov 2005, 20:45
by mitmop
Hiya yes i have checked the 58.38 and that seems about right...i worked the other one out at 14.256 is that right?
Thanks sooo much for this its such a help.
Dawn
Posted: 03 Nov 2005, 21:00
by zenyfish
mitmop wrote:Hiya yes i have checked the 58.38 and that seems about right...i worked the other one out at 14.256 is that right?
Thanks sooo much for this its such a help.
Dawn
Yes, I noticed the error (negative) on my calculation of 101.
Assumming the height (60) is the same:
Vol = 1/2* 28 * 16.9 * 60 = 14196
Posted: 03 Nov 2005, 21:01
by Silurus
The height of the second triangle actually works out to 16.97.
((22)**2 - (28/2)**2))**1/2=16.97
Posted: 03 Nov 2005, 21:33
by mitmop
Thankyou very much for your help you are a life saver I have worked out the tank is about 365 ltr in all.
Thanks again.
Dawn
Posted: 03 Nov 2005, 23:24
by eupterus
The tank sounds like a trigon 350. In jewel tanks the number indicates the litres, Just some info for future use. Save time in the calculation stakes
Posted: 04 Nov 2005, 09:28
by Owch
Everybody is overcomplicating things
83x83x118, surely thats a right angle triangle, therefore:
(83*83)/2 = Area of base = 3444.5cm2
3444.5 * 60 = Volume of tank = 206670cm3
In litres /1000 =
206.7litres
Should havd been a maths teacher, you there quite at the back..... Are you chewing! Oh the power

Posted: 04 Nov 2005, 19:24
by zenyfish
Owch wrote:
83x83x118, surely thats a right angle triangle, therefore:
Actually, two equal side does not imply square.
118**2 = 13924
83**2 + 83**2 = 13778
Not square. BTW, I didn't read the original post close enough and used 82 rather than 83.
Posted: 04 Nov 2005, 21:04
by Owch
I figured that a tank with 2 equal sides was going to be a corner tank

, anyway, I dont think anyone is going to want a tank with an internal angle of 90.607degrees

Posted: 04 Nov 2005, 21:13
by zenyfish
But your method doesn't work so well for 22 X 22 X 28.
Posted: 04 Nov 2005, 21:17
by Owch
zenyfish wrote:But your method doesn't work so well for 22 X 22 X 28.
Its not my method, some guy called Pythagoras made it up
22*22*28 would have 2 equal angles of 50.48deg and 1 of 79.04deg (2dp). And a height of 16.97 (2dp)
Posted: 04 Nov 2005, 21:31
by zenyfish
Owch wrote:
Its not my method, some guy called Pythagoras made it up
I'm referring to assumming 90 degrees?
Posted: 04 Nov 2005, 21:41
by Owch
zenyfish wrote:Owch wrote:
Its not my method, some guy called Pythagoras made it up
I'm referring to assumming 90 degrees?
Well then 22*22*28 cannot be a 90deg triangle
Posted: 04 Nov 2005, 21:56
by zenyfish
Owch wrote:
(83*83)/2 = Area of base = 3444.5cm2
Its not my method, some guy called Pythagoras made it up
I'm afraid I don't see Pythagoras here.
So if it looks close to 90, you'll assume 90? What's the cutoff? At 87, 86, 85? 93, 94, 95?
Posted: 04 Nov 2005, 22:07
by Owch
Things are getting a little out of hand here, like I said earlier, 'overcomplicating things'.
No cut off, just a rough estimate of this ladys tank volume, and I figured the method I used to be fairly accurate, and VERY simple.
And as for looking close to 90deg, well 83*83*118 dimensions of a triangle gives 1 internal angle of 90.6...... and 2 equal internal angles of 44.7, sorry for being pedantic, but thats as near as damn it a right angle triangel

Posted: 04 Nov 2005, 22:15
by zenyfish
Owch wrote:but thats as near as damn it a right angle triangel

Yes, I agree it probably was 90 and her measurement were off a bit or rounded.
But without actually seeing the tank or sketching it out, I would not have assume it.
But you were forced to use 'overcomplicating things' for 22 X 22 X 28?
Posted: 04 Nov 2005, 23:07
by Owch
Your point to this whole debate
I cant see one.
Posted: 04 Nov 2005, 23:44
by zenyfish
Owch wrote:I figured that a tank with 2 equal sides was going to be a corner tank
I would not assume 90, it was coincidental in the case of 83 X 83 X 118. But not so in the case of 22 X 22 X 28.
Maybe you have a better sense of porportions, I cannot look at the numbers 83 X 83 X 118 and say "Oh, that's a right triangle!" How do you know? Is 80 X 80 X 123 a right triangle for instance?
Your method is estimated, subjective, and specific. It relies on what looks like 90 and and only for angles close to it. The method I used is general and works for all cases.
I would argue your method is more 'overcomplicating' since it's subjective. So maybe I used a method not so 'overcomplicating'? I mean, you had to use it also, true?
Posted: 04 Nov 2005, 23:49
by Silurus
OK, this is the Speak Easy forum, but I believe that this post has long outlived its usefulness.
Geometry lesson's over, folks.