Math question -

[Scuba Guy]

Let's see:

If I change out my aquarium water 10% at a time. How many changes would I have to do in order to have 'effectively' replaced all of the remaining water?

I know the answer is not 10 changes.

For you math wizzes out there .....

[henkychif]

Say you have 100 gallons and you do a 10% water change. After the first change, you have 90% of the original water left, or 90 gallons. After the second change, you have 90% of 90 gallons, or 81 gallons. After the third change, you have 90% of 81 gallons left, or 72.9 gallons.... and so on.

I know that's not a formula, but doesn't that describe the situation?

[bielbugs]

Look at it this way, if you were 10 ft from a wall and you move half the distance (50%) closer each time... technically you would NEVER touch the wall. 10, 5, 2.5, 1.25, .625, .3125, .15625, .078125, and so on. Yes at some point you wouldn't be able to "see" any distance between you and that wall but mathmatically you would never touch it.

So technically, by doing partial water changes, you never truely get rid of 100% of the orignial water. :hmm3grin2orange:

[cocoa_pleco]

i actually have an aquarium textbook and it has a huge chart that shows how much old water is remaining and etc.

[henkychif]

There's your answer. An infinite number of water changes are required :-)

[minabird]

Using the logic from the owl in the cartoon commercial for tootsie roll pops....where the kid asks the owl how many licks does it take to get to the center of a tootsie roll pop.....

the answer is.....

3

[Scuba Guy]

Henkychif has it correct.

Using the formulation he described means that it will take about 30 - 5 gallon bucket trips to get about 95% of the water changed for a 50 gallon tank.

It will take half that many or 15 bucket trips for a 10 gallon change at at a time.

It is interesting that to truly change out old water using this partial method requires a lot of water.

And assuming your goal is to get rid of something in the tank (old medication not removable by carbon - or carbon not practical) like salt, etc. it takes massive water changes at one time. 10% at a time won't cut it.

This may explain why we get old tank syndrome and the associated problems when we do partial water changes - even frequently....by the time you finish doing 30 water changes - one per day (10%), a month later you have old water again.

[jeffs99dime]

Using the logic from the owl in the cartoon commercial for tootsie roll pops....where the kid asks the owl how many licks does it take to get to the center of a tootsie roll pop.....

the answer is.....

3

yeah, but that only works if your an owl. lol

[muffin]

ok... this is a very simple mathematical equation... so assuming that you do a 10% water change. Assuming that all of the new water is thoroughly mixed with the old water in the tank and you wish to do another 10% water change of that new mixture.. you would still have 90% of the 90% of the old water still in the tank
also represented as .9 * .9 = 81%...
Following this logic after doing n water changes with these assumptions.. you would still have .9^n of the old water remaining..

i.e. if you did 4 of these changes you would still have .9^4 = .6561 or about 66% of the water left.. as was stated early you can never actually get rid of all your old water doing partial changes...

Sorry this is prolly a little late to comment on this one, but oh well


*** Edit*** Most basic formula...
Let x be the percentage of water change you wish to do (x is between 0 and 1), and n be the number of changes you wish to do.
The percentage of old water in the tank would be (1-x)^n
Alright... there is a mathematical formula that you can use...

[cocoa_pleco]

im too tired to follow your post muffin!

Youre right though with the equation

[muffin]

im too tired to follow your post muffin!

Youre right though with the equation


LOL :22: It's ok..

[Scuba Guy]

Muffin - Thanks for the formula - makes perfect sense.

Interesting though - if I want to change out most of the water in my tank - say 90%. Doing three - 50% water changes would barely get me there.

And doing 10% water changes would take forever to get rid of most of the water (and waste concentration).

What this is telling me is that small water changes are a waste of time.

50% - at least - is the only way to go to lower anything meaningful in the tank and then you have to do it twice in short order.

My strategy will be, when the weather warms and my incoming water temperature is much higher, to change out 60% of the water each week.

Won't be easy.

[Drumachine09]

Have fun with that!:thumb:

[muffin]

What this is telling me is that small water changes are a waste of time.


The problem with making a simple formula to explain something complicated is that information is always left out. Small water changes are beneficial in the sense that you can get ahead of the dissolved waste by sucking out the fish waste before it becomes a problem.

With that said.. good luck with you massive changes. :1: