# Math guru's

Discussion in 'The Guru's Pub' started by dukedave5200, Feb 14, 2012.

1. ### dukedave5200Ancient Guru

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Any math guru's out there? This is probably a very simple formula for you but I could use some verification of my math - or if I'm totally off on how I'm figuring it out.

The long and short of it is that I'll be doing some partial drain and fills of my transmission fluids (let's not get into why or other suggestions about this part, just stick to the math for now.)

So, I can replace/drain 5 quarts out of 12 total each drain. That's around 41.6% new fluids. My goal is to drain and fill as many times as necessary to have at least less than 10% of the old fluid. So, here is how I think the math works.

So first time would be:
5/12 = 41.6% new, 58.4% old

Second time would be:
58.4 * .416 = 24.29% old fluid left

Third time would be:
24.29 * .416 = 14.19% old fluid left

Fourth/last time would be:
14.19 * .416 = 8.29%

So each time I drain my thinking is that I would take the percentage of old fluid remaining and multiply it by the percent I would be replacing. To get to my goal of less than 10% old fluid I would have to drain and fill 4 times.

Is this math correct?

2. ### Matt151bMaster Guru

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Do we assume complete mixing between changes?

1
*5 of 12 new, 7 of 12 old

2
5 of 12 new, 7 of 12 at 5/12th new

5+2.91 new

3
5 of 12 new, 7 at 7.91 /12 new

5+4.61 new

4
5 of 12 new, 7 at 9.61/12 new

5+5.61new

On the fourth change there is only 10.61 qts of new fluid, which means there is still 1.39 qts of old fluid. * *Objective not reached.*

10% of 12 qts is 1.2 qts

(sorry for sloppy, but done on my cell)

3. ### dukedave5200Ancient Guru

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We do assume total mixture after each drain (it will be a week between drains)

If your math is correct I think it'll be close enough to my goal. I didn't mean to say it had to be less than 10% old fluids left, just somewhere close. The stuff is expensive. I could just do a complete flush, but I'm worried about putting too much stress on the transmission, gasgets, etc, because I'm not entirely sure what's in it now. Anyway, I said I didn't want to get into the reasoning for the multiple drain and fills. lol...

Thank you for your reply. I'm no math expert but I think I see where you're going with it. However, could you at some point break it down a little more to help me understand where I'm failing with how I've done it?

4. ### Matt151bMaster Guru

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Back at a a desk. At zero changes you are 100% old stuff, so that is concentration = 1.00.

After your first change, you have the math right and it is 58% old stuff, concentration =0.58.

After your second change you are mixing 5 quarts of new fluid with 7 quarts at concentrtaion =0.58.

5*0.00 + 7*0.58 = 12 * X ; where X is your concentrtaion after you mix the fuilds.

X = .338

Again, 5*0.00 + 7* 0.338 = 12*Y

Y = 0.197

Again...this time for Z

Z = .1149

Z is your fourth change which is close to 10%, but slightly above.

How much is 5 qts? Going from the third to fourth change you don't even reduce it by 10%, so you should decide if the cost of 5 quarts is worth less than 10% new fluid.

5. ### dukedave5200Ancient Guru

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Ah ok, now I see... Yeah, the stuff is around \$10 a quart so good point. I think I will make that determination once I see what the fluid looks like now after the first drain. If it's looking pretty good (not burnt or sludgy at all) I think I will be skipping the last one.

Thank you very much.

6. ### zhengzhoudaveMaster Guru

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Just to interrupt you guys - I made a silly comment about something like this a few weeks ago. I welcome anyone to talk about anything on here but in my ignorance I just don't get it!

Sincere apologies for interrupting!

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(12-5)/12 = .583 remains per change.

(.583 ^ N) = .10

ln(.10) / ln(.583) = 4.2674597 times to get 10%

Last edited: Feb 15, 2012
8. ### NoviceReiAncient Guru

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pardon me (just curious), but why choose to re-fill multiple times if you can simply drain all 12 and do the work once plus save 8 qz transmission fluid?

but to answer your question, i think it would not be linear like the one shown in your equations.

9. ### dukedave5200Ancient Guru

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The majority of the fluid is contained in the torque converter and cannot be drained. It's pretty common... The only way to replace it all at once is to do a complete flush which either requires the use a machine or to remove cooling lines and allow the engine/pump to push it out while adding in new fluid at the same time. My reason has more to do with not shocking the transmission with new fluid as I'm not entirely sure what is in it now - if I did know and my vehicle had far fewer miles I would just do a full flush. Transmissions can be very touchy as they're designed with very specific tolerances which usually must be met with a very specific fluid.

Other's have replied with the correct math. I knew I wasn't doing it exactly correct which is why I made the post knowing that there are several guru's with much higher math skills than I have. So I got my answer.

10. ### NoviceReiAncient Guru

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i see, i got your point. sorry for being too nosy. hehe. hmm.. if i may suggest, i think it would be better if you drain the most that you can the first time then drain/refill less later on. but yeah.. do what you think works for you. and good luck.

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