# Power Steering Fluid

I was replacing the power steering fluid in my car when I stumbled upon some exponential decay math.  First, some background:  There is no drain plug on a power steering system.  You need to siphon out fluid from the reservoir and replace it with new fluid.  This new fluid then mixes into the rest of the system to create a slightly cleaner mixture.  This idea is that if you repeat this a few times, you’ll replace most of the old fluid with new fluid.

So, as you can see, there exists some sort of formula that can determine exactly how many times you need to extract and replace to reach X% replacement.

My car’s entire power steering system holds about 1 liter, and the reservoir itself holds .4 liters of that. So, 40% of the steering fluid is replaced each time I drain and refill the reservoir, and 60% of the old fluid remains elsewhere in the system.  I can do this repeatedly, each time replacing 40% of the “mixed” fluid with brand new fluid.

Let p = Percentage of the system replaced each time you empty the reservoir.
Let n = number of times you empty/fill the reservoir (“flush”)
Let FD = Percentage of dirty fluid in the system.
Let FN = Percentage of new fluid in the system.

If p = percentage of new fluid introduced by a flush, then (1-p) is percentage of old fluid remaining.  (eg: if $p = .40, then (1-.40) = .60)$

$FD_0 = 1$  (initial proportion that is dirty)
$FD_1 = (1-p)$
$FD_2 = (FD_1)(1-p) = (1-p)(1-p)$
$FD_3 = (FD_2)(1-p) = (1-p)(1-p)(1-p)$

$FD_n = (FD_{n-1})(1-p) = (1-p)^n$

$FD = (1-p)^n$
$FN = 1 - FD$

For my car, p=.40 so $FD = (1-.40)^n$

How many times do I need to empty and fill the reservoir to get to 80% clean?
Just set $FN = .8$ and solve for n:

$.80 = 1- (1-.40)^n$
$.80 = 1- (.60)^n$
$.6^n = .2$
$log(.6)^n = log(.2)$
$n*log(.6) = log(.2)$
$n = \frac{log(.2)}{log(.6)}$
$n = 3.15$

With a reservoir that holds 40% of capacity, I need to empty and replace it about 3 times to get 80% of the old fluid replaced.

You can also use the formula to figure out what percentage of the system contains old vs. new fluid, based on the number of flushes you’ve done.  For this, you just plug in n and calculate $FD$  eg:  If you’ve done 5 flushes, $FD = (1-.4)^5 = .08$   So, after 7 refills, 8% is dirty, and 92% is new.

# The older you get, the faster each year passes by. Why?

RE: Birthday Dinner
Yes, life does go by fast.  Strangely, the older you get, the faster it goes.  I do not know why this is.

Ever get an email like this?  Well, as your age varies, the percentage of your life that a single calendar year represents also varies.  As you get older, a year is a smaller percentage of your overall life.  In other words, 1 year represents 50% of a 2 year old’s life.  However, it is only 2% of a 50 year old’s life.  So, perhaps that is why each year seems to go by faster.

Want to see the percentage for every age from 0 to 80?  Yea, so do I.  Let’s make a formula and graph it.  The percentage of your life that a single year represents is just a function of your age:  $f(age) = \frac{1}{age}$  If you graph this on a spreadsheet, you’ll get the following:

How would you interpret this graph?  You’ll notice that once you pass the inflection point, the percentage seems to flatten out.  So, at what point can a person legitimately start saying “Wow, this year really flew by?”  Based on the graph, teenagers might feel this almost as much as middle aged people.

Lastly, do you notice how scaling of the y-axis makes the difference between age 15 and 50 look trivial?  In order to properly display percentage changes, I will scale the y-axis logarithmically.  Here is the result:

With this scaling, you can see there is, indeed, quite a difference between a teenager (~6%) vs. someone in their 50s (~2%)