Finding a Formula That Fits the Data

Date: 01/26/2000 at 21:57:50
From: Kimberley Mullen
Subject: Making a Formula

Dr. Math,

I am a science fair contestant this year. I have tried to focus on 
math and science, but mainly math. I knew that recreating and proving 
an already existing formula would mean null to our society, so I 
decided to make my own formula on the basis of tensile strength.

Tension builds up on a metal strip (in my case, it is sewing thread) 
and stress-relieving signs occur such as unwinding, stretching and 
vibrating, until the thread can take no more and pops. Tensile 
strength is the amount of pressure or stress until it breaks.

I tested, recorded and reported my information and, using a freelance 
spreadsheet, created my sketchy equation with cell numbers. 
Translating that into a mathematical formula is part of the 
concentration and logic score on the project, but I got my ideas 
across and into the formula.

I was wondering if you could take a look at it and tell me what you 
get out of it. The only catch is that I can't use any math higher than 
Algebra I, because the judges will know that there is no way I did the 
math work, much less understand it. Even though I'm only in the 8th 
grade and I have a very good math sense, I am too young to have 
Calculus, the math my father says will work.

Here are my tables and my formula. All I ask of you is to look it over 
and see what you get out of it. I appreciate anything you come up with 
and will be grateful for just your opinion.

                       Table of Information

     # of Threads   Wt. Supported     # of Threads   Wt. Supported
     ------------   -------------     ------------   -------------
           1               1               16              37
           2               3               17              41
           3               6               18              45
           4               7               19              49
           5               9               20              54
           6              11               21              60
           7              13               22              66
           8              16               23              72
           9              20               24              79
          10              20               25              87
          11              23               26              96
          12              25               27             106
          13              28               28             116
          14              31               29             128
          15              34               30             141

(if you need more, e-mail for the rest because I have a ton of string 
and weights supported data.)

Mathematical Formula:

     new weight supported = (previous weight supported * 1.10)
                             + previous weight supported 

or, in the spreadsheet world:

     E35 = (E34 * 1.10) + E34

Can you please help me with an equation that would predict the weight 
supported based upon the number of threads I use? My equation works, 
but there's a fudge factor built in and I don't think its a pure 
mathematical model/formula.

Thanks again,
Kimberley

Date: 01/27/2000 at 13:02:22
From: Doctor Rick
Subject: Re: Making a Formula

Hi, Kimberley. Thanks for writing - you're doing some interesting 
work.

Your data are surprising to me. Not being a materials scientist, I 
would have expected that the weight supported by the threads would be 
a linear function of the number of threads. That is,

     W = aN + b

where N is the number of threads, W is the weight supported, and a and 
b are some numbers. Your results are clearly different from this, and 
it would take someone more knowledgeable than I to understand why this 
is so. I doubt that it's a completely new discovery, but it's new to 
me.

When plotting data and trying to find a formula that fits the data, 
it's very helpful to try plotting different functions of the data to 
see if you can get something like a straight line. For instance, the 
data may be fit by a "power law":

     W = a * N^b

where a and b are some constants and N is raised to the power of b. If 
this is true, then you can plot the logarithm of W versus the 
logarithm of N. If you have learned about logs, you can show that the 
equation above is equivalent to

     log(W) = log(a) + b*log(N)

so that plotting log(W) versus log(N) will give you a straight line 
(like the first equation I wrote.) If you have learned about plotting 
straight lines, then you know that the slope of the line you plot is 
b, and the y-intercept is log(a). You can find a by raising 10 to this 
power.

I tried this, using the LINEST function in Excel to find the best 
linear fit to the log-log data. When I used the values of a and b that 
I found,

     a = 0.991, b = 23.148

I didn't get a very good fit to the W vs. N plot.

What else could we try? The equation that you came up with gave me a 
good idea. I don't think you wrote quite what you mean, but I was able 
to get the idea and rework it a bit. You said

>     new weight supported = (previous weight supported * 1.10)
>                             + previous weight supported 

I think what you are saying is this: the weight supported by N+1 
threads is 1.10 times the weight supported by N threads.

     W(N+1) = 1.10 * W(N)

You didn't want the added term. When you have a sequence of numbers 
and each is a constant multiple of the number before it, we call this 
a geometric progression. The function that fits it is called an 
exponential function:

     W = a * b^N

It looks a lot like the "power law," but the variable N is in the 
exponent this time, and the constant b is the base of the exponent. If 
this is the correct form of the equation, then you will get a straight 
line when you plot log(W) versus N itself (rather than log(N) as 
before).

     log(W) = log(a) + N*log(b)

Try this with your spreadsheet. You will find that above N = 10, it 
fits very well. I don't know why it's different below 10. It appears 
that from N = 3 to 9, you have another straight line with a different 
slope, which corresponds to a different value for b.

I haven't used calculus, but I had to use logarithms and exponents. If 
you can understand what I've said, you will have something very 
interesting to say. Have fun and if I've raised any questions in your 
mind that you think you could understand with a little more 
explanation, by all means ask me.

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/