linearizing with power functions

Linearizing Data

The table below represents water velocity vs. water depth, as inferred from stream range vs. depth by one student:

depth	velocity
0.16 m	1.4 m/s
0.12	1.3
0.08	1
0.04	0.64
0	0.04

To linearize this data we can take different powers of the velocity column and graph them vs. depth.

The most common powers we use include

the second power (the square),

the .5 power (the square root),

their reciprocal powers (i.e., inverse square and inverse square root), and

the -1 power (the reciprocal).

One of these powers linearizes the function.

It is really easy to do this very quickly if you know how to use the fill handle in Excel.

With a small data set like this one it only takes a few minutes to calculate the results by hand, if you don't know how to use the fill handle.

One of the powers mentioned works very well here; none of the others works well at all.

One power that does not work is the -1 power (the reciprocal). Taking the reciprocals of the v column we get

depth	velocity^-1
0.16 m	.7
0.12	.77
0.08	1
0.04	1.53
0	250

This transformation doesn't linearize very well at all, as can be easily seen by graphing y = velocity^-1 vs. depth.

Using a more appropriate power function (which you should find for yourself) the above data gives us the linearized graph below. The y axis isn't labeled, so you will have to find your own linearization.

We can use this linearization to get velocity v as a function of depth h:

If each unit on the y axis is, say, 2 (it isn’t for your data, but let’s use it as a reference example), then the slope of the graph is about slope = rise/run = 2 / .16 = 18 (approx.). In this case the linear function is about y = mx + b = 18 x + 0, or just y = 18 x.

If the function was linearized by taking the reciprocal square root of velocity v vs. depth h (this isn’t the right function for your data), then y in the above graph represents v^-.5 and x represents the depth h.

Substituting y = v^-.5 and x = h, y = 18 x becomes v^-.5 = 18 h.

Solving v^-.5 = 18 h for v we obtain v = 324 h^2.

You should now linearize the given data (or the data you have obtained for the experiment) using an appropriate power of v and using the appropriate units on your graph.

As you saw earlier the power of v shouldn't be -1; and as just mentioned the -.5 power doesn't work either.