Equation of a line

blahblahbla

I have a range of points comming from a proximity sensor, see below.
X values are centimeters and Y values are voltage outputs.
Using Excel I managed to use the threadline fiunction to get the equation of the line that best fits this data.

[CENTER][COLOR=black]y = -0.0035x[/COLOR][COLOR=black]4[/COLOR][COLOR=black] + 0.1849x[/COLOR][COLOR=black]3[/COLOR][COLOR=black] - 2.3623x[/COLOR][COLOR=black]2[/COLOR][COLOR=black] - 17.899x + 614.72[/COLOR][/CENTER]

When I have a value for Y I want to solve for X , so I rewrote the top function as:
594.6436-Y/0.0035

594.6436-Y
X =-----------
   0.0035

This isnt giving me the correct awnser, can anyone spot my error?

      5   596.67 
      6   569.33
      7   542.00   
      8   514.67     
      9   487.33    
     10   460.00   
     11   432.00    
     12   404.00    
     13   376.00    
     14   348.00     
     15   320.00   
     16   306.00  
     17   292.00  
     18   278.00  
     19   264.00   
     20   250.00 
     21   247.70  
     22   245.40   
     23   243.10    
     24   240.80   
     25   238.50   
     26   236.20  
     27   233.90   
     28   231.60   
     29   229.30     
     30   227.00   

Delphi91

You can't do that re-arrangement of the equation to solve for x.

Bear in mind that the equation is a quartic one (highest power of x is 4) so for every value of y there are potentially up to 4 real possible values of x.

Google quartic equations and see the possible ways of handling the equations.

bnt

Since the cubic and quartic coefficients are so small, I'd be inclined to discount them and use just the quadratic equation only: y = 2.3623x² - 17.899x + 614.72.

You could then substitute a y value and subtract it from both sides to get 2.3623x² - 17.899x + (614.72 - y_value ) = 0 . Then use the quadratic formula to find the roots of that formula, and you have two possible answers, only one of which makes sense.

You could use Excel's "Goal Seek" system to find an X value to fit a given Y value using the quartic equation - but since there are four possible Y values, you might need to "seed" the process with something fairly close to the answer.

hivizman

Is it necessary to express the relationship between y and x by a single equation?

If you look closely at the data, you will see that, for values of x between 5 and 10, the value of y falls evenly by 27.33 or 27.34 (assumed to be roundings of 27 1/3 = 82/3) for each unit increase in x. Between 10 and 15, the decrease in y is 28 for each unit increase in x. Between 15 and 20, the decrease in y is 14 for each unit increase in x, while between 20 and 30 the decrease in y is 2.3 for each unit increase in x.

This means that you can express y exactly using the following set of equations, for the range 5 to 30:

y = 624 - (82/3)*(x-4) for x between 5 and 10
y = 460 - 28*(x-10) for x between 10 and 15
y = 320 - 14*(x-15) for x between 15 and 20
y = 250 - 2.3*(x-20) for x between 20 and 30

All the given points lie on the "curve" formed of the four line segments defined by these four equations, and the value of y for non-integer values of x in the given range can be found easily by linear interpolation.

bnt

Yep - just plotted the data and noticed the same thing. You don't have a curve, you have a series of straight lines. :cool: