Strange Math Problem (Resolved)

Greg LaPolla

I found this code in another thread for Steinhart Hart

PRI Calc_StHart (A,B,C,Resistance) : Kelvin | B_x_Rlog , C_x_Rlog , Rcubed, Rlog,  Addit
      
  G.start                                               ' Start math32

  Rlog := G.log(Resistance)                             ' Take the log of the Resistance
  B_x_Rlog := G.FMul(B,Rlog)                            ' Multiply B to log of R
                                                           
  RCubed := G.FMul(Rlog,Rlog)                           ' Cube the log of R by R * R * R 
  RCubed := G.FMul(Rlog,RCubed)                         ' Finish the Cube
     
  C_x_Rlog := G.FMul(RCubed,C)                          ' Multiply C to log the cubed log R

  Addit := G.Fadd(C_x_Rlog,B_x_Rlog)                        
  Addit := G.Fadd(Addit, A)                             ' Add all A,B,C together
  Kelvin := G.Fdiv(one,addit)                           ' Kelvin = 1/ all added

  G.stop

It always returns a negative number.

I put the above problem in mathcad(see attachment) and it calculates properly. Using the same numbers in the prop yields an answer of
-9.549217


Edit - Forgot to attach picture

Post Edited (Greg LaPolla) : 4/2/2009 2:54:06 AM GMT

Mike Green

How are you providing A, B, C, Resistance? If one or more are not floating point values, that could explain what's happening.

Greg LaPolla

Here is how it is done

CON

Coeff1 = 0.0009461261                              ' Steinhart Constant
Coeff2 = 0.00022226704                             ' Steinhart Constant
Coeff3 = 0.000000105745286                         ' Steinhart Constant

R1 = 10000

repeat
    ohms := ((4095 * R1) / ADC.GetData(0)) - R1
    temp := Calc_StHart(Coeff1,Coeff2,Coeff3,ohms)
    updateLcd(ohms, temp)
    waitcnt(clkfreq / 5 + cnt)

Mike Green

Well, there you go ... The whole "ohms" calculation is in integer arithmetic. You may want to do that calculation in floating point or, at least, convert it to floating point.

R1_Float = 10000.0
R1_4095 = 4095.0 * R1_Float

and

ohms := FSub(FDiv(R1_4095,FFloat(ADC.GetData(0))),R1_Float)

Post Edited (Mike Green) : 4/1/2009 11:40:51 PM GMT

Greg LaPolla

Yup I discovered that after I posted the reply. I am reworking the steinhart routing to convert everything to float first.

Thanks for the heads up!


Greg

Greg LaPolla

I have reworked the steinhart and now I have a new strange math problem

PRI Calc_StHart (A,B,C,Resistance) : Fht | B_x_Rlog , C_x_Rlog , Rcubed, Rlog,  Addit, R_Float, Kelvin, Celsius
      
  G.start                                               ' Start math32

  Rlog := G.log(G.FFloat(Resistance))                   ' Take the log of the Resistance
  B_x_Rlog := G.FMul(B,Rlog)                            ' Multiply B to log of R
                                                           
  RCubed := G.FMul(Rlog,Rlog)                           ' Cube the log of R by R * R * R 
  RCubed := G.FMul(Rlog,RCubed)                         ' Finish the Cube
     
  C_x_Rlog := G.FMul(RCubed,C)                          ' Multiply C to log the cubed log R

  Addit := G.Fadd(C_x_Rlog,B_x_Rlog)                        
  Addit := G.Fadd(Addit, A)                             ' Add all A,B,C together

  Kelvin := G.Fdiv(one,addit)                           ' Kelvin = 1/ all added

  Celsius := G.Fsub(Kelvin,273.15)                      ' Convert to Celsius 

  Fht := G.FMul(Celsius,1.8)                            ' Convert to Farenheight
  Fht := G.Fadd(Fht,32)
                                             
  G.stop

The last calculation ( Fht := G.Fadd(Fht,32)) will not run.

TreeLab

Change 32 to 32.0

Cheers!
Paul Rowntree

Greg LaPolla

That did the trick. Thanks