non linear scale applied to data

[leecavturbo]

in this example below i need to work out what the divsor
would be for values inbetween the data?
.

 

[leecavturbo]

or can some sort of averaging formula be used?

 

[leecavturbo]

no one?

 

[mikerickson]

It look like your situation is that you have input values in A and result values in B and C. And you want intermediate result values.

i.e. you would want the values of B and C for A=63?

If that is the case, linear interpolation would be my approach, even though the overall function is not linear.

 

[leecavturbo]

It look like your situation is that you have input values in A and result values in B and C. And you want intermediate result values.

i.e. you would want the values of B and C for A=63?

If that is the case, linear interpolation would be my approach, even though the overall function is not linear.

i'm quite a novice so don't know how to do that
the fixed data is column B. i.e depending on what value
is in A, B should then contain the result after a multipler C.
but as the multiper is non-linear across rows thats my problem

 

[mikerickson]

So B = A * C and C is the input data?

What is the relationship between the columns? Which column holds "the divisor" and how is it obtained? Is it calculated or input by hand?

What do you want to find out?

 

[leecavturbo]

So B = A * C and C is the input data?

What is the relationship between the columns? Which column holds "the divisor" and how is it obtained? Is it calculated or input by hand?

What do you want to find out?

in column A. millivolts from a device.
if 200Mv then the result is 14.9 afr
but if the millivolts was 210 what would the afr be?

 

[mikerickson]

We know that 200Mv yields 14.9 afr
Also 500Mv yields 14.7. Using the general interpolation formula *-below:

(210 -200) / (500-200) = (answer - 14.9) / (14.7 - 14.9)

ans-14.9 = 210-200/500-200 * (14.7-1.49)

ans = 14.9 + ((210-200)/(500-200) * (14.7-14.9)) = 14.9 + (10/300*(-.2)) = 14.89333333

So, 210Mv yields 14.89 afr

In general if D1 contains a value between A3 and A4 then the corresponding result for column B would be

=B3+((B4-B3)*(D1-A3)/(A4-A3)).

Finding the right interval (between A2,A3 or A4,A5) will take more work.

*-General formula: (newValue-lowerValue)/(upperValue-lowerValue) = (newResult-lowerResult)/(upperResult-lowerResult)

 

[leecavturbo]

We know that 200Mv yields 14.9 afr
Also 500Mv yields 14.7

(210 -200) / (500-200) = (answer - 14.9) / (14.7 - 14.9)

ans-14.9 = 210-200/500-200 * (14.7-1.49)

ans = 14.9 + ((210-200)/(500-200) * (14.7-14.9)) = 14.9 + (10/300*(-.2)) = 14.89333333

So, 210Mv yields 14.89 afr

In general if D1 contains a value between A3 and A4 then the corresponding result for column B would be

=B3+((B4-B3)*(D1-A3)/(A4-A3)).

Finding the right interval (between A2,A3 or A4,A5) will take more work.

thats a good start cheers

 

[tusharm]

Given your data set I would think a piecewise linear interpolation would be the 'safest' way to go. For that, see
Interpolation
http://www.tushar-mehta.com/excel/newsgroups/interpolation/

no one?