Interpolation of semi-log paper 1

[RonMB]

How do I do a linear interpolation of values on semi-log paper?

 

[zekeman]

What's the application and exactly what are you trying to do?

 

[RonMB]

I am trying to write a software routine that will determine values from a table to which I have entered points from graphs of data plotted on semi-log paper.

 

[IRstuff]

Well, if it's not a linear curve, you wouldn't necessarily do a linear interpolation.

If you know what function generated the curve, you should do a nonlinear curve fit.

TTFN

 

[amorrison]

Maybe try also

http://groups.google.ca/group/sci.math?lnk=srg&hl=en

http://groups.google.ca/group/sci.math.num-analysis?lnk=srg&hl=en

 

[JStephen]

I ran into this sometime back. The plots I was using had three curves, and I was trying to find values in between the curves. You could find the actual values on each curve, and interpolate between them, or interpolate between the logs of the curves, which gave different numbers. But if you took the top and bottom curve, neither approach would give you values very close to the middle curve. The moral is, uncertainty may be high no matter what you do.

 

[SlideRuleEra]

RonMB - On semilog graphs equal movements represent equal percentage changes in value. For example, the distance from say a value of 20 to 26 (30% increase) is exactly the same distance a change from 50 to 65 (another 30% increase).

For more details, I have posted scans of a couple of pages from my Engineering Graphics textbook on my website (link below). It is near the bottom of the home page. Use of the various types of graphs was emphasized then (1965) since there were few other engineering analysis tools readily available.

As an aside: For the above reason, semilog graphs (both on paper & electronic) are still very popular in study of the stock market. As in the above example, the important issue is the percentage change in value - not whether a stock went up 6 points (first case) or 15 points (second case).

http://www.SlideRuleEra.net

 

[ijzer]

As follows: example

Suppose you have avalue X on the log scale between 30 and 40.
The linear dimension from 30 to 40 is e.g. 20 mm and the point X is at 18 mm from the 30 line .

Than X = 30( 40/30)exp18/20

 

[rb1957]

X = log(x)
two points you want to interploate between ...
(X1,y1), and (X2,y2)

then ...
(y-y1)/(y2-y1) = (X-X1)/(X2-X1)

and, as i imagine you're more interested in x ...

x - 10^X