14 replies to this topic

[marcus]

Hi I'm new and I've got data which needs interpolating. I've tried researching double interpolation but it doesn't seem to work.

My data is 0.9774 g/cm3 at 8.4C (temperature in Celsius)

The table of data I've taken from a methanol/water table is attached below. Can someone please drop a hint. Just stuck on how to get the interpolated wt%.



thanks in advance!

Attached Files

•  table of data.jpg   6.17KB   39 downloads

[marcus]

I think I've found a suitable to method now for my interpolation, from the site below. However can someone please let me know how the order of the "quadrants" in step 2, its a critical step that im unsure of.

http://www.ehow.com/...erpolation.html

thanks

[breizh]

Hi marcus
Let you try LINEST function , excel spreadsheet .
Regards
breizh

[marcus]

Hi marcus
Let you try LINEST function , excel spreadsheet .
Regards
breizh

thanks but im not sure how to do this. i prefer hand calculations, can anyone please help.

thanks

[latexman]

marcus,

First, I suspect an error in the 17%, 10^o C data point. Should it be 0.9739?

I applaud your desire to do this by hand to experience the technical challenge! This method works pretty good as long as the two variables (wt.% and temp.) are independent (i.e. no interactions). Develop a linear equation for density = f(wt.%) for the two data points at 0^o C and another linear equation for density = f(wt.%) for the two data points at 10^o C. They should be very similar. You may want to average the results (i.e. the slopes of the two equations). Create data points using the respective equation(s) for 16% and 17% at 0^o C and 18% and 19% at 10^o C. Now you have one data point at each temperature at all 4 concentrations. Develop 4 linear equations for density = f(temp.) for the two data points at 16%, 17%, 18%, and 19%. They should all be very similar. You may want to average the results (i.e. the slopes of the 4 equations). Now, create one equation that combines the wt.% correction (the slope of the linear equation) and the temp. correction, like density = b + m_wt.%wt.% + m_temp.temp. b is the intercept of the new equation. Use the original 4 data points to calculate what b needs to be and average the results.

I hope you can follow my logic and description.

[kkala]

I think attached interp.xls is what you asked for, subject of course to criticism from other members.
It generally follows proposal of latexman, with some further simplifications. Right sg for 17% w/w content at 10 C seems to be 0.9739 as latexman suggested, not 0.9239.
Extrapolations and interpolations are needed, as explained on the spreadsheet.

Attached Files

•  interp.xls   16KB   50 downloads

[latexman]

kkala,

Looks good to me.

[marcus]

I think attached interp.xls is what you asked for, subject of course to criticism from other members.
It generally follows proposal of latexman, with some further simplifications. Right sg for 17% w/w content at 10 C seems to be 0.9739 as latexman suggested, not 0.9239.
Extrapolations and interpolations are needed, as explained on the spreadsheet.

thanks so much kkala, i never expected the actual calcultion, cant thank you enough

i did try to hand calculate it via ehow's.com step-by-step method but i came out with 26.38wt% and i nearly accepted it as the correct answer.

i guess i'll never understand interpolation properly.

[breizh]

HI Marcus ,
One day you may use linest function .
hope it helps
Breizh

[Padmakar Katre]

Dear,
Even 'spline function' could be used to interpolate the data.

[katmar]

Marcus,

If you have not reached the point in your course where you study interpolation and regression, then a useful technique is to draw a graph on paper and interpolate that way. Do students still draw graphs on paper? I hope so. Doing it this way gives you a good feeling for the numbers and will help avoid the gross errors like you nearly made using a method you did not yet understand.

See the attached (very rough) graph. Luckily no one is grading me on this effort, but it should serve to show you the technique.

Plot your data on the graph. See points A, B, C and D on the graph. Draw the lines through the families of data points and label them. Now measure the distance between the two lines at the target value, ie at density = 0.9774. To interpolate (linearly) between the 0 deg C and 10 deg C lines you simply measure the distance between the lines and then make a mark at 84% of the distance between them. 84% is (100* 8.4 / (10 - 0))

From my rough graph I get 14.35%. kkala got 14.30% by his Excel method. Using Excel I get 14.25%. Within the accuracy that we have here these are all affectively the same number, and much better than the 26.38% the other method gave you.

Attached Files

•  Eng Tips Interp.gif   95.94KB   21 downloads

[marcus]

Marcus,

If you have not reached the point in your course where you study interpolation and regression, then a useful technique is to draw a graph on paper and interpolate that way. Do students still draw graphs on paper? I hope so. Doing it this way gives you a good feeling for the numbers and will help avoid the gross errors like you nearly made using a method you did not yet understand.

See the attached (very rough) graph. Luckily no one is grading me on this effort, but it should serve to show you the technique.

Plot your data on the graph. See points A, B, C and D on the graph. Draw the lines through the families of data points and label them. Now measure the distance between the two lines at the target value, ie at density = 0.9774. To interpolate (linearly) between the 0 deg C and 10 deg C lines you simply measure the distance between the lines and then make a mark at 84% of the distance between them. 84% is (100* 8.4 / (10 - 0))

From my rough graph I get 14.35%. kkala got 14.30% by his Excel method. Using Excel I get 14.25%. Within the accuracy that we have here these are all affectively the same number, and much better than the 26.38% the other method gave you.

Wow there are just so many different ways to interpolate. Thanks for all your help and im pretty sure this has helped so many more people other than just myself.

[Merek Roman]

here is a great webpage for interpolation http://www.ajdesigne...on_equation.php

it even shows the math!

[marcus]

here is a great webpage for interpolation http://www.ajdesigne...on_equation.php

it even shows the math!

hi thanks but i think this interpolation requires double interpolation if im not mistaken?

[kkala]

"a useful technique is to draw a graph on paper and interpolate that way... gives you a good feeling for the numbers and will help avoid the gross errors like you nearly made using a method you did not yet understand.
From my rough graph I get 14.35%. kkala got 14.30% by his Excel method. Using Excel I get 14.25%. Within the accuracy that we have here these are all affectively the same number".

I quite agree with katmar to the usefulness of the graph, especially if numbers for interpolation - extrapolation create a feeling of confusion. A square paper, a pencil and a ruler is what we need for this.
In the attached interp.xls you can also see the three lines to show sg vs wt% in the case of 0 C, 10 C, and 8.4 C. The line of 8.4 C is drawn between the other two, at distance suggested by katmar, and indicates w=14.30% for sg=0.9774. I used excel to draw the lines for clarity, but it is understood these will be hand drawn on the square paper. Note that only two points have been given for the sg - wt% relation (0 C or 10 0C), so resulting lines are straight.
Excel and graph are not actually different methods, only that graph is more intuitive and results can differ somehow from person to person (this may not be annoying for many engineering issues, anyway). The key is to get a physical feeling of the projected results, to avoid going astray. Searching for more data can limit uncertainties, especially in extrapolation.

Basic data for the required projection (interpolation) agree to Perry's 7th ed Table 2-109 (methanol density). Nevertheless Table indicates a more narrow "fence" of basic data, as follows:
For 0 C : 14% sg=0.9792 and 15% sg=0.9780
For 10 C: 14% sg=0.9778 and 15% sg=0.9764
Which results in w=14.46% for sg=0.9774 and 8.4 C (same method as before, linear interpolation, no need for extrapolation). This 14.46% can be considered as a more precise value.
The small deviation is due to extrapolations previously done, e.g. linear extrapolation had given sg=0.9775 for 10 0C & 14%, Table gives 0.9778 (max difference detected). Probably the "exercise" was intentionally made a little more difficult for the students, so that extrapolation should be also applied.

Attached Files

•  interp.xls   21.5KB   10 downloads