Solved: find intercept value of curve - Page 2

ptc-4915476 · ‎Jan 14, 2013

Hello,

totally new to Mathcad, the help files are not of great help at all.

How do I find the Y value of the below curve where it intercepts the Y axis?

The data comes from a CSV where one column is used for X and the other for Y. My goal is to connect the start of this curve to the end. Then I would like to find the maximum difference between the straight line and the actual curve (the X value).

For now I can't even find out how to find the intersection value. I would not like to use a minimum/maximum function as the curve may look completely different for other datasets. The red line is created via the intercept function but it gives me the wrong starting value.

I appreciate the help.

Werner_E · ‎Jan 16, 2013

Some additional questions:

1) Can you guarantee that the data is always sorted in respect to the field values?

2) How do you deal with the situation if 0G is not included in your dataset? Lets say the nearest data point is at (0.2 G / -5 mV). Then, as the worksheet is written, the ideal sensors characteristics will begin at (0/-5) while the real sensors runs through (0.2/-5). Change in my file the 0 in posField to 0.2 and see yourself. In one of the first worksheets I posted, I dealt with that situation by using a spline interpolation, but this was deleted later.

3) Do you expect datasets with neagtive field values? I guess - no.?

ptc-4915476 · ‎Jan 17, 2013

This worked perfectly well. 🙂

1. These are datasets obtained from a DAC system, it just records the magnetic sensor output depending on the applied magnetic field strength.

2. The acquisition always start at 0Gauss.

3. Yes, this is when an inverse magnetic field is applied. It works , I already have it included in the graph. I had to put the correct signs at some points in the formulas.

I don't know yet what a spline interpolation is. 🙂

I'm at the very beginning of learning to work with this software. I must say it's not very intuitive but you have been of great help..

Thanks again.

Werner_E · ‎Jan 17, 2013

Glad you got it working for you!