ACCEPTED SOLUTION
Accepted Solutions
Thank you for your reply, Werner.
Werner Exinger wrote:
Could you elaborate more precise what you try (cycling) between Plot 2 and Plot 3?...
I hope the following screenshot could help:
Step 1:
Step 2:
Step 2 with the table:
Step 3:
Werner Exinger wrote:
How do you define amplitude and minimal amplitude in case of your data? The minimum delta you calculate is only from the difference max_i - min_i, but max_i-min_i+1 (going from a local max down to the next local min) could yield an even smaller result - not relevant?
Amplitude = Max - Min;
min(Amplitude) = Minimum from all set of "Amplitude".
It seems to me that at the end everytime you would arrive at two values which could have been calculated simply by max(maximum) and min(minimum) from the very beginning. Or what am I missing.
Or are you utilizing those intermediate data (the values you cut out of the graph in every step). Yes, that seems to be the point, otherwise it would be too easy.
Some more questions:
1) The cycling ends when ... see later
2) Is guaranteed that the second datapoint is greater than the first? In your file you assumed that as you assumed that the number of maxima is equal or one less than the number of minima.
3) One thing I still don't understand is, why you are interested in the difference (amplitude) of the ascending part of the graph only and not the desvending one.
As far as I understood your algorithm: You look for the ascending line segment with the smallest (vertcal) difference, put the two values in your numbered exit-table and then delete those points from the data set. If there are more than one segments with the same minimal diff you take the first. Sometimes you will have to remove the very first linesegment that way. So every time you remove one minimum and one maximum.
You iterate this until only two minima and one maximum (plus NaN) is left.!? But what, if the number of min and max is equal. When do you stop now?
You are only interested in the minimum of the ascending segment even if a descending segment would have a smaller difference (as in the following example).
Lets see if I got it right. With the data I have chosen, would the following be right or what would be the disired outcome (your routine fails on that data set, probably because the minimum is the very first segment):
Your exit table would end after line 3, does it?
I think a recursive function would do the job, but first I have to be absolutely sure what you want to achieve. On second thought - as we know the number of iterations it could be easier.
Werner,
Some comments on given by you example:
1. Every time we start from left to right and look for the minimum amplitude (Amplitude = Max - Min), as long as there will be three points and then stop ((minimum), (maximum), (minimum); this values of min/max entered last in the exit table). If the amplitudes are equal, then we cut the very first of them (cut one minimum and one maximum) going, with, from left to right.
2. When at some value we started at that value we have to finish (in your example, this is "0") each time the of data reconstruction after cuttting. And in the end we have only 3 points (min(frome all data), max (frome all data), min (from all data)). When our data will have only three values then cycling ends.
3. Amplitude must be between the two lines.
Comment:
Right solution:
Exit table for example must be like this:
So, if we consider your suggestion examples, then first, the data needs to be rearranged so that the cycle starts with the maximum peak or the minimum valley, whichever is greater in absolute magnitude. In example below, the highest peak with a magnitude of 12 occurs first as opposite to the lowest valley and is therefore chosen as the beginning point for the rearranged data. The new data which shown also below is generated by cutting all the points prior to and including the highest peak and by appending these data to the end of the original data. An additional highest peak is included in the new data to close the largest loop for conservatism.
Data:
New rearranged data:
Werner Exinger wrote:
It seems to me that at the end everytime you would arrive at two values which could have been calculated simply by max(maximum) and min(minimum) from the very beginning. Or what am I missing.
Or are you utilizing those intermediate data (the values you cut out of the graph in every step).
Yes, this is correct. Finally we come to the max (maximum) and min (minimum), which can be determined before processing. The condition of the correctness of the final program will be that column for the "delta" should contain the increasing elements and last element in this column must to be the maximum amplitude (max (delta)).
I use these intermediate data (values of min and max, cuted from the graph/our data at each step) for the next calculation of the equivalent load. Initial data represent load-time history. Some methods of processing this data can be found in the following standard: "DDR-Standards TGL 33787 (Schwingfestigkeit. Regellose Zeitfunktion. Statistische Auswertung)". I want to write a program to process the data by using my modified method.
In fact, in the final table, we need only "Max" and "Min", which shall be placed in the table every time we rearrange our data
Ok, thats no so difficult. But I think you did not reveal all rules which have to applied. See the attached file and my annotations.
OK, the new specifications with the splitting and rearranging of the data at the first absolute extrema was easy to implement as was the adressing of the problem with sequences of equal valued data (http://communities.ptc.com/message/195434#195434). Theorde was important, first splt and rearrange, then eliminate the duplicates in series.
Have modified the routine to return for every step the rearranged and processed data and the linesegment to be discarded. Can be used for step by step graphing or making an animation (see attached).
Werner Exinger wrote:
...I think version 4 will be more valuable for you...
But why? I work with version 5 because i would like to see animation step by step. And both worksheets version (4 and 5) give same results in the exit table.
Werner Exinger wrote:
..I posted both somewhere in this thread but I'm beginning loosing track...
They can be found in this post: http://communities.ptc.com/message/195796#195707
So it seems that I don't need the following program line (which showed below) in routine anymore (I added this new line code just for checking)?
Werner Exinger wrote:
ad 1) whats the specific problem in adding additional data? Do you really mean rows or rather columns?
I mean that I want to add some new data in exit table.
This line does not work (the sum of values😞
Connells 70 schrieb:
Werner Exinger wrote:
ad 1) whats the specific problem in adding additional data? Do you really mean rows or rather columns?
I mean that I want to add some new data in exit table.
This line does not work (the sum of values😞
From what I see you are trying to add columns to the exit table, not rows.
The line you marked does not work, because you had chosen the vector sum and the argument is no vector.
As a quick hack I've implemented the first value in a similar way you did after the table.
Hope it helps!
Thought it would be a good idea to introduce some local variables to make the code better manageable. I also got rid of some unnecessary conditions in your if''s (if hi is 0, lo IS neagtive; if lo/hi is negative, hi IS positiv).
Not sure if you are really calculating what you want with the two values "R" and especially "cycle". With "cycle" you may run into problems if "hi" is zero, You may want to use "on error".
