#  Other Applications & Softwares  > Access Tables & Databases >  >  help adding a trend line to a not-so linear graph, and then moving it

## jmarcian

here is what i need to do...

one, make a trend line base on the beginning linear part of the graph
two, move that line a certain amount of units on the graph
three, find the new intersection of the line and the trend line

can it be done...

see attached for graph


thanks for the help in advance!
jared

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## jmarcian

also i am looking to find the area under the curve, is that possible?

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## shg

One, post workbooks instead of pictures, unless you want help with pictures.

Two, be specific.

Three, explain the genesis of the problem so someone has a clue as to what you want to do.

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## jmarcian

ok, the workbook is huge, but i will post it regardless....

here is the deal.. notice the initial linear part of the graph. i need to obtain the slope of that using the initial point (0,0), and the last point of the linear section, not real sure what it would be but you get the jist. Once we find out this line, we are interested in the slope. the slope is an important term called "youngs modulus" anyone familiar with mechanics of materials?

now, the material's yield strength is found by adding a 2% strain (x-value). this is why we want to take that initial line and move it over 0.02 units. after we move the line, we are worried about the y-value of the intersection of the created line and the graphed, this gives the materials yield strength.

following?

lastly, the materials toughness is given by the area under the curve. which is why the area is sought after. of course if the line were given by a simple function, that would be easy, it is simply the integral of the function!

attached is the work book, beware of the size, and sorry if it is not organize hhehe


thanks
jared

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## shg

Young's modulus is the spring coefficient before onset of inelastic deformation. right?

So if the initial slope is indeed linear over some range, you can just use slope  of the first two points, can you not?





> now, the material's yield strength is found by adding a 2% strain (x-value). this is why we want to take that initial line and move it over 0.02 units. after we move the line, we are worried about the y-value of the intersection of the created line and the graphed, this gives the materials yield strength.



That part I don't understand. If you want to explain in the context of Mark's, I have a copy 10 feet behind me.





> ... which is why the area is sought after



So you can do a simple numerical integration of the data you have, right?

You have two curves, and the second is completely unexplained. We are left to guess the relevance?

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