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B How do I linearize this graph?

  1. Jun 16, 2016 #1
    Hello!
    I'm completing my IB Physics IA and after collecting my data and plotting the raw graph my graph seems to be a negative parabola.
    In order to linearize it I though one had to square the x-axis data; however I simply get the same graph shape.
    What am I supposed to do?
    Only when I plot the following I get a straight line [y-axis data = y and x-axis data = x]:
    y/x against 1/x.

    Why is this? How do I linearize this graph?
    Thankyou!
     
  2. jcsd
  3. Jun 16, 2016 #2

    DrClaude

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    Could you post a graph of what the y vs x data look like?
     
  4. Jun 16, 2016 #3

    CWatters

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    or post the raw x and y data.
     
  5. Jun 16, 2016 #4

    CWatters

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    That sounds like it has the form
    y = -xn + C

    If a plot of y vs -x2 is still a parabola then perhaps n is another even number like 4 or 6 etc ?
     
  6. Jun 16, 2016 #5
    If y/x against 1/x is a straight line than y vs. x should be a straight line too. I agree with CWatters that x,y-data would be helpful.

    You should also explain why you want to linearize the data. Maybe there is an entirely different solution for your problem.
     
  7. Jun 17, 2016 #6
    Thank very much for the quick replies, I've attached my raw data graph to this reply.
    I'm attempting to linearize how the heeling angle of a boat affects its velocity.
    Moreover I've already attempted to raise the data to a different power [up to 10] against the other variable however this doesn't seem to work either.
    The data is also on the left hand side if you would like to attempt to try something out.

    Thank you very much in advance!

    upload_2016-6-17_12-33-45.png
     
  8. Jun 17, 2016 #7

    sophiecentaur

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    The data seems to have a nicely behaved maximum and the data points are evenly spread along the curve. What would be the purpose of "linearising" the graph? (Or do you mean something different from what I am thinking?)
     
  9. Jun 17, 2016 #8

    jbriggs444

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    When you attempted to linearize this by squaring the x coordinate, did you square the raw value or did you square its displacement from the midpoint of the fitted parabola?
     
  10. Jun 17, 2016 #9

    DrClaude

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    Also, since the x axis is angle, I would expect something that goes as cos(x) or cos2(x) more than x2.
     
  11. Jun 17, 2016 #10
    I would need to linearize this graph in order to show a proportional or linear relationship between these
    I squared the actual raw data [or coordinates if you will] for example for the first data point 0.21194 I squared that and proceeded to do that for all the other points and then plotted that however I only obtained the same shape of the curve
     
  12. Jun 17, 2016 #11
    I just tried plotting velocity against cos^2(x) and its look more like a scattered graph unfortunately

    upload_2016-6-17_15-49-17.png
     
  13. Jun 17, 2016 #12

    CWatters

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    I've not used LoggerPro but Google found..

    LoggerPro Linearizing a Quadratic Function
     
  14. Jun 17, 2016 #13

    CWatters

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    That doesn't work as your equations is of the form
    y = ax2 + bx + c

    I'm not sure how you linearize a quadratic that has a bx term.
     
  15. Jun 17, 2016 #14
    The plot clearly shows that the relationship is not linear and if you transform it into a linear graph you don't need to show the linearity anymore because it has been designed to be linear. I do not see a reason for a linearization unless you have a corresponding theoretical model.
     
  16. Jun 17, 2016 #15

    jbriggs444

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    In order to get a proper result, you should square the offset of each point from the fitted center of the parabola and re-draw the graph using those results.
     
  17. Jun 17, 2016 #16
    That would be possible with y' = 2ax + b (if the data allow for a numerical differentiation) but the question is why to do something like that.
     
  18. Jun 17, 2016 #17

    sophiecentaur

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    But the relationship clearly goes through a maximum and that, I would imagine, would be worth finding as it would correspond (?) to some optimum design arrangement. It would be relatively easy to identify this maximum. This has been mentioned earlier I think.
    I can't help feeling that someone may have suggested this linearising strategy to you and you feel you need to follow it. It;s an excellent idea when dealing with some types of data. I would have said that the nature of your data is telling you that there is no linear relationship. If it was, indeed, someone else's idea, have they actually seen your data? It would not be the first time on PF that someone wants help with some task that they have been set which is not necessarily appropriate. Ask your teacher / advisor / supervisor and save yourself more angst. :smile:
     
  19. Jun 18, 2016 #18
    Here is an alternative to the quadratic equation.
    I've only spot checked this from 6 to 20 degrees, 5% error. Needs refinement for less than PHIvpk (6 degrees).
    Have fun!

    Vpk = Peak Velocity
    PHI = New Heel Angle
    PHIvpk = Heel Angle at Peak Velocity
    VY = Velocity at PHI

    VY = Vpk x COS15.3(PHI - PHIvpk)
    (Yes, that is the 15.3 power of the COS)
     
  20. Jun 18, 2016 #19
    Thank you very much for the help however when graphing COS^15.3 one can see how some areas of the graph are undefined and as a result I cannot achieve data points for some results
     
  21. Jun 19, 2016 #20
    Which areas?

    If it is areas less than PHIvpk (6 degrees), that is the "needs refinement" area I mentioned. The following should cover that.
    VY = Vpk x COS15.3(ABS(PHI - PHIvpk))
     
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