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Can somebody figure out which unit this should be?

  1. Aug 28, 2014 #1
    Hi guys

    I have a problem - I can´t seem to figure out the correct unit and name (like power or work).
    In the attached file, the blue graph is showing the applied pressure/force on a sensor vs. time.
    The red area is the area below the blue graph.
    I have two of these sensors, and I would like to know if more pressure is applied to one of them over a period of time. I have integrating the blue curve to find the area under the graph, which I guess is the total pressure applied, and the value of this is approx. 40000. But what is the unit and name for this? Should the x-axis be converted to e.g. seconds before it makes sense to talk about the 40000 - i´m a bit confused.

    The green area is just the sensors steady state, and should be substracted.

    Regards - Kasper
     
  2. jcsd
  3. Aug 28, 2014 #2

    nrqed

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    Sounds like the upload of your figure did not work.

    The area under the curve has for units the product of the units of your y axis and of your axis.
    What are these? For the y axis you mention applied pressure/force, what does that mean? Is it applied pressure or force (these are different things)? And your time axis in what units? All you need to do is to figure out those units an multiply them together.
     
  4. Aug 28, 2014 #3

    Thanks for your quick reply. I have attached the graph now.
    Yea, the y-axis is in kPa, and the x-axis is samples. The x-axis could easily be converted to seconds.
    So the unit is kPa*s? That sound weird?
     

    Attached Files:

  5. Aug 28, 2014 #4
    Was just about to hit submit when you edited your original post, so I'll just post here instead :smile:

    If you took into account your sampling rate, such that the time axis would be, for instance, in seconds, the unit of the measure of the area under the graph would just be in kilopascal seconds [kPa*s].

    Maybe the average pressure would be more meaningful to you? You'd just have to divide by the time interval you've integrated over, i.e. the average value of the pressure ##p## is given by:
    [tex]
    \left<p\right> = \frac{1}{T}\int_t^{t + T} p(\tau) \, d\tau
    [/tex]
     
  6. Aug 28, 2014 #5
    Thanks for your reply :)
    Does kPa*s have a name like power or work?
    Actually I´m not that interested in the average pressure as i´m in the total pressure during the period. The sensors are placed in shoes, and I need to know, if more pressure is applied to either sensor. For example if a guy was walking assymetrical, he would put more pressure on one leg, and I need to figure out if this is happening.
    So will it make sense to say, that the total pressure applied is 40000 kPa*s? (If I converted the x-axis)
     
  7. Aug 28, 2014 #6
    When thinking about it - 40000 kPa*s does not indicate how long the time period was, so I dont guess this makes sense?
     
  8. Aug 28, 2014 #7
    Well, coincidentally the pascal-second is also the unit for the measure of dynamic viscosity, but that's a bit of a different concept.

    But what exactly does total pressure mean? If you define it as the area under your graph, then total pressure isn't actually a measure of pressure (it doesn't have a unit of pressure).

    The average pressure would still tell you if the area under one graph is larger than under another since you're just dividing by the same factor in both cases. The average pressure, however, is a measure of pressure, so I thought that might make more sense to you.
     
  9. Aug 28, 2014 #8
    I see what you mean now - I think I´m a bit confused ;)
    So lets say the total area is 40000 kPa. Avg pressure = 40000kPa/1400 samples = 28.6 kPa/sample. That would be the most understandable measure?

    Thanks for your time :)
     
  10. Aug 28, 2014 #9
    Area would be 40000 kPa*sample and average pressure would be 28.6 kPa :smile:
     
  11. Aug 28, 2014 #10
    No, the avg pressure should not be 28.6 kPa/sample, but just 28.6kPa, right?
     
  12. Aug 28, 2014 #11
    I was to slow! Thanks for the help :)
     
  13. Aug 28, 2014 #12
    You're very welcome.
     
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