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jason lake
Joined: 03 Sep 2008 Posts: 3
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Posted: Tue Sep 09, 2008 9:53 am Post subject: new user question |
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Hi there,
I am working working with displacement data and using dplot to differentiate this data to generate velocity, smooth and then differentiate to generate acceleration data.
My question is, is there a way to calculate the objects force from the acceleration curve using the following formula: ((g [9.81]*mass [say 20kg])+(mass[say 20kg]*acceleration)).
If this were possible the velocity curve could then be multiplied by the force data to obtain mechanical power output.
Up until now I have been copying the data over to excel to perform these calculations but was wondering if there was perhaps a quicker way.
Many thanks,
Jason. |
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DPlotAdmin Site Admin
Joined: 24 Jun 2003 Posts: 2310 Location: Vicksburg, Mississippi
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Posted: Tue Sep 09, 2008 10:31 am Post subject: |
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Jason,
You can perform whatever manipulations you want to on your data within DPlot using Edit>Operate on Y. If you want to preserve the curve you'll be operating on then make a copy of it first: Edit>Copy>Data values, Edit>Paste. _________________ Visualize Your Data
support@dplot.com |
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jason lake
Joined: 03 Sep 2008 Posts: 3
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Posted: Tue Sep 09, 2008 11:20 am Post subject: |
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Thank you for your swift response.
That works perfectly in answering my first questioning.
However, I'm a bit stuck with regard to my second... can I multiply one y curve by another.
From your answer it seems that I can but I'm having trouble figuring it out.
Many thanks for your help,
Jason. |
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DPlotAdmin Site Admin
Joined: 24 Jun 2003 Posts: 2310 Location: Vicksburg, Mississippi
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Posted: Tue Sep 09, 2008 12:08 pm Post subject: |
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Yes, use Generate>Y=f(X,Y1,Y2,...)
For example to multiple the amplitudes of the 2nd curve by the amplitudes of the 3rd curve, use Y=Y2*Y3 _________________ Visualize Your Data
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jason lake
Joined: 03 Sep 2008 Posts: 3
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Posted: Tue Sep 09, 2008 12:30 pm Post subject: |
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Hi David...
that's excellent, worked perfectly.
Thank you very much for your help.
Kind regards, Jason. |
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