Geometry Expressions 3.0.8 ML.rar

Geometry Expressions 3.0.8 ML.rar


Geometry Expressions 3.0.8 ML.rar

Hellanzb 3.3 RPM .Q:

Plotting a known progress curve

I am not sure how to ask this question, so the answer may be obvious. I have seen a number of questions on here and the internet, but the answers all seem to use a least squares regression model and I can’t get my head around that. I would like to just know how to create a simple gnuplot curve that shows a trend of how long it takes to complete a process.
I have a very simple process that takes a 3D volume, normalises it then creates a series of cross sections. I have the size of the volume, the duration of the process, and I know I start at the beginning of the volume and it takes 1 second to complete. I know the dimensions of the volume and I know how many cross sections I need. The volume can be defined by:
r, theta, z
where r is the radius of the volume, theta is the angular coordinate in the xy plane and z is the z coordinate.
I need to plot a line that shows the progress as a percentage. I don’t need to know the actual time, only that it takes 1 second to go through the process. If possible, it would be great if I had a simple equation for the curve, but I can work it out in gnuplot if that is harder. Thanks for any help!


Well, since it seems some people want to answer it anyway, I’ll give it a shot…
For a completely arbitrary radial point (r, theta, z) the equation of your interpolated curve is
y(t) = (1 – t)**t * t**(t – 1) * (1 – t)**(1 – t) * (1 – 2*t)**(1 – t) * (1 – 2*t)**t
y(1) = 1 – 1.5 * x + (1.5 – 1.75)*x^2 + (1.75 – 2.25)*x^3 + (2.25 – 3.5)*x^4

or if I don’t assume theta = 0 and z = 0:
y(t) = (1 – t)**t * t**(t – 1) * (1 – t)**(1 – t) * (1 – 2*t)**

Klitora latinica lengtas perminor 1
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Geometry Expressions 3.0.8 ML.rar
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Geometry Expression 3.0.8
Geometry Expressions 3.0.8
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