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Hi everyone.
I'm trying to plot 2 lines and a circle in mathcad prime 4, in the form (x-x0)^2+(y-y0)^2=R^2.
I have values for x0 and y0 and R, but I can't figure out how to plot it. Solving for y gives me complex numbers.
x0 and y0 are defined values of some earlier calculations, so their value can change, so they are not "fixed" numbers.
Any simple way to do this? (Seems all the info on this I can find is pretty old, and I can't open example files and such...)
One way is as follows:
i:=0..360
thetai = i*deg (i as a subscript on theta)
x:=x0+R*cos(theta)
y:=y0+R*sin(theta)
Then simply plot y vs x.
Alan
Ok, that works. Thanks.
- Any way to make it a function of x? That would be great, that way I could check intersection points with the lines..
You can plot versus x, but it's a double-valued function:
You may use your appraoch even though the symbolic result uses the imaginary unit (the values derived are real anyway):
To get a closed curve you may define a range variable for plotting as Fred had shown.
But a better approach IMHO is to use different representations for calculations and for plotting. The parameter representation Alan suggested is best suited for plotting. I prefer a small program which returns a matrix of x and y-values of the points of any circle which can be used easily to plot different circles:
You may use this in the following way. I also plotted the two lines which as it looks are nearly tangents of the circle:
Worksheet in format P5 attached
I used the program that the one person wrote and it worked fine. The challenge for me, not being the sharpest tool in the shed, was to figure out what the parameters were and then to figure out that I had to take the two columns of the resultant matrix separately to get the 2D plot to work. I use PTC MCAD 3.0 PRIME but not professionally.
PDF shows how to plot the circles.
n=100
Using posted techniques.
Plotting of polygons by line is easy for Prime 7. However, plotting of closed polygons by color for Prime 7 needs tips.
Use polar plot
If you consider R to be the square root of the radius of the circle, you're OK. Otherwise:
Success!
Luc