paraboloid from x and y data points

[zerotolerance]

Hello Guys. I have a set of data points in 2D i.e.. (x, y), that produces a parabola. However, I want to draw a paraboloid (similar to the picture below). In the 2D data the height of the parabola is determined from "y" axis, so I was imagining of just rotating this parabola along the "y" axis that can create the impression as seen on the picture. At the moment I'm really confused on how I can generate the "z" axis so that when its all plotted it would produce a paraboloid. 

I've tried various 3D line plots but all that it does is draw a parabola in a 3D axis (but even then I don't think that points in "z" axis were correct). 

Just to give you a background on what I'm doing. I'm designing an off-axis parabolic mirror, which is just a section of a paraboloid. To do this I generated a parabola curve that produces the focal point, diameter, vertex etc. to exactly what I need. But in order to machine the mirror I need 3 axis (i.e.. x, y, z). The parabola that I produce based on the properties that I need only gives me the X and Y points. 

Any help is most appreciated.  

Attached are the data points. 

Thanks a bunch in advanced. 

Kas

[Darin.K · Accepted solution]

Not a lot of time for details now, but something like this?

Pretty easy application for the 3D parametric surface plot.  In case my code is sketchy, you can find more help here.

http://zone.ni.com/devzone/cda/tut/p/id/4984

Message Edited by Darin.K on 08-28-2009 01:00 PM

[Darin.K]

Oops, forgot to save after the final modification.  Now you get the picture and the set of points (which I think is what you really need).  BTW, I just faked the parabola data since I didn't want to deal with importing your file, you can easily substitute your XY data for the output of the Eval vi.

[zerotolerance]

Thank you so much. Its exacly what I needed. Just one thing, I was trying to make the X and Y grids visible rather than just having Z. I tried going into properties and such but I couldn't see an option. Should I just replace the graph or is there an easier way.

Once again thanks for everything. 

Kas

[Darin.K]

Right-click on the graph, select property browser and you should be able to select the grids you want.

[zerotolerance]

Hello Guys, this might be a little more difficult. I'm trying to isolate a set of coordinates that a cylinder would go through the paraboloid. e.g. if I place a cylinder with lets say radious 35 with a certain height (maybe same as the paraboloid), at points ((30, 0) where these coordinates are at the bottom of the cylinder), then the cylinder would intersect a section of the paraboloid as it goes upward. How can I record the coordinate (x, y, z) that are within the surface of the cylinder as it intersects the paraboloid. This is basically the conic section of the paraboloid.
 

I'm not sure if this makes any sence but please let me know if more explanation is needed. Attached is an illustration of what I mean.

 

Thanks you for everything

 

Kas

[elset191]

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Darin.K wrote:
Right-click on the graph, select property browser and you should be able to select the grids you want.

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Additionally you can right click>>CWGraph3D>>Properties and a nice little dialog comes up.

[Darin.K]

You have the set of XYZ points of the paraboloid from my example, and you know the equation of the cylinder (x-xc)^2 + (z-zc)^2 = r^2 (axis parallel to y-axis).  I would just compare each point to the condition (x-xc)^2 + (z-zc)^2 <= r^2, if true the point is inside the cylinder.

[zerotolerance]

The paraboloid parametres are just a, b and c.Ince in my case "c" is always zero, I'm assuming the condition would change to x^2+z^2 <=r^2.

As such, I tried to implement this in the program you provided, but I don't know how to set it so that the condition takes into affect wherever the the center of the cylinder is placed on the x-axis. e.g. if I put the cylinder with radious 20 at (15, 0) then from that point the condition would take affect. 

Attached is the ammended VI. 

Much appriciated Darin. 

Kas

[Darin.K]

Can't open a V8.6vi, at any rate, x^2+z^2<=r^2 is for a cylinder on the y-axis (xc=0), to shift it in x use the condition (x-xc)^2+(z-zc)^2<=r^2 where (xc,zc) is the center of the cylinder, ie. (15,0) in the example you stated.