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BertilRogmark12-Amethyst
12-Amethyst
June 3, 2024
Question

curve equation

  • June 3, 2024
  • 4 replies
  • 9672 views

Is there a way to write the spiral curve equation in a cylindrical coordinate system in  such a way that the pitch is zero in the beginning and the end so that the curve starts and ends tangent to the start and endplanes that are normal to the trajectory center axis?

 

 

4 replies

tbraxton
22-Sapphire II
tbraxton22-Sapphire II
22-Sapphire II
June 3, 2024

You can modify the parameters of an Archimedes spiral here: Archimedean Spiral -- from Wolfram MathWorld

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BertilRogmark12-AmethystAuthor
12-Amethyst
June 4, 2024

That is not what I am looking for.

upload_-aW1hZ2UwMDEucG5n-9110847491704557794..png

The screen dump shows a curve created in a cylindrical coordinate system with the equation:

r=75

theta=t*360

z=t*100

I am looking for an equation for z to make the curve ends parallell to planes FRONT and DTM1 respectively.

Bertil

tbraxton
22-Sapphire II
tbraxton22-Sapphire II
22-Sapphire II
June 4, 2024

The picture clarifies what you need. If you use the parametric equation only to create the curve, I am pretty sure the answer is no as the trajectory of this type is not parallel to the planes at any point along its path.

 

I have had to deal with this type of problem with spring designs and you end up with a discontinuity as seen here to get a flat end on a helix. This is using a helical sweep, but it is the same issue. You will need to add some geometry that is not defined by the curve equation to get the transition to zero pitch. In this case you can see there is a kink at the bottom to go to zero pitch. When these are made it is not a sharp bend as seen in the CAD as the material has a bend radius.

 

tbraxton_0-1717498998893.png

 

kdirth
21-Topaz I
kdirth21-Topaz I
21-Topaz I
June 4, 2024

Wrap may be an easier way to get what you want.

kdirth_0-1717529953380.png

Creo 7.0 attached.

There is always more to learn.
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BertilRogmark12-AmethystAuthor
12-Amethyst
June 5, 2024

Grateful for your suggestion but cannot comment on it since ”wrap” does not exist in Wildfire 3.

Patriot_1776
Patriot_177622-Sapphire II
22-Sapphire II
June 7, 2024

Are you still on Wildfire 3???  Yikes!  Holy outdated software Batman!  Then the graph function is your only option and will do it just fine IF you get the graph and the equation right.  Too bad, that means you can't see my model then since I'm on Creo 8.

 

You still have not defined HOW tangent you want the ends, or how many rotational degrees each transition takes, and what the center section is like as I mentioned before.

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BobNJ4-Participant
4-Participant
June 5, 2024

Kdirth is correct about the graph feature.  Many years ago I used the graph function to do what I think you are wanting to do.  Since I don’t have a way to provide an example, I searched some old sites for a video example for you.  It goes back many years but I can’t tell you if it was available in Wildfire 3.  I haven’t posted for a long time, hope the link works

 

Evalgraph 

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BertilRogmark12-AmethystAuthor
12-Amethyst
June 5, 2024

Thanks, I will investigate this. In the mean time I have found a solution as shown by the attached files.

tbraxton
22-Sapphire II
tbraxton22-Sapphire II
22-Sapphire II
June 5, 2024

I do not see any files attached.

Patriot_1776
Patriot_177622-Sapphire II
22-Sapphire II
June 6, 2024

As Tom mentioned, you can use a graph feature to do this, I've been using graphs like this since 2000, long before I saw anyone else use them in this fashion.  As KDirth said, you can also use a wrap, it just depends on what gives you the best result, in most cases, the end result is pretty much the same, although a wrap gives you the ability to more easily change the number of turns without affecting the radii at the ends that force the tangency.  Fun Fact:  You can also force one or both ends to be tangent to the axis.  The advantage to using a graph and curve by equation, is that you can also vary the radius, unlike with the wrap because a wrap has to be done on a "ruled" surface.

 

Best of luck!

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BertilRogmark12-AmethystAuthor
12-Amethyst
June 6, 2024

Hearing the later suggestions and using a graph like the one shown in the post from  tbraxton would seem to give me what I am looking for, but how do I apply this graph to the r/theta/z-equation that controls the curve?

KenFarley
KenFarley21-Topaz II
21-Topaz II
June 6, 2024

The steps I took to do this are something like:

(1) Create the graph. When I do this I usually try to make the range of X and Y on the graph span zero to 1, with a corresponding "t" range in the curve I'm going to define of zero to 1, also.

(2) Define the necessary references for an equation driven curve (coordinate system, planes, etc.)

(3) Build the curve, using cylindrical coordinates. For the twisted wire I modeled, for  example, the equations looked like this

r = ( diawire + clrwires ) / 2
theta = numcoils * 360 * evalGraph ( "getTheta", t )
z = t * numcoils * pitchcoils

The graph I defined is named "getTheta". The evalgraph call, given a "t" value, returns the "y" defined by the curve.

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