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Rotating Helical Curve from XY plane to XZ plane

ChristianSmith
1-Newbie

Rotating Helical Curve from XY plane to XZ plane

Using the cylindrical helical curve from equation option where you enter the "r, theta, and z" values, Is there any way to rotate the curve so that instead of the curve being in the XY plane it protrudes onto the XZ plane? I've attached an image that shows the curve on the default XY plane, I want the curv to run along the same axis as the threads on the part shown.


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25 REPLIES 25

You can try by creating another co-ordinate system with the desired Z direction. Select this coordinate system for your equation.

That is one way to do it that I hadn't thought about, Thanks

Or just change the equations appropriately; the helix axis doesn't have to be along the Z-axis.

I tried changing the equation by replacing Z = with X = or Y = and it didn't work. Is that the right idea? What would be the correct way to modify the equation?

For example:

y-axis-helix.jpg

x=2*cos(3*t*360)

y=3*t

z=2*sin(3*t*360)

Oops! I just reread your original post more closely; you were using Cylindrical coordinates, and I was thinking Cartesian. If you want to use Cylindrical coordinates, then Srinivasanlyer has given you the correct answer because Z is then the only axis available for cylindrical reference.

You can use D-H parameters (Denavit-Hartenberg aproximation) like if you use a robotic arm.

For example one rotation around X, one translation in X and one rotation of 90 degrees around X transform a single point in a circle.

If you repeat that (6 homogeneus matrix) you have a circular trayectory arround a circle (a torus coil).

x = c1*(l1 + l2*c2)
y = s1*(l1 + l2*c2)
z = l2*s2

can afect z to have an helical coil

helical_coil.bmp

If you repeat it again (9 homogeneus matrix) you can obtain a circle that follow a circle that follow the first circle (a coiled helix halo).

X = c1*(c2*(c3*l3+l2)+l1)-l3*s1*s3
y = c1*l3*s3+(c2*(c3*l3+l2)+l1)*s1
z = (c3*l3+l2)*s2

El mensaje fue editado por: Faustino Garcia

If you repeat it again (9 homogeneus matrix) you can obtain a circle that follow a circle that follow the first circle (a coiled helix halo).

X = c1*(c2*(c3*l3+l2)+l1)-l3*s1*s3
y = c1*l3*s3+(c2*(c3*l3+l2)+l1)*s1
z = (c3*l3+l2)*s2

coiled helical halo.bmp

El mensaje fue editado por: Faustino Garcia

Very interesting, so what then are the variable s and l and how do you define them? I'm in WF 4.0 so the .zip file doesn't work with my version.

s for sin

c for cos

l is length (in this case radio)

These are the equations.

Torus coil.

/* For cartesian coordinate system, enter parametric equation
/* in terms of t (which will vary from 0 to 1) for x, y and z
/* For example: for a circle in x-y plane, centered at origin
/* and radius = 4, the parametric equations will be:
/* x = 4 * cos ( t * 360 )
/* y = 4 * sin ( t * 360 )
/* z = 0
/*-------------------------------------------------------------------

/*coils
n=10

/*torus radio
l1 = 25

/*coil radio
l2 = 5

/*torus total angle
omega1 = 360

/*coils total angle
omega2 = n * omega1

/* torus actual angle
teta1 = omega1*t

/*coils actual angle
teta2 = omega2*t

/* trigonometric simplifications
c1 = cos(teta1)
s1 = sin(teta1)
c2 = cos(teta2)
s2 = sin(teta2)

/* cartesian parameters
x = c1*(l1 + l2*c2)
y = s1*(l1 + l2*c2)
z = l2*s2

Helical coil.

/* For cartesian coordinate system, enter parametric equation
/* in terms of t (which will vary from 0 to 1) for x, y and z
/* For example: for a circle in x-y plane, centered at origin
/* and radius = 4, the parametric equations will be:
/* x = 4 * cos ( t * 360 )
/* y = 4 * sin ( t * 360 )
/* z = 0
/*-------------------------------------------------------------------


/*number of torus
n1 = 2

/*coils per torus
n2 = 40

/*torus radio
l1 = 10

/*coil radio
l2 = 2

/*torus total angle
omega1 = n1 * 360

/*coils total angle
omega2 = n2 * omega1

/* torus actual angle
teta1 = omega1*t

/*coils actual angle
teta2 = omega2*t

/* trigonometric simplifications
c1 = cos(teta1)
s1 = sin(teta1)
c2 = cos(teta2)
s2 = sin(teta2)

/* cartesian parameters
x = c1*(l1 + l2*c2)
y = s1*(l1 + l2*c2)


/* z = l2*s2
/* every torus turn z value afected by 3 coil radus
z = (l2*s2)+3*n1*l2*t

Coiled helical halo.

/* For cartesian coordinate system, enter parametric equation
/* in terms of t (which will vary from 0 to 1) for x, y and z
/* For example: for a circle in x-y plane, centered at origin
/* and radius = 4, the parametric equations will be:
/* x = 4 * cos ( t * 360 )
/* y = 4 * sin ( t * 360 )
/* z = 0
/*-------------------------------------------------------------------

/*helix waves in halo
n2 = 10

/*coils per helix
n3 = 10

/*halo radio
l1 = 36

/*helix radio
l2 = 6

/*coil radio
l3 = 1

/*halo total angle or speed
omega1 = 360

/*helix total angle or speed
omega2 = n2 * omega1

/*coil total angle or speed
omega3 = n3 * omega2


/* halo actual angle
teta1 = omega1*t

/*helix actual angle
teta2 = omega2*t

/*coil actual angle
teta3 = omega3*t

/* trigonometric simplifications
c1 = cos(teta1)
s1 = sin(teta1)
c2 = cos(teta2)
s2 = sin(teta2)
c3 = cos(teta3)
s3 = sin(teta3)

/* cartesian parameters
X = c1*(c2*(c3*l3+l2)+l1)-l3*s1*s3
y = c1*l3*s3+(c2*(c3*l3+l2)+l1)*s1
z = (c3*l3+l2)*s2


Those are the matrix of homogeneus transformation.

/* [wxMaxima batch file version 1] [ DO NOT EDIT BY HAND! ]*/
/* [ Created with wxMaxima version 11.04.0 ] */

/* [wxMaxima: input start ] */
rotz1: matrix(
[c1,-s1,0,0],
[s1,c1,0,0],
[0,0,1,0],
[0,0,0,1]
);
/* [wxMaxima: input end ] */

/* [wxMaxima: input start ] */
trasx1: matrix(
[1,0,0,l1],
[0,1,0,0],
[0,0,1,0],
[0,0,0,1]
);
/* [wxMaxima: input end ] */

/* [wxMaxima: input start ] */
rotx1: matrix(
[1,0,0,0],
[0,0,-1,0],
[0,1,0,0],
[0,0,0,1]
);
/* [wxMaxima: input end ] */

/* [wxMaxima: input start ] */
rotz2: matrix(
[c2,-s2,0,0],
[s2,c2,0,0],
[0,0,1,0],
[0,0,0,1]
);
/* [wxMaxima: input end ] */

/* [wxMaxima: input start ] */
trasx2: matrix(
[1,0,0,l2],
[0,1,0,0],
[0,0,1,0],
[0,0,0,1]
);
/* [wxMaxima: input end ] */

/* [wxMaxima: input start ] */
rotx2: matrix(
[1,0,0,0],
[0,0,1,0],
[0,-1,0,0],
[0,0,0,1]
);
/* [wxMaxima: input end ] */


/* [wxMaxima: input start ] */
p2: matrix(
[0],
[0],
[0],
[1]
);
/* [wxMaxima: input end ] */

/* [wxMaxima: input start ] */
rotz1.trasx1.rotx1.rotz2.trasx2.rotx2.p2;
/* [wxMaxima: input end ] */

/* [wxMaxima: input start ] */
rotz3: matrix(
[c3,-s3,0,0],
[s3,c3,0,0],
[0,0,1,0],
[0,0,0,1]
);
/* [wxMaxima: input end ] */

/* [wxMaxima: input start ] */
trasx3: matrix(
[1,0,0,l3],
[0,1,0,0],
[0,0,1,0],
[0,0,0,1]
);
/* [wxMaxima: input end ] */

/* [wxMaxima: input start ] */
rotx3: matrix(
[1,0,0,0],
[0,0,-1,0],
[0,1,0,0],
[0,0,0,1]
);
/* [wxMaxima: input end ] */

/* [wxMaxima: input start ] */
p3: matrix(
[0],
[0],
[0],
[1]
);
/* [wxMaxima: input end ] */

/* [wxMaxima: input start ] */
rotz1.trasx1.rotx1.rotz2.trasx2.rotx2.rotz3.trasx3.rotx3.p3;
/* [wxMaxima: input end ] */

/* Maxima can't load/batch files which end with a comment! */
"Created with wxMaxima"$

best regards.

I continue experiment and only have 3 solutions til now:

1. SrinivasanIyer solution of use another coordinate system.

2. David Butz solution of use a cartesian system definition, can use homogenius transformation too, is perfect as David Butz says.

3. Make the helical curve (or helical solid) in another part file and assemble both part files, if possible.

Best regards, all of you.

Here is solutions if you want to use other technique: http://communities.ptc.com/blogs/vpalffy/2011/02/09/user-defined-springs

Vladimir

Best Regards,
Vladimir Palffy

I agree with Vladimir.

Use Trajpar and never look back!

That way your helix can adapt to changes in model and you can make basicly any wierd helical thing.

The rounded square section coil looks so PRO!!, the heart is nice too. Excellent command, excellent comment. Thanks.

You are welcome. Thank you too

Vladimir

Best Regards,
Vladimir Palffy

How about this for fun? COILED_SPIRAL-01_01.JPG

beautiful

tell me how make this type item

i wating your answer

Maybe this one?Copy of SPIRA_WAVE_SPRING3.JPG

This?Copy of CURVE-01.JPG

Maybe even this?Copy of ANDYS_COIL-01.JPG

Even this perhaps? Oh, the FUN you can have! Copy of COILSPRING-01.JPG

Hmm... so on this last one Frank, what are you using for the very first and last "hook" at the end of your helical spring? I have my own technique for achieving this, but I'm curious how you built yours.

The end and middle hooks are all cylindrical like the body of the spring itself instead of planar in the above torsional spring. This model was actually the easiest. In the red coil (which is actually the winding of an air-core ferric electrical part) the ends are both forced to be normal to a plane. The example above that has a constantly-changing radius, pitch, diameter, and reverses helical rotation twice. For the Smalley wave spring, their models suck, and i thought I'd make one that represents the true geometry. With the exception of the top coil that spirals inside itself (actually the toughest model to solve though easy to make if that makes any sense) these models were all made at least 12 years ago. I believe my e-mail is listed if you'd like to contact me about them.

Thanks for the extra information Frank.

Personally I love modeling springs and weird helical surfaces with complex transitions. After your explanation and some careful examination, I'm confident I understand how you did them.

I think I was outsmarting myself. The one that reverses rotation twice had me stumped. I can do it in two features easily enough or by merging two trajectories into one and then sweeping the solid. I guess I looked at it and thought you managed a direction-reversing coil all in one feature. Was it just two trajectories back to back or did you indeed figure out some clever mathematical way to reverse direction?

Thanks again!

-Brian

For all those springs, there is only a single continuous trajectory, even for the reversals of rotation, and even for the Smalley spring, which is perfectly cylindrical on it's inner and outer surfaces. The red coil was a trajectory that is the intersection of 2 surfaces, but it is still a single trajectory, even with the ends normal to a plane on the bottom. It was part of a coil and I did it just for fun, I just didn't bother to add the short straight leads for the thru-hole solder joint to the PCB. I have a technique where I can model spring ends normal to or parallel to the cylindrical axis of a spring on one trajectory feature. And, best of all, there is no math involved, just trajpar.

So, how did I do the torsional spring ends?

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