I need some ideas about creating a double helix. Consider the filament in an incandescent light bulb with a strand is coiled and the coil is coiled.
I can think of two potential solutions.
First is to create a complex 3D path and sweep – how would I create that path so I can keep control. By which I mean, I would need to be able to enter the inner, minor coil diameter and the outer, major diameter.
Second would be to create the helix using a coil function, but selecting an axis that is normal to the profile sketch plane and then using the Part Bend function with a Bend Line at an angle.
I’ve tried the latter, but Inv swore and me and then stopped talking to me 😞
Any sensible suggestions welcome
What version of Inventor are you using?
I posted an Equation Curve (produced by one of my students) a few months ago.
Also, I have started experimenting with the Bridge Curve function to tie in ends of curves like this to straight lines.
The CADWhisperer YouTube Channel
Currently using 2012, hoping to move to 2014 very soon
Do you have a link for this equation curve
Two quick searches either produced too many hits or nothing of any relevance.
Using 3ds max to model it is surprisingly easy if you are just looking for a non-parametric model in Inventor.
Attached as an IGES and an Inventor 2103 file (because I don't have 2012 here).
In max you just create one helix, normalize it, then path-deform it around another normalized helix. Export the IGES and bring it into Inventor.
It's actually highlighted a slight problem with Inventor 2014. Importing the IGES on my machine lets me see the model for a split second before crashing 2013 works a treat.
duncan wrote:Do you have a link for this equation curve?
Hi Duncan,
I just posted this. Hope it helps.
Equation for Helix along Helix
http://forums.autodesk.com/t5/Autodesk-Inventor/Equation-for-Helix-along-Helix/td-p/4352925
Glenn
ASM Development
Hi Duncan,
I also noticed a subtle problem with the orientation of the second helix. It's a little slanted in relation to the orientation of the first helix.
Could you try the following equation instead? This may not be what you want, but sweep works better along this path.
x(t): (R + r * sin(turns * t * 1rad)) * cos(t * 1rad)
y(t): (R + r * sin(turns * t * 1rad)) * sin(t * 1rad)
z(t): Z * t + r * cos(turns * t * 1rad)
t_min: 0
t_max: Turns * 2*PI
where
R is the radius of the first helix,
r is the radius of the second helix,
Turns is the number of turns for the first helix,
turns is the number of turns for the second helix (in one turn of the first helix),
Pitch is the pitch of the first helix, and
Z is Pitch / (2*PI)
Glenn
.
I create my own equations that have the over generalised form of,
x(t)=cos{f[t]}+cos({g[t]}+incr)
y(t)=sin{f[t]}+sin({g[t]}+incr)
z(t)=0<t<[distance]
I'll give a better answer to Glenn
.
This is the generalised equation I used
x(t)=(RAD.COS(DIR.2.pi.Z/PITCH)) +(rad.COS((dir.2.pi.Dist/pitch)+(Pass.2.pi/Nos)))
y(t)=(RAD.SIN(DIR.2.pi.Z/PITCH)) +(rad.SIN((dir.2.pi.Dist/pitch)+(Pass.2.pi/Nos)))
z(t)=Dist.[0<t<1]
Where
Dist = length of double helix
RAD = Radius of Major Helix
DIR = Direction of rotation {+=ccw, -=cw}
PITCH = Pitch of Major Helix
rad = radius of minor helix
pitch = pitch of minor helix
Pass = increment number
Nos = number of increments
It isn't possible to use the standard circular pattern/polar array command as the the pattern needs to be revolved around a helix. Therefore, the minor helix needs to be incremented around the major helix mathamatically.
Hi Duncan,
I don't suppose you'd be willing to share the ipt you based that image upon?
I've tried using your equation but I must be missing something in Inventor 2013 as the formula doesn't work for me.
Thanks,
Alex.
.
Sorry, can't share the IPT.
Can you can have the spreadsheets they were based upon.
I was using Inv2012sp2, which meant importing points, conected by a 3D spline, from a spreadsheet.
Four combinations of direction; CW/CW, CW/CCW, CCW/CCW, CCW/CW.
I've had to edit the titles of each input to protect comercial confidentiallity. I might have made a few spelling mistakes, but you should be able to figure.
HTHs
Can I get a file for that, or maybe the equation? I'm young enough that I haven't learned those areas of math.
Up until Inventor 2014, we've used equation curves to model twisted pair cables. Now in Inventor 2015, we can use the new Twist option in the Sweep feature to model them.
From online help for Inventor 2015 > What's New > Part:
Glenn
.
Dogs5336,
If you go back to the message posted 08-14-2013 11:38 AM you'll find the generalised equations.
If you go back to the message posted 10-29-2013 02:29 PM you'll find those equations in four MS Excel spreadsheets.
HTHs