Inventor Distance of Coil Spiral and Pitch Instead of Using Pitch and Revolution

[Anonymous]

Is there anyway to give the distance of the coil spiral and pitch instead of having to use pitch and revolution?

kelly.young has edited your subject line for clarity: Coil Help

[johnsonshiue]

Hi! Are you trying to create a spiral? If yes, did you change the type to Spiral in Helical Curve? If you want to create variable radius and variable pitch helix, you will need Inventor 2019. On 2018 and earlier, you will need to use a workaround involving wrapping a 2D Curve around a cone/cylinder face.

Many thanks!

[Anonymous]

Hello, below is a screenshot of what I'm trying to do. I'm using inventor 2017 and have been using the "spiral" function under coil and it gives me the option to change the revolutions and pitch but in my case of usage, I need to be able to change the length of the spiral part. If you can help its greatly appreciated!

[johnsonshiue]

Hi! I don't believe a spiral can be constrained by length alone. It is because the length is the integral of diameter, pitch, and revolution (angle). Just like volume is the integral of length, width, and height. You cannot constrain a volume. Since for a given volume, there are multiple solutions.

I do understand what you are trying to do though. It looks like you want to vary the spiral in a way animating the deformation. Please note that even if there was a solution, it would not be real. Inventor is a geometric modeler. It is not capable to simulate non-linear material deformation.

Many thanks!

[IgorMir]

Hi aharrington19,

Here is a link to calculating the flat spring:

 http://deepfriedneon.com/tesla_f_calcspiral.html

Using the formulas in there - I think it is possible to represent the inner radius of a flat spiral via the spiral length. Or to have pitch and a length as known parameters and let the radius to be a calculable parameter. In any a case - two parameters have to be defined up front.  Create a flat spiral using 3D Sketch and see, if manipulating the data in parameters DB you can achieve what you are after.

Cheers,

Igor.

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@Anonymous wrote:

Hello, below is a screenshot of what I'm trying to do. I'm using inventor 2017 and have been using the "spiral" function under coil and it gives me the option to change the revolutions and pitch but in my case of usage, I need to be able to change the length of the spiral part. If you can help its greatly appreciated!

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[Anonymous]

Is there anyway that i can solve for the revolution of the spiral when I
have the diameter, pitch and the length though? Im trying to use this
feature so I can size boxes that fit the spirals at different diameters and
different lengths.

Thanks,

[Cris-Ideas]

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@Anonymous wrote:
Is there anyway that i can solve for the revolution of the spiral when I
have the diameter, pitch and the length though?....

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there should be as this data fully defines the solution.

You need to transform the formula given under the link in a post of IgorMir above. and incorporate the solution in  your model.

Cris.

[Cris-Ideas]

I did some transformations of the formula given here (link) while waiting for my FEM to compute and that's what I have:

N^4 *(W+S)+N^3 *2*(W+S)+N^2 *Di-N *6*L*(W+S)+38*L*Di=0

Di- internal diameter

N - number of turns

W- wire diameter

S - turn spacing (S+W) = P itch

L - length (not sure if along central, internal or external surface) (to be found out)

This can be solved analytically so you are able to get algebraic expression for N that you can use in inventor.

You no need to find 4 degree polynomial solving procedure on the internet, follow it and get your result.

I expect, since this is physical problem, that only one of possible solutions will be physical (so no negative, or in other way not applicable)

Wish you success.

Cris.

[TheCADWhisperer]

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@Anonymous wrote:
…. so I can size boxes that fit the spirals at different diameters and
different lengths.

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If you don't want to go to the trouble of figuring out the Equation Curve - you could probably solve by a trial-and-error method pretty quickly for "close enough".  Simply measure the edge length and adjust the other parameters up or down as need by trial.

I haven't tried this yet, but if you have 2019 you might even be able to get the curve length as a driven dimension in a 3D sketch so that you don't have to keep going back and forth with the Measure tool.

[Anonymous]

Thank you!

I'm using an online calculator and plugged in my numbers and I get
imaginary numbers which should be the case. If you could possibly try my
numbers and let me know your results. Any help is greatly appreciated!

P=6 mm
L=250mm
Di=50mm
N=?

[Anonymous]

I tried trial an error but I want to write an equation on Microsoft excel so when the inner diameter changes and the length/pitch all change it will give the amount of rotation every time. It seems so simple but I've been trying to solve this for the past couple of days.

[Cris-Ideas]

Hi,

If you are trying to do it in excel that's should be pretty easy as excel has build in equation solver.

I am not a mathematician and would also have to go threw internet to look for polynomial solution.

If solution with a defined precision is enough for you and you do not chase analytical solution it is even more simple.

And you can do that in inventor directly using iLogic and simple VB loop.

You have to specify:

- what parameters you have in the begging. is it always Di, P and L

- what precision for L1 would you accept (after calculating N with some precision resulting spring will have some length that will come out L1)

I do not have time now to play with it.

Maybe I will try in few days.

In the mean time try excel non linear equation solver or use graph to find a solution.

Cris.

[Anonymous]

I've been running trials with Di= 50mm, P= 6mm, and the L as 250mm, Id
accept a tolerance of + or - 5mm max. I wanted to have an equation id write
in excel that would give the amount of revolutions it would take to reach
the end of the spiral from the initial starting point so when I changed the
parameters of Di, L, and P, it would punch out a new number for revolutions
it would take. If you have any ideas on how to solve this please let me
know as soon as possible. Thank you

[Cris-Ideas]

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N^4 *(W+S)+N^3 *2*(W+S)+N^2 *Di-N *6*L*(W+S)+38*L*Di=0

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Sorry this is totally wrong

I have messed it up totally.

Will get back with proper solution.

[Cris-Ideas]

try

1,36741783902414 turns

I checked this theoretically and the lengths should be

250,174457673522

check if this will be in Inventor

I did not have time to program it.

But I have the solution.

here you have everything you need. If you have no time to wait you have to work it out on your own.

http://mathworld.wolfram.com/ArchimedesSpiral.html

I will be able to make a function in excel in few days

Cris.

[Cris-Ideas]

try

1,36741783902414 turns

I checked this theoretically and the lengths should be

250,174457673522

check if this will be in Inventor

I did not have time to program it.

But I have the solution.

here you have everything you need. If you have no time to wait you have to work it out on your own.

http://mathworld.wolfram.com/ArchimedesSpiral.html

I will be able to make a function in excel in few days

Cris.

[Anonymous]

Thanks for all your help. Let me know if you make any progress on the Excel
function.

[Cris-Ideas · Accepted solution]

Hi,

here you have excel file and pdf with explanation of the procedure so you can code it directly in inventor if you like.

Cris.

[Anonymous]

You're a lifesaver! Thank you so much!

[Cris-Ideas]

accept my post as solution so it is easy to find for others.

Cris.