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Cube root / Power (1/3)

19 REPLIES 19
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Message 1 of 20
nightsheart
2387 Views, 19 Replies

Cube root / Power (1/3)

hello to all,

 

I'm a new french member. I have many problems with parametric equation.

In deed, I would like to make a formula for a dimension with Cube root or Power (1/3).

Is it possible with Inventor ?


Thanks in advance for your helps.

 

19 REPLIES 19
Message 2 of 20
-niels-
in reply to: nightsheart

Welcome to the forum!

It is possible with Inventor to use functions like square root and power.
See this help page for an overview:
http://help.autodesk.com/view/INVNTOR/2014/ENU/?guid=GUID-EE98BCED-5623-4E3B-9E98-432C6738B081

Niels van der Veer
Inventor professional user & 3DS Max enthusiast
Vault professional user/manager
The Netherlands

Message 3 of 20
nightsheart
in reply to: -niels-

Thanks for your answer.

Nevertheless, I have already saw this page.

 

I can make this formula :

(12 nd) ^ ( 1 nd / 3 nd ) * 1 mm

 

But, if I put a variable (quote) for example :

d0 ^ ( 1 nd / 3 nd ) * 1 mm

It's doesn't work.

 

 

 

Message 4 of 20
-niels-
in reply to: nightsheart

Without testing, the formatting seems wrong.
Be sure to follow the formatting shown on the help page.

I think it should be something like:
pow(d0;(1/3))*1

Niels van der Veer
Inventor professional user & 3DS Max enthusiast
Vault professional user/manager
The Netherlands

Message 5 of 20
admaiora
in reply to: nightsheart

The parameter d0 already keep inside the unit mm. Try

 

 

d0 ^ 1/3

 

or the same   (d0)^ 1 u l/ 3ul

 

I hope that it helps you.

 

Welcome in this forum

Admaiora
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Message 6 of 20
nightsheart
in reply to: -niels-

It's look like as Excel but in inventor it doesn't work after testing.

Message 7 of 20
admaiora
in reply to: admaiora

oops ^2    Niels...Smiley Very Happy

Admaiora
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Message 8 of 20
nightsheart
in reply to: admaiora

Hi admaiora.

I test (d0)^ 1 u l/ 3ul but it doesn't work.
If d0 = 12 the result is 3. I test with (d0)^ (1 u l/ 3ul) but error... Inventor doesn't aceept the formula.
Message 9 of 20
-niels-
in reply to: nightsheart

This is how i set it up, though i'm not entirely sure about the resulting value.

Formula.png

It gives a warning, but all that says is that the unit type (2nd column) isn't what the formula gives out. (mm^0,333)

Not sure if this is helpfull...


Niels van der Veer
Inventor professional user & 3DS Max enthusiast
Vault professional user/manager
The Netherlands

Message 10 of 20
admaiora
in reply to: nightsheart

d0 ^ 1/3 

 

neither?

Admaiora
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Message 11 of 20
-niels-
in reply to: admaiora

Tried that as well, still the result remains 10,....
I'm starting to wonder if Inventor is capable of doing this...

Niels van der Veer
Inventor professional user & 3DS Max enthusiast
Vault professional user/manager
The Netherlands

Message 12 of 20
nightsheart
in reply to: admaiora

Non of the solution work... For Niels : I have the same result but 12^(1/3) = 2.89... and not 10.62...

Message 13 of 20
-niels-
in reply to: nightsheart

Yeah, i'm stumped... the result is way off.
Hope someone from Autodesk can shed some light on this.

Niels van der Veer
Inventor professional user & 3DS Max enthusiast
Vault professional user/manager
The Netherlands

Message 14 of 20
admaiora
in reply to: -niels-

Yes, i went on memory, but now that i am tryng  Niels is right.

 

It looks like quite  aproblem with units in parameters..

Admaiora
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Message 15 of 20
WHolzwarth
in reply to: admaiora

Here are my ways, one unitless, one in mm.

Walter

 

Power ul.jpg

 

Power mm.jpg

Walter Holzwarth

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Message 16 of 20
nightsheart
in reply to: WHolzwarth

It's work but the dimension becomes 120 instead of 12.

See picture below :

 

Untitled.jpg

Message 17 of 20
-niels-
in reply to: WHolzwarth

Hmm... interesting, so if you have the starting value as unitless instead of mm you can get the correct value.
Even though the help page states that "expr1" can be "any" type...

 

@nightsheart For this to work, you have to turn your dimension into a unitless value.

I find this to be rather unintuitive, but here's another image:

working_formula.png


Niels van der Veer
Inventor professional user & 3DS Max enthusiast
Vault professional user/manager
The Netherlands

Message 18 of 20
SBix26
in reply to: nightsheart

The problem is the units: the cube root of a millimeter dimension is in mm^(1/3), which is nonsense.  Try this: ( d0 / 1 mm ) ^ ( 1 ul / 3 ul ) * 1 mm



Sam B
Inventor 2012 Certified Professional

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Message 19 of 20
nightsheart
in reply to: -niels-

Thank you to all !!! It's OK !! Thanks thanks !! Cat Very Happy

Message 20 of 20
-niels-
in reply to: nightsheart

Just to add one last thing:

Also_works.png

This also works, it also made me laugh as it just adds to the unit confusion when looking at it.

i just typed it like "pow(d2/1;1/3)*1mm" and inventor added the unit types as shown.

Still, this might be a little better than having to make an extra parameter just to convert the dimension to unitless.


Niels van der Veer
Inventor professional user & 3DS Max enthusiast
Vault professional user/manager
The Netherlands

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