Inventor General

## Inventor General

*Baranowski, John
Message 1 of 5 (79 Views)

# Writing expressions in parametric dim.

79 Views, 4 Replies
11-27-2002 05:11 AM
I have a problem with writing expressions into dimensions in Inventor and
I'm wondering if anyone out there can help? I'm trying to round down a
number in a dimension: meaning if the value of d1=1.3, then d2=1.
I looked up in the help on how to do this and they say to use the "floor"
function. I tryed it and can't get it to work (Example: floor(83 / 2 in) ).
It won't accept it.

Thanks,

John Baranowski
HMS Products
*Dotson, Sean
Message 2 of 5 (79 Views)

# Re: Writing expressions in parametric dim.

11-27-2002 08:20 AM in reply to: *Baranowski, John
floor returns a unitless expression so you'd need to say:

floor (83/2) * 1 in

see my site for a tutorial on Inventor functions

--
--
Sean Dotson, PE
http://www.sdotson.com
Check the Inventor FAQ for most common questions
"John Baranowski" wrote in message
> I have a problem with writing expressions into dimensions in Inventor and
> I'm wondering if anyone out there can help? I'm trying to round down a
> number in a dimension: meaning if the value of d1=1.3, then d2=1.
> I looked up in the help on how to do this and they say to use the "floor"
> function. I tryed it and can't get it to work (Example: floor(83 / 2
in) ).
> It won't accept it.
>
> Thanks,
>
> John Baranowski
> HMS Products
>
>
*AutoCOL
Message 3 of 5 (79 Views)

# Re:

11-27-2002 01:03 PM in reply to: *Baranowski, John
and you wonder why they taught you about the dimensions of equations in
school huh? it was so you could figure out how you have to manipulate your
results from equations in inventor to get it in the units you want!

surely there should be an option to over-ride the resultant dimensions of an
equation with the units you've selected for the parameter in question?
that's how i'd like it to work. say for example you have two lengths and you
want another value (also a length, for me that's mm) to be the result of a
division of the first two.

so your equation is d3 = d1 / d2 but this won't work because the result of
the right hand side of that equation is unitless, so you have to multiply it
by 1mm to get the right answer. i think that's silly and we should just be
able to over-ride the dimensions of the result (in this case ul) with the
selected dimensions of the parameter (in this case d3, and millimetres...)

*end rant*

col.
*Bliss, Charles
Message 4 of 5 (79 Views)

# Re:

11-27-2002 03:10 PM in reply to: *Baranowski, John
The function you are looking for in Inventor is
isolate(Number;SourceType;DestType)
In example, you want to have a linear dimension respond to a degree
rotation. Assume d1 is the angle in degrees and for one revolution of
d1, you want d2 to increase by 36 mm
isolate (d1;deg;ul) /10 ul * 1 mm (or something like that ).

Makes a bit more sense when you are changing betweens systems of
measurement which have no correlation at all (deg -> mm)

AutoCOL wrote:

>and you wonder why they taught you about the dimensions of equations in
>school huh? it was so you could figure out how you have to manipulate your
>results from equations in inventor to get it in the units you want!
>
>
>surely there should be an option to over-ride the resultant dimensions of an
>equation with the units you've selected for the parameter in question?
>that's how i'd like it to work. say for example you have two lengths and you
>want another value (also a length, for me that's mm) to be the result of a
>division of the first two.
>
>so your equation is d3 = d1 / d2 but this won't work because the result of
>the right hand side of that equation is unitless, so you have to multiply it
>by 1mm to get the right answer. i think that's silly and we should just be
>able to over-ride the dimensions of the result (in this case ul) with the
>selected dimensions of the parameter (in this case d3, and millimetres...)
>
>
>*end rant*
>
>col.
>
>
>
>
*Baranowski, John
Message 5 of 5 (79 Views)

# Re:

11-29-2002 07:37 AM in reply to: *Baranowski, John
Thanks for the info. It's seems crazy but I think I can handle it!

John Baranowski
HMS Products

"Charles Bliss" wrote in message
news:3DE5C170.2020207@cbliss.com...
> The function you are looking for in Inventor is
> isolate(Number;SourceType;DestType)
> In example, you want to have a linear dimension respond to a degree
> rotation. Assume d1 is the angle in degrees and for one revolution of
> d1, you want d2 to increase by 36 mm
> isolate (d1;deg;ul) /10 ul * 1 mm (or something like that ).
>
> Makes a bit more sense when you are changing betweens systems of
> measurement which have no correlation at all (deg -> mm)
>
> AutoCOL wrote:
>
> >and you wonder why they taught you about the dimensions of equations in
> >school huh? it was so you could figure out how you have to manipulate
your
> >results from equations in inventor to get it in the units you want!
> >
> >
> >surely there should be an option to over-ride the resultant dimensions of
an
> >equation with the units you've selected for the parameter in question?
> >that's how i'd like it to work. say for example you have two lengths and
you
> >want another value (also a length, for me that's mm) to be the result of
a
> >division of the first two.
> >
> >so your equation is d3 = d1 / d2 but this won't work because the result
of
> >the right hand side of that equation is unitless, so you have to multiply
it
> >by 1mm to get the right answer. i think that's silly and we should just
be
> >able to over-ride the dimensions of the result (in this case ul) with the
> >selected dimensions of the parameter (in this case d3, and
millimetres...)
> >
> >
> >*end rant*
> >
> >col.
> >
> >
> >
> >
>

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