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89 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

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

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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

news:BD64674A43591449415E83FDC6AB04B6@in.WebX.maYIadrTaRb...

> 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

>

>

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"

news:BD64674A43591449415E83FDC6AB04B6@in.WebX.maYI

> 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

>

>

- Mark as New
- Bookmark
- Subscribe
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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.

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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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.

>

>

>

>

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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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.

> >

> >

> >

> >

>

John Baranowski

HMS Products

"Charles Bliss"

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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