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Units are not compatible

ooberandy
2-Explorer

Units are not compatible

I'm trying to create a calc where the answer is a design factor. The calc works fine when just using numbers, but when I try to use a value from earlier in the sheet I get the 'units are not compatible message'. Would appreciate any help on this, thanks.

Capture.JPG

ACCEPTED SOLUTION

Accepted Solutions

You need to give the 50 units also:

 

units.PNG

 

Alan

 

View solution in original post

10 REPLIES 10
vkofanov
12-Amethyst
(To:ooberandy)

1.4 is a dimensionless quantity

a similar question

You need to give the 50 units also:

 

units.PNG

 

Alan

 

What is wrong.jpg

Werner_E
25-Diamond I
(To:II_9800308)

1) Open a new thread for a new question as this increases the likelihood of a quick response and keeps he forum clearer

2) You shouldn't only show a picture but also attach your worksheet. Its much harder to debug a picture than a live worksheet

 

Are you sure about the Unit N/m for G.c, but just N for G.b?

 

The way you define Az both must have the same dimension and the result in either case will be dimensionless and not Newton as you seem to expect.

 

And when you define H, you are trying to add a force (A, Newton) and two lengths (Ab and Ac, meter) . Obviously this can't work and so Prime thankfully refuses to do so

vlehner
14-Alexandrite
(To:ooberandy)

This is the biiiiiiig old problem of "designed equations" from diverse textbooks or scriptums for example. A really big Problem

You have to check out first, which dimension the variable Df must have.

The parameter 1.4 must have the same unit like Df.

If 0.6 is a uitless factor then ok, but the 50 must have the unit of Unit(Tdt)*Unit(Df)²

 

 

Volker
vkofanov
12-Amethyst
(To:vlehner)

If you combine the two solutions: first and second, you can find a way to formalize the empirical expressions

изображение.png


@vlehner wrote:

This is the biiiiiiig old problem of "designed equations" from diverse textbooks or scriptums for example. A really big Problem

You have to check out first, which dimension the variable Df must have.

The parameter 1.4 must have the same unit like Df.

If 0.6 is a uitless factor then ok, but the 50 must have the unit of Unit(Tdt)*Unit(Df)²

 

 


True.  But, in my experience, design factors are usually dimensionless.

 

Alan

Thanks everyone, I've got it working now. All the replies are appreciated 🙂


@AlanStevens wrote:


True.  But, in my experience, design factors are usually dimensionless.

 

Alan


Many of these "design factors" are unit conversions


@Fred_Kohlhepp wrote:

@AlanStevens wrote:


True.  But, in my experience, design factors are usually dimensionless.

 

Alan


Many of these "design factors" are unit conversions


Yes, though these are usually supplied as single numbers rather than expressions that require evaluating, are they not?

 

Alan

 

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