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2-Guest

## 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.

1 ACCEPTED SOLUTION

Accepted Solutions
17-Peridot
(To:ooberandy)

You need to give the 50 units also:

Alan

10 REPLIES 10
10-Marble
(To:ooberandy)

1.4 is a dimensionless quantity

a similar question

17-Peridot
(To:ooberandy)

You need to give the 50 units also:

Alan

1-Newbie
(To:AlanStevens)

24-Ruby V
(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

13-Aquamarine
(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
10-Marble
(To:vlehner)

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

17-Peridot
(To:vlehner)

@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 🙂

23-Emerald I
(To:AlanStevens)

@AlanStevens wrote:

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

Alan

Many of these "design factors" are unit conversions

17-Peridot
(To:Fred_Kohlhepp)

@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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