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Room Tag with Dimension Issue

6 REPLIES 6
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Message 1 of 7
Anonymous
1044 Views, 6 Replies

Room Tag with Dimension Issue

Hello,

I am creating a Room Tag with Dimensions to place on my plans, I have used the formula in this post, the issue is that it always displays the longer side first. (Ex, if a room is 5m on the Y-axis and 3m on the X-axis, it shows 5x3..., if a room is 3m on the Y-axis and 4m on the X-axis, it shows 4x3) I want to be able to have a consistent definition for Length and Width, Length to be on the X-axis and Width to be on the Y-axis so that I can display all rooms in this format X x Y. Is there something I could adjust to just have on annotation family?

 

My temporary solution is that I created 2 families one which shows width x length and one that shows length x width and wherever I want the length on the x-axis to show first I choose the corresponding family. But this is very inefficient and wastes time.   

6 REPLIES 6
Message 2 of 7
lukekvasnicka
in reply to: Anonymous

Can you include the formula you are using for reference?  The post you linked to has a broken link so I can't see the original formula.

 

My first guess would be to split this into two parameters, one which calculates length, one which calculated width.  Then within the tag, you can display "L x W" consistently.

Message 3 of 7
Anonymous
in reply to: lukekvasnicka

Length: if(Area = (Perimeter / 4) ^ 2, Perimeter / 4, ((Perimeter / 2) - sqrt((Perimeter / 2) ^ 2 - 4 * Area)) / 2)

Width: Area/Length

Message 4 of 7
barthbradley
in reply to: Anonymous


@Anonymous wrote:

 

My temporary solution is that I created 2 families 


 

 

Instead of 2 different families, what about 2 additional TYPES within the same Tag Family? 

 

Room Tag 43.jpg

Room Tag 43-2.jpg

Message 5 of 7
Anonymous
in reply to: barthbradley

I can do that, but it still won't solve my problem, I want something to be done with one click. I tried to do something using dynamo, but I can't seem to figure out how to extract the length of the rectangular room boundary in the x and y axis and then input them to the shared parameters and project parameters.
Message 6 of 7
barthbradley
in reply to: Anonymous

I can't imagine how Dynamo would solve. Seems pretty subjective. What if the View was oriented to True North or oriented per a Scope Box - what would the Length be and what would the Width be?  The way I see it, we need to apply the TYPE that "fits" the orientation of the View.   

Message 7 of 7
Anonymous
in reply to: Anonymous

Did you manage to crack this ? I totally get this. have been looking for a solution for a long time myself and the best solution is sadly the one that you have posted in the beginning by creating 2 families. pls do post if you have found a way to get around with this 

 

Thanks 

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