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Thread: Need to schedule # of nested instances within family

  1. #1
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    Need to schedule # of nested instances within family

    Hello All,

    Apologies if the title was unclear. Have a vertical rod stiffener family with nested stiffener family. Depending how long the support of the rack is, (from hanger to deck), depends on how long the rod stiffener will be that wraps around the support. The longer the rod stiffener is, the more stiffeners each support will get. Some have two, some have three, etc. Please refer to the attachment (blue arrows). I know I could create a simplistic schedule to get an overall count but am hoping for something more. My hope is in my schedule next to "LENGTH" of vertical strut, I would like to have "NUMBER OF STIFFENERS". It is my assumption that this would be a conditional formula but I am unsure. I am pretty weak with formulas within families and was hoping someone could help. If there is a simpler way any assistance would be greatly appreciated.

    Thanks in advance
    Rob
    Attached Thumbnails Attached Thumbnails Need to schedule # of nested instances within family-capture.png   Need to schedule # of nested instances within family-capture2.png  

  2. #2
    Moderator Robin Deurloo's Avatar
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    I am assuming that the amount of stiffeners in each rod stiffener is calculated with a formula?
    In that case make that parameter a shared one and you can schedule it.

  3. #3
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    Its not calculated with a formula as of yet. That is partly what I am trying to figure out. Right now I have a visibility paramenter that I switch if its over a certain length. Over a certain distance (i.e. 5'-0") is where I need the 3rd stiffener. Right now I am not sure what to make a parameter. Ive tried looking online formulas on how I can sum the number of stiffeners. Not having much luck. Overall the family I've created is working great and I could create a bulk schedule for the stiffeners by floor but am trying to get that sum for each stiffener on the schedule for pre-fab. Thanks.

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    Moderator Robin Deurloo's Avatar
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    You can make a formula that ticks the right box for you. Put something like IF(Length > 5'-0") in a yes/no parameter and it will turn on when your Length parameter gets over 5'.
    Or you can do something like Length / x to get the number of stiffeners.

    If you put the last formula in a shared parameter you can schedule the outcome.

    With the first one you might need a bit more trickery to get the amount, but very possible as well.
    A with the manual setting of the yes/no parameter you can get it into a schedule.

    I think for each yes/no parameter you can set to make a stiffener show up make another parameter with a number. Put a formula in there that sets either 1 or 0 depending on if the yes/no is set or not IF(ParX,1,0). In a last shared parameter add them all together and schedule that.
    Last edited by Robin Deurloo; December 28th, 2018 at 08:32 PM.
    cadmancan likes this.

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    This is beautiful. Most of iot makes sense. The part that doesn't I am still growing in this area. As I had a bit more time to ponder my issue, I am wondering if I am overthinking this. I've was told this morning that I wont ever need more than 3 stiffeners for (1) Rod Stiffener. Info I am sure that would have been useful yesterday. My most humble apologies. At the end of the day I need two things. An overall, accurate count of stiffeners to order and how many stiffeners each Rod stiffener will get. Right now regardless of each instance of Rod Stiffeners has 2 or 3 stiffeners, the count is coming out as (1). Thank you very much for your time

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    Moderator Robin Deurloo's Avatar
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    What count does is count the number of Rod Stiffeners in your project and NOT the amount of Stiffeners (nested family) in each Rod stiffeners (host family).
    So you still need to setup a parameter that has the number of Stiffeners in each family you can schedule.
    That parameter needs to be a shared parameter or else you can not schedule it, call it something like "number of stiffeners"

    If there is a max or 3 stiffeners you could make it a bit simpler and have the "number of stiffeners" parameter that you fill in for 3 types in the Rod Stiffener family. This will end you up with just 3 types, 1 stiffener, 2 stiffener, 3 stiffener with a fixed number set in the "number of stiffeners" parameter. Users just have to choose the right one.

    If you don't want the users to need to make a choice than you still need the formula, but build one with a max. of 3.

    Give it a go and if you can not get it to work I'll see if I can make a simple example.

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    There will only ever be two or three stiffeners for each Rod Stiffener. My confusion is how to assign values to objects i.e. Stiffener1, Stiffenrs2). I will do some reading and give it a go. If I get stuck I will reach out to you. Really appreciate your time. Thank you so much!
    Last edited by cadmancan; December 31st, 2018 at 02:38 PM.

  8. #8
    Moderator Robin Deurloo's Avatar
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    This is what I would do.


    • A Yes/No parameter that sets the visibility to the 3th stiffener (because the first 2 are always there as you mentioned above)
    • The same Yes/No parameter is used in the calculation for 'calc stiffener 3' to either set that to 0 or 1
    • The 'number' parameter (which is a shared parameter so you can schedule that) has a calculation adding the 2 stiffeners you always have and either the 0 or 1 for the 3th stiffener.
    • Need more stiffeners, just ad the same parameters for number 4 and 5 and 6 etc.





    If you use an array, you can just go with the 'number' parameter and use that for the instances of the array. With some formula trickery you can even prevent people from setting anything other then 2 or 3 in the array as well.
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    Forum Co-Founder Alfredo Medina's Avatar
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    If the schedule is already reporting the length of each strut, is it not just a matter of adding a calculated value to the schedule such as length of the struts dividing by spacing between supports to get the number of supports? Seems simpler than editing each family, creating a shared parameter, etc.

  10. #10
    Moderator Robin Deurloo's Avatar
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    Yeah, but then you have the schedule calculate the number instead of the schedule reporting the actual number, I would always go for the last one. And who is to say that the number of stiffeners is always in relation to the length of the strut, especially because the OP is setting the number manually.

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