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Recit

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Joined: Mon, Oct 30, 2006
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PARAMETER FORMULA

I am trying to work on a parameter formula to allow me to correctly flex a component. See attached image.

Here is what I need the parameter formula to do: 

Length<=4' THEN 2

Length>4' upto and including 8' THEN 3

Length>8' but less than 12' THEN 4

Length=>12' but less than 16' THEN 5

Length=>16' THEN 6

That is it....let me know if you guys have any ideas in mind.

Thanks,

Recit

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TomDorner

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Joined: Sun, Apr 25, 2004
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You can follow these guidelines for some more ideas.

• Use conditionals in the definition of the parameter whose value is conditional.

• In the Formula column next to the appropriate parameter, type the formula for the parameter. Noticethat the formula begins with an equal sign (=).

• IF based on one test statement: =IF (Test Condition, Result if true, Result if false)IF ( Length < 35' , 2'-6", 3'-0" )

• IF based on more than one test: =IF (AND/OR (Test1, Test2) Result if true, Result if false)IF ( AND (x = 1 , y = 2), 8 , 3 )IF ( OR ( A = 1 , B = 3 ) , 8 , 3 )

• IF based on a negative test: =IF ( NOT (Test Condition), Result if true, Result if false)IF ( NOT ( A > B ) , 8 , 3 )

• Simple result if True, another IF if false: =IF (Test Condition, Result if true, IF (Test Condition, Result iftrue, Result if false ))IF ( Length < 35' , 2' 6" , IF ( Length < 45' , 3' , IF ( Length < 55' , 5' , 8' ) ) )

• Set one YES/NO parameter to the opposite of another: =NOT Y/N Parameter

The question though is a complex conditional formula the best solution? Depending on what you are making can the array be driven by a type catalog? Can you do a first/last array and evlauate spacing via a formula that is simpler? Having a multi-nested conditional formula can lead to performace problems within the family. Of course that may not be a problem if you only have a couple of occurances within the project. If you have hundreds or thousands of them, then it is a different story. Without knowing more details it is hard to say what is the best option.

-----------------------------------

Tom

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rkitect

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Joined: Thu, Dec 16, 2004
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Recit,

Take a look at the metalStud family I attached to my post here and see if the parameters in there help any.  I use some parameters to control visibility and when/how an array functions based on the length of the family (it's line based so adapt as necessary).  Let me know if you have any questions about it. 

-----------------------------------

-//------------------------

Carl - rkitecsure[at]gmail.com

Need help? I'm probably in my chat room!

You fell victim to one of the classic blunders. The most famous is never get involved in a land war in asia, but only slightly less well known is this! Never go in against a Sicilian when death is on the line!

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Sumex

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Try this...

 If (Length < 3'11 63/64", 2, (If (and (Length >4' 0",Length < 8'0 1/64", 3, (If (and (Length >8' 0",Length < 12'0", 4, (If (and (Length >11' 11 63/64",Length < 16'0", 5, (If (Length < 16' 0 1/64", 6, 0)))))))))

Length must be a Length Parameter, and the Object parameter to be an Integer.  Less than / greater than or equal to (<=, >=) don't work, so you need to make determinations on tolerances.  I used 1/64th in this case.  You also need to have a "Catch all" which is the final Parameter, for any open loop, or instance not covered.  It acts as a check.  I used Zero (0).

 I hope this helps...

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Sumex

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You know, shortcut smileys shouldn't be active on a site that discusses LISP........

Everywhere you see a smiley in my previous post, assume a close parenthesis.....

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Recit

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THANKS GUYS FOR YOUR HELP. I CAN HANDLE EASY FORMULAS, BUT THE COMPLICATED ONES STILL BEAT ME DOWN!

I'LL LET YOU KNOW IF IT WORKS.

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Recit

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Guys, after all your suggestions, here is the winning formula:

if(Length = 16', 5, if(Length > 16', 6, if(Length = 12', 4, if(Length > 12', 5, if(Length = 8', 3, if(Length > 8', 4, if(Length > 4', 3, 2)))))))

I learned that it makes a difference the sequence of numbers. That is to say that in the above formula begining from the bigger number and ending with a smaller number allows me to end it nicely. the numbering sequence is critical in the formula's behavior. In this case a decreasing sequence works much better than an increasing manner because it closes nicely with the last TRUE,FALSE)

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Sumex

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I have found that the best way to stay organized with nested LISP Statements is to write each requirement on a line of excel and add XXXX under false, and then start at the bottom and enter the line into the previous line's false requirements...  Like this>>>

If (Length <= 4' 0", 2, xxxx)

If (and (Length >4' 0",Length < 8'0", 3, xxxx)

If (and (Length >8' 0",Length < 12'0", 4, xxxx)

If (and (Length >12' 0",Length < 16'0", 5, xxxx)

If (Length <= 16' 0", 6, 0)

Turns into:

If (Length < 3'11 63/64", 2, (If (and (Length >4' 0",Length < 8'0 1/64", 3, (If (and (Length >8' 0",Length < 12'0", 4, (If (and (Length >11' 11 63/64",Length < 16'0", 5, (If (Length < 16' 0 1/64", 6, 0)))))))))

(If (and (Length >4' 0",Length < 8'0 1/64", 3, (If (and (Length >8' 0",Length < 12'0", 4, (If (and (Length >11' 11 63/64",Length < 16'0", 5, (If (Length < 16' 0 1/64", 6, 0))))))))(If (and (Length >8' 0",Length < 12'0", 4, (If (and (Length >11' 11 63/64",Length < 16'0", 5, (If (Length < 16' 0 1/64", 6, 0))))))(If (and (Length >11' 11 63/64",Length < 16'0", 5, (If (Length < 16' 0 1/64", 6, 0))))(If (Length < 16' 0 1/64", 6, 0))

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