relation related inquiry

[megaladon]

i have several parts attached to a skeleton that has a program in that prompts me to enter what mode i want
mode 1 is flat
mode 2 is folded
mode 3 is flexed

i also have several parts that are affected by the mode

when in mode 2 most of the parts are at their longest.

i need to write a relation for a dimension that regenerates or calculates all three modes and returns the longest length to drive my dimension

in the skeleton i have created an analysis feature measuring the curve length and set my dimension to equal it

i have written this but it is not working

if d12:15==1220
x=length:fid_ctr_lp5:23
endif

if d12:15==1554

y=length:fid_ctr_lp5:23

endif

if d12:15==1879

z=length:fid_ctr_lp5:23

endif

d334=max(x,y,z)

d12 is part of the skeleton that is moving but i do not like using this in the relation because the values could change in the future. they currently represent the length of the cylinder extending.

 

[copyboy]

the max() function only takes two arguments i.e. max(x,y). To test three numbers try this:


a=max(x,y)


d334=max(a,z)

 

[megaladon]

if d12:0==2262.82758
length:FID_CTR_FP6:11==x
endif


if d12:0==1377.95
length:FID_CTR_FP6:11==y
endif


if d12:0==2343.15
length:FID_CTR_FP6:11==z
endif


a=max(x,y)
d334 = max(a,z)


i didn't know it only calculated two at a time but i still get this error




errorInvalid left side of assignment

 

[kdem]

Is the error showing up in the IF statements? If you are trying to assign values to x, y, and z those variables can't be on the right side of the equation. You're alsousing a logical expression were one isn't expected. Your equations for x, y, and z from before were okay. You could also write the two relations given earlier using max() as one relation. You probably also want to create a statement that sets d334 afterall modes are run.
Edited by: kdem

 

[megaladon]

you guys are helping tremendously
here's the latest

IF D12:0==2262.82758
LENGTH:FID_CTR_FP6:3==X
ENDIF
<?:namespace prefix = o ns = "urn:schemas-microsoft-comfficeffice" />
IF D12:0==1377.95
LENGTH:FID_CTR_FP6:3==Y
ENDIF

IF D12:0==2343.15
LENGTH:FID_CTR_FP6:3==Z
warning[1]Constraint has been violated
ENDIF

a=max(x,y)
max(a,z)==d340
warning[1]Constraint has been violated

i'm not sure what i'm violating but i think i have to regenerate my model 3 or 4 times in each position to get these to go away. is there a command that will prevent this?

 

[kdem]

The problemis LENGTH:FID_CTR_FP6:3 has a different value than Z and max(a,z) has a different value than d340. The statements you have are comparison statements that compareone value toanother and return a value of TRUE or FALSE not a relation that sets one value equal to another. The messages you are getting mean that when you regenrated the model D12:0 was 2343.15, the value LENGTH:FID_CTR_FP6:3 was compared to Z and the values were not the same so the constraint, LENGTH:FID_CTR_FP6:3==Z,is violated.Similar reasoning is applied to themax(a,z)==d340statement.

 

[megaladon]

i found it

thank you to all for helping me with this!!!!!!

IF D12:0==2262.82758
X=LENGTH:FID_CTR_FP6:5
<?:namespace prefix = o ns = "urn:schemas-microsoft-comfficeffice" />
IF D12:0==1377.95
Y=LENGTH:FID_CTR_FP6:5

IF D12:0==2343.15
Z=LENGTH:FID_CTR_FP6:5

A=max(X,Y)
max(A,Z)==D340

ENDIF
ENDIF
ENDIF

 

[megaladon]

I spoke too soon.


pro/e is not giving me any errors but i can still edit the value of the curve

 

[kdem]

Assuming d340 is the length of the curve try changing the comparison statement max(A,Z)==D340 to the relation D340=max(A,Z).

 

[megaladon]

d340 is the length of the curve. you are exactly right. the below relation works perfectly now


IF D12:0==2262.82758
X=LENGTH:FID_CTR_FP6:11
ELSE


IF D12:0==1377.95
Y=LENGTH:FID_CTR_FP6:11
ELSE


IF D12:0==2343.15
z=LENGTH:FID_CTR_FP6:11


ENDIF
ENDIF
ENDIF


a=MAX(X,Y)
d340=MAX(Z,A)