radice quadrata su formule

[batstar]

Good morning to all,
I have a problem: I should insert the square root into some formulas of inventor parameters but it does not work,
I tried with "sqrt(value)" and "value^0.5" but does not accept them.
I tried with "sqrt(value*1)" and it works, with "sqrt(value/1)" it does not work.... .
How do I do that?

thanks to all

 

[meccanicamg]

the guide qui.

 

[batstar]

I saw her, but it doesn't work.

 

[Peppe]

try to create a parameter and how drive you put on (without units), as an equation test for example 16 (on) ^ 0.5 (on). so it returns you the correct value obviously without unit (on)

 

[batstar]

Nothing to do, it's not okay....proved of everything, I have to include an excel...
... is not allowed on a program for professional use as an inventor!

 

[Peppe]

look in the example I attach you (inv 2019). There are 2 functions inside it and tell me if it works.

 

[Catafratto]

I also confirm that it works: Measuring units must be consistent, explicitly stated.

 

[batstar]

look in the example I attach you (inv 2019). There are 2 functions inside it and tell me if it works.

...not opens it to me

 

[Peppe]

I made screen shot of the functions, I saw now that you have inventor 2018

 

[batstar]

I made screen shot of the functions, I saw now that you have inventor 2018

It also works by me with "on", but I can not use a value already present in mm and derive from that the root... .

 

[Peppe]

It also works by me with "on", but I can not use a value already present in mm and derive from that the root... .

try to make a "support" parameter where you call your value in mm and divide it for 1mm so that you can prove a value without unit.

 

[batstar]

It can be a way, but I don't see why they chose not to add mathematical operators directly in quotas as for basic operations. .

 

[Catafratto]

post the formula

 

[batstar]

post the formula

I have nothing specific but an example could be:
a cylinder with base diameter (d2) equal to the square root of the height (d1), it would be enough to put on the value of d2 "sqrt (d1)". ....
But no, we have to turn around!

 

[Catafratto]

formally you would change the measuring unit from mm to mm^0.5; in practical
d2 = sqrt(d1/1mm) * 1mm,