parameter formulae problems

[batstar]

Good morning to all,
I found a problem in inserting formulas into the parameters in inventor 2014: practically (to explain it simply), I should make a steel cube starting from the weight and what I did was create a user parameter where I insert the weight value and a "side" parameter with formula "(weight/7.85)^(1/3))", (where 7.85 is the specific weight of the steel), I find myself with result a x10 value! the same formula in excel gives correct result, I cannot explain why and at this point I wonder also about the correctness of inventor calculations.
Can anyone explain this error?
Thank you.

 

[Catafratto]

attentive to measuring units, try with:

(weight / 7.85 kg/dm^3 ) ( 1 mm / 3 mm )

specifying in the unit/type column the weight measurement unit (kg, I suppose); otherwise can come pastries like yours

edit: addition

inventor uses as an internal measuring unit the cm, perhaps when not expressly specified there are problems for that. . .

 

[batstar]

attentive to measuring units, try with:

(weight / 7.85 kg/dm^3 ) ( 1 mm / 3 mm )

specifying in the unit/type column the weight measurement unit (kg, I suppose); otherwise can come pastries like yours

edit: addition

inventor uses as an internal measuring unit the cm, perhaps when not expressly specified there are problems for that. . .

I put the weight with unit "g" and side with unit "mm" and formula "(weight/7.85 on)^(1 on /3 on)" and gives the correct value, the formula always appears in red and I know it is not a good signal.
on more complex formulas becomes unmanageable, perhaps it should work on an embedded excel?

 

[Catafratto]

is not unmanageable, indeed, it allows you to mix different units of measurement and have correct results. if the formula is in red means that there is an error, if a formula that is reported as incorrect generates an incorrect result I would say that everything is back. if you read your formula see that you have indicated your specific weight as on, without units. if you read my see that I have indicated the unit of measurement for extended and so does not give you red signal

 

[Catafratto]

By the way, now that I think about it, if you use g and mm, shouldn't density be 0.00785? or am I losing my chicken with the calculations?

 

[Ilario]

in addition it can also be said that if the formula is inserted in the "side" parameter, inventor expects a measure in linear length, then the formula "((/7.85 on)^(1 on /3 on))" is not correct as it returns a mass.

 

[batstar]

By the way, now that I think about it, if you use g and mm, shouldn't density be 0.00785? or am I losing my chicken with the calculations?

Yes it is true...

 

[batstar]

in addition it can also be said that if the formula is inserted in the "side" parameter, inventor expects a measure in linear length, then the formula "((/7.85 on)^(1 on /3 on))" is not correct as it returns a mass.

returns the value in mm but the formula remains red, allego example

 

[Ilario]

no, it returns it to "g" and you bake it in "mm", in fact if the formula writes it so "(weight / 0.00785 g/mm^3 ) ^ ( 1 mm / 3 mm ) )" the formula magically becomes black and no more red

 

[batstar]

It's true, it works as you say,
However I do not understand why it adds the mm in the power elevation since it has no unit of measurement

 

[Ilario]

Yes, in the rush I did not give us weight, but so much "mm"/"mm" is equal to "on", so even if you write:
(weight / 0.00785 g/mm^3 ) ( 1 on / 3 on )
the formula is correct anyway.
the problem was in the "on" side by side with the specific weight

 

[batstar]

adds them by himself inventor the mm in the power elevation, boh, however it works.
Thank you.

 

[Catafratto]

is like an expression, (1*mm)/(3*mm) = (1/3)*on, you may also write 1kg/3kg.
this part of the formulas is not the clearest, but once you understand the concept you go away smoothly. the counterparty is that, specifying expressly (and correctly!) the units of measurement can for example sum inches to millimeters (I do it for the steps of the rollers), type "0.5 in * 15 on + 30mm" or, in your case, you could still write "7.85 kg/cm^3" and the conversion would take place automatically. Finally, in the case of inconsistencies with the units of measurement, it signals them.

 

[Zac69]

I usually use "without unity" formulas (or divide the unit of measurement for itself so as to have a pure number).
then I do all the calculations that go to me and eventually remoltiplico for the unit that I need in the variable with measuring units.

 

[batstar]

I usually use "without unity" formulas (or divide the unit of measurement for itself so as to have a pure number).
then I do all the calculations that go to me and eventually remoltiplico for the unit that I need in the variable with measuring units.

Good, as a system.
I finally incorporated into a excel and there I did everything I wanted, for me it is simpler and faster.

 

[Zac69]

returns the value in mm but the formula remains red, allego example

However for the correct formula you had to divide by 1g in order to make it up and multiply by 1mm to have the side in mm:
(weight / 0.00785 on ) ( 1 on / 3 on ) ) / 1 g * 1 mm

 

[Zac69]

However for the correct formula you had to divide by 1g in order to make it up and multiply by 1mm to have the side in mm:
(weight / 0.00785 on ) ( 1 on / 3 on ) ) / 1 g * 1 mm

wrong, the formula is: (weight / 0.00785 g ) ( 1 on / 3 on ) ) * 1 mm
in this way the grams disappear immediately, all the accounts are made up and in the end gives the length

 

[Ilario]

Yeah, well, here to go from Milan to Rome, we're going through fly. .
the formula is:(weight/specific weight ) (1/3)
weight = "g"
specific weight = "kg/cm^3" which has been converted into "g/mm^3"
1/3 = power elevation = cubic root = pure numbers = "on" for inventor

formula in inventor:(weight / 0.00785 g/mm^3 ) ( 1 on / 3 on )

so the formula is mechanically correct and the result is in "mm" without strange turns.
Everything else is menaced.

 

[Zac69]

Yeah, well, here to go from Milan to Rome, we're going through fly. .
the formula is:(weight/specific weight ) (1/3)
weight = "g"
specific weight = "kg/cm^3" which has been converted into "g/mm^3"
1/3 = power elevation = cubic root = pure numbers = "on" for inventor

formula in inventor:(weight / 0.00785 g/mm^3 ) ( 1 on / 3 on )

so the formula is mechanically correct and the result is in "mm" without strange turns.
Everything else is menaced.

Right. .
I did it more like exercise than anything. I usually said how I act by removing a priori units of measurement

 

[Ilario]

Yes, of course, then everyone is free to do as it is found better, but in my opinion, as from school memories, where not putting the units of measurement, it ran into errors that then made you scare the checks of mechanics, I prefer to put them where they serve, because if then there are errors in the result, proceeding backwards or by degrees you can find out if there was an error of interpretation of units of measurement, otherwise you do not understand what you are
I repeat, I prefer so for my mental order, then if the result is there, everyone is free to do as it is better