Trouble with certification pre-test

[Nhouse1185H6ZG]

Hello, I'm very sorry if this is taboo, but I'm really trying to understand this one question on the Inventor certification pre-test.

Attached is an image of the question. I'm very much inclined to believe the answer is "WIDTH / 4". In fact I don't see how you can relate "SPACE" to "WIDTH" without using the variable "WIDTH".

Despite this, the website says the correct answer is "d4".

Can someone explain what I'm missing, please?

[CGBenner · Accepted solution]

@Nhouse1185H6ZG 

Welcome to the forum.  I'm not 100% certain, but just from a quick glance my take is that since d4 has already been defined, setting the subject dimension to be equal to d4 would ensure equal spacing from the outer edge on both sides of the part.  

[NigelHay · Accepted solution]

D4 & D5 are constants so the spacing will not change with a width change.

WIDTH/4 will place one hole on the R.H. edge of the rectangle.

WIDTH/5 will keep the spacing between the holes & hole to edge distance the same & a function of the width.

I agree that the question seems to be badly worded.

[blandb]

D4

You can test this because if your first dim is "SPACE" then you set the next values equal to space, when you try to place the D4 dim, it will give a references dimension. Then when you edit "SPACE" and make it equal to D4, then all values space out.

[SBix26 · Accepted solution]

Once the SPACE parameter is set equal to d4, the WIDTH is included implicitly, because SPACE + d1 + d2 + d3 + d4 is equal to WIDTH.  There is no need to create an equation that explicitly includes WIDTH when it is included by definition.

Do you have a copy of Inventor available to use?  If so, the easiest way to persuade yourself is simply to create the sketch as shown and try it out.

I will say that this is not a technique that I would ever recommend, but it does illustrate the point that you have asked about.  All the spaces could have been set equal to WIDTH / 5 ul, but it is not necessary to do so, as you can see.

Another method would be to use construction geometry to create the necessary constraints:

Here again there is no explicit use of the WIDTH parameter, and this time not even dimensions-- the construction lines (five individual lines) are set equal to each other.

Hope this helps,

---

Sam B

Inventor Pro 2025.0.1 | Windows 11 Home 23H2

[CGBenner]

I love this exchange of ideas and techniques!  Only.... now I think I need to create this in Inventor myself and play with it.  Darn.  😁

[James_Willo · Accepted solution]

D4 is the answer as all the other dimensions reference 'Space'

You can also use process of elimination. 
It can't be 'Space*5' because the value would reference itself. 

d5 is the vertical distance so this is also wrong

WIDTH/4 as mentioned above is also wrong because it needs 5 dimensions not 4. 

This is technique that I used all the time in industry, but I think maybe the word 'Space' confuses the question as you don't really need to define a parameter name for this.

I think you could also use d4 = Space too.  I think the idea is to make sure you understand how things reference each other and make you actually solve a problem instead of just remembering where a command is located. 

[Nhouse1185H6ZG]

Thank you so much for your response! This makes sense now, I appreciate it!

[Nhouse1185H6ZG]

Thank you for the warm welcome! And from all the help I've received from the forum, I can say you were right on the money with your initial thoughts. Thanks!

[Nhouse1185H6ZG]

Right? Threw me for a loop. Thank you for your response!

[Nhouse1185H6ZG]

It sure does! Thank you very much!

[Nhouse1185H6ZG]

Haha, always a fun time 🙂