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View Full Version : Where is the design tab located?


Abole
28-01-2010, 17:30
I'm trying to find the design tab to help me build a chain for 2 sprockets. I'm following a tutorial to build one and it tells me to click on the desgin tab in the ribbon. Thanks:D

Bob Steele
28-01-2010, 18:16
It might help if you told us what program you were using....

:)

joek
29-01-2010, 11:49
that function is useles for FRC, i've tried it

Allison
29-01-2010, 12:07
Assuming you are using Inventor:

1. Make an assembly file
2. In your assembly file across the top there will be tabs (assemble, design, etc)
3. Under power transmission click the drop down next to v-belts and select roller chains.

I hope this helps

Allison

joek
29-01-2010, 12:23
i tried it, and it didn't work, it wouldn't go from the output sprocket to the sprocket on the mechanum

inventor_phild
03-02-2010, 01:41
All,

Difficult to tell the exact problem, but I suspect that you need to change the options under Sprocket Geometry Option.

1. In the Roller Chains Generator dialog box, click the down-arrow beside Sprocket. There are 3 options, Component, Existing, Vitrual.
2. If the sprockets already exist (and it sounds like they do) select Existing.
3. Keep in mind that it generates a solid representation of the chain or a sketch. We do not need to see actual chain links in a design, the chain data is more important.

Phil

Abole
03-02-2010, 16:29
Sorry, took so long to reply, i found the design tab, just forgot about the assembly part of it from an ipt... thanks though, and it was for making the chains

joek
03-02-2010, 19:00
ok, i used the roller chain generator, and it said this:
"chain drive trajectory is corrupted or data are inconsistent with build in rules. calculation cannot continue"
and
"no sliding sprocket within the chain drive. please set one sprocket as sliding."

Abole
16-02-2010, 17:13
You might need to put a sprocket, so there is a place that the chain can start from. Thats one thing i noticed but it was easier just putting them straight in on the assembly of the whole robot.

joek
17-02-2010, 13:44
i started a thread about this after my post.