Vector Module Chain Tensioning

[ekiss13]

Our team is attempting to design a working vector module during the offseason but we have encountered problems with chain tensioning between the two axels(the live axel and dead axel). We were thinking that the distance between the two would be too small for chain tensioners and instead accurate calculations would have to be used. How exactly would you find the precise distance between the axels so that the chain is perfectly taut?

If you want a better description of what we were planning don't hesitate to ask.

[Aren_Hill]

http://www.wa4dsy.net/robot/sprocket

[AustinSchuh]

Get the two sprockets you want to use, and import them into CAD.

Draw up some "chain", which can be as simple as a block with holes every 0.25".

Mate the "chain" holes concentric to the bottom of the teeth in the sprocket. With a bit of clever mating, you should be able to confine everything exactly, and be in business. I like this method since it's completely general, and lets you also see the chain paths.

[Aren_Hill]

Originally Posted by AustinSchuh

Get the two sprockets you want to use, and import them into CAD.

Draw up some "chain", which can be as simple as a block with holes every 0.25".

Mate the "chain" holes concentric to the bottom of the teeth in the sprocket. With a bit of clever mating, you should be able to confine everything exactly, and be in business. I like this method since it's completely general, and lets you also see the chain paths.

visualizing chain paths is always a good thing


that was a fun to figure out

[ekiss13]

Thanks a lot guys. i haven't tried the cad approach yet, but that website was very helpful.

[Mark McLeod]

Here's the basic math. The center distance sprocket-to-sprocket for a tight chain is determined by:

• c = sprocket center-to-center distance in inches
• L = chain length in pitches (# of links)
• P = pitch of chain or length of a chain link [in inches] (.250" for #25 chain and .375" for #35 chain)
• N = number of teeth in large sprocket
• n = number of teeth in small sprocket

c = P/8 * (2*L - N - n + sqrt((2*L - N - n)^2 - 0.81 * (N - n)^2))

Begin with a chain length approximately the distance you are looking for and solve for it, then pick the nearest even whole number for "L" to get the final distance "c". Using an even number just means you avoid having to use a half-link.


P.S. This is from Machinery's Handbook

[JamesCH95]

+1 to Mark.

Get a machine elements design book, it has that equation and many others that are great to design with.

[viperred396]

Originally Posted by Aren_Hill

visualizing chain paths is always a good thing


that was a fun to figure out

Now i have to ask... what is this for?

[Jared Russell]

For short runs (or heavy duty chain), Mark's math is what you want to do. For long runs of #35 chain or medium-to-long runs of #25 chain under moderate loads, expect measurable stretching over the life of the chain (so make sure you have a reasonable method of compensating).

[Aren_Hill]

Originally Posted by viperred396

Now i have to ask... what is this for?

Dillon our driver and chassis subteam leader was forced to get creative when we decided to construct a swerve with 6 modules