How do teams determine how far apart to place two sprockets to ensure correct chain tension? This year, our robot’s biggest issue was the drivetrain, and specifically, its chain.

As an example, I tried using a calculator, and I input 18 teeth for the sizes of both my driven and drive sprocket, and selected 116 links, and the centre-to-centre distance given was 12.25". Upon making a test piece with the same sprocket sizes to test this number, the chain had significantly less links than 116, and was still very loose. How do I figure out how far apart to place the sprockets for proper tension? Do you use calculators? Do you have a rule of thumb? Did you just eyeball it, or get lucky? Please let me know!

First off, for a long chain run (12.25" certainly qualifies), it’s a good idea to have adjustable tensioning for the chain, both in case you don’t get it quite tight the first time around and to account for the chain stretching as it’s used.

That said, we use a calculator. I typically use this one, but it doesn’t really matter which one you use. Make sure you’re inputting the correct chain pitch. #25 is 0.25" pitch and #35 chain is 0.375" pitch.

The calculator you used appears to be correct, assuming #25 chain and .25" pitch. With same size sprockets, ((2C-C)/pitch)+sprocket teeth = #of links. This is just top length of chain + bottom length of chain + chain wrapped on first sprocket + chain wrapped on other sprocket. The top and bottom lengths = the center-to-center distance, so divide that by pitch to get number of links. The number of links wrapped on 2 halves of equal sprockets is the number of teeth on the sprocket. For your calculation, ((212.25)/.25)+18=116. In theory the chain would’ve been fine. There was some other problem in your setup.

As a rule of thumb, simply make the C-C distance a multiple of chain pitch and you will end up with a nice, even number of links. This is assuming same size sprockets with an even number of teeth, which is what we always use on drivetrains. Every #25 sprocket sold by Vexpro has an even number of teeth anyways. Only when sprocket sizes are different would I turn to the calculator.

As Bazel A suggested, adjustable tensioning is the proper way to make sure the chain continues working even as it stretches. But since FRC robots don’t have to run for all that long, some teams forgo adjustable tensioning with no problems. They just use exact C-C distance, or add .018"-.02" to the C-C distance to account for the chain stretch. Done properly, this setup could last your robot through a full season.

I don’t mind using a tensioner to adjust for chain stretch. We’ve tried using calculators to no avail, the chain is still loose enough to be pulled off by hand. The 12.25" chain run was actually calculated using the exact calculator you linked me to. I am using brand-new unused #25 chain. Are there any other variables that could influence the tension of the chain? What do you view as proper tension?

This may not be your issue, but note that if “116 links” on the chain calculator for two 18 tooth sprockets gave you a C-C of 12.25", the “116” meant “half links”, equivalent to “116 times the chain pitch,” or 29" long if broken and stretched out straight; a bit under 14.5" if the loop were stretched with no sprockets inside, which makes sense if stretched around some sprockets 12.25" apart. Perhaps you thought it meant 116 “whole links”, which would be twice as long?

What are your tolerance’s for the part your used the chain on? That might contribute to the mistake. Also what size chain did you use? I know it is not uncommon for someone to think #35 chain has a .35" pitch instead of .375" which is the real pitch.

Edit just saw that it was #25 chain. Add .018" to each cenet to center distance, but only do center to center if your machines can hold tolerance’s. Otherwise use the versa system with tensioners. I am open to work one on one with anyone needing help with C to C chain designing, message me if you are interested.