Now, before people bite my head off for not searching this topic, I have, but none of them answer my question. In reading CD I have found quite a few threads about what size chain should be used, and the common theme is engineer it properly or don’t do it at all. There has, however, been little discussion as to how to ‘engineer it properly’.

I’m looking for some cold hard math, somewhat like this post, but more coherent. Please, feel free to speak math to me, I’ve completed AP Physics and am currently in Multi-Variable Calculus. If anyone understands how these kind of calculations work and how to explain them, or could point me to a resource where I could teach myself that would be greatly appreciated.

I’m not exactly sure what you are asking for, but I’ll give it my best shot.
I guess one place to start is how much of a load the chain will handle before it breaks. You can find this out on McMastercarr or elsewhere on the net. You also have to see how much torque that your gearbox is going to output to the wheels to see if it is within the load that the chain can transmit, or if it is beyond the breaking point.

ANSI #25 steel chain from McMastercarr has a working load range around only 140 lbs.
ANSI #35 steel chain from the same place has a working load around 480 lbs.

That’s just some help in designing after you design your gearbox. Simply, If your gearbox is outputing more than 200 ftlbs, then you know you should go with the #35 chain instead of the #25 chain unless you like making your pit crew work after every match.:rolleyes:

Also, I’m not sure if this will answer it either, but oh well. The distance between the teeth of a chain sprocket is the pitch, ie, a #25 chain pitch is 1/4" and a #35 chain pitch is 3/8". When you go with two sprockets that have the same teeth count, you can make your life easier by making sure that the distance C-C is divisible by either one that you go with so you don’t need to mess around with tensioning so much.
For different sprockets though, this is a bit trickier. I’m not sure, but I think this only works with even toothed sprockets. (Correct me if I am wrong). You need to think of it as 2 triangles, and a rectangle in between. When you have the pitch diameter or outerdiameter of both gears, since they should have the same difference. You then find the difference in height, when you have the centers of both gears on the same Y axis on a coordinate plane (Note:The difference on the top and bottom of the gears should be the same). Then you just need to make sure the hypotenuse of the two triangles is divisible by whatever pitch you use and you should be fine. Just a bit of geometry, no trig needed there.

I hope that helps… Though that sounds sorta confusing.

One important thing to note is that it doesn’t matter how much torque your gearbox can output, but how much the chain is ever going to see. The force of friction multiplied by the radius of your wheel tells you the torque that that wheel can transfer and the maximum load (theoretically) that the chain will experience. Divide the torque by the radius of your sprocket, and you have the force the chain will experience. Multiply that by at least two or three to maintain a decent safety factor- perhaps more. Compare that to the breaking strength of the chain, and decide if you are too close you are to the rating, and need to bump up the size.

Ex:

Coefficient of friction of wheel (u)- 1.5
Robot weight (N)- 150 lbs
4WD
6" wheels
Single chain run per side
Radius of wheel sprocket- 1.5" Force of friction= uN

Force of friction entire robot can exert- 225 lbs

Force a single side may exert(Ff)- 113 lbs

**Torque on wheel= (Force of Friction)*(Radius) ***We may use the Ff of a side as total torque because we have a single chain per side

Torque- 339 lbs-in

339lbs-in/1.5in= **226 lbs of force on chain

**226 lbs*2 (safety factor)= **452 lbs

**Now, I couldn’t find the ultimate tensile strength of #25 chain, but Mcmaster lists it’s working load at between 140-300 lbs working load. I seem to recall the ultimate tensile strength to be at or over 1000 lbs. In my example, if you had well aligned sprockets and decent chain, #25 would probably be pretty safe, considering the lifespan of a FIRST robot. However, know that you would be operating the system above the recommended working load, and could break chains. If you want the weight savings, bring extra chain, but you would be relatively safe, barring more extreme circumstances.

Unfortunately for those who would prefer a stringently mathematical approach to chain design, no such thing currently exists (in general form). One would need to understand a lot of stress/strain, materials, dynamics and vibration theory to be able to derive working equations for the general case of a chain drive. (And they would be ugly equations.) Engineering practice simplifies this, with tabulated approaches and approximate factors for safety, service class, operational lifetime, etc… Multivariable calculus will not be very useful here, and to be honest, apart from simple concepts like torque and tension, the AP Physics course will probably be of little assistance.

The Tsubaki catalogue also has similar engineering data, which (I believe) was reprinted from Machinery’s Handbook.

You also need to be aware that the published engineering data typically refers to equipment that will be required to withstand continuous operation for several million cycles, with lubrication as indicated, but which may not necessarily be subjected to heavy reversing loads (e.g. the momentum of the entire robot plus the thrust of the robot are going one way, and then your driver reverses the motors). Also, it typically assumes a properly-tensioned chain, and adequately parallel sprockets. Many FIRST robots don’t meet these last two criteria very well, and chain life and efficiency will be reduced because of it.

On the other hand, a FIRST robot doesn’t necessarily need to withstand 10 million cycles; often they’ll only be expected to operate for a few hours (a few hundred thousand cycles, at most). So fatigue performance isn’t something that we would worry about. In principle, we could exceed the working load, and still have an acceptable rate of failure due to fatigue, at the end of the design life.

But while that may be true, we don’t really know enough about the system to be able to set safety factors with a high degree of accuracy. It can therefore be a little misguided to believe that this increase in working load is something which we can actually realize, because other factors may impose limits that are not immediately obvious. For this reason, I would tend to err on the side of caution, or otherwise, have a backup plan for dealing with failures.

That’s from personal experience, especially with the 2003 robot. It was notorious for destroying chains—in fact, in the final match of the 2003 West Michigan Regional, which we won with 245 and 1140, we ended with only one of four wheels driving. It broke at least 13 #25 drive chains over that season. Of course, we erred in design decisions. We didn’t have a proper tensioning system (the chains ran over hard guides), we chose sprockets which were too small (given a set torque, the stress on a chain around a sprocket increases as sprocket diameter decreases, because stress is proportional to force, and force is governed by F=T/r), and we had the most powerful drivetrain in FIRST that year, with over 2100 W (mechanical) available at the motor outputs (and perhaps 1800 W at the wheels).

You can save yourself a lot of unknowns by designing your drives to use an adjustable tensioner, or by using the formula for centre distance between sprockets (from Machinery’s Handbook, 26th edition): Center Distance for a Given Chain Length.—When the distance between the driving and driven sprockets can be varied to suit the length of the chain, this center distance for a tight chain may be determined by the following formula, in which c = center-to-center distance in inches; L = chain length in pitches; P = pitch of chain [in inches]; N = number of teeth in large sprocket; n = number of teeth in small sprocket.

c=P/8*(2L-N-n+sqrt((2L-N-n)^2-0.810*(N-n)^2))

This formula is approximate, but the error is less than the variation in the length of the best chains. The length L in pitches should be an even number for a roller chain, so that the use of an offset connecting link will not be necessary.Even if your sprockets are at the optimum centre distance, it’s not a bad idea to have a place to put a tensioner, just in case you need to change something.

Ensure that you have enough chain wrapping around the driving and driven sprockets. Aim for 180° each, but under no circumstances use less than 135° (for a power transmission system; positioning drives can get away with less, under certain conditions).

In general, read through a few catalogues for chain products, and the appropriate section of Machinery’s Handbook (or similar) to get a feel for what’s out there.

Also, don’t limit yourself to roller chain; belts and silent chain are worth investigating, if you can find a supplier who is willing to produce them cheaply for you.

Just as a clarification; this is an example of using the static load times a safety factor to determine the load rating. It’s a common practice, but because of the difficulty of choosing an accurate safety factor, it’s important that the safety factor be conservatively large, in order to account for the dynamic loads on the chain. It doesn’t directly account for a reversing (dynamic) load, which is rather difficult to predict, without a whole bunch of information about the angular momentum of the gearbox, the linear momentum of the robot, and the rate of change of the speed of the motors—here’s where the calculus starts, if you’re so inclined.

Also, you need to decide whether or not you wish to design for the maximum load that a robot can exert at wheel slip (like in Andrew’s example), or the maximum load that the gearboxes can exert. The latter is only really applicable in a wheels-locked situation, such as if you have a rod jammed through the spokes of your wheel, while it attempts to drive. In practice, it’s just not reasonable to design for this case, unless your drivetrain is sufficiently exposed that you anticipate this happening. And if that’s the case, I’d suggest protecting your drivetrain against jamming with foreign objects, rather than increasing the size of your drive components.

Thank you everyone for your responses. These were exactly what I was looking for, other threads became too much of a debate and less useful for practical information.

One thing Tristan hints at but does not state explicitly: #25 chain, because the sprocket teeth are not as tall as #35, requires the tension to be just right to avoid slippage. This happened to us on our ball conveyer last season. #35 is a lot harder to get to slip, even when the chain is badly under-tensioned.