Tight linear bearings

[pixpop]

I picked up some THK SHS20L linear guides for a machine I'm building. I knew they were overkill for my design, but the price was good. They are brand new.

I've had them for a few months, but didn't open them until last night. They are very tight, which will be a problem for my machine (needs to be fast more than it needs to be precise).

I don't think there's anything wrong with them. Perhaps it's just preload that's making them tight. They do free up a bit if I put my weight on them, but the load they will see in my machine will be much too small to have this effect.

Questions:

1) In the THK literature, they say one should remove the anti-rust oil and feed lubricant before using the bearings. How would I do that, and if I did, are they likely to move any more freely?

2) The literature specifies the 'Seal resistance' at 7 Newton. What is seal resistance?

3) Will they free up if I use them a bit?

My estimate is that I need to apply about a pound of force to get them to move. That's a pound I really need to accelerate my load.

I'm guessing that I should probably sell these, and buy some smaller, lighter ones for my machine. I have used other LM guides before and they've been very free moving, but I've never used brand new ones before.

[pointcloud]

Probablly nothing wrong with them... Got pics? How much are you asking?

Q1 more than likely. mineral spirits
Q2 ? Never heard that term
Q3 They should

[keebler303]

The seal resistance means it takes a force of 7 N (1.57 lbs) to get the bearing block to move. This will definately decrease as the bearing gets worn in. If there is some type surface coating, I would remove it first. If they are still too tight you could try running them in a bit or just swap them for some smaller, more appropriate ones.

Matt

[pixpop]

Marc,

I don't have any pics, but there's lots of info about them on the THK website. Look for the SHS series. Mine are 1060 mm long.

Matt,

1.57 pounds seems about right for what I'm seeing when I try to move them. This Seal Resistance is probably the bulk of the stiffness. In addition, I detect a slight viscous drag that increases with speed of movement. This is probably caused by the grease that's in the bearings.

[davereagan]

How can you be worried about a pound or two of drag? With a decent motor, you should be making forces in the hundreds of pounds. What torque steppers are you using? Ballscrew pitch? If you take the torque of the motor (inch lbs) and multiply by 6.28 and divide by the pitch (inches), you should be making 75-90% that many lbs in force, due to basic losses in the screw, table and bearings. My 40 taper CNC makes ~1400 lbs of force in each axis. This is not off a brochure, but actually measured with a force gauge.

Dave
Dave

[pixpop]

How can you be worried about a pound or two of drag? With a decent motor, you should be making forces in the hundreds of pounds. What torque steppers are you using? Ballscrew pitch?

Not using a ballscrew, using belts. Not using steppers, using servos. This machine is meant to be more like a plotter than a router, and my goal is for it to do a move of 1 meter with an average velocity of 1 meter/sec (i.e. about 4700 ipm peak velocity). To achieve this, I need to eliminate all extraneous drag and mass from the system.

[pointcloud]

LOLOLOL... What are you going to do when it comes time to stop if you think you motors will not pull it without inertia playing any part?

[pixpop]

Originally Posted by pointcloud LOLOLOL... What are you going to do when it comes time to stop if you think you motors will not pull it without inertia playing any part?

I'm not sure I understand your question. Why do you think that I think inertia will play no part?

[davereagan]

Well I must admit, that's a whole lot faster than most guys are talking about on here. I rarely see a mill drill move at more than 50 or maybe 100 ipm, which is crawling. I suspect the drag will lessen with time just like anything else with a seal (bearing, brake caliper, etc). As mentioned above, you might not mind a pound of drag when you are trying to slow things down. You said you are using belts. What are the belts connected to? It pretty much has to be a rack or a screw of some kind? What is the weight of the traveling mass? What is the torque rating of the servo motor? Pitch of the screw or rack? Have you calulated the thrust you expect to generate? A lot of people are not aware of how much force is made with small amounts of torque if the efficiency is decent and the pitch isn't too high.

Dave

[pointcloud]

Maybe you are not seeing:Inertia.... The question IS you have 5 POUNDS of GANTRY moving 1000ipm stopping THE 5 POUNDS will be WAY MORE than just a little ole SINGLE POUND starting OFF..

If your motor will not PULL 1 pound, How will it stop 100?

[davereagan]

I find these equations easier to run in metric units. If the gantry weighs 5 lbs, that's 2.27 kilograms. 4700 inches per minute is 119.4 meters per minute = 1.99 meters per second. Newton's formula says Force = mass * acceleration. What is the acceleration? Well, We have half a meter to get down to zero.
Another formula is S=.5AT*T where S = distance, A = acceleration and T = time. The Time we can solve for. We assume uniform deceleration. So the average speed is 1.99/2 = .995 meters/second. Time =distance/velocity, so T = .5/.995 = .503 seconds, since the distance is .5 meters from the peak velocity to zero at the either end of the travel. So rearranging the formula above to solve for acceleration, A=2*S/(T*T) = (2*.5)/(.503*.503) = 3.95 meters per second. Gravity is 9.81 meters per second, so this is ~.4G (not too shabby). Back to the original formula, Force = Mass * Acceleration, The mass is 2.27 kg and the acceleration is 3.95 meters per second. Force = 8.97 kg*m/second (Newtons). 1 Newton = .224 lbs, so we have a force of 8.97*.224 = 2.01 lbs.
This checks out nicely against our original given weight of 5 lbs multiplied by .4G ( 2lbs). Now I have to admit that the seal drag is very significant compared to this acceleration/deceleration force. Of course, all drag in the screw/belt/bearings would add to the force/torque needed for acceleration, but for deceleration the would help. Having invested this time, I am really curious about the actual gantry weight and the motor torque and rpm ratings, screw/rack pitch, etc.
Being trained as an Engineer and working as a Tool Maker, I love to dust off the books now and then.


Dave

[pixpop]

As I mentioned in post #6 above, there is no screw.

There are two belts, attached to the gantry. Each belt is to be driven by two servos, one at each end. The reason I'm doing it this way is because I got the servos cheap. I don't have official specs on them, but have been told they will do 40 Oz-inch stall torque. Given the small amount of travel, I think the motors will never reach full speed. I'll have a better idea of the actual torque when I construct one bearing assembly, with two motors, a belt and a load, and do some tests.

I reasoned that the customary design, with an idler at one end and motor at the other, just adds to the inertia.. So why not put another motor instead of an idler, and get more torque?

Yes, I know it might be difficult to keep all 4 motors synchronized.

So, four motors gives 160 Oz in. My pulleys are about 1" diameter, which means about 320 Oz of force available (excluding friction). I haven't finished designing the gantry yet, but I'm thinking 5 pounds would be on the heavy side. Each pair of motors sees half the load (mass and friction) of the gantry.

Currently, I'm planning to make the gantry from carbon fiber with a 9mm wide linear bearing. If I can afford the mass, I'll mount a small BLDC servo at each end with a belt running around them, similar to the long axis. But if that makes the gantry too massive to get the speeds I want, then I'll use a cable drive to reduce the moving mass. I prefer the belt drive though, it's very simple.

As I said above.. it's something like a plotter.