[scudzuki]

The 2 nuts are opposed, which means that the load is handled by 1 nut when table moves in 1 direction, and by the other nut when the table reverses. You have to admit that the floating nut is free to move relative to the fixed nut, or there would be no advantage at all to using spring preload to compensate for relatively small deviations in the screw pitch. If the floating nut can move relative to the fixed nut, then obviously it can move relative to the table. Since the preload is created by a spring force that creates the opposition, then when the table moves in the direction that loads the floating nut, if the resistance of the table (bind, tool pressure, table lock) is greater than the amount of force the spring exerts holding said nut in its preloaded position, the spring will compress and the nut moves. Once it moves beyond the amount necessary to remove lash, it is only limited by the spring coil binding (or in this scenario disk binding). It is then anybody's guess what the floating nuts position is relative to the table that it is supposed to be precisely positioning. It's very simple.


As for real VMCs having their screw deviations mapped, yes I'm aware of this. I have a fair amount of manufacturing experience.

Joe

edit; Ah I see what you are saying, that in such a scenario the spring would not coil bind because the lash in the fixed nut would limit the error. Which in the case of my machine would be .01" per axis, which would be unacceptable, but an unlikely deviation while taking a finishing pass on a contour and would never happen on a rapid positioning move. Again, what kind of preload spring rate are we talking about, say, on a 5/8" 5 tpi ballscrew?

[SCzEngrgGroup]

The 2 nuts are opposed, which means that the load is handled by 1 nut when table moves in 1 direction, and by the other nut when the table reverses. You have to admit that the floating nut is free to move relative to the fixed nut, or there would be no advantage at all to using spring preload to compensate for relatively small deviations in the screw pitch. If the floating nut can move relative to the fixed nut, then obviously it can move relative to the table. Since the preload is created by a spring force that creates the opposition, then when the table moves in the direction that loads the floating nut, if the resistance of the table (bind, tool pressure, table lock) is greater than the amount of force the spring exerts holding said nut in its preloaded position, the spring will compress and the nut moves. Once it moves beyond the amount necessary to remove lash, it is only limited by the spring coil binding (or in this scenario disk binding). It is then anybody's guess what the floating nuts position is relative to the table that it is supposed to be precisely positioning. It's very simple.


As for real VMCs having their screw deviations mapped, yes I'm aware of this. I have a fair amount of manufacturing experience.

Joe

edit; Ah I see what you are saying, that in such a scenario the spring would not coil bind because the lash in the fixed nut would limit the error. Which in the case of my machine would be .01" per axis, which would be unacceptable, but an unlikely deviation while taking a finishing pass on a contour and would never happen on a rapid positioning move. Again, what kind of preload spring rate are we talking about, say, on a 5/8" 5 tpi ballscrew?

Joe,

Your edit got it. Pre-load is typically spec'ed as a percentage of maximum static load rating for the screw, which is usually around 1600 pounds for a 5/8" screw. Recommendation is up to 30% for maximum stiffness, but less than that for maximum life. "Typical" runs between 10 and 20%, or around 150-200 pounds. There is also some mechanism (it's mentioned in several manufacturers design guides, but never explained) that makes the actual stiffness greater than the pre-load would imply, supposedly by 2-3X. Even 180# is more than you'll ever see on a small machine, short of a crash.

Regards,
Ray L.

[project5k]

so, let me ask a question, how much spring are we talking about, and we'll have to use some kind of units that i can understand.. are we talking something like a screen door spring, or something like a valve spring out of a car motor?

I have no basis to compare spring stiffness, how thier really measured or any of that.. i just know that a valve spring is pretty darn stiff, and a screen door spring isnt(ok yea i know that one is a pull and one is a push, but you get my point...

how could someone like me know how much force a spring is putting out? is it as simple as just pushing on it on a scale, or pulling on it with a fish scale, soemthing to that effect, or is there a bunch mroe to it?

[SCzEngrgGroup]

so, let me ask a question, how much spring are we talking about, and we'll have to use some kind of units that i can understand.. are we talking something like a screen door spring, or something like a valve spring out of a car motor?

I have no basis to compare spring stiffness, how thier really measured or any of that.. i just know that a valve spring is pretty darn stiff, and a screen door spring isnt(ok yea i know that one is a pull and one is a push, but you get my point...

how could someone like me know how much force a spring is putting out? is it as simple as just pushing on it on a scale, or pulling on it with a fish scale, soemthing to that effect, or is there a bunch mroe to it?

The pre-load is typically provided by Belleville washers (search at www.mcmaster.com to see examples), which can provide a very high force in a very small space. It generally takes a stack of several Bellevilles (I have 7 on my 1" screws) to get enough force. The spec for the Bellevilles tells you what the force is at a given deflection, so you simply design for that deflection, and you know what the force for one Belleville is. Stack as many as you need, and the forces add. So, if one provides 30#, and you want 180#, use 6 of them.

Regards,
Ray L.

[project5k]

well thats easy enough... i think i've seen these before.. these are the ones that you stack like this ()()()() to get more travel, and then like this (()) to add thier forces.. right? so in theory you could end up with something like (())(())(()) for more force and more travel?

[SCzEngrgGroup]

well thats easy enough... i think i've seen these before.. these are the ones that you stack like this ()()()() to get more travel, and then like this (()) to add thier forces.. right? so in theory you could end up with something like (())(())(()) for more force and more travel?

Stack them like this: )))) to increase force per unit displacement. (()) would have 2X the force of a single spring for a given displacement, and twice the total travel. (())(())(()) is equivalent to (((((()))))) or ))))))(((((( and would have 6X the force of a single spring for a given displacement, with twice the travel.

Regards,
Ray L.

[caleb105]

Wouldn't it be 3x the travel?

[philbur]

I thought I understood Belleville washers, but maybe not. Wouldn't (()) give you the same force as a single for a given displacement and twice the travel.

Doesn't (( give half the displacement of a single for a given force then if you add another pair reversed that will give you another half displacement for the given force.

Just trying to check my understanding
Phil

.... (()) would have 2X the force of a single spring for a given displacement, and twice the total travel.

Regards,
Ray L.

[caleb105]

Phil,

We aren't talking about travel for a given force, we are talking about total allowable travel.

Therefore, () would be 2x the distance, b/c you have 2 washers to compress.

)) gives 2 times the force, but still only has the compression distance of 1 washer b/c they compress at the same time.

[SCzEngrgGroup]

I thought I understood Belleville washers, but maybe not. Wouldn't (()) give you the same force as a single for a given displacement and twice the travel.

Doesn't (( give half the displacement of a single for a given force then if you add another pair reversed that will give you another half displacement for the given force.

Just trying to check my understanding
Phil

This is making my head hurt! :-)

If you stack Bellevilles so they "nest" (i.e. -concave sides all pointing the same direction), you will increase the load at a given displacement by the number of springs you've stacked. This is typically referred to as "parallel" stacking. So, if one spring gives 10# when displaced by 0.050", then two springs pointing the same way will give 20# when displaced by 0.050". No matter how many you stack, the maximum displacement of the stack will remain the same as the maximum displacement of a single spring. You can visualize this as being exactly equivalent to putting coil springs between two metal plates. If you put two springs between the plates, placed next to each other, the force required to achieve a given displacement doubles. With three springs, it triples, etc.

If you now reverse one of them them, you increase the available displacement, with no increase in load for a given displacement. This is typically referred to as "series" stacking, in which each spring is facing opposite the direction of its neighbors. This is exactly equivalent to stacking coil springs one on top of another in series between two metal plates. Each spring you add, allows you to increase the maximum displacement by the maximum displacement of that spring. So, if the above spring has a maximum displacement of 0.100" (at which point it's completely flat), then two springs facing opposite directions will still give you 10# when displaced by 0.050", but you will be able to compress the pair by 0.200" before they become flat.

When you combine parallel and series stacking, things get a little more complicated. The coil spring analogy only works well here if you have a symmetrical stack. Suppose you have a stack of two springs pointing up, then two pointing down, then two more pointing up, then two more pointing down. Think of this as two stacks of four springs between the metal plates. You'll have twice the force at a given displacement, and four times the maximum displacement.

When springs are stacked in parallel, you can think of each parallel sub-stack as a single spring having a spring constant equal the n times the spring constant of a single spring, where n is the number of springs in the sub-stack. So, in the above example, you can think instead of a stack of four springs, each with a spring constant equal to twice the constant of a single real spring. Again, this will double the force for a given displacement, and quadruple the maximum displacement.

If you have a parallel/series stack with unequal numbers of springs facing the two directions it gets confusing, so I'm not going to go there.

Regards,
Ray L.

[philbur]

That seems to be a yes!

Phil

This is making my head hurt! :-)

If you stack Bellevilles so they "nest" (i.e. -concave sides all pointing the same direction), you will increase the load at a given displacement by the number of springs you've stacked. This is typically referred to as "parallel" stacking. So, if one spring gives 10# when displaced by 0.050", then two springs pointing the same way will give 20# when displaced by 0.050". No matter how many you stack, the maximum displacement of the stack will remain the same as the maximum displacement of a single spring. You can visualize this as being exactly equivalent to putting coil springs between two metal plates. If you put two springs between the plates, placed next to each other, the force required to achieve a given displacement doubles. With three springs, it triples, etc.

If you now reverse one of them them, you increase the available displacement, with no increase in load for a given displacement. This is typically referred to as "series" stacking, in which each spring is facing opposite the direction of its neighbors. This is exactly equivalent to stacking coil springs one on top of another in series between two metal plates. Each spring you add, allows you to increase the maximum displacement by the maximum displacement of that spring. So, if the above spring has a maximum displacement of 0.100" (at which point it's completely flat), then two springs facing opposite directions will still give you 10# when displaced by 0.050", but you will be able to compress the pair by 0.200" before they become flat.

When you combine parallel and series stacking, things get a little more complicated. The coil spring analogy only works well here if you have a symmetrical stack. Suppose you have a stack of two springs pointing up, then two pointing down, then two more pointing up, then two more pointing down. Think of this as two stacks of four springs between the metal plates. You'll have twice the force at a given displacement, and four times the maximum displacement.

When springs are stacked in parallel, you can think of each parallel sub-stack as a single spring having a spring constant equal the n times the spring constant of a single spring, where n is the number of springs in the sub-stack. So, in the above example, you can think instead of a stack of four springs, each with a spring constant equal to twice the constant of a single real spring. Again, this will double the force for a given displacement, and quadruple the maximum displacement.

If you have a parallel/series stack with unequal numbers of springs facing the two directions it gets confusing, so I'm not going to go there.

Regards,
Ray L.

[caleb105]

Something to ponder...

Your ballnuts already have a maximum amount of backlash before the ballscrew starts riding on the opposite side of the threads. No point in having washers that provide 180lbs of force when fully compressed, if they are only going to be compressed part-way and therefore only provide partial pre-load force.

What you want is a washer that BEGINS to compress at 180lbs. As long as you don't exceed 180lbs of tool force, then you will have NO backlash. This also means that in order to get FULL backlash, you would have to put MORE than 180lbs of tool force against it.

[SCzEngrgGroup]

Something to ponder...

Your ballnuts already have a maximum amount of backlash before the ballscrew starts riding on the opposite side of the threads. No point in having washers that provide 180lbs of force when fully compressed, if they are only going to be compressed part-way and therefore only provide partial pre-load force.

What you want is a washer that BEGINS to compress at 180lbs. As long as you don't exceed 180lbs of tool force, then you will have NO backlash. This also means that in order to get FULL backlash, you would have to put MORE than 180lbs of tool force against it.

What you seem to be describing is exactly how it works. The preload is just that - preload. The springs are compressed sufficiently to provide 180# of static force pushing the nuts away from each other, and into the screw threads when there is zero load on the machine. Unless and until the longitudinal force from machining exceeds 180#, there will be *no* loss of contact between any of the balls in either nut and the screw threads, and hence *no* backlash. The magnitude of the preload will diminish as the machining load increases, but until the preload reaches zero, no backlash can occur.

You *need* partial compression of the springs to allow some movement to accomodate the lead error, variations in ball diameter, and other imperfections in the screws and nuts.

Regards,
Ray L.

[caleb105]

You *need* partial compression of the springs to allow some movement to accomodate the lead error, variations in ball diameter, and other imperfections in the screws and nuts.

Regards,
Ray L.

THIS is what I had not considered. But how do you know what force the spring is exerting at PARTIAL compression? The only values you are given are:

- Force required to START compression

- Force required for COMPLETE compression

Obviously this is not a linear relationship (you can't split the difference and say that happens at 1/2 compression).

[SCzEngrgGroup]

THIS is what I had not considered. But how do you know what force the spring is exerting at PARTIAL compression? The only values you are given are:

- Force required to START compression

- Force required for COMPLETE compression

Obviously this is not a linear relationship (you can't split the difference and say that happens at 1/2 compression).

Not true. It's a spring, and behaves exactly like any coil spring - i.e. - linearly according to: F = k * X. They are spec'ed for a specific force at a specific displacement - generally NOT maximum displacement - and there is a spec for maximum deflection as well (usually just the thickness of the material). You can assume, with reasonable accuracy, that the force *will* vary linearly from zero to maximum.

Regards,
Ray L.

[caleb105]

So it sounds like I need a washer that has an initial compression force below 180lbs, and a full compression force above 180lbs. My problem is that the only washer with an I.D. larger than my ballscrew (.621") takes 410lbs of force just to START compression. (at least at McMaster Carr).

Anybody got any other sources?

Thanks!

-Caleb105-

[Bubba]

So it sounds like I need a washer that has an initial compression force below 180lbs, and a full compression force above 180lbs. My problem is that the only washer with an I.D. larger than my ballscrew (.621") takes 410lbs of force just to START compression. (at least at McMaster Carr).

Anybody got any other sources?

Thanks!

-Caleb105-

This is the exact reason I used "wave washers" also from McMaster. It has lower ratings than "Belleville" as you have found. Yes, I had to stack them to achieve the desired rating.

[SCzEngrgGroup]

So it sounds like I need a washer that has an initial compression force below 180lbs, and a full compression force above 180lbs. My problem is that the only washer with an I.D. larger than my ballscrew (.621") takes 410lbs of force just to START compression. (at least at McMaster Carr).

Anybody got any other sources?

Thanks!

-Caleb105-

Nah! You're not looking in the right place. There are many different kinds of disc springs. Here's one that's 23# at 0.020 deflection:

http://www.mcmaster.com/#94065k54/=qfjtl

Search on "disc spring", and look at ALL the different kinds.

Regards,
Ray L.

[caleb105]

Sweeeeeeeet Himy.

Looks like 8 of those would give me 184 lbs.

(((())))

-Caleb105-

[escott76]

Nah! You're not looking in the right place. There are many different kinds of disc springs. Here's one that's 23# at 0.020 deflection:

http://www.mcmaster.com/#94065k54/=qfjtl

Search on "disc spring", and look at ALL the different kinds.

Regards,
Ray L.

Exactly the ones I'm using for my 5/8" screws. Work perfectly.