[cncfoam]

Has anyone on this list made a spinning nut, verses spinning the screw as normal.

Here is a question to put to the group...

Would it take less torque (or Power) to spin the nut rather than spin the screw. For example.

What if I had an 1 1/2" screw, that was really long. This would weigh alot. I would need power to turn the screw, (because of weight), and more power to push the carriage.

verses, Less power to turn the nut,(weighs less), and the same amount of power to move the carriage as in the example above.

what are your thoughts.

Eric

[ESjaavik]

Yes and no whip, as the screw don't turn. But one problem, The balls will prefer to be in the transfer tubes if the RPM gets too high. The balls entering the tube will have to push harder to get the ones already there out at the other end. But with low speeds the inertia of the screw is no big deal. Nuts meant for this is (even more) expensive than standard ballnuts.

[cncfoam]

Would a better choice be ACME screws and nuts.

Eric

After a few responses, and when I get some time, I was going to draw up a design for this.

[chuckknigh]

Check the open source forum...here's a link to a really good looking setup. It was enough to inspire *me!* But then, it doesn't take much to get my imagination going.

http://www.cnczone.com/forums/showth...?t=4178&page=4

-- Chuck Knight

[duluthboat]

Eric,

I would use a spinning nut where turning the shaft complicates things. On the Z axis, turning the shaft puts the motor up high at the end of the shaft. I would prefer to keep the weight low if possible, so turning the nut makes since here.

I haven’t worked with ball screws but can see if you want to turn a ball screw nut, the transfer tube complicates design of the housing. If I wanted the accuracy of ball screws I would turn the screw and support it along its length as much as possible. The machine I’m working on will be using gears and acme threads.

Sorry I haven’t been much help, I would have to look at this in more detail and my time won’t allow that. If getting to the end is your only goal, than stick with convention, if it’s the adventure of the journey than turn over every rock.

Gary

[arvidb]

Would it take less torque (or Power) to spin the nut rather than spin the screw. For example.

What if I had an 1 1/2" screw, that was really long. This would weigh alot. I would need power to turn the screw, (because of weight), and more power to push the carriage.

verses, Less power to turn the nut,(weighs less), and the same amount of power to move the carriage as in the example above.

what are your thoughts.

Eric

It is not the screw's weight that is a problem, but its rotary inertia. The rotary inertia for a cylinder is proportional to its radius raised to the power of 4, which means a doubling of the radius increases rotary inertia 16 times (2^4).

Acceleration is inversely proportional to rotary inertia and proportional to torque - which means more torque = more acceleration, more rotary inertia = less acceleration.

So by decreasing the rotary inertia (perhaps by using a spinning nut) you can use lower torque motors for the same acceleration (top speed is not affected). Be careful though, since if the radius of the nut is only 2.5 times that of the screw (for instance), its rotary inertia will be equal to a screw that's 40 times longer than the nut.

Arvid

[ESjaavik]

Excellent Arvid! A very good example that common sense does not always make sense. Do the numbers too and you may be surprised!

[cncfoam]

Be careful though, since if the radius of the nut is only 2.5 times that of the screw (for instance), its rotary inertia will be equal to a screw that's 40 times longer than the nut.

Im not catching this, could you use a different explaination.

EZ

Can you get better performance from a spinning nut, or a spinning screw. or is there no difference?

[duluthboat]

Eric,

You have to play by the numbers; it depends on the diameter of the nut, the rpm, in opposition to the length of the screw. In the end if you want to optimize, the engineering gets a little more complicated. There is a point where they will be the same, then go the other way. If I'm reading Avrid's post right.

Gary

[metlmunchr]

there's not much other way to talk about rotational inertia in simple terms than to say it varies as the 4th power of diameter. Past that means getting into the methods by which kinetic energy and polar moments of inertia are calculated to show how its a 4th power function, and then to show how mass distribution can alter the numbers such that its not a true 4th power function at all times. In simple terms, if the nut is 2.5 times the diameter of the screw, then setting the screw diameter to 1 makes its value 1x1x1x1=1, whereas the nut is 2.5x2.5x2.5x2.5=39+, hence Arvid's statement about the nut having rotational inertia equal to a screw 40 times the length of the nut. If you have a copy of Machinery's Handbook, take a look at the section on flywheels and it will give examples of how all this is calculated.

[cncfoam]

Im not an engineer, but logically it seems if you turn the screw, you have to deal with rotory inertia. but if you are just turning the nut, the length of the screw should be irrelevant, but I can see how the diameter of the screw with affect the equation of rotory inertia.

just a thought.

EZ

[InventIt]

How would material factor into this? If the screw is steel (heavy) and the large nut is plastic (light) this should effect the equation.

[jeffs555]

If you are spinning the nut, the length of the screw doesn't matter, however the spinning nut will have rotary inertia also. All the others are trying to say is that if you have a large brass nut, and a short steel screw, then you might have to overcome more inertia by spinning the nut than you would by spinning the screw. And yes, since plastic is much less dense than steel, and rotary inertia is proportional to the density, plastic would have much less rotary inertia.

If you want to work through the numbers, here is a page that has the equations along with some examples. http://www2.danahermotion.com/servic.../pdfs/9011.pdf

[ollopa]

I have to disagree with the physics here.

The moment of inertia for a solid cylinder is 1/2 MR^2. Also note that the nut has some friction on the screw, but it is a ring and not a solid cylinder. It has a moment of inertia of 1/2 M(ri^2 + ro^2). Note the slight difference--the moment of inertia of a nut will depend on both the inner and the outter diameter.

[arvidb]

Yes, the moment of inertia I of a cylinder can be expressed as I = 1/2*mr^2, that is correct. The moment of inertia of a hollow cylinder with outer radius r_o and inner radius r_i is then Ih = 1/2*mr_o^2 - 1/2*mr_i^2 = 1/2*m(r_o^2 - r_i^2).

But observe that this equation contains the mass of the cylinder m. Mass is proportional to volume, and the volume of a cylinder is pi*r^2*L = sectional area * length. Thus its mass is proportional to radius squared.

So, using d = density of the cylinder material and substituting d*volume = d*pi*r^2*L = m, we can write I = 1/2*d*pi*r^2*L*r^2 = 1/2*d*pi*L*r^4.

So yes, if you manage to change the radius of the nut without changing its mass, the change in rotational inertia will be proportional to the factor of change squared.

But much more likely, we are dealing with a constant material that does not change its density depending on how much of it you use , and then the rotational inertia will change as the factor of change raised to the power of four.

Arvid

[ollopa]

OK, so substituting mass in terms of volume and density should lead to something like 1/2 * pi * rho * L * (Ro^4 - Ri^4) for the hollow cylinder, correct?

So you have said that 1/2 pi * rho * 40L * R^4 = 1/2 * pi * rho * L * (2.5^4R^4 - R^4)

40 = 2.5^4

And I find that 40 is approximately equal to 39.0625.

I believe you now. =]

That's probably the most fun I'll have all day :cheers:

[arvidb]

*lol* Yes, mathematics can be fun, can't it?

Actually I disregarded the fact that the nut is hollow:

"So you have said that 1/2 pi * rho * 40L * R^4 = 1/2 * pi * rho * L * (2.5^4R^4 - R^4)"

Dividing out all the uninteresting stuff , this becomes 40*r^4 = 2.5^4*r^4 - r^4 or 41 = 2.5^4. This is "even more approximate", but still valid in the context, I think.

What is really interesting is how little difference it makes that the nut is hollow (from a purely "rotational intertia" perspective, that is - I would hate to have to thread a solid nut onto a screw ). To have the same rotational inertia as a screw with radius r that is 40 times the length of the nut, a "solid nut" would need a radius 2.51r. A hollow nut (where the hole's radius = r) would need a radius of 2.53r. Not much of a difference!

Arvid

[fyffe555]

Which is why flywheels have the majority of their mass on the circumference

[ollopa]

*lol* Yes, mathematics can be fun, can't it?
Dividing out all the uninteresting stuff , this becomes 40*r^4 = 2.5^4*r^4 - r^4 or 41 = 2.5^4. This is "even more approximate", but still valid in the context, I think.
Arvid

On edit, nevermind -- I see the - 1 on the right that I missed =]

[duluthboat]

Before my head explodes, why don’t we throw out a few acceptable rules of thumb that can help guide the DIY’er in making a choice to spin the screw or the nut? Is this a possibility?

Even if the inertia is equal it is easier to control it with a spinning nut. This is offset by the added complexity of turning the nut. Where is the point that it is worth using a spinning nut or changing to a different drive type?

Gary