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  1. #21
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    Default Re: Ballscrew Sizing Options

    Quote Originally Posted by louieatienza View Post
    I think there may be something amiss in your graph. With the 5mm pitch, you show zero cutting forces at about 140IPM, or about 700rpm? That's at the peak of the power curve for the particular stepper. Using CNCRP's torque-speed chart and 48V PSU, at 720rpm there's about 283oz-in of torque. Plugging into Mariss' power equation, that's about 151W power. Or with a 5mm lead ballscrew about 690W at 141IPM. With a 15mm pitch, you'll be spinning at about 235rpm (to get 141IPM). Again, using CNCRP's chart, you have 566oz-in of torque, or about 98W power at 235rpm, or about 164W at 141IPM.
    Can you input an acceleration value into Mariss’ equations? I can tell you that everything changes when you use a different acceleration.

    Let’s elaborate on your example with the equations I used.

    Jmotor = 0.000180 kg*m^2 (assumed value for CNCRP, could not find a published value, so I used this approximation based on other steppers around this size)

    Jballscrew / Length = 0.00000031 kg*m^2 / mm (for a 25mm OD ball screw)
    Length of ballscrew = 1060mm
    Jballscrew = 0.0003286 kg*m^2 = 0.00000031 x 1060

    Mload = 400 lbs = 182 kg (half of 800 lb total for two motors)
    Lead = 5mm
    Efficiency = 0.9
    Jload = (0.0000000255 x Mload) x (1/efficiency) / (1/lead)^2 - You need to look at the links I posted to see where this comes from.
    Jload = (0.0000000255 x 182) x (1/0.9) / (1/5)^2 = 0.0001289 kg*m^2

    Jtotal = 0.000180 kg*m^2 + 0.0003286 kg*m^2 + 0.0001289 kg*m^2 = 0.0006375 kg*m^2

    At 720 RPM there is 2 Nm available from the CNCRP 960 motor. This corresponds to a speed of 141.7 in/min with a 5mm pitch screw.

    Motor Torque = Jtotal x Angular Acceleration
    Motor Torque / Jtotal = Angular Acceleration

    1.965 Nm / 0.0006375 kg*m^2 = 3081.8 Radians / sec^2

    Note there are 2 x pie (3.14) radians in 360 degrees.

    [3081.8 / (2 x 3.14)] x (5mm pitch / 1000) = 2.5 m/s^2 = Linear Acceleration

    Or in other words (rotations / second^2) x (meters / rotation) = meters / second^2

    2.5 m/s^2 / 9.81 m/s^2 = 0.25G linear acceleration

    So we said that there is 2Nm available at 720 RPM.

    2 – 1.965 = 0.035 Nm left over

    Back to the basic equation

    Force = (Torque x 2 x pie x efficiency) / (Lead x 10^-3)

    Where lead is in mm, Force is in N, and Torque is in Nm.
    Force = (0.035 x 2 x 3.14 x 0.9) / (0.005) = 39.9 N = 8.95 lbs

    8.95 lbs x 2 motors = 17.9 lbs Available cutting force. That’s exactly what I have as input for my graph in excel at 141.7 IPM for 5mm pitch.

    The one thing I’ve noticed is between the two links I posted for equations, one of them also divides the ballscrew inertia by the efficiency (not just the load inertia) but the other one does not. I may revise this. It would make the results a tiny bit worse. If you see any errors in the equations I used or in my math, please let me know.

    Also, as I’ve mentioned, this doesn’t include bearing friction or preload drag, probably one or two other things, but I think it's a good approximation to start with.

    Regardless of the acceleration setting used, 5mm pitch is just a really bad choice of ball screw pitch for my application. If I had a servo capable of spinning very fast but delivering low torque and a shorter longest screw length so critical speed wasn’t a factor it could be a different story.



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    Default Re: Ballscrew Sizing Options

    Quote Originally Posted by ger21 View Post
    But you can't find affordable 2515 screws?
    That is correct. And it is the reason I made this thread, to see if there was a supplier I overlooked.

    Quote Originally Posted by ger21 View Post
    And what do you plan on doing at 60rpm? Imo, you always calculate force at max rpm, which is where you need it.
    I just picked a convenient spot on the spreadsheet to use as an example to illustrate the inertia effects. I made the graph to look at the whole speed range.

    Quote Originally Posted by ger21 View Post
    If you run those at 70V, you probably can pick up 20% more speed. As a general rule, speed is proportional to voltage. Going up to 70V flattens out the torque curve.
    That's good to know.

    Quote Originally Posted by louieatienza View Post
    2516 is a common thread lead for rolled ballscrews.
    I found nothing for 2516 ball screw when I searched both ebay and aliexpress. I did find a bunch of 2516 lead screws when I searched through a search engine.

    The more expensive suppliers have pitches between 2510 and 2525, but that is outside of my budget. I did find a Chinese supplier that makes a ground 2016 ball screw. I don't know how much that costs, I need to find out. But with a ground ball screw (20mm OD), that would end machine way better for 15mm ID bearings than a rolled one would.



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    Default Re: Ballscrew Sizing Options

    When I googled 2516 ballscrews, it returns 0.25" screws with 16tpi.

    Gerry

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    Default Re: Ballscrew Sizing Options

    Quote Originally Posted by ger21 View Post
    When I googled 2516 ballscrews, it returns 0.25" screws with 16tpi.
    You appear to be correct. A 0.25" screw, they're just called 2516 lead screws for some reason. I stopped looking when I saw "lead" instead of "ball".



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    Default Re: Ballscrew Sizing Options

    I was in error.... the 16mm leads I saw were for 15mm and 16mm ballscrews. I actually have some NOS ground 1515 screws, C5 accuracy.



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    Quote Originally Posted by NIC 77 View Post
    Can you input an acceleration value into Mariss’ equations? I can tell you that everything changes when you use a different acceleration.

    Let’s elaborate on your example with the equations I used.

    Jmotor = 0.000180 kg*m^2 (assumed value for CNCRP, could not find a published value, so I used this approximation based on other steppers around this size)

    Jballscrew / Length = 0.00000031 kg*m^2 / mm (for a 25mm OD ball screw)
    Length of ballscrew = 1060mm
    Jballscrew = 0.0003286 kg*m^2 = 0.00000031 x 1060

    Mload = 400 lbs = 182 kg (half of 800 lb total for two motors)
    Lead = 5mm
    Efficiency = 0.9
    Jload = (0.0000000255 x Mload) x (1/efficiency) / (1/lead)^2 - You need to look at the links I posted to see where this comes from.
    Jload = (0.0000000255 x 182) x (1/0.9) / (1/5)^2 = 0.0001289 kg*m^2

    Jtotal = 0.000180 kg*m^2 + 0.0003286 kg*m^2 + 0.0001289 kg*m^2 = 0.0006375 kg*m^2

    At 720 RPM there is 2 Nm available from the CNCRP 960 motor. This corresponds to a speed of 141.7 in/min with a 5mm pitch screw.

    Motor Torque = Jtotal x Angular Acceleration
    Motor Torque / Jtotal = Angular Acceleration

    1.965 Nm / 0.0006375 kg*m^2 = 3081.8 Radians / sec^2

    Note there are 2 x pie (3.14) radians in 360 degrees.

    [3081.8 / (2 x 3.14)] x (5mm pitch / 1000) = 2.5 m/s^2 = Linear Acceleration

    Or in other words (rotations / second^2) x (meters / rotation) = meters / second^2

    2.5 m/s^2 / 9.81 m/s^2 = 0.25G linear acceleration

    So we said that there is 2Nm available at 720 RPM.

    2 – 1.965 = 0.035 Nm left over

    Back to the basic equation

    Force = (Torque x 2 x pie x efficiency) / (Lead x 10^-3)

    Where lead is in mm, Force is in N, and Torque is in Nm.
    Force = (0.035 x 2 x 3.14 x 0.9) / (0.005) = 39.9 N = 8.95 lbs

    8.95 lbs x 2 motors = 17.9 lbs Available cutting force. That’s exactly what I have as input for my graph in excel at 141.7 IPM for 5mm pitch.

    The one thing I’ve noticed is between the two links I posted for equations, one of them also divides the ballscrew inertia by the efficiency (not just the load inertia) but the other one does not. I may revise this. It would make the results a tiny bit worse. If you see any errors in the equations I used or in my math, please let me know.

    Also, as I’ve mentioned, this doesn’t include bearing friction or preload drag, probably one or two other things, but I think it's a good approximation to start with.

    Regardless of the acceleration setting used, 5mm pitch is just a really bad choice of ball screw pitch for my application. If I had a servo capable of spinning very fast but delivering low torque and a shorter longest screw length so critical speed wasn’t a factor it could be a different story.
    If we were drag racing identical cars, and I was in first gear, and you were in fifth, who do you tbink would accelerate faster from stand-still?

    I may have caught an error in the math, but I'm at work now and can't sit down and check. what happens when you haven't been in a college physics class in about 25 years. I vaguely remember anything then and have to open my old textbooks.



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    Default Re: Ballscrew Sizing Options

    Quote Originally Posted by louieatienza View Post
    If we were drag racing identical cars, and I was in first gear, and you were in fifth, who do you tbink would accelerate faster from stand-still?
    I think for ballscrew selection we are looking for what performs better at higher speed but still has enough to do what is needed at lower speed.

    Quote Originally Posted by louieatienza View Post
    I may have caught an error in the math, but I'm at work now and can't sit down and check. what happens when you haven't been in a college physics class in about 25 years. I vaguely remember anything then and have to open my old textbooks.
    If you do find one or even if you're unsure if you've found one I'd be happy to hear about it. Better to learn that you've made a mistake in your calculations before you buy stuff . I won't be offended

    The only thing I've found so far is the efficiency question I already mentioned between the two sources, but that amounts to a very small change in the figures. I had a look at a few more sources today. I think some literature doesn't include efficiency in the acceleration equation because they expect you to include friction torque and preload torque values instead of efficiency when calculating acceleration torque.

    Hiwin is an example of this. They also use different units and calculate the ball screw inertia each time (I took mine from a table that lists the inertia values for rolled ball screws of different sizes), but otherwise the equations appear to be the same. Probably the best single source I've found so far.

    http://www.hiwin.com/pdf/ballscrews.pdf

    Pages 26 and 27 have the info.



  8. #28
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    Default Re: Ballscrew Sizing Options

    Quote Originally Posted by Jim Dawson View Post

    My ball nut assembly is about 4 inches long. It's an adjustable double nut system. I haven't actually ever checked the backlash, I should do that one day. I've not had any problems cutting circles, they come out round so that would indicate any backlash is minimal.
    Nic, you asked about backlash. I have had an opportunity to check the backlash and have concluded I need to work on the Y axis. I wrote some new software and had the error correction turned off while re-tuning the steppers to get the steps synced with the encoders. I found that the Y axis has 0.0084 inch backlash. The X is 0.0012 inch. I never noticed because my error correction works, pretty much proves the effectiveness of having linear encoders on the load. With the error correction turned on, can do a 96 inch, 600 IPM rapid to the target position and hit it +/- 0.0002 every time, in either direction. Not bad for an old worn out router.



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    Default Re: Ballscrew Sizing Options

    Quote Originally Posted by Jim Dawson View Post
    Nic, you asked about backlash. I have had an opportunity to check the backlash and have concluded I need to work on the Y axis. I wrote some new software and had the error correction turned off while re-tuning the steppers to get the steps synced with the encoders. I found that the Y axis has 0.0084 inch backlash. The X is 0.0012 inch. I never noticed because my error correction works, pretty much proves the effectiveness of having linear encoders on the load. With the error correction turned on, can do a 96 inch, 600 IPM rapid to the target position and hit it +/- 0.0002 every time, in either direction. Not bad for an old worn out router.
    Ooh, that's interesting. Thank you for the reply! I think that if I use gear reducers and 2525 ball screws that I will need to use backlash correction. It's really hard to tell what things are like if you don't have the opportunity to feel them in your hands.

    How are you going to get rid of that backlash? You mentioned the double nuts you have are adjustable.

    I do wish I had some of your skills to write my own control software. Well, I'm sure I have some of them. But the pieces of knowledge I don't have make all the difference. Of course, if I did, I would be looking to sell it commercially. May I ask, what language it's programmed in? Have you made a nice GUI or is it mostly an interface that only you understand?

    You're a smart cookie, I do appreciate your input, so thanks again for posting in my thread!



  10. #30
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    Default Re: Ballscrew Sizing Options

    Adjusting the backlash out should be pretty straight forward. Getting to the ball nut block is going to be a bit of a pain. It's of course buried under the table, under 3/4 inch thick plastic cover over a 1 inch thick aluminum table. I'm not exactly sure how the plastic is attached to the aluminum. Once I get to the nut it should be pretty easy to adjust. The two ball nuts screw into the housing so just have to loosen the set screws and rotate one of the nuts to set the preload. I suspect I'll also need to replace the bearings, not a huge deal but tedious. I've re-stuffed a few ball nuts. I'll start another thread when I do that and document the process.

    As far as software, I started programming when the Radio Shack TRS80 came out back in the late '70s. I've always found it fun. I started programming industrial systems back in the late '80s The current software is written in VB.net, using Visual Studio 2015. The motion controller software is written in Galil DMC code. It's only taken me about 68 years to acquire my skill set, been at it for awhile.

    Again thank you for the kind words

    Here is the latest screenshot I just grabbed minutes ago. I'll let you decide if you think it's useful.



    Attached Thumbnails Attached Thumbnails Ballscrew Sizing Options-screenshot-jpg  
    Last edited by Jim Dawson; 06-13-2017 at 01:54 PM.


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    Default Re: Ballscrew Sizing Options

    To give you guys a bit of an update,

    Yesterday I went through my ball screw calculations spreadsheet, and added all the specific math for using gear reducers with ball screws. I have done this previously for rack and pinion systems using gear reducers.

    It seems my initial suppositions (this is not the graph I posted, or the sample math I did in post #2, there are no reducers used there) of using planetary gear reducers and higher pitched ball screws as a viable option were erroneous.

    The inertia as seen by the motor, of the ballscrew, and the load, are reduced by (1/Gr)^2 when using a reducer, however, now you have to increase the angular acceleration of the motor by a factor equal to the Gear ratio to achieve the same linear acceleration of the system, and you have to add the inertia of the planetary gear into the equations.

    I just didn't want anyone to read this thread, and think, oh, that's a good idea, based on something I had previously mentioned as a possibility.



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    Default Re: Ballscrew Sizing Options

    Quote Originally Posted by NIC 77 View Post
    To give you guys a bit of an update,

    Yesterday I went through my ball screw calculations spreadsheet, and added all the specific math for using gear reducers with ball screws. I have done this previously for rack and pinion systems using gear reducers.

    It seems my initial suppositions (this is not the graph I posted, or the sample math I did in post #2, there are no reducers used there) of using planetary gear reducers and higher pitched ball screws as a viable option were erroneous.

    The inertia as seen by the motor, of the ballscrew, and the load, are reduced by (1/Gr)^2 when using a reducer, however, now you have to increase the angular acceleration of the motor by a factor equal to the Gear ratio to achieve the same linear acceleration of the system, and you have to add the inertia of the planetary gear into the equations.

    I just didn't want anyone to read this thread, and think, oh, that's a good idea, based on something I had previously mentioned as a possibility.
    Theoretically maximum acceleration is achieved with a 1:1 ratio between motor inertia and load inertia. Thus one can use a reducer to bring the inertia of the load (at the reducer input) closer to that of the motor. Two issues - if the load inertia is close to the motor inertia, adding a reducer could mean more losses from the reducer efficiency and inertia offsetting any gains from using the reducer. The other issue is that the reduction needed could be so large that it would require the motor to spin well past its (or the reducer's) capabilities - which brings the issue of cost in the use of a higher speed motor and capable reducer.

    The acceleration of the motor is already pre-determined by it's torque and inertia. The amount of torque needed to move the load is determined by the final output ratio (disregarding inertia and efficiency for the time being). So the less torque needed to move the load (i.e. the higher the final gear reduction, as the screw acts as a gear reducer), the higher its acceleration should be, at the cost of top speed.



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    Default Re: Ballscrew Sizing Options

    Quote Originally Posted by louieatienza View Post
    Theoretically maximum acceleration is achieved with a 1:1 ratio between motor inertia and load inertia.
    http://www.diequa.com/download/articles/inertia.pdf

    The optimum gear ratio = Sqrt(Jload/Jmotor) is intended for the purposes of limiting overshoot and undershoot in the system. It doesn't give you the maximum possible acceleration, but it may give you the best acceleration in a sense that you eliminate over and undershoot. The inertia mismatch is small in the ballscrew systems I'm looking at, so I don't think it's an important factor.

    Quote Originally Posted by louieatienza View Post
    Thus one can use a reducer to bring the inertia of the load (at the reducer input) closer to that of the motor.
    If your load inertia is high, and you have a significant mismatch, yes, you are correct, you can use a gear reducer to decrease the inertia of the load as seen by the motor and create a system that is less prone to over or undershoot.

    Also, it is assumed that:

    The reducer input inertia is considered to be a part of the motor inertia.

    The reducer output inertia is considered to be a part of the load inertia.

    This is according to the article I posted a link to.

    Usually the planetary manufacturers only give the inertia as seen by the motor which I consider to be the input inertia. For a planetary of multiple stages, I don't know how you'd figure out the output inertia of the gears, as you won't find it in the manufacturer's literature. That being said, the effect is reduced by (1/GR)^2 so it doesn't affect the system as much as the input inertia does. If you are using a belt driven system with two gears for a reducer you can calculate the inertia for each gear and include the gear that is attached to the motor on the input side and the gear that is attached to the load on the output side.

    Quote Originally Posted by louieatienza View Post
    The acceleration of the motor is already pre-determined by it's torque and inertia.
    The maximum acceleration that the motor can achieve is determined by it's available torque at that rotational speed, and it's reflected load inertia.

    For example, the reflected load inertia is

    Jreflected load inertia = Jmotor armature + Jreducer input + (Jreducer output + Jballscrew + Jload)*(1/GR)^2

    Motor Torque Used = Jreflected load inertia x Motor Angular Acceleration Required

    For a rack and pinion system, you'd replace Jballscrew with Jpinion and there is a different calculation for Jload.

    So if you increase any of those values, you have less acceleration possible from the motor for a given torque.

    The thing that you really have to watch out for is that

    Motor Angular Acceleration = Ballscrew Angular Acceleration x The Gear Ratio

    So that your motor must now accelerate faster by a factor equal to the gear ratio in order to achieve the same linear acceleration at the output side.

    How the angular acceleration is converted into linear acceleration depends on the lead of the ball screw. So in essence what I'm saying is that you need to be careful about using planetary reducers on ball screw driven systems if acceleration is important to you. I'm talking about using 3:1 or 5:1 reducers with fairly large input inertia. If you increase the lead or use a rack and pinion system that will have a larger need, this negative effect is diminished. You just need to be cautious and do the specific math for the system to see what will work best. That was my caution.



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