[skray775]

OK guys here we go.

The MD001 should arrive in a few days and I need to start thinking about the servo's drives. I have been really pleased with the DMM Tech AC servo's and I would like to go with them again.

The motors are rated at 3000 RPM and I would like to go with a 2 to 1 gear ratio on the X and Y and maybe 3 or 4 to 1 on the Z if it is needed.

Ball Screws would be from Linearmotinbearings on eBay.
items in linearmotionbearings store on eBay!
RM1605 (16mm) on X and Y
RM2005 (20mm) on the Z
Lead = 5... I assume that is 5mm or .20" per revolution?

Please correct my math if it is wrong.

I am thinking of using:

400W AC brushless servo motor (60EM-DHT-36)
Max 3000rpm
Rated torque: 1.27Nm(179 oz-in)
Peak torque: 3.6 Nm(507.6 oz-in)
Motor

At a 2:1 belt reduction that = 358 oz-in Rated and 1015 oz-in Peak at 1500 RPM. 1500 RPM x .20" = 300ipm

OR

Direct drive
900 Watt brushless AC Servo motor, NEMA34
Max:1080 rpm
Rated torque: 2.9 Nm(409 Oz-in)
Peak torque: 7.2Nm(1015 Oz-in)

1000 rpm x .20" = 200ipm

What do you think?

Edit:
My current machine uses the 300 watt motors at a 2:1 reduction. I can get up to 188ipm in rapids and that is plenty for me and I think it would be fine for the larger table of the MD001.

[skray775]

From the IH web site:

Our Turnkey 3 axis CNC Mill uses 410 in/oz peak motors on the X and Y axis.
The drive ratio is 4:1. So the ballscrew sees 1640 in/oz of torque.
Our math: 410 in/oz x 4 = 1640 in/oz

The Z-Axis uses a 648 in/oz x 4 = 2592 in/oz

So the 507 in-oz motors from DMM at a 3:1 ratio would give me 1521 in-oz.
With a 5mm (.20") lead screw that equals 3000 rpm / 3 = 1000 rpm.

1000 rpm x .20 = 200ipm rapids.

[dwalsh62]

From the IH web site:

Our Turnkey 3 axis CNC Mill uses 410 in/oz peak motors on the X and Y axis.
The drive ratio is 4:1. So the ballscrew sees 1640 in/oz of torque.
Our math: 410 in/oz x 4 = 1640 in/oz

The Z-Axis uses a 648 in/oz x 4 = 2592 in/oz

So the 507 in-oz motors from DMM at a 3:1 ratio would give me 1521 in-oz.
With a 5mm (.20") lead screw that equals 3000 rpm / 3 = 1000 rpm.

1000 rpm x .20 = 200ipm rapids.

Using the theoretical values is not considered good engineering practice, practical values are 0.92 of the theoretical value and is the preferred value that should be used when calculating power.

Also, the 5mm metric screws only have 0.196in lead, not 0.200in lead, a common mistake made by many it seems.


If you want a 0.200in lead screw then I recommend you buy an appropriate screw.

[skray775]

Very interesting... My current machine uses the 300 watt motors rated at 3000RPM and the same 5mm ball screw lead.

Max rapids are 187

200 IPM x.92 = 184

Using the theoretical values is not considered good engineering practice, practical values are 0.92 of the theoretical value and is the preferred value that should be used when calculating power.

Also, the 5mm metric screws only have 0.196in lead, not 0.200in lead, a common mistake made by many it seems.


If you want a 0.200in lead screw then I recommend you buy an appropriate screw.

[dwalsh62]

Very interesting... My current machine uses the 300 watt motors rated at 3000RPM and the same 5mm ball screw lead.

Max rapids are 187

200 IPM x.92 = 184

That is all wrong if you calculated 200IPM for a 5mm lead screw and used 0.200in as the lead value.

If you calculated anything based on 0.200in lead when using a 5mm lead screw your values will be incorrect, a metric screw with a 5mm lead is 0.196in.

You might wonder "Why is a Theoretical Factor (TF) used"?

If you build 5 machines using the same identical parts and then measured IPM and Torque you would find they are all different but should never fall below the advertised values which is calculated based on the Theoretical Factor (TF).

The standard design TF is 0.92% however some manufacturers use a lower value like 0.75% to de-rate the specifications to increase MTBF if the end user never pushes the product beyond the advertised specifications.

If you want to know what your motor torque requirement is you estimate your required torque then calculate it with the following (assuming a reduction of 3:1).
Design Torque Calculation: (Estimated Required Torque / TF) / Ratio
1100 / 0.92 / 3 = 398.5507oz/in

Some more calculations, let's start with a 400oz/in motor and a 3:1 reduction ratio.

Advertised Torque Calculation: (Torque x Ratio) x TF
(400 x 3) x 0.92 = 1104oz/in.
If you get a higher (measured) torque value them your $hit is better quality but worst case scenario will be this value which would be the manufacturers advertised torque specification.

Advertised IPM Calculation: ((RPM / Ratio) x Distance) x TF
((3000 / 3) x 0.196) x 0.92 = 180.23.
If you get a higher (measured) IPM them your $hit is better quality but worst case scenario will be this value which would be the manufacturers advertised IPM specification.

If you only get a max feed rate of 187 using a 3000RPM motor and a 3:1 reduction ration, you are above the theoretical feed rate 180 (always round values down) then I would say your machine is working optimally.

If you want to guarantee a 200IPM feed rate then design for (200 / 0.92 =) 217.3913 (or higher)

[skray775]

The machine was delayed and should arrive in a few days.

DMM has quoted $1100.00 for 3 axis drives, motors, power supplies, cables, estop, limit switches shipped. I will order that soon.

I will use LMB ball screws and order those after I draw up the mounting system.

I am going to use a Fabco 4 stage ait cylinder for a power draw bar but need to figure out the right one. I also need to choose a VFD for the 3 phase motor.

[Starleper1]

It's been discussed and argued about a million times on this forum, but me personally, I'd prefer a slower machine that is reliable, rather than a really fast machine that faults all the time.

I'd take torque over speed any day.

Your math seems right though.

[cyclestart]

Lead = 5... I assume that is 5mm or .20" per revolution?

5mm or roughly 5.08 rev/inch if setting up in standard measure. Always willing to tackle the easy questions LOL.

I have the same machine and may take it cnc someday. Best of luck.

[skray775]

To clarify the motor I listed from DMM is the most powerfull they list. So while I am not looking for 200ipm rapids it looks like with the math I did I have some overhead in case I need to go to 3.5:1 or 4:1 to get the torque I need.

If they had 800 in-oz I would just get those and not sweat it!

[sdubfid]

Ive got the 300w dmms and my friend has the bigger ones with internal gearing. He is really happy with them. So powerful he said he bent his router a bit with them haha.

I was at hui's house last weekend picking up some extra lmit switches and called him with questions a few times and he has always helped.

I used to have an md001 and I would just run the nema 34 1000w motors direct drive. That would be a nice simple setup.

[ninefinger]

To give you an idea of the minimum required to make the MD001 move, I used (and still am using) the 50oz-in continuous rated Nema 23 servos (350oz-in peak) on the X and Y axis with 4:1 reduction (ie 200 oz-in at the screw neglecting friction loss) and 5mm lead ballscrews form linearmotionbearing2000 on ebay. For the Z axis I started out with the same nema 23 servo and a 5:1 reduction (ie 250 oz-in at the screw). I was able to run it this way for a while but I had to have the acceleration dialed back on the Z, mostly due to tight spots at the top and bottom of the travel. I've since changed to a 119 oz-in (600oz-in peak) and I gained a little bit of acceleration.

Personally the 400w servos are what I'd choose, unless going with direct drive and larger servos can save enough money to pay for themselves (1 belt and 2 pulleys per axis plus additional brackets versus a coupler for direct drive). Looks like the encoder is suitable for direct drive (12 bit accuracy ~ 4000 counts/rev @ .2" rev = 0.00005" )
Checking out the pricing it looks like it can - the high torque servo & drives are only $48 more than the 400w servos & drives.

Mike

[jid2]

I've had similar thoughts about belt drive, inside or outside, where to connect the pulley etc. When I get around to it on my machine I plan on modeling the head and then playing around with a couple concepts in CAD to see which one to go with. Simple is typically the best, and also the most difficult to figure out how to execute well.

[skray775]

Here is a rough drawing of the top plate and pulleys'. Excuse the drawing I am not a 3d master! LOL

With the ratios as drawn I would have a 1:1 and 1:1.8.

2HP 3 phase motor.

3" SPINDLE - 3" MOTOR = 1.0 DRIVE RATIO = 1725 RPM
WITH A BELT LENGTH OF 20.16"
@ 50% under speed I would have = 862 RPM
@ 50% over speed I would have = 2587 RPM

2.5" SPINDLE - 4.5" MOTOR = 1.8 DRIVE RATIO = 3107 RPM
WITH A BELT LENGTH OF 21.9"
@ 50% under speed I would have = 1553 RPM
@ 50% over speed I would have = 4660 RPM

I am not that up to speed on VFD so I used a conservative number of 50%.

I am fine with those numbers so if 50% is to low everything above 4660 would be a bonus.

[pete from TN]

Try to make the pulley diameters as large as possible so the belt has the most purchase on the pulley at any one time. That way you can transmit the most power and torque from the motor. Other than that I think it looks pretty good. Personally while I can see the simplicity of using setscrews or whatever to secure the pulley to the spindle shaft using a coupler makes it a simple effort to remove and service the spindle as well as making it also simple to adapt an INTERNAL high speed spindle in the bore such as a pneumatic or electronic spindle. Just a thought.... Good luck with that bad boy!! Peace

Pete

[jid2]

Don't you need some low speed also? I know that I like to do stuff in stainless and you need something around or below 200 RPM it seems like.

[gd.marsh]

I went with 1:2 & 2:1 ratios .. That way you have the exact same belt length so very little motor movement is required to switch.
I also used a 2hp 3ph 1725 rpm motor like you, and I changed out the bearings in the spindle to MUCH higher quality A/C bearings.

At 2:1 (motor to spindle) ratio I get down to 280 rpm with pretty good torque & at the 1:2 ratio I get up to 8000 rpm.
I believe that's running the VFD from 20hz to 130hz if I remember correctly.
If you think about it the motor is only turning at 4000 rpm to produce 8000 at the spindle that way. Standard motor bearings are good to 5500 rpm.
I've been told by motor repair guys that in these general use low hp motors they all have the same bearings regardless of whether their rated at 1725 or 3450.

I just ran a part today using 1/8" ball endmill & the machine cruised along at 8000 rpm for several hours.
Spindle, Motor, & VFD were hardly warm to the touch when done.

[skray775]

I think I found your thread and was reading it last night. Very nice! I think I will use your ratio. So much good info in all the threads but it is hard to find because they are so large!

I went with 1:2 & 2:1 ratios .. That way you have the exact same belt length so very little motor movement is required to switch.
I also used a 2hp 3ph 1725 rpm motor like you, and I changed out the bearings in the spindle to MUCH higher quality A/C bearings.

At 2:1 (motor to spindle) ratio I get down to 280 rpm with pretty good torque & at the 1:2 ratio I get up to 8000 rpm.
I believe that's running the VFD from 20hz to 130hz if I remember correctly.
If you think about it the motor is only turning at 4000 rpm to produce 8000 at the spindle that way. Standard motor bearings are good to 5500 rpm.
I've been told by motor repair guys that in these general use low hp motors they all have the same bearings regardless of whether their rated at 1725 or 3450.

I just ran a part today using 1/8" ball endmill & the machine cruised along at 8000 rpm for several hours.
Spindle, Motor, & VFD were hardly warm to the touch when done.

[skray775]

Recalculated with the same size pulleys on both ends.

4.5" SPINDLE - 2.5" MOTOR = 0.555 DRIVE RATIO = 957 RPM
WITH A BELT LENGTH OF 20.16"
@ 50% under speed I would have = 478 RPM
@ 50% over speed I would have = 1435 RPM

2.5" SPINDLE - 4.5" MOTOR = 1.8 DRIVE RATIO = 3107 RPM
WITH A BELT LENGTH OF 21.9"
@ 50% under speed I would have = 1553 RPM
@ 50% over speed I would have = 4660 RPM

Pete,
Not sure how to keep the spindle so it can slide out quickly if I use a air cylinder type power draw bar. The cylinder pushes down on the draw bar nut but from my understanding you need some type of flange for the air cylinder mount to pull up against so it does not smash down the spindle bearings.

The spindle has tapered roller bearings so I am not sure if they can withstand the 2500 pounds of force needed from the air cylinder? If they could it would greatly simplify the design but my gut feeling it they can't.

My idea is to use the pulley as that flange. The mounting plate for the air cylinder would be something like the Tormach one pictured. Imagine the black plate under the pulley and the top of the air cylinder over the draw bar nut. When engaged it would pull up on the underside of the pulley and push down on the nut.

I will be drawing it up but I need to gather the dimensions of all the parts first.

[gd.marsh]

Please help me understand why matching pulleys require a different belt length? Or is your last post a typo?

[SCzEngrgGroup]

Recalculated with the same size pulleys on both ends.

4.5" SPINDLE - 2.5" MOTOR = 0.555 DRIVE RATIO = 957 RPM
WITH A BELT LENGTH OF 20.16"
@ 50% under speed I would have = 478 RPM
@ 50% over speed I would have = 1435 RPM

2.5" SPINDLE - 4.5" MOTOR = 1.8 DRIVE RATIO = 3107 RPM
WITH A BELT LENGTH OF 21.9"
@ 50% under speed I would have = 1553 RPM
@ 50% over speed I would have = 4660 RPM

You're doing something wrong in your calculations. If you're swapping pulleys between the motor and spindle, the belt length will be identical in either configuration.

Regards,
Ray L.