8/6/2018 9:32 AM  
Posts: 33 Rating: (1) 
Hi experts, I am new to sinamics and struggling with These situtations. I am using a double Motor module with CU 320 for a rotating table application . I have a doubt on the LU implementation. I am using TIA V14 for comissioning the drive. on Motor there are two Revolution data one is n max and n Nominal , which one should I consider while entering the Motor revolutions data for the Drive? and there are two gear boxes one is a closed gear box which has 4500 rev/min and the other is a open gear box section with 24 and 12 teeth , so how do I calculate the total data to be entered? I want the Resolution of 0.1 degree , so please tell me how to calculate LU 360/0.1° ? is this correct Thank you 
8/6/2018 10:18 AM  
Joined: 10/11/2006 Last visit: 11/8/2024 Posts: 5935 Rating: (815)

Hello, first we need to clarify on which system you are working. Do you have a SINAMICS S120 / G120? Do you want to use the function module "basic positioner" (EPOS) on the drive? I am asking, because you just wrote that you use TIA V14 and it could also be that you use technology objects. For the gear ratio: For more than one mechanical gear ratio between motor encoder and load side you need to multiply the two gear factors. The problem is that you wrote the first gear is 4500rpm. This is not a value for a gear ratio. The second value (24 to 12 tooth) is a correct figure for a gear ratio. For example the other gear ratio would be 100 to 12, then the gear ratio in sum would be 2400 to 144 (but this is just an calculation example as we do not know your gear ratio from the first gear!). So you need to find out the correct gear ratio of the first gear. The best figure is to have the exact number of tooth! (sometimes there are rounded gear ratios on the faceplate, which lead to a continuous error in position after a certain number of revolutions) For LU per load revolution setting: If you have a round axis and you want to have your position in 0.1 degree resolution, then you are right and need to set it up as follows: LU per load revolution = 3600LU. It means that 1LU is equal to 0.1 degree. If you wanted to give a position setpoint of 90degree you would need to enter 900LU. Be careful: a resolution of 0.1 degree is not really high. If your encoder is good enough I would recommend to select a resolution of milli degree (LU per load revolution = 360 000LU). Then you have a much higher resolution and a higher positioning accuracy. The rule by thumb is always to select the resolution 5 to 10 times higher than the wanted positioning accuracy. 
With best regards, 

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8/6/2018 10:49 AM  
Posts: 33 Rating: (1) 
Hi expert, Thank you for the quick Reply that clears most of my doubts. yes I would like to use Basic positioner (EPOS). I would check up with the gear Ratio of the first one So when I have to enter in load Revolution as mentioned in the example as 2400/144 = 16,666 and for the Motor Revolution is it Nmax or Nn . thank you for your time. 
8/6/2018 12:33 PM  
Joined: 10/11/2006 Last visit: 11/8/2024 Posts: 5935 Rating: (815)

Hello, don't mix up the setting for LU per load revolution (p2506) and the setting for the gear ratio (p2505 and p2504)! The LU per load revolution just defines your position normalization. In other words: with LU per load revolution you select your user defined position unit (degree or millimeters or inches or whatever you want) and the position resolution (1LU means 1 milli degree or something like that). The gear factor needs to be filled in just that the controller knows from your position how far and how fast the motor has to run because one load revolution is not exactly one motor revolution due to the gear box. You just enter the values for the tooth of the gear in p2504 and p2505. You do not type in a ratio like 16.666! > this is calculated internally by nominator / denominator of the gear ratio. 
With best regards, 

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8/6/2018 4:10 PM  
Posts: 33 Rating: (1) 
Hi, Thanks for the Reply , So I Need a Little bit clarification regarding a typical example I though about example I have a gear box with gear Ratio i = 10.0 and I have other gear box with 25 teeth and by each teeth the load moves a 75mm. so can I say the LU is 75*25*10*10mm (10mm)=1LU 
8/7/2018 7:18 AM  
Joined: 10/11/2006 Last visit: 11/8/2024 Posts: 5935 Rating: (815)

Hello, you say the "other gear" has 25 teeth....but this is not a figure for a gear ratio. A gearbox has a certain number of teeth at the input and at the output stage. I think you have a mechanics that is changing a rotary movement (motor) to a linear movement (load). So it is not a real gearbox, it is a roll with 25 teeth, moving the load forward and backward like a winch. Attached please see the quick drawing. From your description I think you have something similar to that. So for my example I explain how to set up the parameters: First I want to point out that the load moving is not in degree, because it is no round axis. I would say it is a linear movement so we need a positioning unit in mm, µm, inch or something like that. In this case you say the load is moving 75mm each tooth and there are 25 teeth on the roll. So the load is moving 75mm * 25 teeth = 1875mm per load revolution. To have a good positioning accuracy I suggest to select µm as positioning unit. So we calculate the LU per load revolution p2506 = 1875mm * 1000 = 1 875 000 LU per load revolution. 1LU equals 1µm of load movement. Then we take into account the gear ratio of i=10 for p2504 and p2505: p2504 = 10 p2505 = 1 (take care that the gear ratio is not rounded! the most exact way is always to know the number of teeth for gearbox input and output. Values on faceplate can be rounded, e.g. 17,53 which is not exact in the end and will lead to continuous position deviation after a certain number of rounds). Hope this comes close to your mechanics and I am not totally wrong. Let me know if this helps. 
With best regards, 

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8/7/2018 7:54 AM  
Posts: 33 Rating: (1) 
That is exactly what I am looking for. Thank you 
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