5/26/2011 12:24 AM | |
Posts: 20 Rating: (0) |
Dear Sirs: Sorry to interrupt you, butI need to know where can I find some help about Chain calibration in the BW500. I have BW500 with three idlers (6 load cells), each load cell 50 lb. I tried to do the chain calibration, but i could not, so I just did the manual span adjust (test material). I bought the chain thinking that this would be much more simple to calibrate, but it did not work. The chain specifucation says: 7,62 Kg/m, 2,134 mt. If you could help em in this, I will appreciuate your help. Best regards. hp. Splitted from Advice for Siemens Load Cell Calibration |
Last edited by: O_Moderator at: 5/26/2011 11:55 AMnew subject after splitting |
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5/26/2011 9:34 AM | |
Joined: 3/21/2006 Last visit: 9/10/2024 Posts: 10228 Rating: (1181) |
Hello, Did you check in manual the necessaries step for its calibration? http://support.automation.siemens.com/WW/view/en/18252870And then to discuss eventually if some step is not proper done or is in fault. Best regards, Hristo MIhalev |
5/27/2011 5:06 PM | |
Joined: 5/28/2008 Last visit: 9/21/2024 Posts: 4515 Rating: (839) |
Hello hhpp, Please have a look on this video: Milltronics BW500 configurationHope it helps. |
Ayman Elotaify |
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5/30/2011 11:21 PM | |
Posts: 74 Rating: (3) |
Hello hhpp Re test chain - you may be just able to purchase the extra length and connect is. I think it will be hard to return the piece you have. You'll have to remove the shackles for one end of the existing piece and one end of the new one. Please make sure your wiring is as the drawing attached shows. Confirm that the BW500 indicates the actual speed of the belt. If in error adjust the speed constant so it reads correctly. Make sure the shipping stops are removed.Now you will do a zero. Stop the belt. Apply the correct length test chain making sure it is secure. With the belt running perform a series of span adjustments.There should be very very small or no errors between the span checks. Remove the chain. Recheck the zero. There should be small or no error. Any large variations in error would most likely be due to conveyor, idlers or alignment. To prove the calibration, you should then perform at least three material tests. Each test should be about 6 minutes in length. Catch the material and compare. Tests 1 and two to check for consistency. Then make the adjustment. Test three to verify the adjustment. Then factor the test chain value to the correction. Did you get test weights with your scales? Regards AttachmentA5E0237471901.pdf (483 Downloads) |
Last edited by: Toppie at: 6/2/2011 3:05 AMI should add that most clients will use a test chain when they can''t perform material tests. The test chain will compensate for the initial belt tension, where a test weight calibration will not. If you can perform a material test a test weight calibration with material test would almost always be a better choice that a test chain calibration without a material test. The material test compensates for the tension of the loaded belt. Of course it is important to factor (adjust) the value of the calibration standard (weights or chain) to the tests. Also, it is imperative to begin with a good installation on a good, well maintained conveyor in good condition. [ |:) |
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5/31/2011 2:18 AM | |
Joined: 5/28/2008 Last visit: 9/21/2024 Posts: 4515 Rating: (839) |
Hello hhpp, As you have 3 MSI idlers, so you need a very high precsion. From data sheet it is +/- 0.125%. So, there are some points we have to take into account to achieve your target. 1. Installation
2. Wiring
3.Paramteterization
4. Balancing
5. Zero
6. Span
7. Material test
Hope this helps. |
Last edited by: Ayman Elotaify at: 5/31/2011 2:24 AMAyman Elotaify |
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6/2/2011 6:28 PM | |
Posts: 20 Rating: (0) |
Hello partners, and thanks again.
First of all the wiring is according to the drawing (Siemens dealer gave me the same drawing). The parameters speed are as follows: P14= 1,156 m/s; P15-01=236, 1952; P18 -01 oscilating between 1,103 to t,176 m/s, I think tha´s fine. Stops removed I do not know what to think, the test chain is really expensive, test weights are much cheaper ( test chain like 2000 $, test weight 60$ each). If I decide to buy test weight, what must be the weight?, how many pieces? At the beginning I used the test chain (the smaller one) for span calibration with the P017=7,62 Kg/m. In those time I think that everything was fine, I did three times span and the counts were similar. Then I decided to run the conveyor with material to the mill, the results were really bad because it was reading much more rate than the real rate ( for instance reading 50 t/h when there were just 20 t/h). After this problem the only Think I could do was the test material P019. During the test I found out that the BW500 was reading more than the real rate. So doing and doing the material test I finally found that the P017 was really high. With P017=4,47 the BW500 was just fine ( for instance 500 Kg load(real), ( first test totalizer 502, totalizer=409Kg and so on). Since then I have been working like that, but I do not know if What I did was really right.Is it really necessary to do 6 minutes of test?, what weight should I use?I been doing the test with 400 Kg (blinker), I mean putting load over the belt and catching to compare the total weight with the totalizer ( 5 or 6 tests) . the problem is that I have been doing this in T units, so the totalizer shows 0,400 t. Is it possible to do the test material in Kg units, and then change to T? I do not remember that I said 60 tph (the max. is 90 tph) please give a light. You mean that I could do the balancing with my test chain? How? Yesterday I saw another problem, Let´s say that the belt conveyor is running with 40 tph, then suddenly the feeder stops and the belt conveyor is running empty, but the BW500 does not show “0” immediately , it takes like 5 minutes go to “0”. How could I solve this problem? Thanks again Hp. |
6/2/2011 8:14 PM | |
Posts: 74 Rating: (3) |
Hi hhpp Please check your private messages. I understand your comments about the high cost of the belt scale (and storage reel) and the difficulty of doing long material tests. You have to consider the value of your material being weighed to the importance and cost of an error. A test chain and real may pay for itself in a month if it provides an improvement of only 1/2% depending on the value of your material and the production. You also have to ask youself if you need absolute accuracy or just repeatability. For most processes, repeatability is absolutely necessary whereas accuracy is only secondary. Repeatability is delievered through good equipment and good installation. Accuracy is determiend by calibration and verification. I did a quick calculation with values taken and inferred from this thread. My calculation says that you can calibrate with 3 -18 lb (8.16 kg) test weights, one for each suspension. You cam make your own and determine the starting value as follows. Assume your full scale loading (Max Rate/Belt Speed)is 13.89 kg/m and you have a weighlength (idler spacing) of 1028mm. This represents a weight of 13.89 X 1.028 m = 14.28kg. You would calibrate at about 60%, so a test weight of 14.28 X .6 = 8.57kg would allow you to calibrate at 8.57 / 1.028 = 8.34kg/m Or a 10 kg weight would allow you to claibrate at 10kg / 1.028 m= 9.73 kg/m. Note you would neet 1 weight per scale and the weights should be identical. This calibration (if a good installation) should give you stated repeatability. To confirm the calibration, you can run representative material over the scale and weigh it. If six minutes is not practical, 1 minute will have to suffice. If you cannot do this and you have small sized material, you can run the conveyor and while it is showing a constant Load, stop it and clean all of the material off the belt for (as an example) 10 m and weight it. The mass /10 will give you the loading in kg/m. Compare. Yes If doing the running material tests you should adjust your totalizer resolution to as fine as allowed. BTW I did all the calculations at a full scale of 60 tph since you were running at 40 tph. For 90, the same loadcells are calculated. Your Design Load 100% will become 20.83 kg per m. and using 2 8.2 kg weights per suspension will allow you to calibrate at 15.89 kg/m. I hope this helps Toppie [ |:) AttachmentCalcs.zip (452 Downloads) |
Last edited by: Toppie at: 6/2/2011 8:21 PMSorry - To balance the loadcells you need equal size test weights. One for each suspension Balancing is recommended for maximum accuracy as it aligns the gain curves of each load cell. They are pretty close out of the box. If you are taking 5 minutes for the Rate to go to zero on rate, you probably have your display damping P080 set way too high or if viewing Rate on mA P220. A value of 10 is a god starting point. Note that if viewing load, Load will report anytime there is something on the belt. |
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6/4/2011 3:18 PM | |
Posts: 20 Rating: (0) |
Hello and thanks again. I think there is a missunderstanding, sorry. I did not say taht the belt scale is expensive, what i meant was that the test chain is more expensive than a test weight. I think the best way to adjust the scale is with atest material. Now I just can do the SPAN with the test chain or the test weight, am i right? If i am rightthe best solution is buy the test weightsdoes not it?. Or may be there is another option to do the correct span? I already check theP080, P220, you were right it had a numerical value of 90, so I change it. My question now, during this time that the scale without load gotofrom 45 to 0 (belt scale empty) the totalizer is still totalizing? During my manual SPAN adjusnt with test material I did like 10 tests with an error of 1,5% average. there is repeatibility ( 404, 3998, 402, 402, 404 and so on), so I am fine with that, but I would like to improve that errorand be sure that the calibraation is well done. Now I amgoing to change the units to Kg, just for calibration to improve the calibration, thenI will chagethe unis to tph fro normal operation. Best regards. hp. |
6/5/2011 4:46 PM | |
Joined: 5/28/2008 Last visit: 9/21/2024 Posts: 4515 Rating: (839) |
Hello hhpp, For some applications, you do not have all choices for calibration procedures. We can calibrate using:
Not all calibration have spce enough to install the chain storage wheel for chain calibration as you know. This introduction just to tell that we can not say that test weight is always the best way for calibration. Because in many cases it is the only way if we excluded the E-Cal.
Did you made a three succeful zero calibration without diviation? Please tell. And please check the allignment. Also please measure the belt length and enter an accurite value to P016. I mean by belt length a complete one revolution of the belt. When you enter this value accuite, and start running the belt, then start zeroing, the zero calibration start measuring the load cell count based on forces appliend on them for a complete one revolution of belt scale. If there's extra force applied due to any reason, this force will be repeated each cycle. So when you make zeroing, you tell the integrator that this force will come to you each cycle, so it will be neglected during normal operation. For this reason, belt length is critical.
Value of 3998 is far awy of other values!!! I think it is 399.8 Am I right?
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Ayman Elotaify |
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6/6/2011 4:07 PM | |
Posts: 74 Rating: (3) |
Hello hhpp My apologies. I meant to say I understand your comments about the high cost of the test chain... rather than"I understand your comments about the high cost of the belt scale (and storage reel) and the difficulty of doing long material tests." Your data supplied says your Design Belt Speed (P014)is 1.156 m/min. Should this be 1.156 m/sec? The variation in your Rate display from 0 - 45 tph is a concern because this represents 50% of your design capacity. When I see this I would normally suspect a conveyor with substandard idlers, a belt pieced together with old and new pieces and a bad alignment. Amplifying the heck out of the loadcell signals can also do this which can happen when the shipping stops are not released from the scale suspensions. So too, can locatingthe scalein a curve. The repeatability of your tests looks not too bad if I assume the 3998 is in fact 399.Your resolution now is 1/400 but you are only able to achieve 5/400 accuracy.With such a scale, I would expect much better. How are you doing these material tests and are the samples representative of normal loading? Remember, because of (changing) belt tension effects (with loading), a belt scale is linearover the range of.25 to 1.25 of design. The scale should continue to be repeatable but the accuracy at low loading tends to error negative. That is, the scale will weigh light (due to the belt holdingup material). In case you haven't seen this video, it's worthwhile watching. http://www.youtube.com/watch?v=8EXnV8zM88E&feature=channel_video_title As well, our installation guidlines: http://support.automation.siemens.com/WW/view/en/18246952 Cheers [ |:) |
6/15/2011 5:50 PM | |
Posts: 20 Rating: (0) |
Hello and thanks again. I had a problem with the speed sensor last days so i been working on that. Now itis working fine. Going back to the the post: Actually I made like four zero calibration, two of them with 0 %deviation, the other two asmall diference. I put the leght asaccurate as i could.. What do you mean by "neglected during normal operation"? Actually the real value was 398 cause the units (tn). It was my mistake when i wrote 1.156 m/min, Actually taht value is 1.156 m/sec. the shiipins stops are out. The scale is not instalating in a curve. the material test: P019 option 2. My samples are bags of 20 and 30 Kg. I run the belt then 4 idlres before the scale I put the bags one by one as fast as i can, then for idlres after the scale i take out the bags as fast as i can. Finally I check the totalizer if the deviation is smaller than before I accept otherwise not. Please tell me how do you know that the scale is lineaer from 0.25 to 1.25?. Thanks agains. hp. |
6/15/2011 6:55 PM | |
Posts: 74 Rating: (3) |
Hello hhpp I was wondering where you had been? I shall embed my responses into your edited message below:
Good luck and tell the boss you will need a long shower after doing those belt cuts. Toppie [ |:) I should have added an important point. A belt cut is only practical if the material size is suitbly small. I would say the maximum practical material size would be about 1/2" (12 mm) due to the difficulty of handling. |
Last edited by: Toppie at: 6/16/2011 6:22 PM |
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6/15/2011 7:47 PM | |
Joined: 5/28/2008 Last visit: 9/21/2024 Posts: 4515 Rating: (839) |
Dear hhpp, When you run the belt conveyor for one revolution on the belt scale, if you monitor the mV output signal measured by the load cell, it will not be constant. Thus due to many factors.
As this forces happen when the belt is clean, and as you have entered the length accuritlly as you could, so, when you make zero calibration, the integrator measure the zero signal that read by the loadcells during complete one revolution. It is like as you draw a treand and take the average of it. Then each time there is a material comes over the beltscale, then the load that will appear on the integrator is the current load sensed by the loadcells - the average value of zero calibration which calculated during zeroing. So all un-expected forces will be taken into consideration whether it was positive or negative.
Please explain your question.. |
Last edited by: Ayman Elotaify at: 6/15/2011 7:47 PMAyman Elotaify |
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6/16/2011 11:36 PM | |
Posts: 20 Rating: (0) |
Hello : First of all. just summarizing. WhatI understand so far, is that I need to buy the test weights, because the test chain is not prácical. You mean that with the test weight the calibration would be more repeteable and acurate. Application: Feed load to the SAG mill. What I am worry about is that last months there is a big difference between what the total BW reports and the operations guys (like 15 tons in a month), i think that there is anothjer problem (samples), but I want to be sure that I am doing the job the best I can. I think that in the future I will need to demostrate that the BW is working fine. I do not understand "do a belt cut" How can read load from the display? Measure 6 to 10 idlers centered or contered over the belt scale? Substract from what? Factor test weight to do this correction? the material that we feed has sizes from 10 mm to 250 mm (fine and coarse) there is no homogeneous load. (The counterwight of the belt is not what I want, but have to deal with it). Another test That I been doing so far is run the belt at normal rate, then stop the conveyor, take out like 10 meters of load from the conveyor. measure the load in a scale . With thelenght (10 meters) and the weight I can get the rate, so i compare this rate with theBw rate. I would like do what you suggested, please explain that to me. I am not sure what you mean load or rate(rate is common i think, but load??). I want you to ex´plaim to me the the linearity. Best regards. hp. |
6/17/2011 3:13 AM | |
Posts: 74 Rating: (3) |
Hi again hhpp Actually, If you spend the money on the correct length test chain, you would save yourself allot of work and allot of argument with operations and the metallurgical guys. What I'm hoping to do is prove your test weights so that you can get some sort of calibration. With your test chain so short, you are not loading all the suspensions equally so you are not getting an accurate calibration. The fastest thing you can do is make some test weights. If I assume your test chain is a correct load but incorrect length you will need test weights weighing 7.62 kg/m X 1.2 m (idler spacing) = 9.144 kg. Not practical, eh? So get the machine shop to make you 3 weights, each weighing 10 kg. Or 6 weights each weighing 5 kg, Have them weld a little hook on each so you can hang the weight from the test weight bar on the MSI. The weights need to be all the same weight, within 1%. Assume you will hang 10 kg from each suspension and your idler spacing is 1.2 m. Your Test Weight value will be 10 kg / 1.2m = 8.33 kg/m (See how the units work out?). A Belt Cut is just as I described. You cut a representative sample of material over a set length. Weigh it and divide by the length of the cut. You get a result of kg/m. This is referred to a Belt Loading. The BW500 has two displays Rate and Total is one. Load and Speed is the other. When you view Load and Speed, the Load is reporting what the scale sees. Rate = kg/hr or tonnes per hour. Load = kg/m. So a belt cut gives you a load that you compare with the Load display on the BW. Most likely, it will not read perfectly zero after you remove all of the material. Variations in belt thickness and idler roundness will cause this. In practicality, your material sizes are too large for an accurate belt cut. My recommendation is to get the test weights made and do the calibration to the test weights. Then collect data for a few weeks and compare to what the operations/metallurgical guys are telling you. There should be an average error in what they tell you. Input this as the correction. But remeber, the Belt Scale is monitoring continuously. The metalurgical balance is calculate based on samples taken what, 1,2 maybe 4 times a shift max. I don't know what your production is so 15t in a monyh maight be large or might be small as a percentage. So if you have now made a correction to the test weight calibration and then you calibrate to the test weights again, you will re-introduce that same error. You need to adjust the value of the test weights to the correction. This is called factoring and is explained in the BW manual. You may still want to consider the test chain option after you get a good calibration with test weights. Many sites used test weights for periodic checking and use a chain for example, once a quarter. Linearity and Turndown. A loadcell is pretty much perfectly linear. Load it with 1% of its rate and it will report that. Load it with 100%% and anything in bwteen and it will report with near perfect accuracy. We can demostrate the same linearity with our bet scale. It will weigh a coin or you. Now do a perfect installation into a conveyor. If you place a coin directly over the scale, the output of the scale won't move. Why? The belt is supporting the weight of the coin. Place 10 coins and maybe 10% will be sensed. 100 coins, maybe 50%. At light loading, the belt supports some of the load. If the belt tension is high, it supports more than if the tension is low. It takes some weight to overcome this. The scale will be repeatable (okay probably not for the coin) at light loads but will not be accurate. Generally minimum loading should be around 25% of conveyor design. Here's another. Take a piece of steel flatbar 6mm thick. It's easy to bend. Now form the flatbar into the shape of an angle iron. It becomes very rigid. So does a belt. It transmits load better with 20 dgree trough idlers than with 45 degree. So becaue of the 2 above points, a conveyor needs to be loaded to about 25% of design for the scale to report as accurately as if it is loaded to 100%. It's a little worse for 45 degree trough idlers. Conveyors are generally designed for a max Rate (tonnes per hour) and have an overrange capability of 25% or 1.25 of design. A properly sized belt scale should meet this design mainly to handle surges. Therfore, a scale is linear in a range of .25 to 1.25 of design (which should match conveyor design). Below that, it's repeatable but will weigh progressively lighter with lighter loading. BTW. My spelling is not perfect and I don't see a spell check in this blog. As I already said in my PM to you. I have a team of experts available and they can be reached through www.siemens.com/automation/support-request or by clicking on mySupport above. I believe you can use your same log-on and response is supposed to be very quick. Cheers |
7/4/2011 9:20 PM | |
Posts: 74 Rating: (3) |
Hello hhpp How are you making out on this issue? Toppie |
7/11/2011 4:04 PM | |
Posts: 20 Rating: (0) |
Hello Toppie, i am really sorry, but I have been very busy here so many problems in a short time. Electric lines fall down becuase of the wind (100 k/h) and the snow storm. Anyway I finally will get the test weights in a few days, i am really ancoius. Thanks. hp. |
7/11/2011 4:39 PM | |
Posts: 74 Rating: (3) |
Snowstorm? What is your location and elevation? I come from the mining industry too and have expereinced these unplanned disasters so I know what you are going through. Good luck and I hope to hear from you soon. Cheers |
8/3/2011 10:34 PM | |
Posts: 74 Rating: (3) |
Hello Hector Any updates for me? Toppie |
8/7/2011 6:40 PM | |
Posts: 20 Rating: (0) |
Hi Toppie, finaly I have news. I worked yesterday the whole afternoon, first checking the instalation. to summarize: There is a litle slope in the conveyor ( really, small). The instrumentwas installed ina placewhere the conveyor is at the almost the same level (horizontally). Two idlers before the instrument there is a missaliagmente of 3 milimetersand two idlres after the istrument there is misaligment of 8 milimiters. Next shut down I will fix this and put the instrument as aligment as i can, in the same place because it is thebestplaceto install because of the slope that is really small. During the test: I put a weightof 10.1734 Kg on each idleronce at a timewith the conveyor stopped, the instrument read this values: 1st idler: 1,9 Kg 2 d idler: 2.01 Kg 3 th idler:1.96kg Next I started the conveyor, and i do not know exactly why the load increased, but theload showed in the instrument was going up and down between this values 2.25 kg to 2.45 kg (More weight ) The rate showed by the instruement was: 1st idler: 9.29 to 9.95 t/h 2didler : 9.32 to 10.04 t/h 3 th idler: 9:29 to 9.95 t/h this values looks fine if we check the load while the conveyor is runnig, but not agaisnt the load while the conveyor is stopped. Now if we do the theorical calcs the real Kg/m with a load of 10.1734 Kg should be 2.826 (k/m), and the t/h should be 11.76 (t/h). I think this difference due to last adjust, where I did it a litle smaller, for isntance if the test weight was 400 kg, the test gave us 396, 398, 396, and I took 396 kg because of repeatebility. Next time I will fis teh isntalation and i will do a calibration, so verything must be fine till then. Mean while i neeed to know waht do you think, am i fine, I need to take something else into account?, my calcs are fine? Thanks. |
8/9/2011 2:59 AM | |
Posts: 74 Rating: (3) |
Hi Hector Some of your variations may be coming from the misalignment. This can cause the belt to bounce a bit, especially when it's unloaded. Please don't do this but if the belt were lifted off of the idlers, I'm sure you would see the same delta on all three scales with the same test weight. I don't understand your theoretical calcs. If you have 10.1734 kg and an idler spacing of (example 1.2 m) your Load = 8.48 kg/m. Your calculation yields an idler spacing of 3.6 m. (Or is it 3.6 feet? Careful not to mix engineering units) You should also make sure there are no self aligning idlers within 12 m of the belt scale. these are the type with little rollers on the wings and they serve to 'bump' the belt back into alignment. You could help me by confirming a few details. I looked through this whole thread and I can't see where you had provided me: Design Rate Design Speed (This I have at 1.156 m/s) Idler Spacing at scale (Approach idler to scale = Scale Idler 1 to SI2 = SI2 to SI3 = SI3 to Retreat Idler 1). Cheers Toppie [ |:)
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8/9/2011 11:01 PM | |
Posts: 20 Rating: (0) |
I do not understand what you meant with this words "Don´t do this but..." About the calcs, I put the weight over the whole scale (10.1734/3.6 = 2.82 kg/m) that´s the reason so if you see I have a big difference between the theorical rate and what is showing the instrument. according to this load ( 2.82 kg/m) my calcs are fine right? there are selg aligning idlers, but i think i can take outthem and put somewhere else on the conveyor. You say 12 meters, 6 before the first sclae idler and 6 after the last idler? Desing rate = 90 t/h Speed rate = 1.156 m/s AI to SI1 = 121 cm SI1 to SI2 = 119.7 cm SI2 to SI3= 122.2 cm SI3 to RI = 118.2 cm Best regards |
8/10/2011 10:04 PM | |
Posts: 74 Rating: (3) |
Hector. Re: Don't do this...I don't want you to lift the conveyor belt off of the idlers, at least until you are ready to do an alignment. The belt has weight and is supported by the idlers but the weight can vary so this can throw your tests off, especially with a light weight. Re Applying weights over whole scale: The weight is quite light compared to toatl loading so some of the effects you are seeing is from the structural support of the belt. The weight needs to be hung from the test weight bar of each scale, preferably as close to the center of the suspension as possible. Re Sel Aligning Idler Location: I made a little erroe working from memory. I suggested 12 m before the first scale idler. I should have said approx 9m. The S-A idlers have a different profile, sit higher plus have those wing rollers. It's covered in this application guide http://support.automation.siemens.com/WW/view/en/18246952page 13. Thanks for the data. Cheers |
8/27/2011 8:05 PM | |
Posts: 20 Rating: (0) |
hi Toppie, yesterday I finaly made the aligment two idlers befor and two idlers after. All rollers are working, the belt is fully supported by the rollers. |
8/27/2011 11:30 PM | |
Joined: 5/28/2008 Last visit: 9/21/2024 Posts: 4515 Rating: (839) |
Hello, I recommend to make first balancing before zeroing of the belt scale. If you have "Calibration out of range" massage, then initial zero should be applied first. Please see the attached picture. Please revise my previour post in this thread (post # 8)
Hope this helps.
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Last edited by: Ayman Elotaify at: 8/27/2011 11:33 PMAyman Elotaify |
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8/29/2011 2:56 PM | |
Posts: 74 Rating: (3) |
Hello Hector So what's happened is you have changed to zero point by more than is allowed by the BW500. By changing the alignment, you have either added or removed considerable belt weight. This needs to be reset by performing an initial zero and then an initial span. See the information in the manual about P377 and P388. Hector. If you would like me to call you so you can ask any other questions, let me know your phone number and a contact tme. Cheers Toppie. |
9/20/2011 10:45 PM | |
Posts: 20 Rating: (0) |
Hello everybody. Thank you very much for everything, yesterday I finally made the calibration with the test weights (6 pieces of 6 kg each). the span is ok, not perfect but ok. there is a little difference but i think this because of the speed is not constant, belt and so on. Now how I make the factoring, because I need to decrease the span 5% according to the data that i got last month. Factoring accoridng to the manual is for know the value of a new weigth or unknown weight. Best Regards. |
9/21/2011 6:07 PM | |
Posts: 74 Rating: (3) |
Hi Hector It sounds like very good news! So I will guess that the scale isreporting about 5% less than the actual weight. This is good because this is in the range of what I would expect with a test weight calibration. This variation iscaused bybelt tension. The belt can structurally support some of the load due to how tightly it must be pulled. This is a limitation on all conveyors andiswhy a test chain calibration is somewhat better if you can't do a material test. It should not be caused by speed variation because the speed sensor will compensate for this. So nowwe must assume that the error is constant and future results will prove this.Now, you would manually adjust your span and adjust the values of the test weight by the errors reported. Since you have calculated your data, I would suggest you do as follows: 1) Perform a zero calibration 2) Access P019 and enter the Edit Mode. 3) Enter the calculated/inferred error completewith the direction (eg -4.9%) 4)The original test load value will be ajusted to compensate for this correction. Record this value (P017) since this is you new calibrationtest load. Your done. Nice job. |
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