Hi fritzuki,

The apparent weight reduction on a belt weigher weighbridge is roughly proportional to the cosine of the angle of the conveyor so:

cosine of 6.8 degrees = 0.99296

If the t/hr was 100 then the apparent weight on the weigher would be:

0.99296 x 100 = 99.29 t/hr.

This error is usually eliminated when performing a weighed load check. If you are using test weights then this error has to be catered for.

Jon ■

Hi fritzuki,

Not sure of the 3000 series weigher, but it probably uses the same calibration technique as the old Microtech.

so as I don't have your exact data, here is a sample calc into which you can substitute your own values:

Test weight 30kg

Weigh length 1.2 m

belt speed 1.5 m/sec

conveyor angle 6.8 deg - (cos 0.992)

equivalent weight per metre - test weight / weigh length

30/1.2 = 25 kg/m

equiv Kg/sec = 25kg/m * belt speed = 25 * 1.5 = 37.5 kg/sec

equiv kg/h = 37.5 * 3600 = 135000

equiv t/hr = 135000/1000 = 135

modified by the cos of the conveyor angle = 135 * 0.992

Calibration Rate = 134.0 t/hr

You can then work out integrated tonnes over the test period from the rate value.

Hope this helps

Jon ■

Mr Fritzuki,

I wanted to make some comments about weighing on an angle, it is a simple question with a simple answer, but its easy to get confused.

1. In the normal course of events, on a fully suspended weigh frame (Ramsey 10-14 and maybe IDEA, but not 10-17 or 10-20) when you apply a static calibration mass, the weigh frame sees less of it during calibration. Thats OK though because it also sees less of the real material and the affects cancel. This is the normal situation.

2. The Ramsey product has a lot of pre-packaged calibration method in it, and it would not be too hard to get confused. You need to have selected the correct weigh frame type, and if it is a pivoted type weigh frame, this could also involve the entry of the angle. The best way to get around this would be to apply the same mass to each idler set (usually hung on both sides of each set) and then you can treat the weigher as if it is a fully suspended unit, in which case do not use the angle entry in the calibration process.

3. Ramsey also like electronic calibration, this needs an angle input in the process. They introduce a mV signal (with a resistor across the LC bridge) and this with the loadcell characteristic and the angle of inclination is used to simulate weight for calibration. This is fine, but at least on some models, the angle input should not be used by the calibrating engineer because it leads to errors - when angle input is used when not needed. This question has led to a lot of eroneous calibrations in Ramsey units.

What to do, either get a fully suspended weigh frame or apply the weights evenly at the idler locations only and do not consider the angle in the calibration process.

If you want to put your calibration calculations in a message I am sure you will get plenty of help to work it out. ■

There appears to be a problem here.

When you place a test weight on an idler set, the whole weight is reflected down through the idler. The incline makes little or no difference if the weigh carriage is designed correctly. The angle compensenation is required to make it appear as if it is material on an inclined belt.

It is the spread of material on the weigh section that makes the apparent weight change proportional to the angle.

Jon ■

Gentlemen,

If I may. I've set up static weights on Ramsey 10-20's, 10-17's, 10-14's, 10-30's,and 10-101 (idea) carriages. In the formula that Ramsey uses, you need to know the distance from pivot to loadcell, the distance from pivot to idler, the distance from pivot to test weight. You also need the hieght from the top of the carriage to the pivot, hieght to top of the roll, and hieght to to test weight. This hieght measurement comes into play with pivoted carriages such as the 10-17, 10-20, 10-22, and I bevieve some of the older 10-11's and 10-12's but those are for the most part obsolete. Test weights used on an incline conveyor on non pivoted carriages are not calculated back based on the angle. The loadcell does see less weight, but the weight of material is proportional to the static test weight. So a 100 lb test weight on an incline might only register 98 lbs. at the load cell, but 100 lbs of material on the same incline will also look like 98 lbs. The scale should still be set to display 100 lbs. On the pivoted carriages the angle is used to compensate more for the hieght differences as this changes the relationship from pivot to the test weight

Attachments

pivot scale (JPG)

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Gentlemen,

If I may. I've set up static weights on Ramsey 10-20's, 10-17's, 10-14's, 10-30's,and 10-101 (idea) carriages. In the formula that Ramsey uses, you need to know the distance from pivot to loadcell, the distance from pivot to idler, the distance from pivot to test weight. You also need the hieght from the top of the carriage to the pivot, hieght to top of the roll, and hieght to to test weight. This hieght measurement comes into play with pivoted carriages such as the 10-17, 10-20, 10-22, and I bevieve some of the older 10-11's and 10-12's but those are for the most part obsolete. Test weights used on an incline conveyor on non pivoted carriages are not calculated back based on the angle. The loadcell does see less weight, but the weight of material is proportional to the static test weight. So a 100 lb test weight on an incline might only register 98 lbs. at the load cell, but 100 lbs of material on the same incline will also look like 98 lbs. The scale should still be set to display 100 lbs. On the pivoted carriages the angle is used to compensate more for the hieght differences as this changes the relationship from pivot to the test weight

Attachments

pivot scale (JPG)

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You may use the formula

Tonnes/Revolution of belt = Belt Load x Belt length / 1000.

In your case you are using 30kg Load. multiply this by belt length and divide by 1000 will give you the totals reading per revolution of belt.

Multiply this with 5 and take reading for 5 revolutions. If there is an error that nmeeds to be fed into the controller.

Sundar

newtonws@gmail.com ■

Mr. Sundar,

I don't often reply to posts that are quite this old, but in your reply, I think you might have ommited some information. Unless I've been doing the math wrong for a long time, the 30 kg. wieght applied to a scale will not necessarily represent 30 kg. of belt loading. You must also allow for idler spacing. If the idlers are spaced at 1.5 m intervals and the full weight of the carriage is carried by the load sensor then 30 kg would apply a loading of 20 kg/m registered by the scale. If the idlers are spaced at 1 meter intervals then 30 kg would represent 30 kg/m, or in Mr. Scarrott's example 1.2 m spacing = 25 kg/m.

This next part is getting off topic...

I believe you are correct in not factoring the weight for the COS of the Angle. The reason that the angle is needed in the Ramsey scale is for the electronic calibration. The electronic method of calibration applys a resistor across a strain gauge loadcell and forces an electrical offset. This offset must also take into account the angle of incline to give a representation of live load. I can get the formula if you are interested.

If you take a strain guage loadcell and apply enough weight to see 10mV of change, then tilt the loadcell and load 45 deg. the loadcell output will change to 5 mV. Loadcells measure force in the vertical. If you turn the loadcell on it's side to 90 deg. the loadcell will not register any load at all. If you take the same loadcell and short a resistor across to make 10 mv of change you can turn it any direction you like and you will still have 10 mV of signal. ■