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Thread: Belt Friction Factors

  1. #1
    Paul Allardice - Downer EDI Engineering, Australia Guest

    Belt friction factors

    I am trying to estimate the force required to pull off and replace a steelcord belt from a conveyor.

    I understand that GoodYear recommend a friction value between 0.02 - 0.03 be used, but this is for a conveyor where the take-up tension is applied, hence would not be relevant for this situation.

    Has there been any tests carried out on determining belt friction forces over rollers when the take-up mass is removed from the conveyor system?

  2. #2
    Kent_Rieske Guest
    Hi Paul,

    Any belt conveyor design program can be used to determine the force to pull off the old conveyor belt. These programs give the running force between any two points which includes friction and lift forces. The slow velocity pull off friction would be a little higher than normal running forces.

    We market our PRO-BELTcomputer software for the design of belt conveyors, belt feeders and pulley shafts. Our software includes an idler friction factor, Kx, MULTIPLIER which can be used to adjust the running friction. I suggest a multiplier of 1.25 for pulling off the belt. You can visit our web site at:


    Kent R. Rieske, President
    Professional Designers & Engineers, Inc.
    PO Box 11380
    Boulder, CO 80301
    Tel: 303-530-1551
    Fax: 303-530-3775

  3. #3
    Lawrence K. Nordell

    Lawrence K. Nordell

    President and CEO

    Conveyor Dynamics Inc.

    Conveyor Dynamics Inc.

    Professional Experience 59 Years / 10 Month Lawrence K. Nordell has 59 Years and 10 Month professional experience

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    Rolling Resistance to Belt Pulling


    If you want to make a rough guess, use the CEMA method. Kent sells Probelt. Conveyor Dynamics, Inc. sells BELTSTAT at These can give a order of magnitude estimate to the true value. They would be lucky to get within 30%. They can be too low or too high. It is usually academic, except when the installation is on a slight incline or decline (1%) and threaded opposite to the positive slope. In this case, the sag resistance may dominate depending if you use the counterweight or a resistive brake or not.

    BELTSTAT computes the ky rolling resistance depending on tension up to an idler spacing of 3m (10 ft.). You must put in the equivalent lowest sag tension and allow the program to close the loop. Individual tension values can be read around the belt as though you were threading it.

    The true rolling friction resistance, per unit length, is fairly complex to evaluate. First, it continually changes as the belt tension changes. As the tension changes, the belt sag shape changes. More sag = more resistance due to belt compound bending flexure between idlers and bending over idler rolls.
    It also depends on how the belt is being threaded - uphill; downhill... It varies with the ambient temperature. It varies with the belt construction and mass, fabric, steel cord, cover gauge, rubber mfg's. compound rolling efficiency... It depends on the idler roll configuration, diameter, spacing, and drag properties (new vs run-in), and so on.

    Yes, it has been measured on specific projects that are equipped with tractor-mounted load cells (or equivalent) pull-bars obtaining force measurement on tractor or the like. The number is meaningless without other knowledge as noted above.

    A rough friction number in DIN 22101 f= 0.04 or CEMA ky=0.03 plus the "new" idler kx term. Big belt may require a little more.

    Wishing you success in your quest.

    Lawrence Nordell
    Conveyor Dynamics, Inc.

  4. #4
    Lawrence K. Nordell

    Lawrence K. Nordell

    President and CEO

    Conveyor Dynamics Inc.

    Conveyor Dynamics Inc.

    Professional Experience 59 Years / 10 Month Lawrence K. Nordell has 59 Years and 10 Month professional experience

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    Belt Pulling Resistance


    Sorry about the first rsponse. It was made in haste. I did not get the bit about installing new belt into old. In this case, the takeup is often used to control the threading rate and pulling dynamics. Typical event: cut belt and splice in new from a prefabbed run of belt the is layed flat on prepared ground. As the new blet is pulled, usually by tractor, the old belt is removed in large sections or threaded onto reels. In this case, the threading resistance can vary from a low of f = .013 up to .03 again depending on many of the previously spoken parameters. With long belts, the friction resistance value will be lower and trend to .013-.015. The value is best determined by getting the old belt empty resistance before replacing it. Depending if you replace it with a better rolling loss compound, you could see a lowering of the old belts resistance, even with the gain in weight.

    Yes, it to has been measured. In some cases, the drive system is used in the threading process. This is designed into the conveyor systems configuration as new.

    It is still a rough number.


  5. #5
    Paul Allardice - Downer EDI Engineering, Australia Guest
    Larry & Kent

    Thank you for yor replies and input.


    As you mentioned this calculation depends on many variables and factors which can vary greatly depending on the design of the conveyor system.

    The belt change re-placement procedure that I am looking at involves lapping out the fully spliced new belt alongside the exisitng conveyor, cutting the old belt and splicing the new belt to the old belt, then wind the old belt off onto empty belt reels. As the old belt is wound off the new belt is pulled from the lapping station onto the conveyor.

    I intend on taking a conservative approach and use a high range frictional resistance factor, f = 0.03.


  6. Dear Paul,

    As I understand, you intend to calculate pull which is needed to drag the empty belt on idlers, while laying it out during erection / installation. In this context, following information will be of use to you.
    Widely used value of f = 0.02 to 0.03 for ‘conveying artificial friction coefficient’ is valid for conveyor operating under sufficient load say approximately 70 % to 110 % design capacity (full available capacity potential); and belt is held tight by take-up device so that maximum sag at any point is not exceeding say 1 % or 2 % or 3 % (1 % is standard value for design). Therefore, the aforesaid value range is not valid in this case. Most of the investigation has been done based on above conditions, and hence only option is to project your solution.

    To solve your problem we have to go into constituents of ‘ f ’. The ‘ f ’ is made up of :
    1) Idlers turning resistance 2) Belt denting flexure resistance 3) Belt Bending flexure resistance and 4) Material flexure resistance.

    The important judgement here is the % of sag value which can be utilised in calculation. The free end will have unlimited sag, but as we advance toward pulling device, the sag value will go on diminishing. Also, for empty belt, the sag value will be very much dependent on stiffness of belt. As per my assessment, the sag maximum value will not be more than 10 % in 80 % of belt length, when belt is being pulled. The 10 % sag implies the sag numerical value of 120 mm (5 inch) for carrying idler pitch of 1.2 metre (4 feet)
    Idlers turning is app 10 % of 0.02 = 0.002
    Belt denting flexure is 50 % of 0.02 = 0.010
    Belt bending flexure is (15 % of 0.02 ) x 10 = 0.030
    Material flexure resistance = Nil
    Total of above, f = 0.042

    The aforesaid idler turning resistance, as a ratio of load, do not give true picture of seal resistance when material load is absent. Therefore, we have to separately add for rollers seals (grease shearing) resistance. The test value of seal resistance is 1 N to 4 N per roller (one can consider 1 N for 500 mm belt and up to 4 N for 2000 mm belt, an approximation for the purpose). The formula for approximate calculation of belt drag force will be as below
    Drag force = L. (Mb + Mi) x 9.81 x 0.042 + Mb.H x 9.81 + (1 to 4).Ni Newton

    Here Mb is belt mass kg/m, Mi is idler rotating mass kg per meter of belt length, H is lift when length of belt being pulled becomes L and Ni is number of rollers supporting L.

    The above calculation is for the purpose to decide capacity of pulling device for belt piece of length L (when it becomes L ). The L to be of sufficient length say not less than 100 m (to uphold the sag condition).
    One can generate formula / solution appropriate to specific case, using the described method. This is reasonable judgement. For example ‘ f ’ value will be comparatively less for uplifting as sag will be less. Hence, decide issue with open mind. Seal resistance 1 to 4 N is per roller (for idler set of 3 rollers, its value is 3 to 12 N per idler set). Reverse check will generally show sag value within assumed condition.
    For more information about the ‘ f ’ etc, please refer my book on belt conveyors, which provides detailed information for this crucial parameter.

    I G Mulani
    Author - Book - 'Engineering Science and Application Design for Belt Conveyors'.
    Advisor / Consultant for Bulk Material Handling.

  7. Friction factor

    Dear Paul,
    You want to replace the old belt by new.In most of the cases I have seen the old belt is cut & then dragged by BULLDOZER. But before that take-up is removed.I have seen that the belt is sliding over the structure.So the friction factor may be as high as 0.35 to 0.4.There is no proven test for the same.
    Although my experience is limited to the installation of India only.

  8. Hello Paul..
    My goodness what a wild range of theoretical values you have had for the running friction factor.
    I did some extensive tests on a 4km undulating curved overland conveyor that I put in as well as a 1.2km dead straight one, and a 3.4 Megawatt shaft conveyor. Up to 3000t/h on steelcord belts, to see the exact value of "f" for running full, running empty, and coasting full, and coasting empty.
    The results are quite surprising, give a huge range over the different scenarios, and have proven to be very usefull to me.
    If you contact me at I can enlighten you.
    It will give a good indication of the pull in value which, being empty, is actually reasonably low.

    LSL Tekpro
    Graham Spriggs

  9. #9
    Ray Latchford Guest

    Friction factors for belt pulling

    We are using .04 to .05 for inclined or horizontal sections and .005 to .01 for regenerative sections plus 2.5 to 5kN to pull the slack out and tension the system to get started.

    Our objective is NOT to predict exactly what forces are involved but to get ourselves close enough to the mark that we are able to select appropriate size equipment and also be sure that we have an adequate safety margin. There is a lot more at stake than a prediction of motor size or belt tension.

    Safety is paramount and pulling can be a dangerous task if not approached with appropriate caution. Situations can be very difficult and awkward.

    We have done enough load cell testing to satisfy ourselves that these factors are on the mark, and all our pulling equipment (winches, cables, belt clamps), are rated and certified.


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