Friction value

Posted in: , on 24. Jun. 2003 - 04:42

Need to know what fricition value to apply so the correct winch can be selected for pulling on new 1800mm wide PN2000/UG ply belting - mass of the belt is 48.6 kg/metre - total beling 2060 metres.

There is only a 10 metre rise over 1000 metres



Re: Friction Value

Posted on 25. Jun. 2003 - 12:35

Previous Answers Posted Oct. 25, 02:


Paul Allardice

Registered: Oct 02


Posts: 2

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?


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.


See other posting on the responses to 12-02 for more information.

Lawrence Nordell

Lawrence Nordell Conveyor Dynamics, Inc. website, email & phone contacts: phone: USA 360-671-2200 fax: USA 360-671-8450

More Information On Responses

Posted on 25. Jun. 2003 - 12:44

Title: Belt friction factors

see: Conveyor Forum, second page near mid point

Author: Paul Allardice 5 434

Last posting 12-08-02 15:39

by I G Mulani

L Nordell

Lawrence Nordell Conveyor Dynamics, Inc. website, email & phone contacts: phone: USA 360-671-2200 fax: USA 360-671-8450

Re: Friction Value

Posted on 25. Jun. 2003 - 05:17

Previous Answer Posted on 9th Dec 02 in :

Forum : Belt Conveyor Technology

Thread : Belt Friction Factors


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'.


PS: To investigate your specific problem, please indicate roller rotating mass per meter length of conveyor and idler pitch and also approximate location of rise (whether at tail or at head or in-between)