Belt Pull and Power for Inclined Conveyors

[@somedas1]

Superb Explanation, please continue to teach us sir.

[@rulmecacorporation4596]

It is our pleasure to offer these videos.

[@f3nunez]

Thank you very much for this video.

[@jhonnyrestrepo6780]

Thanks for sharing this information. Excellent

[@rulmecacorporation4596]

You are very welcome. If there is any way our company can help you, please let me know. My email address is [e-mail]

[@adelciosouza9018]

Thank you very much. Great explanation. Congrartulation.

[@rulmecacorporation4596]

we are happy you found it useful. we will make a “metric version” with SI units soon.

[@Begem1S]

Thank you for this video, it will be great in metric units also.

[@rulmecacorporation4596]

I am working on the metric version right now (working from home because of COVID-19). Did did you see the metric version on the "horizontal conveyor" lecture?

[@jacobthomas785]

Hello. I have a bidirectional incline/decline conveyor. I can't track the belt in one direction.

[@nishantdhengre8569]

Sir is this calculation will be same for drag chain conveyor

[@mikegawinski3450]

The calculation is not valid for drag chain conveyors. You should contact a drag chain conveyor manufacturer for assistance

[@osh-1336]

Hello Sir, thank you very much for explanation. I would like to ask, what if there is no slider bed?

[@rulmecacorporation4596]

Your question could have several answers. Slider beds can have many different frictional coefficients, from low to high. Rollers may be used beneath conveyors instead of slider beds. One rule of thumb (if rollers are used) is to substitute a coefficient of 0.1 instead of the frictional coefficient. I suggest you visit our website and download our free power calculation software. Just let me know if you have any problem finding it.

[@billwells8054]

Thanks. How/why do you have gravity pull as (load/ft) * h? Hmmm... I'm calculating.... I would use (sin 17.46) (250) = 75#. Same as you. But why is it also (load/ft) * h? Appreciate your help. EDIT: And if you don't mind, another question. Let's say the conveyor stops fully loaded. What is power required to get it going again? I'm assuming there must be a ramp-up time to go from 0 to 50 ft/min. So this would be the power to accelerate the total mass? I'm writing a continuing education engineering course. I studied this subject 60 years ago! Bill

[@rulmecacorporation4596]

Hi Bill. You could use trigonometry, but the result will be exactly the same since the belt pull required to overcome gravity is linearly proportional to the change in elevation. Rather than provide a complete "derivation" I simply showed the accurate short cut method we use to calculate the force.

[@jacobthomas785]

The conveyor is centre drive with rubberized drive pulley and tensioning pulley. Also head and tail pulleys are crowned pulleys. Still the tracking in one direction is difficult. Can you please guide?

[@rulmecacorporation4596]

Dear Jacob, We will be happy to give you some suggestions. You can obtain our contact info from our website rulmecacorp.com/contact-us/

[@souravkumardas3618]

Sir, Why we are considering belt weight 3 x 10 = 30. Should we didn't require to calculate the lower part weight i.e 30 x 2 = 60 ?

[@rulmecacorporation4596]

To simplify the explanation, we assumed there would be no slider bed carrying the return strand. No lower slider bed implies no friction, which implies no belt pull required to move the belt from the head pulley to the head pulley. If there is a slider bed for the return strand, then you should add belt pull to overcome that friction force. Note also (on the inclined conveyor) that we ignored the effect of gravity on the belt because gravity's effect on the top strand is offset by gravity's effect on the bottom strand.

[@rulmecacorporation4596]

Correction: my previous remarks should have stated ""implies no belt pull required to move the belt from the head pulley to the tail pulley"

[@ShafiqulIslam-vy8xy]

Hello Sir, Thanks for your video. However, what about for inclined conveyor up to downward.

[@mikegawinski3450]

Thank you. We do not understand your question. Do you mean “How do I calculate belt pull and required power for a declined conveyor?”

[@rayzjr3762]

Is this applicable to inclined slat conveyors?

[@mikegawinski3450]

unfortunately, it only applies to flat rubber belt conveyors.

[@wycliffeapindi3270]

A very informative lecture. am still enquiring on the pulley drum weights why have you not considered them in determining the power?

[@rulmecacorporation4596]

Drum pulley mass is irrelevant to the power required to keep moving the belt continuously in stead state. However, you are correct that drum pulley mass should be considered when determining "required start up power" because the conveyor drive should have enough "available start up power" to overcome inertia. If you send me your email address I will send you a link to a more comprehensive program which allows the user to specify drum pulley mass and calculate required start up power. Since most conveyor drive manufacturers supply "Design C" motors (which are able to supply 200% of running torque at start up), required start up power is not calculated unless the conveyor is very long (i.e. > 300m).

[@wycliffeapindi3270]

@@rulmecacorporation4596 , very well of you, my email address is:[e-mail], I kindly look forward to the link since am currently working on an Inclined conveyor belt project .thanks

[@wycliffeapindi3270]

@@rulmecacorporation4596 Hi, hope your well. just a reminder am still waiting for the link you promised to send me kindly.

[@mikegawinski3450]

@@wycliffeapindi3270 I am very sorry that I missed your recent message. I will send you links via email within five minutes.

[@madhudanduboyina5848]

good explanation sir please do more videos regarding motor selction for linear moments and rotary applications tank you sir once again nice exlanation

[@rulmecacorporation4596]

We are working on a metric version of the "inclined plane" video now. We do not understand what you mean by "linear moments and rotary applications". We manufacture Motorized Pulleys to drive conveyor belts. So, our training is geared toward helping engineers with those considerations. You may find our webinar playlist useful because it is more comprehensive

[@faizahmedbudihal6285]

Thank you so much Sir These are amazing. Can you please do the same for roller chain conveyors or chain conveyors

[@rulmecacorporation4596]

Our specialized expertise is with driving belt conveyors because we manufacture Motorized Pulleys to drive conveyor belts. You would be best served soliciting help from manufacturers of chain conveyor drives.

[@faizahmedbudihal6285]

@@rulmecacorporation4596 Thank you for the reply, Can we suggest any book for chain drive systems

[@rulmecacorporation4596]

@@faizahmedbudihal6285 Here is something I found on the internet. It may help you. www.renold.com/upload/renoldswitzerland/conveyor_chain_-_designer_guide.pdf

[@sardurithesecond520]

Hello Sir, Thank you for the amazing video. However, I have 2 questions I guess I must be missing something. 1) shouldn't the total weight (Material and belt) be resolved on the inclined plane to calculate the friction pull and the force required to overcome gravity? [ie for friction: f = Mue * N = Mue * W * (H/L) ] 2) Shouldn't we factor in the required acceleration force to accelerate the newly fed weight to the belt velocity? assuming material is fed with a mass flow rate m_dot so, F_ac = m_dot * V_belt. Thank you in advance for your clarification.

[@rulmecacorporation4596]

Thanks for your comments. The video you watched shows a very simplified technique to calculate required power for handling packages. Our bulk handling belt pull video elaborates on the factors you mention in your comments. An explanation of the belt pull required to accelerate material is definitely included in that video. Also, package handling conveyors typically have low mass and slow belt speeds. So, acceleration of small mass from 0 fpm to 50 fpm is negligible whereas bulk handling conveyors with high mass and belt speeds => 300 fpm need momentum to be considered.

[@JiteshKumar-yq8cl]

I think angles are really less for inclined conveyor so they did not take cos theta component ;because for less angles that will be equal to 1

[@patelghanshyam1986]

For gravity pull why don't we consider coefficient friction ?

[@rulmecacorporation4596]

Friction is included in the "friction pull" calculation. The "gravity pull" calculation only considers the power required to overcome gravity since friction is already accounted for. We do not need to add friction twice.

[@kiranmathure880]

Sir what about gear box,how to select gear box on the basis of torque for similar application pls reply

[@mikegawinski3450]

We have a video which addresses gear box torque, etc. would you like the link to the video?

[@rulmecacorporation4596]

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[@rulmecacorporation4596]

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[@rulmecacorporation4596]

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[@neelchauhan9326]

hellow sir i want to calculate torque of dc geared motor which is used in side gripe conveyor which is inclined at 30 degree to pick a pvc pipe of 0.335 kg in total there are 6 pvc pipe to gripe by conveyor can u tell me how to do and also can u make a video on side grip conveyor belt which is inclided and can u give any id so i can reach u as i have some other problem related to side grip conveyor as i am making an project so

[@rulmecacorporation4596]

Dear sir: You may reach me at [e-mail].

[@RajeshtopTechnical]

Friction=0.5 ? I can't understand. constant for every inclined conveyor .

[@rulmecacorporation4596]

Here is a link to our power calculation program which will enable you to calculate conveyor power for a variety of frictional coefficients from 0.2 to 0.65 rulmecacorp.com/unit-handling-power-calculation-program/