Re: Wagon Tippler Drive
Torque is force multiplied by distance. T=FxD
The distance is measured from the point of application of the force to the pivot point. That distance can be expressed as a radius with the pivot at the center. If the force is not perpendicular to the radius, then only the component of the force which is perpendicular to the radius is used for torque.
Your problem is going to be more of holding the load from free wheeling away on you as it turns more so than the actual torque required to turn it. ■
Gary Blenkhorn
President - Bulk Handlng Technology Inc.
Email: garyblenkhorn@gmail.com
Linkedin Profile: http://www.linkedin.com/in/gary-blenkhorn-6286954b
Offering Conveyor Design Services, Conveyor Transfer Design Services and SolidWorks Design Services for equipment layouts.
Torqueing Is Cheap
Talk to the Experts...We're always saying it....
You need to examine the roll axis, load centre, car geometry and so on before you can start to anaylse the equilibrium of the waggon. There is no simple formula and the polar monment of inertia, and its gradual decline, must be considered for you to have any neaningful control over the machine, like Gary said.
Better to talk to Heyl & Patterson, PO Box 15, Pittsburg, Pennsylvania. How can you forget such an address? PO Box 15, Pittsburg! They know what they're doing. ■
Control Is (Quite) Everything
Hello Nash,
you study a quite complex machine. So, if you do research or study or a paper to some qualification, try to have a look at an existing installation, search the libraries for substantial input and then do a sketch as near to the truth as possible. As Gary points out, there's the always urgent problem of keeping the wagon(s) under control, during long time operation under very demanding circumstances. Also, the structures must be sufficiently rigid, which counteracts to the demand of torque = power due to weight. Then, speed is of the essence, meaning that the share of dynamic torque dyn T=J*e at acceleration and deceleration must be considered due to cycle time. And this is just the beginning, you see..?
If this is for design reasons, your questions might be understood in a way that i personally would strongly support Mr. Panjangs option, however there's suppliers all around the globe and certainly somewhere near you. In this case, experience, read references, of being in the market say 10 ys. + and under a reputable flag, is everything. There must not have been incidents of any grave nature, i personally would check on that.
Have a nice sunday
Regards
R. ■
Re: Wagon Tippler Drive
Hello Nash,
Deciding the wagon tippler drive is a complex subject, and it can best be done by ‘wagon tippler designer’ who has specialised for the subject (i.e. mechanical engineer who is well versed in electric motors and drive and whose work consistently concerns with the wagon tippler). Also the subject is too big to come under category of casual discussion in forum. However some brief information is as below which would be of interest to the readers:
1) Tippler drive can only be decided after concluding (or assuming) practically complete construction aspects of wagon tippler. It is presumed that you have the wagon tippler G.A. and constructional data and possibly you are making some calculation as a buyer. The nature of query implies this.
2) Side discharge wagon tippler is considered here and its rotation is zero to 150 degree and back to zero degree.
3) Wagon load torque (resisting torque) is not constant but varies from negative to positive during forward rotation, as well as during backward rotation.
4) Wagon/s to be tipped, their mass and C.G. location, when empty and loaded are known.
5) Wagon tippler load torque (resisting torque) is decided considering
- No wagon on tippler platform (rare occurrence, but it becomes convenient basis for calculation).
- Empty wagon/s on platform.
- Loaded wagon/s on platform, with material of high density, low C.G. situation. Loaded wagon with material of lowest density i.e. high C.G. situation.
6) Consider wagon tippler platform standing at rotation by 10, 20, 30, ---- 150 degree and find the resistance torque due to moment of the various gravity forces. Then add motional frictional resistance torque in above for dynamic condition.
7) Using the values as per sr. no. 6 plot the graphs. All the above graphs will have positive torque (power) for some zone and negative torque (power) for some zone.
8) Select the drive motor which has rated operational torque (positive as well as negative situation) more than the resisting torque. The drive motor takes care of torque values without use of brake, except for stopping or holding. Tippler rotates at fixed speed. It rotational speed and torque provides the power.
9) The exercise is done many times by changing / adjusting magnitude of counter weight (concrete mass) to have either least size of drive unit or least power consumption kWh which has relevance with respect to wagon type applicable for most of the operational time.
10) Rotary wagon tippler will also be on similar line.
11) The information is for readers at large and is not addressed to experts on the subject.
Regards,
Ishwar G. Mulani
Author of Book : Engineering Science And Application Design For Belt Conveyors (new print November, 2012)
Author of Book : Belt Feeder Design And Hopper Bin Silo
Advisor / Consultant for Bulk Material Handling System & Issues.
Pune, India.
Tel.: 0091 (0)20 25871916
Email: conveyor.ishwar.mulani@gmail.com
Website: www.conveyor.ishwarmulani.com ■
Wagon Tippler Drive
Hello,
I am trying to design a drive system for a Wagon tippler (also called Car dumper).
I have the data regarding the capacity (tons per hour) and the number of cycles per hour (this can also be read as number of cars to be tippled per hour) for the tippler.
I would like to know a basic calculation method by which i can figure out the torque on drive shaft required to tilt the wagon.
Best Regards
Nash ■