5/10/2012 3:29 PM | |
Posts: 767 Rating:
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Hello Karnan, Static torquefor a servo motorin an actual application could best be described as the amount of torque necessary to maintain motor at zero rpm. This static torquewould be for instance on a vertical lift application where a load is raised to position y and held (by the static torque) at that height for a short amount of time while a machine operation is being performed. Then while load is moved to next position a dynamic torque would be necessary to be applied by the motor to move load. In description of your application it appears that the motor wil be operating at a constant speed. Therefore it would be technically most correct to select motor which could continuouslysupply the amount of torque (i.e. dynamic torque) at that speed. A servo motor is designed and often documented for its torque capabilities in catalogs and some manufacturers do not supply all information. Attached is an example catalog page and documentation for a servo motor 1FK7083-5AF71-1TA0 where its capability of supplying toque continuously is per below. Static Torque 16 Nm (@ or near zero speed) Rated Torque 10.5 Nm Rated Speed 3000 rpm Rated Power 3.3 kW Manymanufacturers only display in catalogsthe static torque rating of a motor, rated speed in rpm, and a power rating. If rated torque is not listed in catalog this can be very misleading and can result in selection of a motor which is undersized since aservo motor will not be able supply as much torque continuously at rated speed as at zero speed. When a manufacturer does not display rated torque in catalog pages for selection thenthe rated torque must be calculated or to be more exactrefer to manufactuerstorque / speed curves for that motor for the speed of operation range of application. Calculation of Rated Torquewould be performed as follows: Power (kW) = Torque(rated Nm) * Rated Speed (rpm) / 9550 Using above catalog example: Power (kW) = 10.5 * 3000 / 9550= 3.3 kW Deputy AttachmentCatalog Page 1FK7.pdf (682 Downloads) |
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5/11/2012 6:56 AM | |
Posts: 664 Rating:
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Hai Deputy, Thanks for your Valuable information.. Please tell me how we defined the consatnt 9550 for the calculation ??/ Thank&Regards, RAj
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5/11/2012 2:55 PM | |
Posts: 767 Rating:
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Hello Karnan, Wikipedia is one source where the value 9550 is rounded slightly from what is documented on Wikipedia per below as 9,554. The below copy and paste is better viewed from website, but if you take the following constants in the formula 60,000 / 2 * Pi 60,000 / 2 * 3.14 9,554 http://en.wikipedia.org/wiki/Torque Conversion to other units A conversion factor may be necessary when using different units of power, torque, or angular speed. For example, if rotational speed (revolutions per time) is used in place of angular speed (radians per time), we multiply by a factor of 2π radians per revolution. Adding units: Dividing on the left by 60 seconds per minute and by 1000 watts per kilowatt gives us the following. where rotational speed is in revolutions per minute (rpm). Some people (e.g. American automotive engineers) use horsepower (imperial mechanical) for power, foot-pounds (lbf·ft) for torque and rpm for rotational speed. This results in the formula changing to: The constant below in, ft·lbf/min, changes with the definition of the horsepower; for example, using metric horsepower, it becomes ~32,550. Use of other units (e.g. BTU/h for power) would require a different custom conversion factor. AttachmentKilowatt formula.pdf (633 Downloads) |
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5/18/2012 7:09 AM | |
Posts: 165 Rating:
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hi Deputy Nice explanation!!! Regards Mothilal |
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