Power Required To Rotate Load
Power Required To Rotate Load
(OP)
I have a D - shaped support attached at Point A, and a load attached at Point B (See attached). The distance from attachment points is 8 inches. The load is 197.5 lbs
I need to rotate this support around point A using a rotational actuator, but I am having trouble calculating the power required.
Could somebody possibly point me in the right direction? The Point A attachment is secured with a bolt, and must rotate 180 degrees.
If more information is required for this problem, I will post it up. Thanks again for your help. It will be appreciated
I need to rotate this support around point A using a rotational actuator, but I am having trouble calculating the power required.
Could somebody possibly point me in the right direction? The Point A attachment is secured with a bolt, and must rotate 180 degrees.
If more information is required for this problem, I will post it up. Thanks again for your help. It will be appreciated





RE: Power Required To Rotate Load
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Which direction is down, direction of gravity?
Ted
RE: Power Required To Rotate Load
Thanks for the site referral!
RE: Power Required To Rotate Load
2) how is it supported before you rotate.
Friction, load inertia and time to rotate will determine the power.
Absent friction and very slow time to rotate, the answer is nearly zero.
RE: Power Required To Rotate Load
The torque at A needed to support your load at any given angle is solvable with high school trigonometry. You need to know the mass and centre of gravity of your D_shaped link. You need to know the friction in your joints. You need to know how fast you need to accelerate your load when the system starts. Power is a function of how fast this thing moves. This is Newtonian physics, which you should have learned in first year engineering.
Are you qualified to do this?
RE: Power Required To Rotate Load
Tobalcane
"If you avoid failure, you also avoid success."
RE: Power Required To Rotate Load
How do you plan on stopping this load?
How precisely is stop position?
can it overshoot the stop position?
Sometimes a mechanism can do the job without too much difficulty.
Give us what you can of the specs and I think we can help you.
And don't worry about power equations just yet.
RE: Power Required To Rotate Load
There will be a rubberized stop at the end of the 180 degrees to limit the movement of the support.
I have the bending moment of the support, and the angular velocity. (180 degrees over 5 seconds ... 0.63 rad/s, or 6 RPM, correct?)
The stop position needs to be 180 degrees from the starting point, but like I stated, there will be a stop limiting the range of motion from 0 - 180.
The confusion arises for me where the rotation occurs about the y axis, but the moment occurs about the x axis, and where the weight of the load is off center for the rotation.
Thanks again for the advice. I think it is great that most of the contributors on this site are so helpful.
RE: Power Required To Rotate Load
Your calculation of speed appears to completely ignore acceleration. It's the acceleration that requires the torque, and you could potentially be accelerating and decelerating at all times. This would give you the smallest torque requirement. Assuming 180 total motion and 5 seconds total timeline, ignoring settling time, we get an acceleration/deceleration of 14 deg/s^2. This multiplied by your moment of inertia results in the required torque.
TTFN
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RE: Power Required To Rotate Load
Given that you have to lift the 1600 lb weight 8 inches in 5 seconds, I get
1600*8//5=2560 inch-lb/second or
2560/12= 213 ftlb/sec
Since 1 HP=550 ftlb/sec,I get
213/550=.387 HP
without considering friction or inefficiency in the motion.
To get this done ideally, you need to control the motion, either by servo ( positio/velocity control, bang- bang syatem (acceleration control + for 1/2 stroke and - for the remainder).These will yield almost no velocity at the ens of the stroke but are not inexpensive solutions.
If cost is an issue there are simpler methods but will involve significant impact at the end of the cycle.Also, if possible, you could significantly reduce the HP if youi could add a counterweight so that there is no unbalanced static torque.
Then you could use Irstuff's calculation of 14 deg/sec^2 to get a much smaller motor.
RE: Power Required To Rotate Load
Although, Zekeman, where exactly did you get the 1600 lb value for the weight?
I get about a 1600lb moment on the end of the D shape.
Just to be sure everyone understands what I am trying to get at, I made a small animation in Solidworks. See attached.
Thanks again for all the help so far. Well appreciated.
-Adam
RE: Power Required To Rotate Load