Power Required To Rotate Load 2

[AdamDrover]

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

 

[hydtools]

You could start here:

http://www.engineeringtoolbox.com/motion-formulas-d_941.html

Which direction is down, direction of gravity?

Ted

 

[AdamDrover]

Sorry for the confusion. The left picture is the top view, and the right picture is the side view. (On left picture, gravity is in the down direction)
Thanks for the site referral!

 

[zekeman]

1)You need to give us the time you need to rotate
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.

 

[drawoh]

AdamDrover,

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?

JHG

 

[Twoballcane]

I'm not near my books, but I think there is an equations to go from torque to horsepower (also in SI units)...

Tobalcane
"If you avoid failure, you also avoid success."

 

[zekeman]

The power is the least of your problems.
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.

 

[AdamDrover]

The time to rotate the support 180 degrees is 5 seconds.
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.

 

[IRstuff]

Your question seems a bit nebulous. "Power" for the rotation at constant speed is the work per unit time required to overcome the friction of the bearings involved. The big question is how you intend to start and stop, and how fast you do want to get up to speed?

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

FAQ731-376

 

[zekeman]

The minimum HP motor you need is about .4 HP.
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.

 

[AdamDrover]

Thank you guys, that was exactly what I was looking for.

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

 

[zekeman]

Sorry,with 200 lb, the the result should be reduced by the factor of 8,or about 1/20 HP without friction. The additional torque needed for the acceleration can add an additional .025HP.