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Topic: Rotating a box (Read 2739 times)previous topic - next topic

Kahlid74

Feb 15, 2012, 03:34 pm
I posted in the motor section looking for help on understanding how to move a box and now I have a good idea of kg*cm and lbs*in for torque.  I am however still lost when it comes to understanding the gear box reduction ratio.  I posted in that thread but I thought I would post in here since it's really more of a project.

So if I need 675 lbs*in (22.5" radius on a box weighing 30 lbs) torque and I have a motor that gives me 10Lbs*in torque I would need a gear reduction of 67:1 right?  Below is a picture of what I'm trying to recreate.  The black platform on the gear shaft would actually be 45" by 40" instead of the smaller one shown in the picture.  Any help on understanding the gears would be awesome.  Thanks for your time.

johnwasser

#1
Feb 15, 2012, 05:01 pm

So if I need 675 lbs*in (22.5" radius on a box weighing 30 lbs) torque and I have a motor that gives me 10Lbs*in torque I would need a gear reduction of 67:1 right?

67.5:1 or higher. 70:1 or 80:1 would allow for more margin of error

Any help on understanding the gears would be awesome.

You need a big gear with 70 or 80 times the number of teeth as the small gear.  What other information do you need?
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Kahlid74

#2
Feb 15, 2012, 07:38 pm

You need a big gear with 70 or 80 times the number of teeth as the small gear.  What other information do you need?

Thank you!  That exactly what I needed.  My understanding of the Gear ratio was limited to size based on inches/cm but what you're saying is that that ratio is more along the lines of teeth.  This makes perfect sense now.  I'm a dolt.

Jantje

#3
Feb 15, 2012, 09:58 pm
Are you sure you need 675 lbs*in?
You need 675 lbs*in  when you want to lift a box weighing 30 lbs and the center of the box is 22.5" horizontally away from the axis.

If I look at the picture it looks like your square (box?)is connected to the axis in the center of the square so the distance is 0 and the torque is 0*675 =0.
There is no torque in your setup. This is proven by the fact that the box doesn't change position by itself when turned in any position. This means that your engine only has to deliver the power to rotate the square and this is small torque. Conclusion you only need a very small engine to drive it.
The only reason for using a gear is to have "the right turning speed".
You want to reduce speed as most motors run really fast (unless you go to a stepper or a already geared motor) and touching a box that is spinning at 3000rpm hurts. I guess you don't want to hurt people.

I hope this helps
Best regards
Jantje
Do not PM me a question unless you are prepared to pay for consultancy.
Nederlandse sectie - http://arduino.cc/forum/index.php/board,77.0.html -

PeterH

#4
Feb 16, 2012, 12:25 am
It would be worth revisiting your mechanical design to understand what forces, torques and speeds you need to generate. You're describing a system with rather high torque requirements but it may be possible to reduce these substantially by counter-balancing, moving pivot points etc.
I only provide help via the forum - please do not contact me for private consultancy.

Kahlid74

#5
Feb 16, 2012, 02:45 pm
Thanks for the replies.  The picture above was from someone else who completed a similar project.  I was planning to use his same design but retrofit with a stronger motor etc.  I think since I'm not a mechanical engineer, the best way for me to truly experiment is to grab a motor/drive controller/Arduino and build my box and try turning it.

Jantje

#6
Feb 16, 2012, 03:57 pm
Kahlid74
It is a valid way of thinking when you are in software. It is a "expensive" way of thinking when you deal with hardware.
Note that components you buy are seldom reused (I have plenty of "misbought" items at home)
The torque you are describing is not needed. For this setup a modded cheap servo would do.
With the external gear which has at least 1 to 100 it should easily turn the box whatever material it is.

Quote
I was planning to use his same design but retrofit with a stronger motor

I'm not sure why you want a stronger motor. Do you know which motor was used? Was this  motor insufficient?
Best regards
Jantje
Do not PM me a question unless you are prepared to pay for consultancy.
Nederlandse sectie - http://arduino.cc/forum/index.php/board,77.0.html -

Kahlid74

#7
Feb 16, 2012, 04:17 pm
He used a 311oz*in motor for a 19 inch monitor.

I plan to do the same but with a 42 inch TV.  My TV is an LED (super light) but I want a box to hide the bezel of the TV.  I plan to make the box out of Sanded oak, which is pretty damn light and strong.

Here's the motor/Driver Controller I'm leaning towards: http://www.ebay.com/itm/130646106715
Here's the power Supply I'm looking at: http://www.ebay.com/itm/280819612259

I found a post on these forums about controlling a stepping motor with Audrino and Easy driver (http://arduino.cc/forum/index.php/topic,68555.0.html) which is perfect because I could basically replace the easy driver with the DQ542MA but it seems kind of complicated code wise in Audrino.  The part I really struggle with is looking at Dan's code I can't really see the area where you tell the pulse pin how to ramp up/down.  I see where you tell it to go but don't you ramp up/down with stepping motors?

My apologies if I'm missing obvious points here.  I'm new to all this stuff.

PeterH

#8
Feb 16, 2012, 04:41 pm
If you're just wanting to move and tilt a specific monitor then design the mounting to be neutrally balanced and you will reduce the forces required to negligibly small values. Having a servo mount system which is directly supporting large weights and moments is not the smart approach.
I only provide help via the forum - please do not contact me for private consultancy.

Kahlid74

#9
Feb 16, 2012, 04:44 pm

If you're just wanting to move and tilt a specific monitor then design the mounting to be neutrally balanced and you will reduce the forces required to negligibly small values. Having a servo mount system which is directly supporting large weights and moments is not the smart approach.

Does the first picture I posed do this?  Or does that design introduce weight in places that aren't naturally balanced?

Simpson_Jr

#10
Feb 16, 2012, 05:48 pmLast Edit: Feb 16, 2012, 05:53 pm by Simpson_Jr Reason: 1

Does the first picture I posed do this?  Or does that design introduce weight in places that aren't naturally balanced?

It depends...

If the rotating plate is mounted to the middle of the TV, the weight above/below/left of/right of the middle (and torque needed) will be roughly the same, depending on how weight is distributed inside the TV.
After rotating the TV, you can simply turn the motor off if it's balanced enough. Torque needed will also be constant during rotation and as low as possible.

Once the TV is mounted off-centre it requires a lot more torque which... by itself also differs a lot during rotation. Heaviest part mounted to the rotating plate will also be pulled downwards by gravity which may require you to add a brake or keep the motor powered to prevent it from rotating back after you've turned it the way you want.

Kahlid74

#11
Feb 16, 2012, 05:52 pm
Makes sense.  I'll have to play with that and make sure it's mounted in the middle.

jwatte

#12
Feb 16, 2012, 06:18 pm
Quote
you will reduce the forces required to negligibly small values.

That's only true for more or less continuous movement. However, because there is oscillating movement involved, you have to overcome the inertia (angular momentum) of the moving object.
When you want to control a large inertia like that, you have two options:
1) Specify strong enough motors that you can step them at a known rate and not "miss" steps because of overload, and run as open-loop.
2) Use a feedback system (servo, or high-resolution position encoder) in closed-loop

PeterH

#13
Feb 17, 2012, 02:25 am

Does the first picture I posed do this?  Or does that design introduce weight in places that aren't naturally balanced?

It is hard to say, since you don't show what is attached to it or what range of movement you require.
I only provide help via the forum - please do not contact me for private consultancy.

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