Submitted by moptop on Tue, 08/30/2011 - 21:38
Posted in
I just placed an order for some new toys but I just received an email from my supplier about the stepper motors I ordered. He now has me second guessing my order. I pretty much ordered these:
http://www.pololu.com/catalog/product/1207/specs
Now he has me worried about the torque. I would like to know if thats powerful enough for pan/tilt and linear(lets say Nikon d300 w/17-35 or 40d w/equivalent lens). Which motors would you buy. I been looking through out this forum and there seems to be alot of answers and not one main answer.
I see that sparkfun only sells this motor:
http://www.sparkfun.com/products/9238
Should I just buy that one? Thanks in advance for the help.
There isn't one main answer,
There isn't one main answer, because it's not possible to have one =)
Total torque required is a function of load weight, the length of the lever which the load is on, and the length of the lever which is driving the lever which the load is on! *phew!*
But, here's what I do, in a nutshell, to simplify it - please note this is NOT exact, and may be way off. You can always research how to calculate torque for a more exact answer! =)
Establish maximum expected load weight. (e.g.: my camera with biggest lens I'd use for TL is 6 lbs)
Establish force against the drive train. This is easy if you're dealing with purely radial loads: e.g.: pan, dolly. You can ignore this for these loads, and just use maximum load weight. For tilt (axial) loads, I use a napkin calculation that considers the distance from the center of the mass of the camera to the and simply presumes that much weight per inch of lever (again, napkin calculations!). e.g.: 6 lbs, where the center is 2" from output shaft mounting point = 12 lbs of load.
So, we need to get to 12lbs of torque to hold that load in a position, how do we get there?
Gear train.
You could, of course, get a 4000 oz/in motor, and with no gearing, drive the output directly - but as we understand in basic circles, the lever length of a circular gear is its radius. Therefore, with no additional gear, presuming the shaft thickness is 3/8" (pretty standard) we get a 3/16" lever. So, the actual torque elicited on the load is 4000 * (3/16) / 16 = 46 lbs. We're golden! Right? Well, not really - that 4,000 oz/in motor is probably NEMA 34 (big!) and probably consumes several amps of power. Not a good choice for a portable rig...
So, go back to gear trains. We could take that 9 oz/in motor. How can we get it to move 12 lbs? Gear reduction! We'll both get more output torque and higher resolution.
Let's assume we use a 100:1 worm gear train (just worm and wheel), and it has a hypothetical efficiency of 60% and a worm diameter of 1/2" and a wheel diameter of 3", our output torque (again, napkin calculation) becomes:
(I * r1) * R * r2 * e / 16
Where I is the motor torque, r1 is the radius of the worm, R is the reduction ratio (fractional for multiplying ratios rather than reducing ratios), r2 is the radius of the wheel, and e is the efficiency - 16 oz's are in a lb, so... that gives us:
(9 * 0.25) * 100 * 1.5 * .60 / 16 = 12.65lbs
The short answer as to why it's not an easy answer? Output torque is a system variable, generated from numerous factors, and not a single-factor component. Meaning, I can't give you any answer without knowing every piece of information I just went through - and that's only applicable to a single-transition worm gear train, whereas adding in more transitions complicates it further, and mixing gear types also complicates it further. But, I would resist the idea that you should not use a gear reduction of some sort. You really should be considering gear reduction, even if you're just using a pre-built gear box.
You can always modify these values around, if you can't get or fit a 100:1 worm gear ratio, but you can get/fit 50:1, you'll need to multiply your motor torque at least 2x, and so one and so forth.
Consider this article as well: http://openmoco.org/node/243
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Thank you for the reply that
Thank you for the reply that was great info.
Basic info:
Pan/tilt
Linear(dolly)
Focus(at a later date)
Price- Cheaper then a moco commercial unit
Distance of slider is 3'
Alright lets say with pan/tilt head, camera, lens, follow focus and the three motors the whole thing weight 12 pounds(Camera 6 and pan/tilt head 6). So if I got this right the 44oz*in stepper motor with no gear is:
44 x 0.1875 /16= 0.515625 lbs not any where near the 12 pounds.
Now I need gears my plan was to run a timing belt with two 10 tooth pulleys which have a face diameter of .60 or .6 so it would be:
44 x 0.6 /16= 1.65 lbs?
Am I doing that correctly? I just want to make sure I got the math correct.
You also need gears to get a
You also need gears to get a higher resolution. 200 steps per revolution will not cut it, especially for a pan/tilt head.
Moptop, two ten tooth pulley
Moptop, two ten tooth pulley gears will get you no reduction.
Generally, the best reduction ratio you'll get out of a reasonable pulley set is 4:1. So, again, napkin calculations (very simplified) result in an output of 11 lbs, still short of your target. Thus, you either need to use a different motor if you're going to use timing pulleys, or different gear setup.
As pointed out a couple of times, consider gear reduction on your resolution as well. 200 steps/revolution is 1.8' per move, which is too rough for smooth timelapse moves. Yes, you can microstep, but you could microstep through a gear train for better resolution too =)
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Alright so this is becoming
Alright so this is becoming alot harder then I originally planned. So I ordered a kessler crane w/motion control to cover me while I build this thing.
I swear I am not trying to sound stupid. I am actually doing research on this and trying my best to understand the physics behind all this. I do thank you all for the help If I could I would buy you all beers. Maybe ill paypal you a 6 pack. If that makes sense.
Alright so lets break this down. Ill worry about the gear ratio for the dolly movement. (This I am guessing will be the same gearing for the pan) I can also buy a premade pan/tilt head and mount it to the dolly.
Now my biggest concern is I need to make this as portable as possible. Thats why I didn't want to buy the kessler crane. I will be using a 4 foot track that can be broken down into 2 parts. Also the motor and everything will need to come off. This is going to be for a base jumping documentary so after I climb up the mountain I either paraglide down or I base jump. Btw I need both for timelapse and video.
So with that said I can't have this huge setup.. So I stayed up looking how to calculate gear ratios and what stepper motors are out there.
I found this:
http://www.robotshop.com/geared-bipolar-stepper-motor-3v-416-oz-in.html
Its a geared stepper motor which is 26:1, 3v, 416oz*in and a 8mm shaft. I dont know how many steps to full rev.
So if the maths the same:
416 x 0.3125 / 16 = 8 lbs
or they have
http://www.robotshop.com/geared-bipolar-stepper-motor-3v-555-oz-in.html
which is 555oz*in, 3v, 8mm shaft.
555 x 0.3125 / 16= 10.8 lbs
So if I wasn't going to do the pan system this would work perfectly. But because I am going to do the pan tilt I need another gearing ratio. Which I have not found a good calculation for determining a gear ratio. Which is where I am stuck.
It's not that hard I think.
It's not that hard I think. At least I hope so, because otherwise my own calculations are wrong :-).
I am using a 20:1 gear ratio for my dolly (check: http://openmoco.org/node/380). That gives me a lot of steps in combination with the 8 micro steps of the stepper driver. The stepper had 200 steps per revolution. This gives me 20 * 200 * 8 = 32000 steps for a full revolution of the stepper motor. I use a small pulley on the dolly and that gives more than enough resolution (over 320.000 steps to travel 6 feet).
The gearing for a pan/tilt head is different. I see no use to turn the camera more than 360 degrees. So, there is no use to turn the axis of the stepper more than one full revolution. I still have to build my pan/tilt head, but I want to use a 120:1 ratio for it. That gives 120 * 200 * 8 = 192.000 steps for a full 360 degrees travel of an axis.
Quote:416 x 0.3125 / 16 = 8
Quote:416 x 0.3125 / 16 = 8 lbs
The math is a bit off, I believe there... According to the page, the max achievable (without breaking the gearbox) torque is 30Kg per CM, so that would be 66 lbs/0.3937 inches, or, given the 4mm radius (not 8mm diameter) of the shaft:
66 * (0.15748 / 0.3937) = 26.4 lbs of torque, perfectly within your requirements.
But, you have figured out one important part of the whole equation:
Cost is relative to requirements, and can be paid in either money, time, or effort. A system perfect for your requirements will rarely cost less than a professional system in total loaded cost. (That is, calculating how many hours you spend, etc.) For some people, time is freely available, so the cost per unit of time isn't calculated - but it really is a cost.
The real beauty of building your own system is not so much saving money (especially when one has strict requirements), but the experience and knowledge the quest provides. =)
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It's true, though, that
It's true, though, that choosing the right gear setup will be somewhat difficult, but it's not rocket science.
You have to consider:
* Torque requirements
* Resolution requirements (when using steppers)
* Cost and implementation issues
Torque is the easiest you can ballpark it and be safe.
Resolution is really easy - I tell you this: you will be wholly unsatisfied in the long run with your TL shots if your minimum resolution is greater than 0.08' - this is an empirical value I have arrived at through experimentation =) Best results are had once you get down to 0.01' or less as a minimum resolution. (That's 1/100th of a degree.)
A 200 step motor has 1.8' of resolution. So, we can easily calculate resolution:
360 / ( steps * usteps * gear reduction ratio)
Thus, with a 200 step motor, running in 8x microstep mode, and 24:1 gear reduction ratio, we get:
360 / ( 200 * 8 * 24 ) = 0.009375' per step.
There is a drawback to using microsteps: you'll be forced to maintain power to maintain holding position, even a motor brake, or braking effect in the gear train will not matter - full current must go through the poles in the motor to hold that position.
Cost and implementation are often the hardest part. You have a budget, you have a skill level. Figuring out that you can get 100:1 by using a 25:1 worm gear setup, chained into a 2:1 spur gear setup, leading to a 2:1 pulley set certainly works, but can you build it with your tools, and did you just lose all gains due to ineffeciencies in your implementation?
The fewer parts you use, with wider tolerances, the easier it is to build.
As for your dolly, say you take that motor and attach a 3" pulley to it and use an Omega drive configuration, generally speaking, you would get the following resolution:
circumference of pulley / ( steps * usteps * gear ratio ) - e.g.:
(3 * pi) / ( 200 * 1 * 24 ) = 0.049" per step.
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Price and time somewhat dont
Price and time somewhat dont matter right now. I would like to have the dolly motion done by middle of September. I bought the kesller crane/slider and that will do for the red epic and my other cameras. If I need to ill buy another camera to which is lighter to make this work.
My only worry is the size. This one because I have to hike up huge mountains and jump off those mountains it has to be small and lite.
I will be buying a pre built pan/tilt head and then swap the dc motors for stepper motors and just try my best to match the ratio. All I need to worry about is dolly/linear motion.
Sorry my math is off I have so many numbers on my mind I am going crazy.
So after shutterdrone math the http://www.robotshop.com/geared-bipolar-stepper-motor-3v-416-oz-in.html stepper motor will fit my requirements.
If I did a direct drive with a 20 tooth timing belt pulley do you think I can get some more rpm out of the motor? I know bigger pulley faster. Smaller more power.
Right now I am going to get the arduino uno with the lcd keyboard and install minimal engine code just to test it. The motor will be running off the easy driver controller.
As for your pulley, the teeth
As for your pulley, the teeth count doesn't help me much (it could mean anything - there are a lot of different pitch belts out there) - what is the pitch diameter of your pulley? (What I accidentally called 'circumference' in the formula above.) Your pitch diameter should be listed on the pulley.
EDIT: I had an arithmetic error which led me to the wrong conclusion =) The next paragraph that follows is modified
The full speed of 23RPM on that motor gear combination is 110,400 steps per minute, which works out to 1,840 steps per second. This should be do-able in full steps easily using blocking code: it's 1840 uS between steps, so subtract the step time its self from 1840 and then step high, low, delay, and so on. Doing it using non-blocking code, where you can talk to the device and do other things while the motor is stepping is easy as well, until you start trying to add in more complex features like ramping, velocity smoothing, and so forth. So, take the possibility of reaching the maximum speed a question of the implementation you are doing.
But if you had a pulley with a pitch diameter of 3", the max speed would be (3 * pi) * 23, or 226 inches per minute. Do you really need to go this fast? At half that speed you'd still cover your entire dolly length in about 20 seconds. So, if 20 seconds end-to-end is ok, then get a pulley with 1.5" pitch diameter or run your stepper at half-speed.
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