running motors in pairs

hi all..

if i run say 2 nema 17's, one on each X axis and one on each Y axis in 'pairs', would that in effect double the torque/power?

or is it better to just run larger nema 23's.. as in does two cheepies perform as well as the bigger ones?

which leeds to... conecting them up? whats the difference between parraleling the pair and series? pro's cons?


(i've read a few places its better to run in pairs as its more accurate..)

so it looks like parralell for sure for conecting up. series tends to draw power to all the other motors in the curcuit when only one is taxed, verses parallell they tray and maintain the same speed but draw diferent powers to each irrispective of the others needs.

so, that leaves what to power them with, what system? x y and z 6 motors 2 per axis.(posably two on the extruder...??)

any one?

Edited 2 time(s). Last edit at 11/15/2014 05:21PM by munchit1.

no takers? hey-ulp? hey-lup??

hmm on me tod.. billy no mates boo hoo!

Whether two NEMA17 have more torque than a single NEMA23 is like asking how long is piece of string. You need to look at the specs of the motors. In general bigger motors give more torque per amp than smaller ones.

If you wire them in parallel you need a driver that can handle twice the current, or use two drivers with the step and direction inputs linked.

If you wire them in series the current is the same but you need to double the supply voltage to get the same top speed. Also they stay in sync better if you wire them in parallel.

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[www.hydraraptor.blogspot.com]

i was thinking two 1.7A nema 17's versas an equivelent of the two....hmm i see what you mean now i've started typing.

ty.

nema23 stats..

Specifications
Size: 56.4 mm square × 56 mm, not including the shaft (NEMA 23)
Weight: 0.7 kg (25 oz)
Shaft diameter: 6.35 mm (0.25″) “D”
Steps per revolution: 200
Current rating: 1 A per coil
Voltage rating: 7.4 V
Resistance: 7.4 Ω per coil
Holding torque: 9 kg-cm (125 oz-in)
Inductance: 10 mH per coil
Lead length: 30 cm (12″)

nema 17 stats

Specifications
Size: 42.3 mm square × 38 mm, not including the shaft (NEMA 17)
Weight: 285 g (10 oz)
Shaft diameter: 5 mm “D”
Steps per revolution: 200
Current rating: 1.68 A per coil
Voltage rating: 2.8 V
Resistance: 1.65 Ω per coil
Holding torque: 3.7 kg-cm (51 oz-in)
Inductance: 3.2 mH per coil
Lead length: 30 cm (12″)

the 17 producing 51 onz-in, so for 2 it would be 102 onz-inch-ish for 3.4amp, verses the 23's 125 onz-inch for 1Amp..

the answer being no...all bar the very low end 23's with higher rated 17's.

i thinks it all needs the calculasioms doing being even thinking about buying a motor.

Edited 4 time(s). Last edit at 11/16/2014 04:08PM by munchit1.

You are comparing apples with pears because the NEMA23 is higher voltage and inductance so it will have more torque but be slower. A better comparison would be with a lower voltage NEMA23.

Also the NEMA17 you picked is only 38mm long. A 47mm NEMA17 would have a torque of about 4.2kg.cm for the same current and voltage.

It also depends on what driver you intend to use as that is often the limiting factor rather than the motor.

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[www.hydraraptor.blogspot.com]

Quote
nophead
You are comparing apples with pears because the NEMA23 is higher voltage and inductance so it will have more torque but be slower. A better comparison would be with a lower voltage NEMA23.

Also the NEMA17 you picked is only 38mm long. A 47mm NEMA17 would have a torque of about 4.2kg.cm for the same current and voltage.

It also depends on what driver you intend to use as that is often the limiting factor rather than the motor.

need the math doing on a build before looking doesnt it? or its like the question 'how longs a peace of string?.. from one end to the other'' ok its answered but no ones any the wiser lol.

lets start with?

the gravity thing.. 9.8, the gravivity number is 9.8 (summed up to one decimal place). 'g'.

so mass 'm' is the weight divided by the accelerasion of gravety, which is in the case of a printer, the weight it everything on that one axis, the print head the bars the bearings etc, weigh it or what ever...i'm in kilograms at the mo.(the numbers work out so much easier lol).

mass = weight/9.8 (m = w / g)


the answers in newtons. now then we're working in kilograms so is that newton kilograms?

and gravety is 9.8 m/s squared... what does the squared mean in turms of calculating?

Edited 2 time(s). Last edit at 11/18/2014 05:17AM by munchit1.

Mass is measured in kg and force is measured in Newtons. The force on an object is mass times acceleration F = ma, so the force due to gravity is F = mg.

Weight is also expressed in kg but is usually a measure of the force due to gravity with the result displayed by dividing by 9.8. Weighing scales that do that would give the wrong answer on the moon.

To work out the force to accelerate your axis in Newtons it is mass in Kg times the acceleration in metres per second. To convert that to motor torque you multiply by the pulley radius. g does not come into it unless your axis is vertical, in which case you would add it to a. I.e. F = m(g + a).

Just to add confusion motor data sheets often show torque in Kg metres and convert Newtons to Kg by dividing by g. Not technically correct but force is often expressed in kg outside the scientific world just as weight is.

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[www.hydraraptor.blogspot.com]

ahaaa...thank you very muchly.


just for by the by.. for any one starting or wandering how to play, you'll need some transposition... http://www.youtube.com/watch?v=UsMv7_Dg9-U

Edited 1 time(s). Last edit at 11/18/2014 05:53AM by munchit1.

so to run through..

the X axis (printer head bars mounts etc total weight) say its 1kg, for ease..
mass would be = 1kg x 9.8 = 9.8

and the accelerasion i presume you choose one just to get started?

so 9.8 x 0.5 m/s/s = 4.9 so the F = 4.9 n

pully (on the motor spindle?) has a radiuos of 5mm i'll asume for the maths its in meters?

4.9n x 0.005m = 0.0245 torque.. in what denominasion?

Edited 1 time(s). Last edit at 11/18/2014 06:33AM by munchit1.

Quote
munchit1
so to run through..

the X axis (printer head bars mounts etc total weight) say its 1kg, for ease..
mass would be = 1kg x 9.8 = 9.8

and the accelerasion i presume you choose one just to get started?

so 9.8 x 0.5 m/s/s = 4.9 so the F = 4.9 n

pully (on the motor spindle?) has a radiuos of 5mm i'll asume for the maths its in meters?

4.9n x 0.005m = 0.0245 torque.. in what denominasion?

No if the weight is 1kg then the mass is 1kg.

Weight is how much force an object experiences due to gravity. It should be expressed in Newtons but is actually displayed in kg by dividing by g and assumed to also be a measure of the mass. If you take an object up a very tall mountain or to the moon it will weigh less but its mass will be the same. So your weighing scales make an assumption about gravity. If they displayed the result in Newtons they would always be correct but people want to know the mass more often than the force, so the scales do a handy conversion.

So to accelerate 1kg at 0.5 m/s^2 the force is 0.5N. With a 5mm radius pulley the torque is 0.5 * 0.005 = 0.0025 Nm.

To convert that to Kg.cm divide by 9.8 and multiply by 100.

There is also the force to overcome friction and the torque to accelerate the motor's rotor inertia.

When comparing with the motor's data sheet torque you have to note that it is the holding torque (i.e. stationary) with two coils on at the stated current and a temperature rise of 80C. We never run motors that hard as it would melt the plastic brackets. Also Allegro motor drivers quote the current as the peak current into one coil. So when people typically set VREF to get 1A that is only 0.7A when both coils are on. So with a 1.7A motor the torque is only 0.7/1.7 times the data sheet value. Fortunately the temperature rise is a square law so it will only be 80 * (0.7/1.7)^2 = 13C above ambient.

Also note that the torque produced by the motor reduces with speed depending on its inductance and the supply voltage.

Having just enough torque for the acceleration is not the best situation either as excess torque gives better positional accuracy because the motor will come to rest with an offset proportional to the static friction divided by the holding torque.

Edited 3 time(s). Last edit at 11/18/2014 07:44AM by nophead.

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[www.hydraraptor.blogspot.com]

hmm, so i'd used forse as the mass..

and i'm asuming a nema 17 with 28Ncm holding torque is way ample for 1 kg on a pully system horozontaly?.............(asuming the ''So to accelerate 1kg at 0.5 m/s^2 the force is 0.5N. With a 5mm radius pulley the torque is 0.5 * 0.005 = 0.0025 Nm'' is A, the 0.0025 bit into Ncm is 0.25Ncm, and listed 28 holding talk are of diferent decimal places).

http://www.aliexpress.com/item/5-pcs-lots-Nema17-17HS3401-4-lead-Nema-17-Stepper-Motor-42-motor-42BYGH-1-3A/1887674673.html?s=p

i've been pulling some really strange numbers out of my calcs lol.

another asumption i'm using would be, if the holding torque will accelerate the object (printer head) then after it starts moving the required torque lessons to continue moving? (so the same applied 'forse'would actualy accelerate up to a speed over a longer distance, like a terminal type velocity after say a second or over 10cms type of thing?

Edited 2 time(s). Last edit at 11/18/2014 08:14AM by munchit1.

and part 2.

''Having just enough torque for the acceleration is not the best situation either as excess torque gives better positional accuracy because the motor will come to rest with an offset proportional to the static friction divided by the holding torque''

is this like the momentum builds up as it travels being greater than its start momentum, in affect stopping the preverbial car at 80mph is way heavier than at 5mph?

Your figures show much less torque than the motors have because 0.5m/s^2 is a low acceleration for a Reprap machine and you haven't factored in the fact that the torque of the motor is much less at high speed. Typically machines don't start losing steps if they move very slowly but do when they attempt to go fast.

Yes the carriage only accelerates and decelerates in at the start and end of the travel, typically less than 1mm. In between you only have friction to overcome.

Momentum is simply mass times velocity and is nothing to do with friction. When an objects slides over another the coefficient of friction is a bit lower after it has started moving.

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[www.hydraraptor.blogspot.com]

thank you nophead, thats given me stuff to think on..