gearing question again

[timmyb199]

hi guys thanks for all the help so far

ok so i am using chain and sproket and am confused abit

i put my gears on and bacame dumbfounded maybe i am tired maybe not but here we go

gear on stepper 13 tooth

gear on axle 40 tooth

equalling about 3 to one

but gear on x axis drive is same as on stepper so if i go from 13-40-13 am i actually 1 to 1 again or am i going to deep into it

and how do i use all this and figure out how many steps my motors will turn per inch

thanks tim

[timmyb199]

plus should i just go one to one i thought at first 3 to one would be better, this is a plasma table but i would like to do engraving also if possible

thanks tim

[timmyb199]

sorry again

here are stepper specs i am using the xylotex kit with 497 oz nema 23 steppers

1.8 degreee angle

[Torchhead]

Engraving AND plasma????......wow


So you are going from a 13 to a 40 then back to a 13 at the drive shaft? (confused look here....) Sounds like it's the same as a direct connection!
If you used a jackshaft and went from 13 to 40 and then 13 to 40 again you would be a 9:1 reduction. As is is you are just wasting chain and sprockets (:_)

[timmyb199]

yeah i have a jackshaft so i need a 13 tooth on stepper, a 40tooth on jackshaft and 40 tooth on both ends of jackshaft to gear it down right??

[Al_The_Man]

For a few extra $ I would have gone with Timing belt and pulleys, Rather than chain.
Resolution is (steps/motor rev) x reduction ratio x (final reduced element turn/inch).
Al.

[handlewanker]

Hi Timmy,in order to calculate gear and pulley ratios the basic formular is:- number of teeth on driving gear divided by number of teeth on driven gear will give you a ratio.
EG. If the driving motor has a gear of 20 teeth and the driven gear has 40 teeth then 20/40= 1/2, which means your output speed is half your input speed, but direction of rotation is reversed.
This is for a "simple gear train" which means there are only one driver and one driven.
If you introduce another gear between the driver and the driven gear, E.G. 20 driver, 50 intermediate gear, and 40 driven gear, you will still get 1/2 the output speed but now the output is in the same direction as the input.
This is still called a "simple train" The intermediate gear does not change the ratio.
The second method is if you have 20 teeth on the input driver gear and a driven gear of 40 teeth, then you add a second gear of 20 teeth to the shaft with the 40 tooth gear and mesh this with another driven gear of 40 teeth, you will get 1/4 of the inut speed. The formular is always drivers/ driven, which means that we have 20X20 = 400 = 4 1/4
40X40 1600 16
This is called a compound gear train.This also applies to pulley drives, only now you take the OUTSIDE diametre of each pulley instead of gear teeth.
To get the actual speeds involved you calculate the ratio as just shown, Eg if the input or motor speed is 3,000 rpm then the output, as in the last example will be 1/4 of this at 750 rpm. If you get an odd ratio e.g. 4/7 then multiply by the top number and divide by the bottom number, so if we have 3000rpm then 3000X4 /7 = 1714.3 rpm.
I hope this helps,
Ian.

[timmyb199]

thanks ian, i am still grabbing all this.

[Al_The_Man]

Originally Posted by timmyb199 and how do i use all this and figure out how many steps my motors will turn per inch

Going back to your original question.
if your steppers are 1.8°/step then this will be 200 steps/rev.
if your reduction ratio is 9:1 then this results in (9x200) 1800 steps for one revolution of the output shaft, if the output shaft requires 2 revolutions to move an inch, then you have a resolution of 1800x2=3600 equal to 1/3600 = .000277"/step.
Al.

[timmyb199]

ok i think i am getting this , but if i have 13 teeth on stepper and 40 on axle and 13 on final drive then i am still 1 to 1 right

i apologize for this ahead of time

[project5k]

ok, let me see if i get this right... you have a 13t on the motor, and a 40t on the jackshaft, then also on this same jackshaft there is another 13t that actually moves the chain that is tied to the gantry and moves it... yes??
the you do not have 1:1 you have 3.076:1 meaning that your motor will turn a little over 3 times for each rotation of the jackshaft... so then you take your motor rpm, lets just say 600 rpm for round math... so then the motor is spinning at 600, devide that by the 3 we got for reduction, and that means that the jackshaft is spinning 200 rpm... then to figure the travel rate, you would multiply the circumfrence of the other 13t on the jackshaft by the 200 rpm, so just as an example, if the circumfrence is 5 inches, then your traveling at 1000 inches per min.... also, if your motors are 1.8 deg, and with this gearing, you would take the 200 steps per rev, times your reduction, so thats 200*3=600 steps per rev of the jackshaft, and then devide that by the circumfrence of the 13t on the shaft and thats 120 steps per inch travel... does any of this make sence, or did i totally miss the question...

[handlewanker]

Hi P5K, you're right up to the end of the jackshaft. From here on one rev of the shaft is 13 teeth of the gear on it and this travels 13 X the pitch of the chain. If the pitch of the chain is for example 1/4", then we have:- 13x1/4"=3 1/4"
This is because the gear contacts the chain at it's pitch circle, not at the tip. If you look at spur gearing you will see that two gears meshed together roll on a pitch circle as if it were friction drive, and have protuberances called teeth to reduce the slip.
So for every turn of the jackshaft gear you travel 3 1/4" and your speed of travel is the jack shaft speed (200 rpm) times this, which is 650 inches per min. This is based on a motor speed of 600 rpm.
Working backward, for every step of the motor you will travel:-
3 1/4" divided by 600 = 5.4 thou or .0054" which is your table resolution per step.
This is based upon a motor of 200 steps per rev, and a jackshaft reduction of 1:3
Ian.