Jon, Here's a screen shot (assuming it makes it through). I used your 100 oz-in motor and 0.2" pitch screw. I set the RPM to 3000 but with a 3:1 reduction. Not sure If it would be better to just add two boxes for the reduction so that 0.25 is really 4:1 in two boxes. Using your table weight of 200 lbs and the resulting Continuous Linear force of almost 200 I get approximately 1 G as expected. Now the ideal would be that the last box there would tell us what acceleration value we want to enter into the INI file. So would you multiply the 0.98G by 32 Ft/Sec/Sec to get 31.36 or turn it into inches 376 Inch/sec/sec but since our table speed is in inches per minute divide by 60 or by 3600? For inch/min/min?
Oh and of course the checkmark will convert the numbers and units indication to N.m etc. Doesn't yet but that's just a bit of math and text output. John Dammeyer > -----Original Message----- > From: Jon Elson [mailto:[email protected]] > Sent: July-21-20 1:52 PM > To: Enhanced Machine Controller (EMC) > Subject: Re: [Emc-users] Need help with Bostomatic BD18-2 to linuxcnc > > On 07/21/2020 02:54 PM, John Dammeyer wrote: > > Hi Jon, > > How did you come up with the constant 0.0318? > Long story. There's complicated formula to compute it from > the helix angle of the screw. > Too much fiddling around. So, if you had a drum with a > string wrapped around it that advanced the > axis as much as the leadscrew, it should be equivalent > (ignoring friction and diameter of the string). So, how big > would such a drum be? The circumference should be equal to > the pitch of the screw, in this case 0.2". So, what is the > radius of a circle with a 0.2" circumference? 2 Pi R = > Circumf. so > R = circumf./2 Pi so, 0.2 / (2 Pi) = 0.0318 > > Now, if you apply 10 in-Lb torque to a 5 TPI leadscrew, you > get 10 / 0.0318 = 314 pounds force. > > > > " So, that 100 oz-in motor (0.52 lb-ft) would > > produce 0.52/0.0318 = 16.35 lbs of linear force (neglecting > > friction)." > And, this was wrong due to a inch/foot mixup! > > 0.52 lb-ft is 6.24 in-Lb, and the linear force would be 196 Lbs. > > And how did you work out the 5G? > > > > "So, if your machine has a 200 Lb table, and the leadscrew > > were to produce 1000 Lbs linear force, > > it would accelerate at 5 G? > > > If you dropped the table on your toe (don't do this at home, > kids!) it would accelerate at 1 G. > if you push on the table with a force equal to 5 times it's > weight, that will accelerate at 5 G. > > Jon > > > _______________________________________________ > Emc-users mailing list > <mailto:[email protected]> [email protected] > <https://lists.sourceforge.net/lists/listinfo/emc-users> > https://lists.sourceforge.net/lists/listinfo/emc-users
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