Thanks.
I'll whip up a quick app for this.  Nothing special.  Could be done in as 
spreadsheet but where's the fun in that?
John


> -----Original Message-----
> From: Jon Elson [mailto:el...@pico-systems.com]
> 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
> 
> 
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