RE: [CAD_CAM_EDM_DRO] Does this make sense? wasRe:BentNookballscrews ?????

Kim -

Just to clear this up -

The coefficient I used of 0.2 is not the efficiency. It is a
coefficient that takes into account lead angle, pitch circle
diameter, friction angle, and friction coefficient. The range
they give for the friction coefficient is from 0.003 to 0.010.

BTW, you're applying the thrust force equation to a
preload-torque-drag problem. Apples and oranges.

Regards,

Carlos Guillermo
VERVE Engineering & Design

-----Original Message-----
From: Kim Lux [mailto:lux@...]
Sent: Tuesday, August 12, 2003 2:25 PM
To: CAD_CAM_EDM_DRO@yahoogroups.com
Subject: RE: [CAD_CAM_EDM_DRO] Does this make sense?
wasRe:BentNookballscrews ?????

See below.

On Tue, 2003-08-12 at 15:13, Carlos Guillermo wrote:
> Kim -
>
> So then by your wheel bearing example, if I adjust my wheel
> bearings for 50 lbs of axial preload, then it would take an
extra
> 50 lbs per wheel to roll the car?

By my calcs, it would take 500 lbs/31 inlb per lb = 16 in lb. The
ratio
of force to torque for a 5 TPI ball screw is about 31 lbs / in lb.

> I'm not disputing that the drag
> increases, but it is certainly not 1-to-1, and you must take
into
> account the moment arm length. You hit it on the head, though:
> The drag force is roughly PROPORTIONAL to the preload, and the
> proportion would be a function of the efficiency of the rolling
> elements.
>
> According to the formula that you read:
>
> Td = (Kp*P*L) / (2*3.14),
>
> where:
> Td = preload drag torque (Kgf-mm)
> Kp = preload torque coefficient = typically between 0.1 and 0.3,
> I'll use 0.2

The docs that I've read use a ball screw efficiency of 0.9 or 90%.
Here
is the difference between our arguments.

torque = force/(2 pi x TPI x eff) = force/ (31.4 x .9) for a 5 TPI
ball
screw.

The difference between our equations is that I am using 0.9 and
you are
using 0.1 to 0.3 ???? We both agree there is drag due to the
preload
and we are using the same equation, with different co efficients.


> P = preload force (Kgf)
> L = lead (mm)
>
> Then:
> For a 5mm lead screw with a 250 Kgf preload between ballnuts:
>
> Td = (0.2*250*5) / (2*3.14)
> = 39.8Kgf-mm
> = 0.390 N-m
> = 55.3 in-oz
> = 3.45 in-lb
>
> This torque would be a small fraction of the peak torque
available
> from the motors of a medium-sized mill.

Depends on the motor. The Camtronics are rated at 600 oz in peak.
I
call that 120 oz in continuous. 3.45 in lb = 55.2 oz in, which is
46%
of their continuous capacity.