Re: Timing Belt Sizing

Crew:

I've gotten a few comments back regarding my selecting a
timing belt 2:1 reduction in attaching steppers to my
Bridgeport size machine. As I suspected everyone is
using smaller belts than I specified.

However, I think I'm starting to understand the difficulty.
I think timing belt selection takes into account overload,
i.e. an electric motor will produce a torque 2 or 3 times
the running torque during start-up. In contrast to this a
stepper rating is the maximum torque seen, static holding
torque. If true the belt capacity need be only about half
what standard selection procedures indicate. Any thoughts?

Also, you want to stay way under the maximum torque a
stepper can apply. This because it is such a disaster
if a step is lost. Anyone have a rule of thumb here, I
just heard something like 10 to 1?

Let's also consider this from the other end. (Anyway,
someone asked about this earlier, motor sizing) SO,
what is the force on the table when taking a "Big"
cut? Let's consider a one inch four flute HSS end
mill taking a full width cut in 1018 mild steel.
From Machinery's Handbook I find this cut requires
a cutting speed of 110 fpm and a feed rate of .003
inches/tooth. From this I calculate a spindle speed
of 380 rpm and a feed of 4.6 in/min.

Now consider power. Let's say the mill has a 1.5 HP
motor. Again from Machinery's Handbook I find for
mild steel a power constant Kp of .74 HP/(in^3/min).
In one minute this cut removes 1" wide x 4.6" long x
d" depth.

.74 = 1.5/(1x4.6xd) for d = .44"

To me this is a Big cut.

We don't need all these number to estimate table force
but I wanted to see what kind of cut we're talking
about. Now, let's assume all the force is taken
perpendicular to tool travel. This isn't quite true
but I think reasonable for ball park figures and will
give a slightly larger force than actually exists.

Then the torque for this 1.5 HP mill running at 380 rpm:

T = 63000 x P / N

Where T is torque (in-lb), P is power (HP) and N in rpm.
For 1.5 HP at 380 rpm the torque T is 247 in-lb. The
moment arm is .5" for this 1" end mill so the force of
the cutter on the table is 495 lb.

To move the table against this we need to also overcome
friction in the ways. Let's assume this is 50 lb,
seems reasonable. The maximum force to move the table
is then about 550 lb. Let's not consider acceleration
here. This 550 lb is the biggest force we can generate
with a 1.5 HP motor so I think it's unlikely any
acceleration force would exceed this. We could however
figure what acceleration this force would give and see
if it's adequate. Anyone?

Now for a ballscrew with a 0.2 in/rev lead we can find what
torque is required to move it against this 550 lb linear
force. Assuming there is no friction in the ballscrew
(reasonable ball park estimate) the equation for a power
screw becomes:

T = F L / 2 pi

T is torque (in-lb), F is force (lb), L is lead (inches), and
pi is 3.14159. This gives a lead screw torque of 17.5 in-lb.
With a 2:1 timing belt reduction this gives a motor torque of
about 8.75 in-lb or 140 oz-in.

Now if this is the max torque actually seen, or anywhere near
it, the belts are smaller. With this torque and a service
factor of 1.7 I could use a 1/2" wide L belt 10T (1.2" diam)
or bigger. XL is still too light.

Unless something else comes up I now plan to use this size
belt. L's don't readily come in 1/2" so I'd use a 3/4" wide
belt with a small pulley of 10 or 12 teeth. Some of the
metric sized belts may match better though, don't have
catalogs.

Anyone have a good source for timing belts and pulleys?

Whew! Sorry for the length of this post. If you're still
with me, thanks for listening. Any comments or suggestions
would be appreciated. Or if I'm wrong please do let me
know before I spend $. :-)

Thank you.

Hugh Currin
Klamath Falls, OR

Hugh Currin, PE
M&M Dept.
voice: 885-1649
fax: 885-1855