Re: Timing Belt Sizing

Nice post. But i disagree with your conclusion. It looks to me like
you calculated the minimum belt width needed with the simplest
loads. What if the table has to accelerate a several hundred pound
part and start the heavy cut at the same time. Jon Elson posted that
he snapped his belt by accelerating to quickly. Acceleration is
obviously a big concern.

I think what you should do is use your calculated number as a minimum
belt width and use it as a referance point. Then do what i said in
my earler post to you and use the maximum HP number your drives can
put out or you hope they will put out if you go to bigger drives.
Recalculate belt size with this new HP rating. Now look at the
differance in these 2 numbers i would guess a couple of belt sizes.
Now look at the differance in price for sprockets for these 2 belt
sizes, maybe a few dollars. You can spend about $400.00 in sprockets
and belts for 3 axis and have a marginal system that snaps belts or
spend $450.00 and run for 20 years without a problem.

Wally K.

--- In CAD_CAM_EDM_DRO@egroups.com, Hugh Currin <currinh@O...> wrote:
> 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