Differences between HTD, GT, AT and T pulleys?

[peteeng]

Hi Mactec - The working strength of the belt does not really matter in this application, they are way strong enough. The GT belt does not state what reinforcement it uses. The T uses steel and its no surprise if the GT uses fibreglass that its much stronger. But designers still need to see stiffness figures to make a clear decision between different belts or different widths. For this application stiffness is the key you can either have a bigger belt or a stiffer belt but if it's stretchy and very strong then it's not as good as stiffer and strong enough. So I'm still looking for a Gates belt stiffness chart....

The mectrol belts table provided earlier quotes their AT5 working strength as >1000N with a SF=4 whereas the table you have shows a GT as 712N so if this is the deciding factor use a mectrol belt as this is clearly not whimpy. (They quote their T5 at 625N which is >>182N a per the Gates data I presume) ) Peter

again a note on the ATL profile - see attached. Its profile is slightly different to the AT profile to minimise backlash and to correct for the extra pretension required in synchronise or positional applications. Its up to sploo to get what he can now... I'm not hung up on the profile or the manufacturer if someone comes in and says I've got a wizzbang belt and its twice as stiff as others & same cost then I'm in. Just need the evidence...

[sploo]

Having run a few more numbers from the Optibelt catalog it looks as though I might have a problem with the proposed design. The 25mm AT5 belt itself is fine, but a 20 tooth to 60 tooth combination appears to have insufficient tooth engagement to carry the torque.

I don't completely understand the metric, but the belts have a W/mm rating based on the small pulley rpm (where the figure is higher for higher rpm figures). I'm guessing this is torque related (as the same wattage at a higher rpm requires less torque, vs that wattage at a lower rpm).

The doc suggests I should probably consider my power requirement to be around 0.126kW, but the above configuration has a nominal power capacity of just under 0.1kW.

Moving to a 30 tooth pulley for the smaller side results in more tooth engagement, and a nominal power capacity of 0.25kW (so well above 0.126kW). That obviously only gives 2:1 reduction, and there doesn't appear to be an AT5 pulley larger than 60 tooth.

I could use a 20 tooth small gear, but the figures suggest I'd need a 66mm wide belt! Further thinking required...

[mactec54]

Having run a few more numbers from the Optibelt catalog it looks as though I might have a problem with the proposed design. The 25mm AT5 belt itself is fine, but a 20 tooth to 60 tooth combination appears to have insufficient tooth engagement to carry the torque.

I don't completely understand the metric, but the belts have a W/mm rating based on the small pulley rpm (where the figure is higher for higher rpm figures). I'm guessing this is torque related (as the same wattage at a higher rpm requires less torque, vs that wattage at a lower rpm).

The doc suggests I should probably consider my power requirement to be around 0.126kW, but the above configuration has a nominal power capacity of just under 0.1kW.

Moving to a 30 tooth pulley for the smaller side results in more tooth engagement, and a nominal power capacity of 0.25kW (so well above 0.126kW). That obviously only gives 2:1 reduction, and there doesn't appear to be an AT5 pulley larger than 60 tooth.

I could use a 20 tooth small gear, but the figures suggest I'd need a 66mm wide belt! Further thinking required...

You can gain more tooth engagement by using 2 idlers and then a longer belt the 2 idlers will give you more tooth engagement like this snip this is a AT5 belt being used like this

[sploo]

You can gain more tooth engagement by using 2 idlers and then a longer belt the 2 idlers will give you more tooth engagement like this snip this is a AT5 belt being used like this

That's a great idea - thanks!

My calculations tell me I'd get 7 teeth engaged on a 20 tooth pulley. Getting that up to 9 would take the nominal power calculation to exactly 0.126kW, so that might well be a solution. I'll look into it.

[peteeng]

Hi Sploo - Publish your calcs and I'll review them if you like. Peter

[sploo]

Greatly appreciated; thanks. Table and page references are to the Optibelt Timing Belts document you posted earlier...

My input is likely to be a 0.06kW motor, 60rpm, calculated torque ~= 10Nm

Initial design:

• Small pulley (Zk) = 20 tooth AT5, 31.83mm dia pitch
• Large pulley (Zg) = 60 tooth AT5, 95.49mm dia pitch
• Drive centres from page 48; 0.7 * (95.49 + 31.83) = 89.124mm
• AT5 Belt length from https://www.bbman.com/belt-length-calculator/ = 390mm

Calculations:

• Table 2.2.1; c0 of 1.8 (wood processing, constantly running, up to 16h)
• Table 2.2.2; c6 of 0 (no idlers), c8 of 0.3 (will be starting under load)
• Table 2.2.3; c3 of 0.8 (AT5, <=440mm belt)
• With 2 idlers c6 = 0.4 and c3 probably 0.9 due to slightly longer belt
• Page 45; c2 = c0 + c6 + c8 = 1.8 + 0 + 0.3 = 2.1
• Page 46; design power of Pinput * c2 = 0.06kW * 2.1 = 0.126kW
• Page 49; engaged teeth on small pulley; Ze = 20/6 * (3-(95.49-31.83)/89.124) ~= 7 teeth
• PNSpec for 25mm ALPHA TORQUE AT5 - ST belt at 60rpm input = 0.017 W/mm
• Nominal power (PN) from above data sheet = 0.017 * Zk * Ze * belt width / 10^3 = 0.017 * 20 * 7 * 25 / 1000 = 0.0595kW

The nominal power calculation effectively seems to be torque related (W/mm for a particular rpm) so I'm assuming it's an indication of whether the belt will likely slip due to insufficient teeth engagement with the chosen pulley and belt width. The calculated PN of 0.0595kW is well below the "design power" of 0.126kW so I assume this design isn't sufficient.

If I instead use a 30 tooth small pulley, 47.75mm dia pitch, I get 12 teeth engaged:

• Ze = 30/6 * (3-(95.49-47.75)/100.268) ~= 12
• That gives a PNSpec of 0.017 * 30 * 12 * 25 / 1000 = 0.153kW
• That's above 0.126kW so should be OK, but only gives me 2:1 reduction

Returning to the 20 tooth small pulley, but assuming I could use idlers to get 10 teeth engaged:

• PNSpec = 0.017 * 20 * 10 * 25 / 1000 = 0.085kW

That's still less than 0.126kW, and would need 2 idlers, so I assume c2 = 1.8 + 2 * 0.2 + 0.3 = 2.5, so the design power is now 0.06kW * 2.5 = 0.15kW


A UK supplier has a "GEN III" AT5 belt claimed to be 25% stronger, but I assume that's irrelevant for PNSpec as it's about tooth engagement vs torque, not the actual strength of the belt.

A 24 tooth pulley with idlers (for 12 teeth engaged) gets me pretty close, but I was hoping to get 3:1 reduction, and my supplier's biggest AT5 pulley is 60 tooth.

Working backwards with the PNSpec calculation, I believe I'd need >37mm wide belt for the 20 tooth pulley with idlers, and >52mm with no idlers.

Moving to AT10 gets pretty pricey due to the cost of the larger pulley, so it looks as though a 30 tooth small pulley may be the only feasible option (if my calculations and assumptions above are correct). Adding idlers (therefore assuming 15 teeth engaged) should be worth it as it gives even more nominal power vs the increase in the design power due to the addition of idlers.

[peteeng]

Hi Sploo, is this a "turning" axis ie rotary or a trunnion? Peter

Edit- My first look is that you are considering this as a power drive and its not. It's a positional drive and the inertial conditions are different to a power drive. So have a look at the linear drive section. Plus describe in a bit more detail what your 4th axis will be doing. If you look at P44 at 60rpm the AT5 is good for 0.3kW >> 0.06kW. I've looked through your numbers and the maths is correct but C3 is for power drives. So I'll look thru the linear drive process and see how that fits.

Vmax for the AT5 is 80m/s and your doing 2.8m/s so the considerations of a power drive are diminished considerably. These considerations evolve around belts and teeth trying to fly off the pulley which they don't do in a slow positional drive with lots of pretension. Peter

[sploo]

Thanks Peter - to confirm, it is a rotary lathe axis.

Without going into lots of dull detail, I'm essentially looking at turning the exterior of large (maybe 50cm / 20" diameter) hollow wooden spheres. Rough calculations for an object on the lathe are maybe 12kg (26lb) in mass, with a moment of inertia of 0.5kgm^2.

With the 60W motor ultimately geared to a ~30rpm output (~20Nm of torque), and accelerating from rest to full speed at 0.25s, I make that ~12 radians/s^2 acceleration, with under 6Nm of starting torque required.

The objects will be mostly pretty symmetric around the turning axis (i.e. I'm not trying to spin an eccentrically weighted object), but my main concern is the surface speed at the largest diameter. My CNC machine is capable of cutting at around 3000mm per minute, but with a 500mm dia sphere having a circumference of ~1500mm, I need to be able to spin it as slowly as 3000 / 1500 = 2 rpm. Plus there's going to be a not insignificant torque load with a milling cutter ~250mm from the lathe axis - thus I'm very much after low rpm and high torque.

It's for wood, so I'm not trying to hold for thousandths of an inch accuracy, but I would like to be within 0.5mm on the surface of a 500mm dia sphere.

[peteeng]

Hi Sploo - That's an interesting application. Are you turning the timber with a lathe tool or a a spindle? Is the drive motor a stepper? Peter

[sploo]

I'm setting it up so that the CNC 4th axis will use an identical spindle to what's in my lathe (so I can share chucks and accessories). I may well rough turn an object on the lathe, then move it to the CNC to do the intricate stuff that can't be done on a lathe. For that I'll be using a spindle; no plans for using lathe type tools on the CNC; specifically because the point of the CNC will be to mill the off axis and non-symmetric (is that a word?) stuff.

[peteeng]

non-symmetric or asymmetric will do working some numbers... Peter

[peteeng]

Hi Sploo - Instead of working with their service factors on power I have worked with the belt strength figures from the timing section. I do this as for power applications belt life (and its service factors) are based on fatigue and in power applications its the fatigue of the steel wires that runs the show as they run fast for very long periods of time and clock up huge stress cycles. In this sort of application you won't fatigue the wires. So the calcs show that if you run at 20-40rpm you are way way below the 80m/s max velocity of the belt, 2.8m/s which shows that the belt is not influenced by centrifugal and long term fatigue effects. You do fatigue the belt however and only your application after a long time will know how the belt fails.

First comment is that the 20x60 combo only has 7 maybe 8 teeth engaged. The nominal allowable torque with 7 teeth is 9.75Nm so your cool at 10Nm with a SF=2.0. The nominal working tension is 613N which is well below the nominal 1560N max which has a SF=4.0. This number is to ensure the wires and cover are not stretched too much to start to pull apart. This is a bonding endurance failure. So in short it will work by the numbers. If you can get the pinion teeth count up its all the better. This is via an idler. A note on idlers, if you use an idler that bends the belt in the opposite direction to the other pulleys then you fatigue the belt faster. If possible use an idler that bends the belt in the same direction as the pulleys and this will then not affect the fatigue life. The smallest diameter pulley does this. In this case the smallest dia recommended is 15T and you have 20T so expect a good life.

So the life of a belt is determined by:
1) fatigue of wires - smallest pulley diameter and number of cycles at load
2) If very highly loaded and small pulley the wires disbond from the cover, another form of fatigue
3) The teeth fatigue due to high load, number of cycles and if very high speed they do not engage smoothly so there is an imposed improper fit issue that fails them over time.

I could not find a minimum teeth engaged recommendation only the specific teeth load calc. which in your case is good. Hope this helps. I'm sure if the starting data is good then the 3:1 straight with correct preload will be fine and simplest is best. Regards Peter S

calcs are based on table on Page 93 for AT5 belt. If you get ATL belts I think they have an extra 50% margin in strength from memory...lets check (look up metrol doc) 2560N vs 1560N so 2560/1560= 55% stronger more margin although teeth strength is same I believe...

[sploo]

Many thanks Peter.

I've reproduced your FN(spec) calculation (though shouldn't it be 3.5 x 20 = 70 for a 20T small pulley)? Meaning Mn would be approx 7.8Nm; though it would rise to approx 11Nm with idlers if that gave me 10 teeth engaged.

I can't find the formula for the Mn calculation though, and there's no mention of Dp in the document. There's MN = FN x dk / 2 x 10^3 but there's also no obvious explanation of dk.

I assume SF means safety factor, but that's not obvious from the doc either. Can I ask where you got that, and does SF=2.0 mean that the system would be safe for twice your computed 9.75Nm?

In summary though, I'm getting the message that the belt moving away from the pulleys due to centrifugal forces won't be an issue, the strength of the belt itself won't be an issue, and the torque that I'd be putting on the smaller pulley probably wouldn't be an issue (from the point of view of the belt slipping due to insufficient tooth engagement)?

It's likely that this system would run for at most a few hours per week (maybe 4-10), so fatigue due to long term use should be low.

[peteeng]

Hi Sploo - Sorry Table is on P93 with formulas. Thks for checking my numbers. Dp is pitch diameter which is my usual description not Optibelts. They use dw1 for the driver diameter. FN is per tooth engaged not # teeth on pulley.

MN =FN x dw1 / 2000 dividing by 1000 is to correct mm to m and the 2 is to get the radius. Last night I assumed the 2 was a hidden Service factor, sorry . See calcs attached in Optibelt nomenclature. The nominal power calc falls short of your 0.06kW but this is a fatigue consideration and in your application this can be discounted I feel.

If you run 20Nm all the time you will be up against their recommended max service tension and you may get problems over time. But also they are conservative so suck it and see? Belts are made to a normal distribution and they have a spread of performance so you may buy a great belt that lasts forever, you may get a lemon that fails early.....Peter

By the way I calculated the 7 teeth engaged by drawing the pitch diameters at your 89.12mm spacing and calculating the tangent angle hence the contact distance. In CAD. Then divided the contact distance by 5 to get the 7.7 teeth. The minimum clamp teeth is recommended at 6 which indicates 7 engaged would be fine. I'm an optimist when doing these things sometimes...

I tension the belt with a tensioner at the crown or the pinion. I find it's cheaper for me to do this then build an idler and it becomes more compact.

[peteeng]

Hi Sploo - Looking closer at the Optibelt calcs I see for an Alpha V belt its Zeb teeth max is 6 whereas a Alpha linear is 12. This means the Alpha V is stiffer and can only distribute load over 6 teeth. Can you get The Alpha V belt ? In which case 7 meshed is greater then 6 so all's good?

Then the FN=3.5x6x25= 525N and MN=10.2Nm which is on target.

I take the specific tooth load at the rpm to mean that if you have this load or less the tooth will last forever. Reading further there is a discount on short belts due to the join, so I think your failure mode will be the join sometime in the future. Some of the short belts are not joined so then this element disappears. Getting a bit myoptic now. Peter

Edit - I had a further look and the Alpha V is a welded belt ( edit - in that example) and they have a further SF=2.0 for the join so the recommended max working tension is 780N not 1560N but that's Ok 525N<780N. So it seems in your case the belt join is the limiting factor. 20Nm takes you to 1225N so every time you go over 780N Optibelt is saying you chew into the join life. 780/525=14.8Nm. So <14Nm infinite life over that, life is reduced? Then there's the preload which imposes a further tension all the time...

Edit - I think the short belt you have will be cast so no join... Carry on as intended Peter

[mactec54]

Hi Sploo - Looking closer at the Optibelt calcs I see for an Alpha V belt its Zeb teeth max is 6 whereas a Alpha linear is 12. This means the Alpha V is stiffer and can only distribute load over 6 teeth. Can you get The Alpha V belt ? In which case 7 meshed is greater then 6 so all's good?

Then the FN=3.5x6x25= 525N and MN=10.2Nm which is on target.

I take the specific tooth load at the rpm to mean that if you have this load or less the tooth will last forever. Reading further there is a discount on short belts due to the join, so I think your failure mode will be the join sometime in the future. Some of the short belts are not joined so then this element disappears. Getting a bit myoptic now. Peter

Edit - I had a further look and the Alpha V is a welded belt ( edit - in that example) and they have a further SF=2.0 for the join so the recommended max working tension is 780N not 1560N but that's Ok 525N<780N. So it seems in your case the belt join is the limiting factor. 20Nm takes you to 1225N so every time you go over 780N Optibelt is saying you chew into the join life. 780/525=14.8Nm. So <14Nm infinite life over that, life is reduced? Then there's the preload which imposes a further tension all the time...

Edit - I think the short belt you have will be cast so no join... Carry on as intended Peter

I think you are confusing yourself

Optibelt ALPHA V are joined endless timing belts manufactured from Optibelt ALPHA linear belting.

The belt join means that the tension cord is not endless.

They are primarily used in transport/conveyor drives but they can also be used for normal drives provided a reduction in power transmission
capability is accepted.

As you can see a joined belt like this has limited use and not what he should be using

[peteeng]

Hi Mactec - The numbers indicate that an Alpha V will work in the given conditions and the consideration is the join life. It may last a month, a year or forever in Spoos application. An application or design engineer from the company would be able to help from experience to choose a belt, there are many out there and people have favorites. If a jointless belt is available then its clear thats preferable. All belts should be inspected regularly and replaced when deemed needed. We aim at infinite life but in reality this is not achieved.

The exercise has been worthwhile to gain a better understanding of the belt design process (for me at least as I had not thought much about joins before) and gives Sploo the information to talk to a belt rep in confidence. Thanks for umpiring. Peter

[sploo]

Peter/Mactec - thanks for the extra input.

I've reproduced Peter's figures (I also realised the "25" in one of Peter's calculations is of course the belt width, not the number of teeth on the pulley).

In terms of belt choice, my supplier has the Conti Synchroflex belts, and although I couldn't immediately find the same level of spec detail as the Optibelt catalogue, the supplier claims that the Optibelt AT5 is essentially a copy of the Synchroflex belt and "from what I can work out they just used the Synchroflex catalogue figures".

That (hopefully) being true, our calculations should be applicable to the Conti belts. They also have a GEN III belt, which is rated as 25% better than the standard Synchroflex AT5. The only thing not clear to me is whether that translates to a 25% increase in running load. If true, that's great (as it takes the MN [load] and PN [power] figures of the belt above what the motor can provide). If the calculations are about whether the belt will slip on the small pulley then I guess that 25% improvement doesn't apply - but if I understand Peter correctly it is about fatigue on the belt due to load, so probably the GEN III will help.

In short, with 7 teeth engaged on the 20 tooth pulley I get an MN of 9.46Nm, and a PN of 0.0595Kw. If I use idlers to get 10 teeth engaged that goes up to 13.5Nm and 0.085kW respectively. If the GEN III belt would increase those figures by 25%, that becomes 16.9Nm and 0.1kW (vs an input to the small pulley of 10Nm and 0.06kW). Should be OK?

EDIT: Having just found a Conti Syncroflex data sheet with the figures; it appears my supplier is correct; their basic AT5 belt looks pretty much identical to the Optibelt in performance terms, with the GEN III having roughly 25% higher W/mm figures (though they use W/cm instead). I guess that bodes well for the GEN III for my application.


PS As an aside, last night I knocked up a terrible hack with a smaller motor I have; ultimately it would have been providing around 2.25Nm to the small pulley; which was knocked up out of plywood on my lathe for use with an old SPZ belt. The ratio of the larger (also plywood special) pulley on the lathe spindle resulted in a further reduction of about 1.5x; for a "massive" 3.5Nm of torque at the lathe spindle. Even given that low torque figure it wasn't particularly pleasant trying to grab the chuck on the spindle to stop it, and the planned setup should result in something nearer 10 times the torque. As such, if the belt will hold up to it I'm hopeful it should do the job.

[peteeng]

Hi Sploo - How did the mod work out? Peter

[hanermo]

Fwiw..
Even with HTD-8 profiles, about 3x and HTD-5, and 30 mm wide belts,
the results are not good for 4th axis apps.

Positioning, yes.
Angular holding, no.
With 10.000 count ac industrial 220V servos, 1:3 belts, 10 Nm cont, 30 Nm peak, 90 Nm at spindle.