Planetary reduction 1

[jucasan]

I'm working in a reduction system and would like some help to be sure this is going to work properly.

The electric motor (green) is rated to 1.4HP @ 7500 RPM and the shaft torque = 187 in oz. The belt reduction is 2:1 and the planetary system info is

Rated Output Torque (in-lbs):177
Max Output Torque (in-lbs):283
Ratio:32:1

I'm trying to make a mid drive electric system for a bicycle and my concern is if the planetary reduction is strong enough.

Any advice would be welcome.
Thanks

 

[BrianPetersen]

I hate old English units but can work with them enough to calculate that 1.4 hp at 7500 rpm requires very close to 1.0 lb.ft of torque. (This doesn't imply that this is the maximum torque at any conceivable rpm ... only that it's that torque at 7500 rpm at the stated power output, which is all we've got to work with.)

This will then give 2.0 lb.ft of torque at 3750 rpm at the INput of the gearbox.

And at that ... your gearbox rating concerns the OUTput condition of the gearbox, not the input.

How the input condition relates to the output condition, depends rather strongly on what the internal reduction ratio inside the planetary gearbox is ... which you haven't told us

 

[jucasan]

ups, sorry the planetary reduction is a 32:1

 

[BrianPetersen]

2.0 lb.ft input x 32:1 ratio gives 640 lb.ft output, i.e. 7680 in.lb. It will be somewhat less in reality by the efficiency of the gearbox.

Given that this exceeds all ratings given for the gearbox by a factor of around 25 or 30 (can't be bothered to do the exact math, but it's irrelevant at this point) it's safe to say that it's gonna break!

(Do you really need 640 lb.ft of torque to move a bicycle??)

 

[jucasan]

Are you sugesting a smaller motor? That is the one I have. I'm probably going to need a stronger planetary reduction. What kind of output do I need. Like 640lb-ft output? Sorry but as you can tell this is not my field of expertice.

Thanks Brian

 

[handleman]

What is your field of experience?

I think Brian may have hit an extra zero somewhere... 2.0 x 32 = 64, not 640. So you're only at 2 to 3 times your max torque.

You need to approach this from the other end. What portion of the bicycle are you trying to drive? How much torque you need depends on where you're trying to drive it from, and the performance requirement of the bicycle.

 

[jucasan]

What is my field of experience??? I'm a certified tarod reader. And what I see right now Handleman is a great sucess in your life. Tons of money...it is a little bit blurred but I think I see a Ferrari BUT ONLY IF YOU HELP ME. If you don't.....sad to say.

I know the system looks more complicated than it should. I'm doing a system that is going to allow to engage or de-engage the motor and pedal with freewheels but you don't have to worry about it.

As I mentioned before the motor charasteristics are:
1.4 HP @7500RPM. Shaft torque 187 in oz . The primary reduction with a ratio of 2:1 and the planetary with a ratio of +/-30:1. As you can see there is a lot of room for fine tuning by changing the chain rings so there is no need for precision. I need somewhere close to 100 RPM on the output shaft of the planetary.

These are my gears options.
Link

Help please. Thanks

 

[3DDave]

I love the smell of bridges being set afire.

Edit - and that's a 1:2 step up at the start. So you have half the torque at the input to the planetary as produced by the motor.

 

[BrianPetersen]

Busted on the extra zero. It's only 2 or 3 times not strong enough!

Just pick a planetary reducer with a smaller reduction ratio and spin the motor at a slower speed.

 

[handleman]

OK... The image helps. Don't put such a big dang chainring on your gearhead. A small sprocket there instead of a big chainring will give you another great opportunity to gain torque using components designed to handle it. I'm not sure what sort of control you have on your motor and whether or not you can intentionally limit the torque output to stay below the reducer limit.

Typical crank arm is, what, neighborhood of 7 inches? And say you have a 180lb rider. So, max (intermittent) torque is around 1260 in-lb. You have a max of 283. So you need about a 4.5:1 ratio from the output of the gearhead to the chainring to duplicate a 180lb rider's max torque. According to a quick Google search, typical highest ratio from chainring to cassette is about 4:1, so you're somewhere in the realm of feasibility. Now get rid of your reduction between the motor and gearhead so that you don't put in more torque than your planetary can handle. You probably also need to check the spec for max radial load on the planetary's shaft. Picturing a small sprocket, they're what, 2inch (or less) Pitch Circle Diameter? That means at max output torque the radial shaft load is 283lb. Bearings may not be able to take that.

 

[jucasan]

Thanks Handleman,

The planetary gearbox is rated up to 18000 RPM. If I make a primary gear ratio 2:1 with the motor at 7500 RPM the input in the planetary would go to 15000 RPM (still in the limits) and decrease the torque. Would this help?

 

[BrianPetersen]

Pick a gearbox with a more favorable reduction ratio, and quit struggling to find a way to use one that's wrong.

 

[handleman]

What is your plan for a bearing to mount the reducer input pulley? The coupling and input of the planetary reducer you show is designed to be directly attached to the motor, with a pure torque input. Neither the coupling nor the input bearings are designed for the radial and moment load of a belt and pulley. You need to add a shaft/bearing assembly that withstands moment and radial load.

I assume you are a student or a hobbyist who's found a mismatched motor and gearhead, but that's all you have and you need to make it work? Do you have or have access to appropriate machining and welding capability to get this stuff mounted square and securely?

 

[drawoh]

jucasan,

1.4HP [×] 550 ft.lb/sec = 770ft.lb/sec.

T = P/[α] = 770ft.lb/sec / (7500rev/min [×] 2[π] rad/rev [×] 1/60 min/sec) = 0.98ft.lb, or about 1ft.lb torque.

T = 0.98ft.lb [×] 12in/ft * 16oz/lb = 188oz.in.

You are putting 2ft.lb torque into your gear reducer, and 64ft.lb out of it. Bicycle wheels are around 26" diameter.

T = 64ft.lb [×] 12in/lb = 770in.lb.

F = T/R = 770in.lb / 0.5[×]26in = 15lb 59lb.[ ]

You are propelling your bicycle up a hill with 15lb 59lbforce.

If your rider and bike weigh 200lb, your maximum slope is

[θ]_max = asin (15lb59lb/200lb) = 4.3[°] 17[°] (draw a free body diagram)

Of course, I did not account for rolling resistance or air resistance. If this is a permanent magnet DC[ ]motor, the stall torque will be very much higher than the torque at 7500rpm. Can your gear reducer manage this?

More editing: If we assume this is a PM motor with power quoted at half of no-load speed, then the maximum torque and the force are doubled, and the bike will climb almost 36[°].

This was not very complicated. Except the arithmetic.[ ]

What side load can your reducer take? Interesting things happen when people spin wheels attached to highly speed reduced motors.

--
JHG

 

[BrianPetersen]

F = T/R = 770 in.lb / ( 0.5 x 26 ) = 59 lbs

Missed that critical set of brackets ...

In the diagram a couple of posts up, I spy what appears to be a chain speed-reduction on the output of this gearbox. I'll guess that this is about a 4:1 reduction (can't see all the teeth to count them) and then there is the chain speed-increase to the drive wheel whose ratio will vary.

Figure out what overall drive ratio that you need, and then do what's needed in the simplest manner possible.

 

[rb1957]

a clear description of the physics, but ...
"F = T/R = 770in.lb / 0.5×26in = 15lb." ... 770/13 = 59 lbs.



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