calculating RA torque

[alistairsam]

Hi,

can anyone help with calculating torque required at the RA shaft if the only parameters known are the weight and length of the scope?

I know torque = force x distance and
force = torque / distance
but how do you calculate force?

Reason I wanted to calculate is to make sure I use the right gearbox and mechanical components with the correct torque rating.

taking a simple example, a scope of 800mm length, with midpoint at the RA axis, RA shaft of 50mm, scope weight -> 7kg, ra shaft axis is parallel to ground at 0 deg.

is it possible to calculate torque required to overcome inertia and rotate the shaft with that scope?

[jenchris (Jennifer)]

Inertia is a strange thing - the way a bearing is tightened will determine the point at which friction is overcome and must be determined by experiment.
The system should be balanced anyway so the mass inertia will be almost zero as there is no acceleration involved (not at 15 degree an hour at any rate.)
All you are overcoming is the stiction between the bearing surfaces and the grease.
I reckon a small scope like that will come in at 20 watts no problem - depending on your mount size

[alistairsam]

Hi,

thanks for the replies.
the stepper motor i'm using is able to rotate the load with a total reduction ratio of around 670.

But I'm changing the gear train and am using a geared stepper with a reduction of 125:1, and the output torque rating of the gearbox is 100Ncm or 1000mNm.
This will further be reduced by a ratio of 4:1 with timing pulleys at the output stage.

so I just wanted to get a rough idea of what the torque required would be for my load or how to calculate it.
just wanted to understand the concept if we ignore friction at the bearings, mass of the gear, etc. just the basic calculation first and then consider friction and other factors.

or, since moment of inertia is dependant on mass and distance of the mass from the axis of rotation, how do you calculate force required to overcome the moment of inertia.

in practical terms, if the gearbox output torque rating as above at 100Ncm, how does the subsequent reduction ratio alter the final output torque. eg. if the reduction stage after the gearbox is 4:1, what would the final output torque rating be after reduction.

just wanted to get my head around the math involved or parameters to be considered.
another reason is that if i replace the stepper that came with the gearbox with one that has a higher holding torque, how do I calculate if the gearbox can handle it.

[alistairsam]

Originally Posted by jenchris

Inertia is a strange thing - the way a bearing is tightened will determine the point at which friction is overcome and must be determined by experiment.
The system should be balanced anyway so the mass inertia will be almost zero as there is no acceleration involved (not at 15 degree an hour at any rate.)
All you are overcoming is the stiction between the bearing surfaces and the grease.
I reckon a small scope like that will come in at 20 watts no problem - depending on your mount size

Hi,
Even if the load is balanced correctly, won't the moment of inertia be a factor to get it rotating, and won't that increase with the weight of the ota?

eg. there would be a significant difference when you try and turn the shaft by hand between a perfectly balanced ota that weighs 10kg and another balanced ota that's say 30kg on the same shaft?
how do you measure that?

[torana68 (Roger)]

[QUOTE=alistairsam;766030]Hi,
Even if the load is balanced correctly, won't the moment of inertia be a factor to get it rotating, and won't that increase with the weight of the ota?

yes there is a force involved in putting to motion that is higher than normal running , you would need to know friction of the bearings to be exact (operating temp, adjusted to manufacturers spec and or material etc these can be found on makers sites usually) but I dont think you need to be exact. You can apply a fudge factor to cover all issues, maybe 20% above normal running??
have a look at this
http://bftgu.solarbotics.net/tutorials_mech_torque.html

eg. there would be a significant difference when you try and turn the shaft by hand between a perfectly balanced ota that weighs 10kg and another balanced ota that's say 30kg on the same shaft?
how do you measure that?

torque.....the twisting force, you should consider an overload clutch as well or over engineer the drive to compensate.

[alistairsam]

[QUOTE=torana68;774147]

Originally Posted by alistairsam

Hi,

have a look at this
http://bftgu.solarbotics.net/tutorials_mech_torque.html

torque.....the twisting force, you should consider an overload clutch as well or over engineer the drive to compensate.

Thanks for the link. very useful, especially torque in geared systems.
I will over engineer by around 10 to 20% and use a clutch for safety, but i just wanted to know if the torque rating of the gearbox i've chosen is within my load requirements as there are multiple types of gearboxes with the same ratios but different torque ratings due to materials used and construction.

I could use a torque wrench to turn the RA shaft with full load and work out what the required torque is and work backwards along the gear train.

[torana68 (Roger)]

[QUOTE=alistairsam;774211][ multiple types of gearboxes with the same ratios but different torque ratings due to materials used and construction.

can you ask the manufacturer for load limits? if not what sort of gears are they and tooth width (im guessing spur but helical would be better or maybe worm?)

[alistairsam]

Hi,

its one of these which is similar to the ones used in the losmandy mounts.
an IIS user has replaced his mounts gearbox with this one.
difference is that the mount uses a worm wheel reduction stage after the gearbox, I'm using timing pulleys.
http://australia.rs-online.com/web/p/gearboxes/0336416/
as per the datasheet, the output torque is 1 Nm.

But there are several similar ones with different ratings, some with meta spur gears.

[jenchris (Jennifer)]

Yes there'd be starting torque inertia and other things trying to stop it starting.
The biggest part of that would be the bearing stiction, because nothing esle is stopping it moving except the bearing and Newtons 2nd law.
That's the reason for careful balancing - there's less likelyhood of the motor getting overloaded if there's no 'weight' on the system.
The motors for the keck scope are about 1/2hp - and the thing weighs tons but it is balanced.
Your motor loading is likely to be rated on wind resistance and interference with external parts like collision with the mount.

[alistairsam]

Hi

won't the moment of inertia of the whole mass be the biggest element that is to be overcome to get the load rotating since the body is at rest?
bearing stiction will also be involved, but if we assume zero friction at the bearings, there would be a reasonable amount of torque required to start the load rotating.
I've tried rotating the RA shaft by hand with a fully balanced load, and its a fair bit. just wish I knew how to calculate it.

as per the earlier link, we can calculate torque required for a weight suspended by a string over a gear wheel. but a telescope load would be quite different.

I will have to try some practical tests.

[AndrewJ]

Gday

won't the moment of inertia of the whole mass be the biggest element that is to be overcome to get the load rotating since the body is at rest?

Nope ( well partly no )
As someone noted earlier, a small motor can get a large inertial mass moving as long as its balanced and has low friction, but it takes time.
Where the effect of the inertial mass comes into play is
in respect to angular acceleration, ie "how fast" you can
"change the angular velocity" of the mass on its axle.
This will control how long it takes to get up to max speed,
for a given driving torque.
If you just google "Angular acceleration Torque"
you will get a lot of good basic descriptions on how it all ties together.

Andrew

[bojan]

Actually, with worm gear reduction, the main issue that will determine the required motor torque is the friction between worm and worm gear.
This friction will be highly variable, and is dependent (among other causes) on the pressure the worm teeth have to exercise on worm gear to achieve the required angular acceleration rate of the the scope/counterweight assembly.
As Andrew, mentioned, it could be surprisingly small at motor shaft.
But, this is not easy to calculate - so the experiment may be the only final judge here.

I have similar issue at the moment - recently I modified my eq 6 such that I have only timing belt with 12/60 reduction between motors (from old 5.25 floppy drive) and worm gears, and I am keen to find out how this arrangement performs with my Bartels motor driver. If the problem arises, I have the alternative ready: stepper with 0.48 Nm holding torque, nominal.

[joncarruthers (John)]

Hi Bojan, have you seen Alan's (AWR) 'How To..@ on torque ?
http://www.awrtech.co.uk/oddtopic.htm#TORQUE

[alistairsam]

Originally Posted by joncarruthers

Hi Bojan, have you seen Alan's (AWR) 'How To..@ on torque ?
http://www.awrtech.co.uk/oddtopic.htm#TORQUE

Thanks. thats a very good article. pretty much what I was looking for.

[bojan]

Originally Posted by joncarruthers

Hi Bojan, have you seen Alan's (AWR) 'How To..@ on torque ?
http://www.awrtech.co.uk/oddtopic.htm#TORQUE

Hi John,
Thanks for posting - no, I haven't seen this page earlier, however I was familiar with the stuff discussed there from other sources.
Very good reference