More maths..

[hotponyshoes]

So to work out how much motor power would be needed to turn a rotary table / weld positioner,

In a "worse case" the table was mounted vertically (like a lathe chuck) as a rotator with a 100kg load at 1m away from the center,

9.8nm = 1kgm so 9.8x100 = 980nm.

If the table has a 90:1 reduction then
980/90 = 10.8

So only 11nm of torque required to lift a 100kg load? Sounds low to me but then it is a 90:1 ratio..

Obviously not accounting for any drive losses or the fact that a 100kg item may well be larger than 1m but presuming the table will be used horizontally and only rotating a load rather than lifting it then the actual input torque needed will be in the region of sub a couple of nm or so?

 

[Aff]

Your gearbox should have an efficiency rating on the spec sheet. Depending on what type it is it could be anywhere from 60-90% so can have a large effect on required motor power.

 

[mtt.tr]

Use a spring balance to get a direct reading

 

[Brad93]

As said you’ve calculated a system with 100% efficiency.

 

[8ob]

As said you’ve calculated a system with 100% efficiency.

There is friction and sticktion to take into consideration, never underestimate sticktion

 

[hotponyshoes]

But aside from the losses (which I'm sure may be considerable) is that about right?

 

[Brad93]

I can’t imagine it’s much more. If anything i think it would be less for a uniformly distributed mass

 

[mtt.tr]

I can’t imagine it’s much more. If anything i think it would be less for a uniformly distributed mass

Doesn't make much difference according to coefficient of friction.

 

[zzr1200]

If the load is a horizontal cylinder then there's a part load, that is a balancing load moving down on one side of vertical, once the correct motive speed is attained, you only need to overcome the resistance forces in the positioner to maintain rotation speed...

 

[daedalusminos]

...and don't forget inertia.

 

[MoreWellie]

Assuming the table is horizontal I would say your calculations are correct but they certainly don't take account of the worst case vertical table scenario

you also have to look at what happens when the mass moves to the "downwards" side of the table in that case and it is going to need quite a reverse force to hold it up or else balance the load around the centre point

100kg on the end of a 1 metre arm is going to take a lot of effort to lift even through a 90:1 gearbox

 

[hotponyshoes]

100kg on the end of a 1 metre arm is going to take a lot of effort to lift even through a 90:1 gearbox

It's going to take 980nm of force to lift it if my maths is correct?
Divide that by the 90:1 reduction of the gear box and it needs 11nm.

As above, not allowing for any losses, but then he table will be used for turning not even lifting.

 

[daedalusminos]

As above, not allowing for any losses, but then he table will be used for turning not even lifting.

So doesn't that make your maths totally irrelevant?

If you have the table and mass then you can measure the torque required to turn it using a fish scale and lever.

 

[MoreWellie]

It's going to take 980nm of force to lift it if my maths is correct?

I am hung up on the situation when the table is vertical and the weight is on the end of a 1m arm, it seems to me the rotational force at the centre should be multiplied but I don't know how to calculate it

 

[daedalusminos]

I am hung up on the situation when the table is vertical and the weight is on the end of a 1m arm, it seems to me the rotational force at the centre should be multiplied but I don't know how to calculate it

I believe you find the centre of gravity of the mass then use the position of that as your moment

 

[MoreWellie]

my brain hurts

assuming you have the table mounted vertically, a 1cm dia shaft and a 1m arm I think you will need 2,200nm of turning force when the weight is horizontally out to the left (also assuming clockwise rotation)

Although why you would want to swing a 100kg weight around in a vertical 2m diameter circle escapes me, but I am very glad you are in Somerset

 

[hotponyshoes]

my brain hurts

assuming you have the table mounted vertically, a 1cm dia shaft and a 1m arm I think you will need 2,200nm of turning force when the weight is horizontally out to the left (also assuming clockwise rotation)

Although why you would want to swing a 100kg weight around in a vertical 2m diameter circle escapes me, but I am very glad you are in Somerset

How are you working that out?

 

[Pigeon_Droppings2]

Just the ratio of the radius of the mass to the radius of the force applied to the shaft I would guess.....not allowing for gearing?

It's just like a kids seesaw if you ignore the gearbox I guess?

 

[MoreWellie]

yes, as @Pigeon_Droppings2 said I used a 1cm diam shaft and 100cm arm which gave me a ratio of 200 times

 

[hotponyshoes]

Humm..

Where does the 200 ratio come from? Is 1cm to 1m not 100:1 or are you taking half the shaft, 0.5cm?

My working was that a mass of 100kg =980n
So at 1m away you would need 980nm to lift it.

Or more precisely, 980nm to hold it, 981+ to lift it.