Calculating Torque with Auger 1

[DEVELONET]

I am developing an application with an auger in which this auger will be pushing balls up a tube. I am assuming this is working much like a worm gear. The weight of the balls equals about .5 lbs but I would like to have 2.5lbs of pressure applied to them. The auger will be powered by a DC motor and I need to figure out the torque rating needed for this motor to push this kind of weight in inch pounds. The motor needs to spin at 1500-1800 RPM if that is helpful. Does anyone have a formula or formulas for calculating something like this? The auger spec drawing is located at

http://www.develonet.com/auger1.GIF

Please detail your formula if you have one.
Thanks a ton!

 

[aviat]

There is a formula in Marks Handbook for Mechanical Engineers for screw converyers you could look at.

 

[EnglishMuffin]

How about this :

T = n*Wb*f*p/(2*pi)

Where T = torque applied
p = screw pitch
n = number of balls present in 1 turn of screw (probably non-integer)
Wb = weight of a ball
f = desired weight multiplier factor (5 as specified)

This totally ignores any friction effects which might be present. Hopefully, your 5x multiplier will take care of that. Figuring out n is an exercise in geometry.

 

[Bgoldstein]

Or how about treating this as a power screw? Then could use this equation (from Shigley and Mischke):

T=F (d_m/2) [sin(a) + u cos (a)] / [[cos (a) - u sin (a)]

F= force/load being moved
d_m = mean diameter
u = coef of friction
a = lead angle of auger

 

[EnglishMuffin]

Bgoldstein :
Since my equation is for a frictionless power screw, your equation should reduce to the same thing as mine for u = 0, and indeed it does.

Using your nomenclature, without friction your equation would be:

T = F*d_m/2*tan(a)

Since tan(a) = p/(pi*d_m), we obtain

T = F*p/(2*pi)

which is the same as my equation where F = n*Wb*f

Exactly how one should incorporate friction seems somewhat arguable, (possible extra friction on side walls, unknown entry and exit conditions etc), but since DEVELONET is using such a large safety factor, it would not seem terribly important, at least to me.

 

[Bgoldstein]

E_Muff,
I'll buy your argument, since we are talking about balls which are probably rolling, so negligible friction. (Especially with your fudge factor, f)

 

[EnglishMuffin]

Well, friction wouldn't necessarily be negligable because they are rolling, since they will rub against each other and the side walls, but we seem to agree in principle.

 

[EnglishMuffin]

Actually, one other point is that my equation gives the torque required to lift only the balls contained in one pitch. You need to multiply by the number of pitches in the screw to get the total torque.

 

[Spurs]

Follow the attached link.

There are two pieces of software that will help you with your problem.

One deals with worm drives where you can calculate torques for a given axial load

The other has to do with matching a DC motor with a gear box to achieve a desired speed of operation.

http://members.rogers.com/webgear/home.htm