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bb99
10-26-2003, 04:22 PM
Hi All,

I'm in the beginning phases of my first machine design. I have several designs floating around in my head (and on paper). I would like to determine the size of motors I need for some of these designs. For example, let's say I have a gantry with a router motor attached and it all weighs some number of lbs. How strong of a motor do I need to properly move the gantry? I understand that motors are described in ounces/inch, but how do I determine if motor Y will hold up my Z axis that weighs X lbs?

Thanks,
bb99

Mariss Freimanis
10-26-2003, 09:52 PM
bb99,

Here's where you start. By the way, I think you are smart to ask that question first rather than getting a couple of motors first, then asking "How can I make them do what I need?'

First, how fast do you want it to go? Second, how many lbs of "push" do you need at that speed? Most people have a good grip on the speed but freeze on the "lbs" part.

If you are a machinist, you probably are familiar with manual machines. Ask yourself how many lbs of force you would apply to the manual crank before you figured you were doing something very wrong.

Example: Say your crank diameter is 4", your leadscrew pitch is 5 TPI and you say "10 lbs is plenty". The lbs of push would be:

Lbs = pi/8 times TPI times in-oz

Rearrainge it to sovle for in-oz of torque.

Leaving that aside, figure how fast you want to (IPM), multiply it by the Lbs af force you want and divide the result by 531. The answer will be in Watts mechanical to do what you need.

If the answer is 100W or less, think steppers. If it's 200W or more, think servos. If it's in between, either will do.

Mariss

bb99
10-26-2003, 11:16 PM
Hi Mariss,

Thanks for the quick response. I'm a computer programmer by trade. I have no experience in mechanical or electrical engineering so excuse me for asking lots of questions (I plan on asking tons of them!)

Here are my design goals for the CNC machine:
1 Woodworking projects- e.g. cutting out the parts for a grandfather clock
2 Woodcarving projects- e.g. sculpting the face of the grandfather clock
3 Milling of aluminum- once I get the xyz axes working I plan to have the machine mill the aluminum for its 4th and 5th axes
4 Tile cutting- haven't seen this addressed here at CNCZone.com, but I plan on doing some tile work in my bathroom
5 Cutting of brick/stone- this is more of a wish then a requirement
6 More to come I'm sure!

Assumptions:
1 The system will need a good sized work area (TBD)
2 Since this is for my own weekend use, cutting speed is nice but not critical
3 Price will factor into my final design, but since all this is costing me right now is some brain power, I'm ignoring this for now (I'm sure I'll pay close attention to it once my credit card is pulled (no pun intended!))

With that said, do the following answers to your questions seem reasonable?

Cutting speed = max out at 30 IPM
Push = 10 lbs (I have no real idea on what this number should be based on the design requirements above, so if anyone has a better number, please feel free to correct me)
Threads Per Inch = 5 (I'll use your number) BTW what if I'm using a rack and pinion system? Is TPI really "turns of the motor" per inch?

Now to the math!

Solving for Watts:
W = (30 inches * 10 lbs) / 531 = 0.565 W <- is that correct? Does the weight of the gantry need to be factored in some how?

Here's what I have for the other equation:
10 lbs = (3.142 / 8) * 5 TPI * in-oz = 81.49 in-oz (I converted the 10 lbs to 160 oz, so I think this is correct if I don't have to factor in the weight of the gantry)

It's nice to see how the TPI relates to the in-oz needed. The more turns, the fewer in-oz but the longer it will take!

bb99

ger21
10-27-2003, 07:20 AM
If you download theie catalogs from here:

http://www.nookindustries.com/Catalog/CatalogDownload.cfm

The Acme catalog lists torque required to lift 1 lb for all their Acme leadscrews. You'll need to account for friction and other inefficiencies, but it might be a good start.

Gerry

Mariss Freimanis
10-27-2003, 11:36 AM
You might want to check your math; I think you zigged when you should have zagged. I get:

(8 * Lbs) / (TPI * pi) = in-oz = 80/15.7 = 5.1 in-oz

Your result was 16 times too big.

About weight:

The mechanism can lift 10 lb mass if oriented vertically. That same 10 lbs force applied horizontally to a 10 lb mass would accelerate it at 1G if it is on a frictionless surface.

1G = 32ft / sec / sec, or 6,144 IPM / sec. This means you would reach your 30 IPM in 30/6,144 sec or about 0.005 seconds.

About rack and pinion:

in-oz = 32 * Lbs / diameter of pinion

Mariss

balsaman
10-27-2003, 12:48 PM
Mariss,

I am the guy you mentioned earlier. I buy something, bolt it all together, and hope it works! (I hope I threw some common sense into the equation at the time). So far I am two for two!

It's the TLAR method. :)

Eric

bb99
10-27-2003, 08:42 PM
Thanks Guys!

I'll review my math. Don't want to send my gantry into space!

ger21 –great site, thanks.

bb99

cnczane
12-04-2003, 06:32 PM
I missed something in Mariiss's original response (post #2).

"Say your crank diameter is 4"..."

but then crank diameter doesn't come into the equation:
Lbs = pi/8 x TPI x oz-in

I think it's in that '8' in the denominator.