motor sizing/design help needed 3

[mcfridge]

Hi, I cross posted this here because I didn't know how many people viewed different forums:

http://www.eng-tips.com/viewthread.cfm?qid=106072

Hi, I'm new here, but I really need some help on something I am working on. I did a search but couldn't find exactly what I was looking for so I thought I would ask for help:

I am doing a simple design where I use a motor, coupled to a set of gears (probably spur gears), which rotates another shaft, which raises and lowers an amp in the trunk of a car (maybe by means of a screw jack, not quite sure yet). What my question is this: I'm thinking that I have to size the motor first before I can design everything else, but how do I do that? I need the motor to run off 12V DC because I want to run it off the car battery (if possible). I hate asking because I don't want the work done for me, but I needed someone to explain it on kind of a step by step basis (what measurements I need to take, what formulas to use, etc) because I don't have a formal engineering background. Thanks a lot if you can provide any help.

 

[JStephen]

The size (ie power) of the motor could be anything you wanted to make it, provided you used the proper gearing. A more powerful motor would raise it faster, of course.

You could get a sort of rough idea using energy considerations, without fooling with the gearing in between.

Suppose the amp weighs 50 lbs. Suppose you want to raise it 12" in one minute. The rate of work done is then 50 lbs x 12" / 1 minute = 600 in-lbs/minute, or 5/6 ft-lbs/second. One kilowatt = 737.56 ft-lb/sec, so you get 0.0011kw or 1.1 watt. This is the energy actually put into the raising of the weight. With a small gear system and screw jack, you probably would have very low overall efficiency. Suppose 10% efficiency overall, and you'd need an 11 watt motor. With 12 vdc, that would be about a 1-amp motor.

If this is a home-made type thing, you might check into salvaging the motor and gear train off one of those rechargeable cars that are made for little kids. I'm not sure if they are 6 or 12v, but they do come with a handy (although noisy) gear train.

 

[drawoh]

mcfridge,

You need to work out everything simultaneously. Your motor affects your drive. Your drive affects your motor.

A crude first pass at your investigation is to work out the motor power required. Do not bother with motor torque because your drive controls this. JStephen is worked out much of this, above.

As yourself the following questions...

- How much torque do you need at the output?

- Does the system have to remain stationary with the power off?

- How efficient must your system be? If you are running off the battery with your engine off, power consumption could be an issue, although I suspect your amp will be involved here too.

- How must can this cost?

You have spur gears, worm gears, harmonic drive, off-the-shelf drives, screw jacks, ball screws, scissor lifts and who knows what else.

I always punch my information into a spreadsheet. It makes it so much easier to try different setups.

JHG

 

[amorrison]

From an automotive scrap dealer - a car window raise/lowering motor/mechanism combo.

http://hardwareus.dunmarsh.com/hardware/motor.html

 

[loquillo]

The gear reduction would do the trick. After you calculate the speed you want the load to move (work output) and the motor parameter you decided to use you just have to play with the gearbox until you converge with the inputs and outputs. For that you should know a little about gear reduction calculations which can be found in Desing of Machinery references (author recommended Juvinall).

 

[mcfridge]

Thanks a lot for the help you guys. I just have one more quick question: I found one book (Machine Elements in Mechanical design by Robert Mott) and it walks you through the gear design step by step, but it first gives you the power transmitted to the pinion at a certain speed (mainly 1750 rpm), and then specifies how fast the gear must rotate. I was just wondering how I determine how fast I want the gear to rotate? I'm sorry if this question seems redundant or anything but I really appreciate any help you can provide.

 

[pso311]

I'm a fan of the Mott book. The 1750 rpm is pretty common for motors, but what you're looking for is the tooth-to-tooth relationship for speed reduction. It is a direct ratio, number of teeth on the gear divided by number of teeth on pinion.

ex. 118 (teeth on gear)/ 12 (teeth on pinion) = 9.83

This is your speed reduction, so your gear (if pinion was at 1750) is 1750/9.83 = 178 rpm.

You find the rpm you need by how much torque is required to raise the amp. (see above posts)

Remember the lower the rpm, the higher the torque. Unfortunately, I don't have my Mott book here, or I help further.

-Scott

 

[loquillo]

If you have the RPM of one of the shafts you can reduce or increase the speed of the output shaft by playing with the diameters of the gears. After you establis what kind of motor you are going to use and how fast you want your outut to be, you just got to play with the gear ratio until you converge to your specifications.

 

[plasgears]

The motor you choose should match the power required in the candidate motors, whether they are large slow motors or small high speed motors. I usually start with:

T1N1 = T2N2/eff

where, at the design point:
T1 = motor torque,
N1 = motor speed,
T2 = output torque,
N2 = output speed.
eff = mech efficiency, say .95 per gear stage. Number of stages could be estimated about (gear ratio)/4.

Knowing the proposed function of the machine, you should know the output parameters. Solving for T1N1 gives you a measure of motor power. The candidate motor should be able to meet T1N1 at or near the max eff point shown in the catalog. Choose a motor with a little more power than you need. Now,

N1/N2 = gear ratio

Tentatively assign gear pitch and tooth numbers to meet the target gear ratio, and make a ballpark estimate of stresses using the Lewis equation. With acceptable stresses (y.s./3), now you can wrap a gearbox around the gears and have a sense of how big this assy is.

 

[MintJulep]

What you asked was "how do I size the motor?"

Power is force time distance divided by time. How much do you want to move, how far and how fast.

For reference, 1 HP = 550 foot-pounds/second. Since you won't need anything near a HP, lets multiply by 12 to work in inch-pounds/second.

1 HP = 6600 inch-pounds/second.

If you want to lift 6600 pounds a distance of one inch in one second, you need 1 HP. Got 2 seconds, then you only need 1/2 HP. Likewise, 3300 pounds, one inch in one second needs 1/2 HP.

How heavy is your amp, how far do you need to lift it, and how fast? Thats your starting point. Add to that the force needed to overcome the friction in your mechanism. Add another 20% for good measure (gear drive efficiency and such).

Thats it. You can trade RPM for torque using plasgear's formula.

Now, having said all that, I agree with amorrison. Go to a scrap yard and get a power window operator. If your amp is really heavy, go for a tail gate window operator, or a power seat mechanism.

 

[automatic2]

construct your mechanical linkage first, then you can take off torque requirements, then add desired speed.

 

[mcfridge]

Okay, I swear this is the last stupid question you're going to get out of me. I was arguing with my friend that the lower the rpm, the higher torque I will get. He seems to think that we need to be running the other shaft at like 100rpm, and I keep getting frustrated on this. Now, correct me if I'm wrong, but say I found a motor that outputs 2.1 in*lbs @ 800 rpm (0.027 hp). Now, say I have an amp that weighs 25 lbs, and we want to raise it only 4 inches. Wouldn't the torque required to raise that be at least (25lbs)(4") = 100 in*lbs ?? And then, with

T1N1 = T2N2/eff

where T1 = 2.1 in*lbs, N1=800 rpm, and T2=100 in*lbs, wouldn't the rpm I need be about 16 rpm (if we are just assuming an efficiency of 100%)? Thanks again.

 

[MintJulep]

"Now, say I have an amp that weighs 25 lbs, and we want to raise it only 4 inches. Wouldn't the torque required to raise that be at least (25lbs)(4") = 100 in*lbs ?"

No. This is the work required.

 

[Iskit4iam]

Unless you're planning on making hundreds of these and must have a proprietary design I'd suggest a search for "linear actuator" on google. I know there are 50 pound force 12v actuators available from Danaher.