Problem summary:

I need to produce about 0.289 Nm with a rack and pinion system and compress 24 mm of a 110 mm x 21.6 mm, 0.99 N/mm steel spring.

I thought about using with a 10mm pitch diameter gear with ten 1mm gear modules at an angular velocity of about 15rpm (or 1.571 rad/s). Ignoring the efficiency losses, and the friction of the gear system a 12 V DC motor consuming 37.8 mA would do the job:

$$E \cdot w=0.454 W$$

$$ V=12 V \quad I =\frac{(E \cdot w)}{V}=37.837 \mathrm{~mA}$$

(I assumed the torque on the gear to be the same as the work done by compressing the spring and set this to be the electrical power consumed.)

However, in the real world, things are more complicated and I have been oriented here. To choose a motor expecting a final efficiency of 25% and, thus, expect using about 4 times more current.

Motor specs:

After some research I found a DC motor with the following specifications:

Supply Voltage                  12 V dc
DC Motor Type                   Geared
Output Speed                    14 rpm
Shaft Diameter                  6mm
Maximum Output Torque           59 Ncm
Gear Ratio                      300:1
Dimensions                      39 (Dia.) x 83.2 mm
Current Rating                  180 mA 

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My questions:

Because the output speed is about the same as my rack and pinion uses, I imagine my operating conditions will be the same as the graph Gear ratio 1/300. Is that correct?

Thinking back now, I assumed the torque on the gear to be the same as the work done by compressing the spring and set this to be the electrical power consumed.That's wrong, isn't it?

$$Tangential force \cdot \frac{( pitch diameter)}{2} = torque$$

so if I need about 24N of linear force on my rack, the torque will actually be about 13 Ncm instead of 29 Ncm like I first thought. In that case, is this motor I found still a good choice?

Any corrections, comments and/or suggestions are welcome!
For a more thorough description of my problem go to here


1 Answer 1


On the motor spec sheet you are correct to used the "Gear ratio 1/300" graph because that is the motor+gearbox that you intend to purchase. Those other graphs are for different models that the company offers for sale.

I did not have time to dive into all your previous questions. So I will just explain what this motor can do and you can decide if it is satisfactory for your application or not.

Per the 1:300 graph in your post we can identify many different operating conditions. (you can also reference the 1:300 portion of the table):

Since you are only concerned with the torque and RPM you just need to look at the T-N line, the other lines are for figuring out the electrical current (amps) required. Also note that [min-1] = RPM = Revolutions per minute

No load, no resistance = 0Ncm @ 17RPM
Rated load = 98Ncm @ 14RPM
Another possible operating mode = 196Ncm @ 10RPM
Startup, Stalled rotor = 490Ncm @ 0RPM

Torque and RPM are inversely proportional. As the load torque goes up the RPM goes down. gear ratio graph


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