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Linear actuator & motor pairing (check my math)

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  • Linear actuator & motor pairing (check my math)

    I want to see if this linear actuator can lift 50 pounds
    https://www.servocity.com/linear-act...stroke-6-0-sec

    They suggest pairing it with the goBILDA 5202 5.2:1 ratio motor
    https://www.gobilda.com/5202-series-...-3-5v-encoder/

    And they say it will lift 50 pounds. I believe it will, but my math shows otherwise. Specifically, I think the motor is WAY overpowered. Unless I am reading the specs wrong.

    Math time!
    The screw in the linear actuator has a 2mm pitch. That means every revolution of the screw equals 2mm distance traveled.

    The gear driving it is 1.5in in diameter, or 38.1mm. The gear coming off the motor and the gear attached to the linear actuator are both the same size, so no mechanical advantage/disadvantage there.

    The circumference of the gear is 38.1mm * pi = 119.7mm (call it 120mm).

    That makes an overall mechanical advantage for the linear actuator is 120/2 or 60:1. Instead of 50lbs, we feel 0.8333lbs at the gear teeth (13.333oz).

    The motor and gearbox are rated for 1,150 rpm and a stall torque of 7.9kg-cm (109oz-in). Again, the output gear is 1.5in in diameter (0.75in radius). The motor can handle a force of 109/0.75 = 145oz force at the gear teeth. Way way way more than the force I am calculating we will need. This motor seems to be more than ten times as strong as we need.

    I take servocity's word that this is a good pair, but I'd still like the math to work out. Or is my math correct and in general I should always plan on getting motors at least ten times as strong as I need?

  • #2
    I think there are 4 threads in the lead screw, so although the pitch is 2mm, every rotation is 8mm traveled. If the rest of your math is right, that would put the force on the gear at 4x13.33oz = 52 oz, or around 1/3 of the 145 oz of stall force. Since peak lifting speed happens at 1/2 the stall torque, or half the stall force using your derivation, this is a pretty good match (and also has a little overhead for battery voltages dropping, friction causing less than 100% efficiency, etc.)

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    • #3
      Oh, it sure does say "4 start", which means 8mm per rotation. Great catch!! It all makes sense now.

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      • #4
        We had that motor on the same set up with a 22 lb bot. One of the mentors had the kids switch it for a slightly slower/more torquey motor. Then we found that the early season safety-minded code had the motor power limited to 0.3. So, had we commanded more power from the motor, it is possible that it wouldn't have gotten so hot. There were situations where I think there was too much friction on the lander side and it couldn't get started. If we nudged it, it would break through. More applied power would likely have gotten past this. We never went back, however.

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        • #5
          Assuming no friction (which is far from correct), the torque required to lift using this screw mechanism would be:

          Torque = weight * lead / (2 * PI)

          where the weight in this case is 50 lbs, and the screw lead is 8 mm (0.315 inch).

          So, the required torque is 2.5 in-lbs, or 40 in-oz.

          The added torque needed to overcome friction is often greater than the torque needed to overcome gravity.

          So that motor isn't giving you way more torque than you need.

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          • #6
            We made a presentation for this:
            https://docs.google.com/presentation...it?usp=sharing

            Comment


            • #7
              That is an awesome presentation. Thank you for putting it up. I suggest you put your team name & number on it so people know who to thank.

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              • #8
                Excellent presentation! I'll also suggest adding team name and number.

                My only suggestion: on slide 16 you say (essentially) that peak efficiency is achieved at 1/2 the stall torque. Technically, peak power is achieved at 1/2 the stall torque, power being energy / time. In this particular application, that's the same as peak lifting speed for a given robot weight. Efficiency is different: that's measuring the percent of electrical power converted to mechanical power. Peak efficiency is typically achieved at much less than 1/2 the stall torque.

                Rev has a great overview of motors that explains a lot of this. See Figure 2 in this doc to see the difference between power and efficiency. I might humbly suggest that adding a figure like Figure 2 from the Rev doc would also be useful in your presentation.

                http://www.revrobotics.com/content/docs/Motor-Guide.pdf

                Thanks again! Nice work!

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                • #9
                  Oops. OK, one more suggestion.

                  Also on slide 16, you calculate the speed, but using the free speed of the motor (1150 RPM). The motor only achieves this speed when there is no load on the motor. The speed decreases linearly as load is applied until the speed reaches zero at the stall-torque. Again, see Figure 2 in the Rev document. Since you're operating at about 1/3 of stall torque, you'll be getting about 2/3 of peak speed, and the motor will really be spinning closer to 750-800 RPM at that load.

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                  • #10
                    Great feedback! Thanks. I have incorporated all of your recommendations.

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