I'm designing a machine that I need to spin a horizontal disk at low RPM using a motor and gearbox. I'm not good at physics so I need help to find a motor/gearbox to fulfill my needs.

It's a horizontal disk, a shaft is connected at the middle of the disk, connected to the gearbox on top of disk.

There are wheels under the disk to lower the friction and handle weight.

On top of the disk, there are meats.

On the disk there is a blade. When the disk spins, cutted meats comes out of disk from bellow.

Disk and shaft's weight is 5kg, meat is 25kg (max), motor and gearbox are secured to the machine, Disk's friction to the machine is negligible.

Required gearbox speed is 30 RPM.

How should I decide which motor/gearbox is good for this project? I know the speed (30 RMP), how to calculate the needed Torque?

Thank you.

  • 3
    $\begingroup$ Torque depends on the resistance the disk gets from the meat being cut. Simplest way to find that out is to measure it. $\endgroup$ Sep 4, 2019 at 13:39
  • $\begingroup$ ^we need to know both how much force it takes to push the blade through the meat, and also the radius at which this happens on the disc $\endgroup$ Sep 4, 2019 at 14:03
  • $\begingroup$ @JonathanRSwift Radius is 15cm, but the blade is 10 cm in length (secured to one side of the disk). I don't know about the needed force right now, because the needed blade is not manufactured yet, can we put a variable like "CF" for cutting force in the equation? Thank you. $\endgroup$
    – amil.eddie
    Sep 4, 2019 at 15:36
  • $\begingroup$ Thanks amil - best practice is to edit your question so that important details are not buried in the comments. If you could also include a diagram (is ComradeH's suggestion correct?) that would really help us help you. $\endgroup$ Sep 5, 2019 at 7:51
  • $\begingroup$ Jonathan is right on,.Without knowing the exact force required of cutting the meat, you can calculate a range of force that would be available using the equation for power, 2pitorquerotational speed (rpm), and the equation for torque, F*r. $\endgroup$
    – r13
    Sep 24, 2021 at 0:28

1 Answer 1


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Base on your description, is tis the system you are talking about?

First thing: Having just a single blade is just a waste, I recommend having a few more blades on it (more "cutting" per revolutions of the disc, or more "cutting" in the same time frame)

Now, you need to find the first thing: cutting force, for example, based on experience, experiments or simply a value out of nowhere (it is acceptable for freshman projects, as long as it "feels right"). For example, you know that you can cut a slash of beef with a force of 100N. Due to your design, the blade is 0.1m from the center of the rotation (or from the centre of the shaft). Therefore, the torque is

$T=F*r=100*0.1=10 Nm$

Converting 30 RPM into rad per second (basic unit), we have

$\omega = \pi rad/s$, or approximatly 3.14159 rad/s

The power is equal to the product of torque and angular velocity, or

$P=\omega*T = 3.14159 * 10 = 31.4159W$

This is the power output of the motor (with the gearbox included). The mechanical efficiency of the gearbox should be high, say 95%, which means the actual output of the motor (before the gearbox) is 33.069W, or just 33W because I hate long decimals. Continue to extrapolate the input of the motor, again we assume a 95% efficiency, leads to a 34.8W motor.

Now you have it, a motor of 33W is "just enough" to cut the meat as you want. Now, you have to look for the technical data online, seeing which is the best motor for you (power is the closest match, range, size, weight...). You check for its speed limit, say it delivers 35W at 1600RPM. Now, you want to have an actual output of 30RPM (after gearbox). Therefore, the ratio must be


So, for a summary, with a crapload of assumption, you need a ~35W motor and a gearbox with 53.33 ratio.

Note, a 53.33 decreasing ratio means that gonna be huge! Still, I make a lot of assumption here, which means "making an ass out of u and me".

The main idea is just pure Math (secondary school math), so it should not be too hard... Hope it helps

  • $\begingroup$ Thank you for the help, your time and suggestions. What about the weight of meats on top of the disk? Doesn't the weight add friction like a brake? Also as I may look for Stepping Motors (to control speed and time of cutting), should I look for a motor with > 10 N.M motor, right? It seems a lot of torque, or am I missing something? Thank you. $\endgroup$
    – amil.eddie
    Sep 5, 2019 at 10:36
  • $\begingroup$ Because you have "wheels at the bottom" (the ones that I did not draw in the figure), the coefficient of friction would be fairly low - Actually, it is very low according to this link. For specific type of motors, again, it depends greatly on your situation. But for the torque value of 10Nm, I suggest you to ignore the value (it is a number gained from random values, completely untrustworthy. Take some research online, find some numbers and then try to work out the problem like I did. $\endgroup$
    – ComradeH
    Sep 5, 2019 at 13:51
  • 1
    $\begingroup$ Take this advice, ‚multiply by ten and forget about it‘ is the rule of thumb i use when it comes to laying out electric motors. The cost to buy a slightly larger motor than needed is rarely worth the extra time it takes to perform more than back of the envelope calculations. $\endgroup$ Sep 24, 2021 at 9:33
  • $\begingroup$ @amil.eddie A stepper motor to control this seems like a waste of money. I'd simply use a 'clearly big enough' DC motor, control the voltage and be done with it. $\endgroup$
    – Chris_abc
    Jan 17, 2023 at 15:09

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