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I have a weight of 1 kg and I want to properly choose 2 DC motors for them in order to be able to move the 1 kg of weight with variable speeds in the range of 0 to 1 m/s with an increment of 0.05 m/s. How can I do this?2 DC motors

Theodor
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  • "*... with an increment of 0.05 m/s*". Do you mean "with an acceleration of 0.05 m/s²"? If the drive is low friction it is usually the acceleration that determines the power required. – Transistor May 16 '22 at 21:25
  • No, the speed’s increment should be 0.05 m/s. I will monitor it with a shaft encoder – Theodor May 16 '22 at 22:09
  • "*... the speed’s increment should be 0.05 m/s.*" I don't understand that sentence. I think there may be a language problem here. – Transistor May 16 '22 at 22:22
  • kg is a unit of mass, weight is in Newtons (N). velocity is in m/s and acceleration is m/s^2. Please check your units. – Solar Mike May 17 '22 at 05:40
  • @Transistor well I will be controlling the speed with a joystick. When I will slightly incline it forward the speed of the platform should be 0.05 m/s, then when I incline it slightly more it should go at 0.1 m/s, then 0.15 m/s …. Then finally 1 m/s. Keeping the joystick in the same position should maintain the speed of the platform. – Theodor May 17 '22 at 06:13
  • @SolarMike well yes, sorry, the Mass is 1 kg – Theodor May 17 '22 at 06:18
  • @Theodor, the 0.05 m/s is not relevant. The **acceleration** is what is relevant since force required is given by ***F = ma***. – Transistor May 17 '22 at 06:28
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    Does this answer your question? [How can I calculate the power and torque required for the motor on a wheeled robot/vehicle?](https://engineering.stackexchange.com/questions/31501/how-can-i-calculate-the-power-and-torque-required-for-the-motor-on-a-wheeled-rob) – DKNguyen Oct 14 '22 at 13:54

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one kilogram meters per second is 9.8 Watts. Ignoring the friction, this is the power you need. then divide by two and that's each motor.

Then you need a motor drive controller which is available in many electronic online stores or on Amazon for approximately $10.

Edit

After OP's comment,

Moving a mass horizontally theoretically does not need any force. only friction or if you need a preset acceleration in between the steps.

Let's say you require acceleration of $ \alpha$ in between the steps and again ignore the friction,

$$F=1kg*\alpha= \alpha watts$$

DC driver

kamran
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  • "*one kilogram meters per second is 9.8 Watts*" Isn't the watt defined as $1\ \mathrm{kg⋅m^2⋅s^{−3}}$. "*Ignoring the friction,*" the power required for constant speed is zero. – Transistor May 17 '22 at 02:25
  • @Transistor, OP is asking to lift 1kg.m/s not horizontal moving. lifting has kg. g per sec .it is power meaning work per second. factor: g has s^2 in the denominator. – kamran May 17 '22 at 03:12
  • @Theodor, can you make clear: is the weight moving horizontally or vertically? – Transistor May 17 '22 at 06:27
  • @Transistor the weight is moving horizontally – Theodor May 17 '22 at 06:28