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I am currently building a bldc motor and am trying to gauge the amount of force the coils in my motor would need to generate. In order to move a desired load - to a particular speed.

Without taking losses or friction into consideration. Can I use the moment of inertia to ball park the amount of force required by the coils to achieve this? For example - I have a 16inch diameter Rotor which will have a 7kg load uniformly distributed to it. The RPM I would like the motor to operate at with this load is 3000 rpm. I am also not to worried if this takes 1second or 60seconds to achieve.

Can I Use... ( 1/2 * m * r^2 ) * angular acceleration = to find this force?

Secondly - if I can use this formula. Would this be the force needed by a coil - a phase - or in sum overall from all coils?

Or is this the wrong formula/way of approaching this type of problem?

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  • \$\begingroup\$ What is a 7kg load? Just to rotate something doesn't take any force at all if you don't care how long it takes. Is there any upper limit on how long it takes to come up to speed? Is there any load other than windage and bearing friction? \$\endgroup\$ Commented Mar 17, 2021 at 21:37
  • \$\begingroup\$ You're actually overthinking it a bit as far as the math goes. If I told you it was a linear system instead of a rotating system what would you do? Same thing. Except you use rotational instead of linear equations. \$\endgroup\$
    – DKNguyen
    Commented Mar 17, 2021 at 22:02
  • \$\begingroup\$ Sorry...I really meant anywhere between 1second to a minute. For arbitrary purposes let’s say in 5 seconds. No other loads simply bearings and windage. The load is uniform...it’s another disc on top of the rotor with the same diameter as the rotor. \$\endgroup\$
    – JPH
    Commented Mar 17, 2021 at 22:08

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Without friction, anything eventually gets up to speed if you provide a force or torque. What mass and moment of inertia determine is how long it takes for that to happen. In practice (while still neglecting friction) you are probably limited by how long you can take to spool up rotate due to heating form stall currents. How quickly you can spool up is probably limited by core saturation, then heating.

So you need to decide how quickly you need your load to accelerate.

Then you either use all the \$F=ma\$ and \$v = at\$ from high school physics or their angular equivalents which are \$ \omega = I\alpha\$ and \$ \omega = \alpha t\$ depending on whether your 7kg mass is being moved linearly by the motor or being spun by the motor.

If the load mass is being translated by the motor then you need to use the lever arm (aka wheel/gear radius) to convert back and forth between linear force and torque.

If the load is being rotated by the motor, then you need the dimensions and geometry of the 7kg mass to calculate its moment of inertia through calculus or a table. Moment inertia being the equivalent of mass in an rotating system.

If the inertia of the load is much larger than the inertia of the motor's rotor, you can neglect it.

If you can't neglect rotor inertia then:

  • For rotational movement you just add it to the load moment of inertia.
  • For linear movement, you will have to use the lever arm to convert the load mass to a moment of inertia and add it to the motor's inertia.
  • Alternatively, for linear movement you can convert the rotor inertia to a linear equivalent mass using the lever arm and do all your calculations in terms of linear forces and masses rather than torques and moments of inertia. Then at the end you can convert the linear force into a torque again.
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  • \$\begingroup\$ Thank you - I guess the confusion for me. Was if I know how much force is required to accelerate a mass rotationally to a given speed within a given time frame. In terms of a motor which is moving in phases or steps how does this translate. Is it the sum force produced across all coils in the same given time frame what matters? or is it every single time each phase takes place ? It must be able to produce this amount of force. Over and over again. \$\endgroup\$
    – JPH
    Commented Mar 17, 2021 at 23:20
  • \$\begingroup\$ There are always two phases active a time. \$\endgroup\$
    – DKNguyen
    Commented Mar 17, 2021 at 23:21

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