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Physics has never been my strong point... I have a task to size a motor. The question is: Based on information below, how do I calculate required motor torque to accelerate a mass with the information below? Please explain the formulas and where they originate from. I understand how to calculate the force required. I don't understand how I compute the required torque to achieve this force.

Variables:

Motion Direction:  Horizontal - Perpendicular to gravity
Mechanics:  Ballscrew & Linear Slide mechanism.

(W) Weight of Load to move:           100KG
(D) Distance load should move:        1 m
(A) Speed load should move:          .04 m/s
(C) Coef of Friction of Mechanism:    .003
(p) all Screw Pitch:                 2.75 mm
(d) Ball Screw Dia:                   20mm
(T) Min motor torque required:         ?

*Please arbitrarily choose variables for any missing information and declare them in your post.

Thank you for your time.

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assuming constant acceleration, a, we need to cover 1 meter with an average speed of $$04m/s =\frac{0 +V_{final}}{2} \\ V_{final}= 0.8m/s $$ Force, F, needed for this acceleration is:

$V_{final}= \sqrt{2a*x}: 0.8 = \sqrt{2a*1}\\ a=0.8^2/2= 0.32 m/s^2 $

$$ F = ma =100*0.32m/s^2 =32kgm/s^2 \\ and\ 32* 1.003 = 32.09kgm/s^2$$ This is force demand increased by friction coefficient.

and the torque is $$\tau = 32.09kgm/s^2/(mechanical\ advantage\ of\ screw)=\frac{32.19}{(\pi*0.02 )/0.00275)}\\= 1.4kgm*m = 1.4* 9.8 =13.76Nm$$

We have assumed $sin( small\ angle\ \alpha)=\alpha$ on the screw pitch.

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  • $\begingroup$ Hi Thanks for reply. When I try and convert to SAE units, I do not get correct results and the reason I belive is in the acceleration calculation. The value becomes drastic. This is not correct. .... Example::: wt(lbs) = 45...... Distance (in) = 39.37...... Desired Velocity (in/min) = 1183 ........ Friction = .003...... Pitch (in) = .109....... Dia (in) = .789...... ----- VFinal = 1183*2 = 2366 in/min....... a = VFinal^2/2 = 2798978 in/min...... F = 126331872 in/lbs...... Mech Adv = 22.844..... T = 12801xxxxxxx in-lbs..... What am I missing? Thanks very much, $\endgroup$ – ExcelMania Jan 17 at 1:52
  • $\begingroup$ @ExcelMania, I will check it later, you check for the difference between lbs weight and lbs force, if I remember correctly you need to divide by 32.2. $\endgroup$ – kamran Jan 17 at 4:52
  • $\begingroup$ Hi kamran. Thanks for your help... 1 lbf = 1lb weight. So I am confused. $\endgroup$ – ExcelMania Jan 18 at 1:17
  • $\begingroup$ Go metric, man. Go metric. It's all much simpler and it's international as is this site. $\endgroup$ – Transistor Oct 13 at 10:16

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