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request help calculation and sizing torque motor opening door swing

  • Thread starter Thread starter Enry774
  • Start date Start date

Enry774

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Good morning to all,
I have a problem that I'm facing these days, but since I'm unfortunately not very hard on the size, I'm not sure I can solve the right way.
I would like to be able to calculate the necessary torque that an electric motor applied to the rotation axis of a swing door, should provide to open it and close it. leaving out the frictions of the carts.
I have all the data, dimensions and weight, so I can calculate the moment of inertia with the formula i=1/3m(l^2+s^2)
then I think I have to calculate the angle of the door with the formula m=i*alfa where alfa=accel. angle.
below the data:
weight port = 40kg
l (width)=0,6m
(thickness)=0,03m
================================================================================================================================================================================================================================================================

hypothesized to open the door of 90° in 3 secs, the angle light is 0.173 rad/sec^2

so the angle moment turns out 0.00075 n.m
If I didn't miss the calculations, it seems to me too low

If the value was correct, would this be the pair that an electric motor should apply to open the door?

Thank you so far who will give me a hand...

Hello, everyone!
 
Hello exxon,

thanks for your valuable contribution. I ask first of all vein for the error in calculating the moment of inertia, I confirm 4.8 kg m2.

I apologize for the lack of familiarity with the formulas, but as I wrote I am not very hardwareized in the measurements and in this type of calculations.

I am not, however, with angular acceleration. .
ω = 90° /3s =30° /s =0.52rad /s
α = (30°/s −0) / (3s −0) =10°/s2 =0.174rad /s2

Am I wrong with the calculation of angular acceleration?

Thank you in advance!
 
to have π/2 rad (90°) in 3 s, you must have an average speed of (π/2 rad) / 3 s = 0.52 rad/s.
in order to have that average speed with constant acceleration, you have to accelerate in the first half of the race and choose in the second half. This implies that the maximum speed reaches it halfway and that this speed will be double compared to the average speed.
from the above, α = (2 x 0.52 rad/s) / 1.5 s = 0.7 rad/s2
 

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