Ask Me Help Desk - Need to work out what torque I need to rotate a door from stand still to 90 degrees

I have a s/s door weighing 900kg and it is hinged on one size with bearings to overcome friction issues, the door measures 2200mm wide so from shaft to edge of door is 2200 + shaft radius (around 30mm).

I need to find how much torque I require to move the door from standing position to 90 degree shift and closed position so that I can size the correct motor and gear sizing for this application.

Cheers any help would be great.

This is not a tax question - would be better to post this question in either the engineering or physics forums. [**EDIT - I see the mods have moved it - thanks!] But to help you along:

Any discussion of sizing of a motor based on torque calculations requires that you specify how quickly the door is to be opened. In theory you could do this with a toy motor and a 1-1/2 volt D cell battery IF it was acceptable to require all day to open or close the door.

Hi Ebaines,

Obviously I know that was not a tax question as I posted it in the physics section but I gather it got crossed over somehow.

In regards to the door it will be shutting / opening with a time of 15 seconds. Meaning 15 seconds to open and 15 seconds to close.

Just trying to get a torque figure on this operation so I can look at a motor and gearbox to suit.

Cheers.

The formula for calculating torque is

$ \tau = I \alpha $

where $\tau$ = torque, $I$ = the object's moment of inertia, and $\alpha$ = rotational acceleration. For a hinged door the moment of inertia is:

$ I = \frac {mw^2} {3} = \frac {900 Kg (2.03m)^2}{3} = 1236\ Kg-m^2 $

The average rotational velocity you want to achieve is 90 degrees in 15 seconds, or 1 RPM. If we assume that the door starts at rest and accelerates to a top speed at time 7.5 seconds when the door is half open/closed, and that it then decellerates to 0 in the next 7.5 seconds (i.e. the door doesn't slam open or closed but glides smoothly into place) then the required angular acceleration can be found from

$ \theta = \frac 1 2 \alpha t^2,\\ \frac { \pi} 4 = \frac 1 2 \alpha (7.5s)^2 \\ \alpha = \frac {\pi} {2 (7.5s)^2} = 0.028 radians/s^2 $

So
$ \tau = I \alpha = 1236 Kg-m^2 \times \frac {0.028} {s^2} = 34.5 N-m. $

This can be reduced a bit if you're willing for the door to slam open or shut, as the rotational acceleration could be made smaller. Note that this does not take into account any frictional forces in the hinge or gearing.