Calculating Force for 500lb Electric Actuator Door

[RLCOS]

I have a door that weighs 500 lbs. It raises up on an arc. It uses an electric actuator to pull on the lever to raise the door and push to lower. I am modifying the door and do not know how to calculate the force in pounds to make sure I have the right actuator. I currently use a 1500 lb actuator. I need assistance with the equation/formula to come up with an answer. Forgive the typo.

 

[AJ Bentley]

Rather a strange door! If I read this correctly it will move quite a lot backwards and forwards as it lifts.
The force will also vary substantially over the lift.

The maximum force will be required at the top of the lift as the short lever arm becomes almost straight-on to the actuator - in fact at that position the force required is theoretically infinite because it's a lever with a very short (hardly any) leverage. Whatever actuator you use if will simply run out of 'grunt' at some point near the top.

It might be an idea to re-design the mechanism to make it more efficient in the use of leverage.

 

[RLCOS]

The door actually goes out on the arc and straight up. In the up position the bottom of the door is just above the top of where the door was when in the down position. I think this more than a simple type 1 lever calculation, yes?

 

[AJ Bentley]

From the drawing, it looks to me like a simple lever. Maybe I'm wrong to see it that way, my interpretation suggests it's about the worst possible design for mechanical advantage.

 

[RLCOS]

Thanks A.J. A simple lever calc shows me with about 636 lb force required to balance the load. So my 1500 lb actuator will work. The design actually works pretty well. However your comments give me food for thought and I will certainly look a different design.

I was most likely trying to over think the calc. I do that a lot.

 

[Dhanesh]

Hello,

Have u succeeded with this design concept?
If yes, please help me by posting the calculations and design.
I will be very thankful to you.

Regards,
Dhanesh