Actuator Installation Query

[Oryen]

Hi

I have two engine room doors that I would like to fit electronic actuators to open. The doors weigh 50kg's and are 1460 mm (from hinge) by 1465 mm.

I have attached a sketch of the installation I Would like to do. I plan on using one actuator per door.

How do I calculate the minimum pushing force of the actuators? As you can see the actuator will be installed very close to the hinge so I figured it would need more lifting force than the weight of the door.

I have had my eye on a a set of actuators that each have a lifting force of 120 kg is this sufficient for this installation that is so close to the hinge (i.e one actuator for one door)? Or do I need to upsize the actuator to open this? What is the formula for calculating this?

I would appreciate any help or advice

Thanks
View attachment 21627

 

[vas]

Hi

I have two engine room doors that I would like to fit electronic actuators to open. The doors weigh 50kg's and are 1460 mm (from hinge) by 1465 mm.

I have attached a sketch of the installation I Would like to do. I plan on using one actuator per door.

How do I calculate the minimum pushing force of the actuators? As you can see the actuator will be installed very close to the hinge so I figured it would need more lifting force than the weight of the door.

I have had my eye on a a set of actuators that each have a lifting force of 120 kg is this sufficient for this installation that is so close to the hinge (i.e one actuator for one door)? Or do I need to upsize the actuator to open this? What is the formula for calculating this?

I would appreciate any help or advice

Thanks
View attachment 21627

Hi,

I hope someone with less rusty physics will come soon and tell you, however, I'd like to know why you don't "move" the mounting on the door a bit to the centre and the lower mount on the boat a little bit outside (well not more than the axis of the door on open position as it obviously wont work. what I'm trying to make is to INCREASE the travel of the actuator thus reducing the force it will be called to provide. Again you have to be right in not overcoming the length of the piston but I think the closer you get to fully extended position the softer on the forces (and the cheaper the ram you'll need)

OK, looking at the diagram again, force is 50kg effective length 1400, so a 50+ ram would move the door if it was mounted smack in the centre (where the center of gravity is) The further you go towards the hinge the worse the situation. I'd guess that moving the ram mounting point to more or less 1/3 of the effective length, would be ok for the 120kg actuator. A bit of trial and error would help, but obviously only if you can shift the mounting point that far to the centre...

cheers

V.

 

[jfm]

You haven't given enough data or dimensions (algebraic if you wish) to write out the forumla (and ironically most of the dims you have stated are irrelevant to the calculations! ) but in that picture (if it is to scale) the actuators need a theoretical force of 4x the weight of the door ie 2000 newtons. If you allow for inefficiencies, friction, safety factor, etc etc you need to buy 3000 or 4000 newton actuators. If you move the actuator attachment away from the hinge and fit a longer stroke actuator, you can reduce the newtons. Also if you fit a gas strut to help take some of the weight, you can reduce the newtons

 

[Spi D]

The maths are the same for gas springs. If you Google that you'll find on-line services like this

http://www.bansbach.de/jgctest3/e-berechnung1.htm