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

 

[Georgesalmon]

Re: Actuator Installation Query

I'm sure someone with a brain will come to help you but in the mean time,,,I remember that old saying,"If I had a lever long enough I could, etc, etc. But, no matter where you put the actuator it can never have to lift more than the full weight of the hatch, right? So, you can lessen the amount of force needed to raise the hatch by moving farther from the hinge point using the lever anology, but the force needed can never increase beyond the total hatch weight of 50kg. Where would the extra weight come from? I think you'll be fine.

 

[Bondo]

Re: Actuator Installation Query

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 173026

Ayuh,... Welcome Aboard,.... I ain't no mathamagician, hopefully, 1 will be along...

I'd think you'd want longer travel actuators, at a better angle myself,...
Everything I've seen like it, were that way...

The way you've got it in yer picture, I see short lived hinges...

 

[TerryMSU]

Re: Actuator Installation Query

I believe that your answer that it can never be more than the weight of the door is incorrect. For a thought experiment, picture a 12 ft long 2by4 weighing 10 pounds. You can easily lift that board with one hand. You and a friend can easily lift the board with 5 pounds on each end. Now place a hinge at one end. Lift the other end of the board way out at the end. The hinge supports 5 pounds, you support the other. Take the same board and lift it at the middle. Now your friend at the end lifts zero pounds. That is the point where the actuator (you) will lift exactly the weight of the door. Now move closer to your friend (say maybe one foot from him) and lift the board. He will need to push down on the board with some force to keep the board in balance. Where did that down force come from? (I see Bondo mentioned short lived hinges. That is a direct result of this extra load at the hinge end.) The board still weighs only 10 pounds. The answer is that you must lift with exactly the total of his down force and the weight of the board.

The good news is that for the OP, the ideal spot is to put the actuator at the end of the door oposite the hinge end. If the door is uniform from one end to the other he will need to lift only 25 kilograms. The negative is that the farther he gets from the hinge, the longer the actuator will need to be.

As a general rule, think of this as torque. The longer the wrench is, the less force you need on the wrench. Force times length = torque. You will need 36.5 kilograms at one meter from the hinge, and at one half meter, you will need 73 7ilograms.

 

[Grandad]

Re: Actuator Installation Query

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'm no mathematician either, but I enjoy puzzling this sort of thing through using my form of logic. So here's my thinking
1. The doors are the heaviest when fully closed.
2. The 50 kG weight of the doors could be represented by a downward vector at the center of the door (730 mm).
3. The upward push by the actuator is 200 mm from the hinge.
4. You have a mechanical disadvantage of a lever ratio 730mm/200mm or 3.65.
5. Therefor you need an actuator that can provide 3.65 x 50 kG or 182.5 kG.
- Grandad

 

[Georgesalmon]

Re: Actuator Installation Query

Told you people with a brain would come along, lol

 

[glenn property of pam]

Re: Actuator Installation Query

i think grandad is correct but i was wondering if you could just use bathroom scales and a jack. open one door set scales under other door then jack at 200mm until door rises 1/2" then weigh the jack and minus that. grandad has already confirmed that the load should be reduced as the door opens so the first 1/2" will be the weight of the door at a mechanic dis-advantage

 

[UncleWillie]

Re: Actuator Installation Query

Are these Electric/Hydraulic Actuators where you expect to push a button and the dooors open unassisted?

Or are these Gas Springs that you want to assist in opening the doors by hand and keep the doors in the open position until closed by hand.
Much like the back hatch on an SUV.

If the gas springs; Your plan will be fine.
They will make it easier to open.

If they open to full straight up, it will take almost nothing to keep them there.

 

[UncleWillie]

Re: Actuator Installation Query

Lever Theory

In your example The center of mass is 730 mm from the hinge (Fulcrum).
The center of force is 200mm from the hinge.
730/200 = 3.65, 3.65 x 50Kg = 182.5kg force at the Center of Force (Actuator Mounting Point).

182.5Kg - 120Kg actuators = 62.5Kg remaining at the Center of Force.

62.5/3.65 = 17Kg at the center of mass (center of door) ; And 8.5Kg (<20lbs) at the far edge of the door (1460mm from the hinge.) to start to open the doors.
Note: As the doors raise the actuators become more effective and will open themselves at some point.

It is going to take 120/3.65/2 = 16.5Kg (>35lbs) to start to close them from straight up.

Move the Actuators out to 304mm and the doors will weight Zero and open themselves in a stiff breeze.

EDIT:
I just noticed you stated ELECTRIC Actuators.
The 120Kg actuators will need to be MORE than the 304mm from the hinge to be within rating.
Or be upgraded to more than 183Kg to keep them at the 200mm locations.
The previous was assuming Gas Springs.