Current versus dynamic force on linear actuators

[Apple&Orange]

G'day guys

So I currently have a 2x 8" stoke 150lbs dynamic force actuators with a maximum current draw of 5A.
The exact one is the FA-PO-150-12-XX in the link below:

https://www.firgelliauto.com/products/feedback-rod-actuator

The load I have on it causes the actuators to draw almost its' maximum limit (i.e. 4.5A - 4.8A).

To minimize the current draw, I was thinking of using actuators with a higher force rating (i.e. the 200lbs dynamic force actuator in the link above). However, one of my colleagues believes that current draw will still be the same because it shares the same motor specifications. The only difference is the gear ratio, which will result in a slower stroke speed, but still draw the same amount of current.

His conviction has left me double thinking myself, and am wondering if he is correct. Could someone put my doubts to rest?

 

[anorlunda]

Conservation of energy is a powerful tool. Ignoring efficiency, electric power = mechanical power.

To minimize power, make it move slower, or use a lever or gears.

 

[CWatters]

Take a look at the specification tab. The speeds and gear ratio are different for each version.

35 lb model: 35lb at 2"/S, 5:1
150 lb model: 150lb at 0.5"/S, 20:1
200 lb model: 200lb at 0.3"/S, 30:1

The actual current draw will depend on how the load behaves at different speeds. What is the load?

 

[CWatters]

It's quite interesting to calculate the power required to move the max load...

For the 150lb model..

150lb = 68kg
0.5"/S = 0.0127m/S

The power required should be..
= Force * Velocity
= 68*0.0127
= 0.86W

Yet you say it's drawing 5A at 12V = 60W?

Have I made a mistake?

 

[Apple&Orange]

Conservation of energy is a powerful tool. Ignoring efficiency, electric power = mechanical power.

To minimize power, make it move slower, or use a lever or gears.

So the 200lbs actuator should draw less current then, given it has a higher gearing?

It's quite interesting to calculate the power required to move the max load...

For the 150lb model..

150lb = 68kg
0.5"/S = 0.0127m/S

The power required should be..
= Force * Velocity
= 68*0.0127
= 0.86W

Yet you say it's drawing 5A at 12V = 60W?

Have I made a mistake?

The actuators will need to lift around 105Kg of vertical load. However, I have both of them opening at an angle of 27.72 given the design constraints.
The power of the motor is seems correct, because that is what my colleague has calculated as well.

I'm not an expert when it comes calculations...but for using the power formula above, shouldn't the mass be multiplied by acceleration first?

 

[CWatters]

shouldn't the mass be multiplied by acceleration first?

Darn it you are right.

The actuators will need to lift around 105Kg of vertical load. However, I have both of them opening at an angle of 27.72 given the design constraints.

So is the vertical load effectively 105*9.81/Sin(27.7) = 1030/0.46 = 2215N ?

Either way the fact that it would move slower with the 200lb model would reduce the power consumption and hence the current.

 

[Asymptotic]

Looking at the spec sheet there's another thing to keep in mind ... duty cycle. These actuators are rated 20% duty cycle at 100% load, 50% duty cycle at 25% load, and a maximum of 5 minutes continuous operation.

 

[Apple&Orange]

Hi guys, I've had another thought.

Assuming same actuator specifications, which of the three scenarios would draw more current from the actuators?

Based on my understanding of basic physics, I am inclined to say Scenario B, mainly because the vertical force component of the actuator relative to gravity has decreased, or even gone into the negative. Scenario C would draw the least amount.

As you can tell, these actuators will be used to lift a lid with mass on top. These lids are attached to trailers, and the trailer may be parked at an X incline.

 

[CWatters]

Can you put some dimensions on one of the drawings?