Linear Actuator Sizing for Angular Force (BMX starting gate)

[Bodine1187]

I am sizing a linear actuator to build a BMX gate. Please see the drawing below. I have a metal gate that is 20" tall and 9.5' wide. It will be steel and weigh approximately 150lbs.

I am mounting the rear of the linear actuator on a clevis 8" off the ground. The front will be mounted with another clevis 6" above the pivot point of the gate. I am trying to understand the amount of force I need to pull the gate up into a position perpendicular to where the bikes sit on the platform. The actuator length, fully extended will be 24.25". The force used to pull the gate into a perpendicular position will be at a 19.26-degree angle to the gate.

I've also included a picture of a bmx gate to give a better idea of what I'm building. I am only building the moving gate in the front, not the platform piece.

I do understand that the force needed will decrease as the gate moves through its motion, but I'm just worried about the gate force needed to start the motion.

 

[JBA]

Assuming the 150 lb weight of the gate is uniformly distributed across it s area, gthen its CG (Center of Gravity) will be 20 in. / 2 = 10 in. from the pivot point and Fv (the vertical force required to lift the gate) at the 6 in connection point will be: Fv = 150 lbs x 10 in / 6 in. = 250 lbs; and, due to the low 19.26° angle of the cylinder rod, the resulting tension force on the cylinder rod to start the lift of the gate will be: 250 lbs / sin (19.26°) = 758 lbs.

Form a rod loading perspective, apart from any other issues it might create, a better arrangement would be to place the 8 in. vertical stand connection on the gate at the rod end and the cylinder connection on the face of the ramp. This wold reduce the initial rod tension force at the start of the lift to: (250 x 10 / 8) / cos (19.26°) = 331 lb.

 

[berkeman]

Looks like a fun project. Could you consider adding counter-weights on arms at the two sides of the gate to make it easier to raise it up?

 

[Bodine1187]

Looks like a fun project. Could you consider adding counter-weights on arms at the two sides of the gate to make it easier to raise it up?

No, because the counterweights would effect the speed by which the gate drops which is a critical variable.

 

[berkeman]

No, because the counterweights would effect the speed by which the gate drops which is a critical variable.

Makes sense, but the counterweights can be released once the gate is in the upright and locked position...

 

[Bodine1187]

Assuming the 150 lb weight of the gate is uniformly distributed across it s area, gthen its CG (Center of Gravity) will be 20 in. / 2 = 10 in. from the pivot point and Fv (the vertical force required to lift the gate) at the 6 in connection point will be: Fv = 150 lbs x 10 in / 6 in. = 250 lbs; and, due to the low 19.26° angle of the cylinder rod, the resulting tension force on the cylinder rod to start the lift of the gate will be: 250 lbs / sin (19.26°) = 758 lbs.

Form a rod loading perspective, apart from any other issues it might create, a better arrangement would be to place the 8 in. vertical stand connection on the gate at the rod end and the cylinder connection on the face of the ramp. This wold reduce the initial rod tension force at the start of the lift to: (250 x 10 / 8) / cos (19.26°) = 331 lb.

JBA, thanks for the input. I'm struggling to understand the basis for the 250lb load input. Also, reversing the action and changing it to a 8 inch mount on the gate connection would only bring the total load down to 724 lbs. Because that would extend the length of the ram to nearly 31 inches and change the angle of the mount point (150x10/8)/sin (15degrees) = 724 lbs.

 

[Bodine1187]

Makes sense, but the counterweights can be released once the gate is in the upright and locked position...

Not enough room within the design, plus the added cost and engineering. A good ram/linear actuator will do the job just fine on it's own. I just want to size it appropriately so that it works and doesn't cause premature failure.

 

[JBA]

(150x10/8)/sin (15degrees) = 724 lbs.

The lifting load for the 6" distance from the pivot to the connection point with the center of gravity at 10" from the pivot is : 150 x 10 /6 = 250 lbs vertical.

Also, the divisor in the above equation should be "cos .." not "sin ..." for the revised configuration.

As a result the actual revised load for the new configuration is: (250 x 10 / 8) / cos 15° = 323 lbs.

As for the required stroke length, the hypotenuse length from the pivot to the top of the 8" connection 6" from the pivot = sqrt(6^2 + 8^2) = 10 in (which represents an arm directly from the pivot to the arm connection point) and the arc length to rotate that hypotenuse arm, and the gate, 90° = 2πr/4 = 2 x 10 x π /4 = 15.7 in. and the circle segment for the 90° enclosed right triangle = sqrt (10^2 + 10^2) = 14.14 in. which will be the required stroke length for the actuator.

 

[Bodine1187]

The lifting load for the 6" distance from the pivot to the connection point with the center of gravity at 10" from the pivot is : 150 x 10 /6 = 250 lbs vertical.

Also, the divisor in the above equation should be "cos .." not "sin ..." for the revised configuration.

As a result the actual revised load for the new configuration is: (250 x 10 / 8) / cos 15° = 323 lbs.

As for the required stroke length, the hypotenuse length from the pivot to the top of the 8" connection 6" from the pivot = sqrt(6^2 + 8^2) = 10 in (which represents an arm directly from the pivot to the arm connection point) and the arc length to rotate that hypotenuse arm, and the gate, 90° = 2πr/4 = 2 x 10 x π /4 = 15.7 in. and the circle segment for the 90° enclosed right triangle = sqrt (10^2 + 10^2) = 14.14 in. which will be the required stroke length for the actuator.

JBA, does the possibility of the flat horizontal platform, or bottom of the triangle in the drawing, being on a 17 degree decline make any difference?

 

[JBA]

If the platform is elevated and horizontal and the gate alone dropped down at an angle of 17°, then that would be a different matter and require a longer stroke to lift the gate to a position perpendicular to the platform because the gate would need to swing through a: 17 + 90 = 107° arc and the force required would be considerably higher at the start of the lift (and would take a bit of work to calculate) because the top of 8 in. high end connection on the gate would be lower and the angle of the rod above platform significantly reduced as a result . The rod stroke would be: 2 x 10 x sin (107° / 2) = 16.08 in.

While doing all of this keep in mind that the cylinder and rod load required is not only going to have to handled by the cylinder and rod but also by your deck and gate connection structures and the deck and gate structures. For example, for the 45° strap you have from the top of your post to deck (or for the post to the gate for my recommended configuration) the tension on that strap and it connections will 1.4 x the cylinder and rod load.