Parallel movement for the hard of thinking

[Turbocharger]

I'm struggling with brain fade so I need a second opinion.

I'm making a parallel movement linkage. The left side is anchored and the right side can rise and fall. I'll use a diagonal gas ram to assist the lift but I want to know how much it'll lift.

Am I correct to consider it as torques at the right-hand side? Eg F y = L x

Or am I completely on the wrong track?

[dirtyninety]

I *think* thats right, or atleast very close.

for example

F = 150N

y = 120mm

x = 800mm

150 x 120 = 18000

18000/ 800 = 22.5N

22.5N = 2.29kg

Correct me please, ive lay in bed now for 30 minutes reading up on four bar linkage and reading GCSE physics websites..

Good website for struts aswell http://www.sgs-engineering.com/gas-struts/range/nitride/fixed/14mm-rod

[ian_s]

i think you need to know the angles between the ram and the linkages to be able to work it out accurately, and as the angles change the resultant lifting force will change too

[sean f]

I *think* thats right, or atleast very close.

for example

F = 150N

y = 120mm

x = 800mm

150 x 120 = 18000

18000/ 800 = 22.5N

22.5N = 2.29kg

Correct me please, ive lay in bed now for 30 minutes reading up on four bar linkage and reading GCSE physics websites..

Good website for struts aswell http://www.sgs-engin.../fixed/14mm-rod

From my look at it the for the force value it should be the angle of the ram so the "y" is the measurement from the base of the ram not the one from the already inclined position. Also you need to look at it as a right angle triangle so first look at the ration of the 2 sides and calculate a length

Using the above numbers.

f*f = x*x + y*y (Pythagerous) f*f= 120mm *120mm + 800mm * 800mm f*f =654400, f=809mm

so for a force of 150N

F = 150/809*120, F = 22.2N

With a Horivontal force of 148N

So at that angle horrible inefficient.

In simplified form to calculate the amount of lift available or the force needed in the ram think of a right angle triangle with the ram as the long side. The lengths of the sides are the forces so at a shallow angle as in the diagram most of the force will be trying to rip the lift apart with only a small amount supplying lift, as the angle increases the amount of lift for the same ram force increase. This ignores mechanical friction etc. For the most lift you want the ram to be as close to vertical as possible for the required amount of travel, or use some form of fulcrum and lever design.

If maths isn't your strong point draw it as accurately as you can with the required load as the y value length and the angle correct, measure the F value length and this is probable accurate enough. Also bear in mind the physical length of the ram will obviously change as it moves, just to make it more complicated.

The same holds true for forces in a lifting or towing bridle if the 2 legs are at 45deg angles the tension in the bridle is 140% of the lifting load i.e. for a 1000kg load the tension would be 1400kg, if the angle is increased to 60deg then the load doubles. The rest of the force is going to be trying to force the two lift or towing points together. Simple lesson if you use a towing bridle keep the legs as long as possible and the angles as shallow as possible.

Just finished a 13hour night shift so feel free to tell me where I have gone wrong as my brain has died. One day off then 3 more weeks of 13 hour nights to go.....

[simonr]

If you assume that the pivot of the ram on the left hand side is coincident with the pivot of the lever - such that X, Y, F forms a right angle triangle.

If Y is zero, the force in direction Y is zero and if X is zero, the force in the direction Y is the same as the force F. The relationship is sinusoidal (because the lever and ram are moving in a circular path) so:

Call the Force in Y direction = fY

fY = F x Sin(Angle XF)

Angle XF is given by ArcTan(Y/X)

so fY = F Sin(ArcTan(Y/X)

Using DirtyNinety's figures:

F = 150N

y = 120mm

x = 800mm

fY = 150 x Sin(ArcTan(0.15)

fY = 22.25N

Trying the two extremes:

F=150, X=0.001 (it cannot be 0 as 800/0 = Infinity), Y=800

fY = 150N

F=150, X=800, Y=0

fY=0N

Which seems to work!

Si

fY =

[fozsug]

^^^^^^^^^My brain hurts^^^^^^^^^^

[simonr]

Even the above is not right for the full arc of movement - ill have,another look at it in the morning! It is right for the position as shown, with the three links forming a right angle triangle, but not quite when they don't.

Si

[Turbocharger]

Which is about where I got to - it needs to travel from horizontal (as shown, assisting the full weight of the mechanism) to vertical, where the "weight" will be much smaller as it goes vertical, so I need a lower force - I think it's going to work.

I know the ram is much stronger than I need - hence the shallow angles to tone it down a bit because I don't want to catapult the whole truck around the garage when I release a catch. I could just drill several holes and try the ram in different positions... but I can't do that until the bits arrive and "scribbly maths" curiosity got the better of me in the meantime.

The real measurements are 530mm horizontally, lifting about 9kg at the worst case (horizontally) and the ram is 350N (35kg) ish. The ram position is adjustable to taste.

[discomikey]

right, so im doing off road vehicle design at uni. This is EXACTLY what i had to do in my exam less than 2 months ago

but i dont have a frigging clue!!!!!! haha. sorry i cant help

[Troddenmasses]

I think that the issue is that the force is proportional to the angle between the ram and the linkage (constant of proportionality being sin). In the down position where greatest force is required, the least is available. As the angle rises and the mass decreases, the force increases. Unless you are trying to make a robot wars style 'flipper', I think that a slight redesign is in order.

[Turbocharger]

In the down position where greatest force is required, the least is available.

I don't think that's true - when the links are vertical, the ram will be vertical too (because its mounts are immediately above & below the 'lower' link bar) so there won't be a resultant.

I've played with the maths on a whiteboard today and I think the best method is vector diagrams, and the answer is "it'll be reet".

[Turbocharger]

I built a model in Excel last night, so (if my maths are right):

The vertical ram force is the cosine of its angle, so greatest when the ram is near-vertical.

The tension in the links also rises (and becomes more evenly shared / less biased towards the lower link) when the mechanism is near vertical.

The lifting force is roughly (12-13kg) constant throughout the movement, until the ram goes over-centre near the top.

The lifting torque will decrease as the mechanism rises, but then (because of the load) the load should reduce too.

End result - I think it'll be ok. I should find out at the weekend, then I'll post some pics.

[Turbocharger]

In the end I simplified the parallel part, just because that damn Pythagoras meant the ram wasn't long enough to reach the far end anyway. Here's the end result (though if I'm honest it wasn't worth the effort). Should make the truck slightly more secure on holiday though.

[Turbocharger]

I've finished the moving parcelshelf cover thing tonight. It has a parallel movement so it can cover the whole area but still lift vertical to allow access, it locks to the tailgate to make the whole vehicle more secure, and now it's well-lit for camping trips (and ready for winter!).

I bought some adhesive strips with 3528 LEDs (not the brighter 5050 LEDs, but it doesn't seem to be struggling for light), and I've put 2x 50cm under the shelf, and a metre on the rearmost hoop that holds the tilt, to complement the 50cm of "ice white" 3528s that I've had on the ceiling just behind the seats for years. Overall, i'm quite pleased with the whole thing, it doesn't even rattle!