scissor jack mechanism
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Posted by: w i l l

02/21/2006, 06:39:05

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Hi,

I'm trying to design a device for lifting my own weight, which I've bascially modelled using 3D CAD, shown below.

This is a scissor jack mechanism. The user sits on the seat (shown at bottom of mechanism) and pulls themself up using the handles at the top. The idea is that I can adjust the fulcrum point of the bars and therefore change the mechanical advantage and weight that I am lifting (in the range 25% to 75% of my body weight).

Could anyone give me any links to basic/in depth mechanical formulae to calculate this or any insight into whether this can actually be achieved? (PS I'm not an engineer so that basics maybe better to begin with).

Your help would be very much appreciated.


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Re: scissor jack mechanism
Re: scissor jack mechanism -- w i l l Post Reply Top of thread Forum
Posted by: w i l l

02/21/2006, 10:08:08

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ok I've calculated a force required to lift someone sat on the seat... but that would be assuming that someone else was pulling down on the handles to lift that person.

I'm thinking that the force, for the person sat on the seat, to lift their own body weight must be this force minus the force that they are applying upwards to their body to lift their own weight (E.g. if the mechanism didnt move and the handles were static the force that they would apply upwards to lift their body would be the weight of their body)...what I'm struggling with is how to work out the force that they are applying in lifting their own weight...as this is a moving mechanism.

Basically, what I am trying to say is that the user is helping to lift themselves by pulling on the handles... how can I calculate this lifting force?








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Re: scissor jack mechanism
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Posted by: zekeman

02/21/2006, 12:01:06

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You can't do what you are trying to do, since there is no external force on the closed system. If you look at the proposed system from afar you would see an acceleration of person sitting on the lift and since there are obviously no external forces, you can't do that. However, if you could figure out a mechanism whereby an equivalent mass would drop as your body was lifted, you might have something.







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Re: scissor jack mechanism
Re: Re: scissor jack mechanism -- zekeman Post Reply Top of thread Forum
Posted by: w i l l

02/21/2006, 12:07:53

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sorry, I dont understand what you mean by that. The main vertical part is fixed. The user pulls on the handles to lift their own weight and by varying the mechanical advantage they can vary the force needed to lift their weight... does this not work?







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Re: scissor jack mechanism
Re: Re: scissor jack mechanism -- w i l l Post Reply Top of thread Forum
Posted by: zekeman

02/21/2006, 14:53:35

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You're right, I didn't see the support. The device will work but you will have varying forces during the lift . If you look at the two sketches you provided, the left one in the extended position would require considerably more force to pick up the man than the right one occurring after some motion, so that you have varying mechanical advantages. Also, you have a stability problem in that there is no means for the unit eliminate the rotation it will experience without some means of holding the mechanism symmetric. You cannot count on the man to stabilize it( or can't you) But you do have the beginning of a possible working unit.







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Re: scissor jack mechanism
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Posted by: newpunkrkr

02/21/2006, 13:49:47

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Seems possible to me. Like throwing a rope with a swing over a tree limb and pulling yourself up.

I don't think your force required to lift the person on the seat is changed by them pushing down on the handles vs someone else pushing on the handles. This is because in order to lift yourself off of the seat you would have to apply a force to the handles greater than your weight. And since your creating leverage with the handles, you would apply enough pressure to lift yourself in the seat before you could lift yourself out of the seat.

Don't know if that makes sense, but that's my thought





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Re: scissor jack mechanism
Re: Re: scissor jack mechanism -- newpunkrkr Post Reply Top of thread Forum
Posted by: ChrisMEngr

02/21/2006, 19:14:22

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Lets say you apply 25 lbs with each hand..... Pull down on each handle with a force of 25 lbs that is. So total, you hands are pulling down a total of 50 lbs.

Think of the free body diagram, you have a 200 lb person. Lets take the torso, 200 lbs are pulling dowon. But look at the load on the shoulders (50 lb load from the hands transfers to the wrist, to the arm, to the shoulder) you have 50 lbs acting in the opposite direction of the of the 200 lbs which is acting downwards.

This means that the seat only has a load of 150 lbs sitting on it when really a 200 lb person is in the seat.

But as zeekman mentioned, this can be a complicated problem. But I think rough estimates can be made with some fun geometry and mechanics... only if you assume VERY slow movement.

Unfortunately, the further you pull your handles down, the larger the moment arm you create and the easier it is to lift.

As far as the origional question, what force must be pulled down to lift the body. If you assume VERY slow movement, you can get some ballpark numbers but it all depends on the weight of the person, the force they apply to the handles, and the geometry of the aparatus. Even with this you are looking at some fun geometry and mechanics work.

Exactly what do you need here? The maximum load the person must pull down? How heavy of a person can sit on this? etc...?








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Re: scissor jack mechanism
Re: Re: scissor jack mechanism -- ChrisMEngr Post Reply Top of thread Forum
Posted by: w i l l

02/22/2006, 10:37:16

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answer to your original question... I need to design this so that the user is pulling 75% or 25% of their own body weight, but different people will use it (different weights).







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Re: scissor jack mechanism
Re: Re: scissor jack mechanism -- ChrisMEngr Post Reply Top of thread Forum
Posted by: w i l l

02/22/2006, 06:37:59

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As i've modelled this on CAD I was thinking of measuring the total vertical distance moved downwards by the handle and comparing it to the total upwards movement by the seat and by comparing the ratio I could calculate the force required for the handle (by using the weight of the user).

From what you just mentioned I think this method is inaccurate as with this method you are assuming that someone else (not the user sat on the seat) is pulling the handles to lift the weight.

I'm assuming I'd have to take into account the force that is lifting the user created by pulling on the handles in the first place (as you mentioned).

What if have done is;

1. (I am assuming the user wants to apply a total force of 600N to the handles) so I calculated the upwards force lifted the user off the seat that this would create.

2. I then subtracted this force from the force needed to lift the body weight and the result is the force applied to the handles to simulate 600N.

Is this a suitable way of calculating this?







Modified by w i l l at Wed, Feb 22, 2006, 07:36:34


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