I have created a toggle link force calculator in Inventor. I got the equations from and used the arms of UNEQUAL lengths set up. Now, the issue we are are trying to figure out is what is the actual force we can apply without stalling the cylinder. I have attached a picture of the model. Because of how the equations for toggle links work, as your angle approaches zero between the links(top dead center - TDC - as we are calling it), the force goes to infinity...which we also know is not the case. So, is there a way to determine that if I have cylinder appliing say, 600 lbf on the bottom of the link, what is the true amount of force it can push against, before it can't go past TDC?
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Finding Stall Force on a Toggle Joint Assembly
- Thread starter RNDInov8r
- Start date
- Thread starter
- #3
So, that .ipt file is the simplified version of this. Ultimatley, that calculator is for us to size linkage assemblies/cylinders. This attached image shows what I am really doing. So, what I am trying to figure out, is what is the point at which the cylidner can't push past TDC (after the lower plate contacts the upper thin [red]gladden plate.
Oh wait...I think I figured it out...it's the X component of the "triangle" in the lingake...when that force exceeds the cylinder, or mechanical advantage of the lever...the toggle is stalled....correct?
Oh wait...I think I figured it out...it's the X component of the "triangle" in the lingake...when that force exceeds the cylinder, or mechanical advantage of the lever...the toggle is stalled....correct?
Hi RNDinov8r
Could you provide some more information like link dimensions and cylinder stroke, cylinder pressure and the resultant force of the device its driving, there are some numbers on the first drawing but reading them isn’t easy
“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein
Could you provide some more information like link dimensions and cylinder stroke, cylinder pressure and the resultant force of the device its driving, there are some numbers on the first drawing but reading them isn’t easy
“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein
yes, I think it's the dynamics of the link (it's movement in the x direction) that keeps it moving thru TDC.
Now a crankshaft may work 'cause there's more than 1 cylinder (link) so there's always something driving the motion.
When you have a single crank then I can see stalling the mechanism (at BDC) by slowing down the load application, slowing the movement of the link, ...
another day in paradise, or is paradise one day closer ?
Now a crankshaft may work 'cause there's more than 1 cylinder (link) so there's always something driving the motion.
When you have a single crank then I can see stalling the mechanism (at BDC) by slowing down the load application, slowing the movement of the link, ...
another day in paradise, or is paradise one day closer ?
- Thread starter
- #7
So, RB1957 pushed me in a direction I think that solved my issue. Where i have ended up is this.
1) Determine how much compression force is require on the assembly.
2) Determine how much vertical compression of the o-ring is required.
3) Determine what total angle the linkage has to move thru to accomplish the above vertical movement.
4) Using that fact that the vertical load builds to my required compression force from zero, and the angle which the linkage moves thru is small, I made a chart looking at the resultant "X" direction force from time of engagement to max resultant force. This shows me my maximum force in the X direction that I have to overcome, my stall force.
So my overall numbers were that I needed 200000 lbf for compression.
That force creates 1.5 mm of vertical movement.
My linkage was 70mm, so my angle at time of contact is 12° (I assumed this angle was small enough to consider my building ocmpression load to be linear)
Since the X-Direction is Compression force * Tan(theta) and I assumed the compression force increased linearly, I found 6° to be the maximum resisting force in the X direction...it comes out 1073 lbf.
My cylinder produces 600 lbf on a lever arm of 140 mm, so that means i am exerting 1200 lbf at the linkage, therefore, My cylinder will not stall while locking into position.
1) Determine how much compression force is require on the assembly.
2) Determine how much vertical compression of the o-ring is required.
3) Determine what total angle the linkage has to move thru to accomplish the above vertical movement.
4) Using that fact that the vertical load builds to my required compression force from zero, and the angle which the linkage moves thru is small, I made a chart looking at the resultant "X" direction force from time of engagement to max resultant force. This shows me my maximum force in the X direction that I have to overcome, my stall force.
So my overall numbers were that I needed 200000 lbf for compression.
That force creates 1.5 mm of vertical movement.
My linkage was 70mm, so my angle at time of contact is 12° (I assumed this angle was small enough to consider my building ocmpression load to be linear)
Since the X-Direction is Compression force * Tan(theta) and I assumed the compression force increased linearly, I found 6° to be the maximum resisting force in the X direction...it comes out 1073 lbf.
My cylinder produces 600 lbf on a lever arm of 140 mm, so that means i am exerting 1200 lbf at the linkage, therefore, My cylinder will not stall while locking into position.
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