Reply With QuoteAm I not getting a response because this is super easy and you all are laughing at me for not figuring it out?
I've seen several questions about the equations for a scissor lift but I need to see how the equation might be modified for a slightly different "scissor" design.
Let's use the red colored FBD in this scissor equation but change it. Instead of an X shape, lets have the legs of the scissor meet at the top under the "W" so it forms a sort of triangle. We'll call that point "A" where they meet and we'll leave points "B" and "C" the same name. However, "B" and "C" are mounted on lead screw and the thread direction is such that the "B" and "C" move towards each other as the lead screw is turned. The turning of the screw moves "B" and "C" towards each other and "A" moves upward. What does the equations look like for this scenario? It seems like it would be easier since there is no pivot point ("E" from the above FBD) in my scissor design.
In addition to the equation to determine the force needed to move "B" and "C" towards each other to lift "A", I need help with the equation to determine the required torque on the lead screw. I can find my pitch, diameter, and RPM of the lead screw, but I need help putting that all together.
My ultimate goal is to find out how much torque I will need to turn that screw to lift the weight at point "A".
Am I not getting a response because this is super easy and you all are laughing at me for not figuring it out?
Hi,
OK, this is gonna sound a little harsh, but bear with me. I intend no disrespect, just hopefully edification.
I suspect it is because no one can understand exactly what you are asking. While the concept may be clear in your head because you have been working on it day night for a week, it is far less obvious to us, well certainly me at least.
Also, in order to asses what you might be seeking it would mean that I should figure out what you are talking about and develop a sketch or two. That's a lot of work to expect someone to go to for some free answers.
You need to make it simple for us to understand. Most of the regulars here work a day job and have a life outside of that as well. Spending half an hour just to get started resolving your question is far from enticing to them so they move on to the next request for help. I know that's what I did.
Give it another try with some more detail.
None taken, I should have provided more info or at least a sketch.
Hopefully this link will work.
XZA6PoF.jpg
The weight is on top at "W", the three points are marked as described in the original post, and the red line is the lead screw.
Last edited by Kelly_Bramble; 01-25-2013 at 01:05 PM. Reason: Attached image directly to Engineers Edge Forum
Your problem is that you no longer have a lift of any kind. Where is this thing anchored?
At C? If so, then E will rotate about C and you lose the vertical motion you need there.
On the other hand,if you have a partitioned jack screw with left and right handed threads, you could effectively have E take a vertical path.
Is that what you envision?
The equations a real easy and if you would still be interested they will be forthcoming.
Last edited by zeke; 01-25-2013 at 09:38 PM. Reason: Added a comment
I think I need to give a little more info on what I want and why "C" isn't fixed in my design and how that works. The sizes of the arrows in my picture are not representative of the force acting at that location, I just can't use Paint very well.
The red line in my picture is a lead screw. This lead screw has a right hand thread on one end and a left hand thread on the other. It is not in the picture but imagine a gear or sprocket located in the middle of the lead screw so that when it turns, point B and point C move inward or outward (depends on the spinning direction of the lead screw). This action will make A move up or down but not move sideways.
I was able to figure out that I can find the force in the X direction at point B (and C) by using Fx=(W/2)/Tan(θ). From that equation, I was able to determine the force required to hold point B (or C) in place at a given angle. Sorry if that sounded confusing.
Ok, so now I need help with the next part. Because the lead screw has a beveled face, how do I apply that to my scenario. If I put 72Nm of torque on the screw and the diameter of the screw is 24mm with a 3mm pitch, then how do I figure out how much force I am applying to point B (or C)? I'll add friction to my scenario at the end, I just don't want to clutter what I've done until I know that I get the equations right.
The lead screw torque, T is found from the energy equation
T*6.28=F*p
p= pitch
Since,as you found
F=W/tan(theta), you end up with
T=p/6.28*W/tan(theta)
Thank you. What is the 6.28 and how did you get that?
Ok, the 6.28 is 2PI radians, right? This takes my pitch (linear movement) and puts it over rotation, right?
Am I wrong in thinking the threads are helping me "push" the weight upward? If I went from the ACME thread in my design to a thread an square thread, how would I apply this to my situation?
Last edited by aakrusen; 01-28-2013 at 06:00 PM. Reason: I'm an idiot
Ummm, yes, the lead-screw is sucking Gravity from the mechanism and the load is floating upwards. Much the same way a suction cup behaves on glass when the air inside the cup is removed.Originally Posted by aakrusen
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You do know I am joking -- right?
I didn't think the lead screw had enough mass to cause a gravity sucking pull on the scissor system. While the screw will be spinning, is it spinning fast enough to to create an effective gravity field?
Ok, back to work.
I was able to get an equation from Wiki where it includes frictional loss:
T=((F*Pitch Dia.)/2)*TAN(µ+lead angle)
Wiki says the lead angle is the TAN(pitch/(PI*pitch dia.)). I might be missing it but I don't see where the beveled face of the screw is coming into play. Anyone?
What do you mean by "beveled face of the screw?" A screw and nut are in total contact (theoretically) for the entire surface of tooth/teeth engagement.
Are you asking or suggesting that friction will change depending on where you are radially on a thread-tooth? I am too confused to suggest that I am confused.
The face of the thread is at 14.5° from the perpendicular of the lead screw. The block that is sliding along the threads has a mating surface that is also at 14.5°. I don't know how to word it but the normal force of the threads on the block is at 14.5° from the horizontal movement. Again, I'm most likely butchering the description. So are you saying that because the block also has a angled face that they cancel each other out?
Dave,
He means the screw profile angle , like 60 degrees for a standard thread'
The answer is to replace
TAN(µ+lead angle)
with
(mu*sec(beta)+tan(alpha))/(1-mu*sec(beta)*tan(alpha)
beta=screw angle
alpha= helix angle
If you need a derivation consult any good mechanical engineering book
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