Time for an index plunger to fully release 2

[mymagicpie]

Hi,

I need to calculate the time for an index plunger to fully release.

The basic setup is simple. It pretty much identical to a collapsible loft ladder. Within a sliding mechanism, we have a fixed plate and a sliding plate. A plunger is housed within the fixed plate. This plunger is compressed as it is pressed against the sliding plate. The sliding plate has a through hole in it. As this sliding plate slides across the fixed plate the through hole will at some point align with the spring plunger. At which the plunger will be free to release.

The challenge is the moving plate moves very quickly. I need to ensure that the spring plunger can release quickly enough to guarantee the plunger engages into the sliding plate. Can anybody suggest how I can calculate the speed for the plunger to release.

Thanks.

 

[MintJulep]

Fundamentally, I would apply Newton's Laws of Motion to the problem.

 

[desertfox]

Hi

First work out your average acceleration of your plunger from F=m*a

Then use S= u*t + 0.5*a*t^2

S= plunger travel

U= initial plunger velocity = zero

a= average accel from the first part

t= average time for plunger travel

 

[Jboggs]

Plus friction! Could be a significant factor with all this lateral motion going on.

 

[mymagicpie]

Ok some numbers.

Well, Pre plunger release the spring is providing a 24N force onto the plate. After release (AKA bottoming of the plunger) the plunger is providing 9N force. The mass of the plunger is 0.012kg. Therefore

F = Ma

So:

a = F / M = 24 / 0.012 = 2000 m/s^2

and

9 / 0.012 = 750 m/s^2

(2000 + 750) / 2 = 1375 m/s^2

Well at least I have done that much! But I am Struggling to rearrange this formula to find t. However the values I know:

S= 0.003m (known stroke of plunger)

U= 0s (from static)

a= 1375 m/s^2 (as above)

t= ? (help?)

 

[desertfox]

t=square root (S/(.5*a))

I get about 2 mili seconds

 

[mymagicpie]

Ah ok I think I see it now.

S = u*t + 0.5*a*t^2

So as the inital velocity is = 0, the u*t can be cancelled out, and then it is just a case of rearranging the 0.5*a*t^2?

So:

t = SQRT (s/(0.5*a)) = SQRT (0.003/(0.5*1375)) = 0.0021 s.

Is that correct?

 

[MintJulep]

Then look at the linear velocity of the slide and the diameters of shaft and hole.

Are they lined up long enough?

 

[desertfox]

Yes that's what I got but bear in mind that's an approximate or average time.
Also as mentioned by others friction might play a big part in changing that time

 

[racookpe1978]

Good news and bad news. Your problem is both much more complex than you think - certainly more complex than you have simplified your calculation to plan for! - but the first part of your solution is also much easier than what you are planning.

To stop the moving plate, "all" you have to do is engage that pin into the "moving" hole. I suspect this is a homework problem, since it makes no sense realistically; but assuming it is a real-world problem, I also suspect you plan on drilling the hole in the moving plate the same diameter as the pin. Don't do that.

Your hole MUST be large enough so that the pin ALWAYS can go into the hole BEFORE the hole "goes past" the top of the pin. Now, to do that, you need to know EXACTLY how much side-to-side tolerance you MAXIMUM can expect your sliding plate to be able to move + the mis-location tolerance of the hole in the moving plate + the tolerance of the fixed plate + the tolerance of the hole in the fixed plate + the tolerance UP + DOWN as well as left + right of the mechanism of the pin and spring + the left + right + up and down maximum tolerance of the screw holes in the fixed plate for the pin and spring assembly!

That is, your mechanism will fail if the worse case position of ALL of the components puts the pin out of position of ANY future moving position of the hole as it slides past.

You know your mechanisms and assembly tolerances: Maybe your pin is 1/4 inch dia (6 mm if metric) and your maximum out-of-position tolerances require NOT a 1/4 inch hole but a 5/16 or even 3/8 dia hole. A very fast moving plate might need even a 1/2 inch hole.

Or you can specify a sharp=pointed pin. That reduces the need for larger "moving" hole, but requires you get the entire length of the pin into the hole before the hole "goes by" the entering pin. If you do not get the entire length of the pin not merely started "up" into the hole as it goes by, BUT ENTIRELY THROUGH THE LENGTH of the hole before the hole goes by, the sliding plate is not stopped, but jammed into the side of the pin before it is far enough in to be strong enough the stop the plate.

If only the "tip" of the pin gets in the hole before impacting the side of the hole, you break off the pin. Or bend it - same level of problem. If the pin is bent, it can't retract, and you can't use the mechanism twice.

But even if you do that calculation to get the pin engaged safely into the moving hole, then you need to calculate NOT the speed to get the pin INTO the hole, but rather the speed to get the pin ALL THE WAY THROUGH the hole so it doesn't bind half way down the hole as the plate slides by.

Now, your real problem is NOT how fast the pin needs to be moving when the pin engages the hole, but how you are going to stop the sliding plate assembly when the pin engages. Do you seriously think you are going to drive a small diameter (light weight!) pin into a hole in a moving plate, engage that spring-driven pin entirely up and through that hole in that moving plate, and then stop that entire plate from moving when the pin engages that hole more than once?

In a way, the mere fact that you never mentioned the thickness of the sliding plate nor the weight of the entire sliding plate assembly tells me you don't have a realistic problem.

Above, we discussed four possible easy solutions to getting the pin engaged positively into that moving hole: locating the pin and spring assembly as accurately as possible in the fixed plate, making the hole in the moving plate as large as possible (or a slot if need be, though that is much, much more expensive to machine), and making all tolerances in the moving plate and fixed plate as close as practical to ensure the movement is actually going to take place over the range that can be accommodated by the pin and spring assembly.

Now, again, how are yo going to stop that plate once the pin engages the hole?

This is not a "braking" stop. This is a sudden impact that must be totally taken up by the pin, spring mechanism, and the screws holding that pin and spring onto the moving plate. Further, once the impact has be absorbed by the spring and plunger, then the fixed plate and ITS mounting mechanism must absorb the crash. At the same time, the mounting brackets and slides of the moving plate need to absorb that impact of the crashed sliding plate.

Now, assuming your crashed piece of machinery survived each impact, you also to plan a way to take the weight of the piece OFF of the pin to pull the pin back and release the sliding plate to move again while holding the sliding plate in position with some other mechanism.

 

[ornerynorsk]

Assuming your pin is round-nosed, you have the advantage of additional reaction time because it begins to move at-or-near r distance from the centerline of the hole. Racookpe1978's idea of a sharp pointed pin is even better, but may be prone to damage if the pin is constantly loaded as you have suggested. You may buy some additional time by ramping an entry to the hole so the pin has less distance to travel once it has reached centerline. Just a few thoughts. Is this a one-time-use safety or arresting arrangement, or meant to be used repeatedly? Repeated use poses a larger problem than just getting the pin into the hole, as others have mentioned.

It is better to have enough ideas for some of them to be wrong, than to be always right by having no ideas at all.

 

[tbuelna]

As racookpe1978 points out, the situation is quite complex indeed. As the old saying goes, it's not the velocity or distance traveled that poses a problem. It's the sudden stop at the end.

A pin popping into a hole in a sliding plate with minimal clearance will produce huge dynamic contact forces as the sliding plate is brought to an abrupt stop.