# Spring+mass to do known job. How dimension? Impulse? Energy?

Hello there,

I'm working on a design project where I have come upon a mechanical problem that I'm having trouble with. Basically I'm making a kind of specialized stapler (at least I think that's a good translation...), and I want it to clamp the staplers using a mass accelerated by a spring, that strikes a surface on the tool that does the stapling.

I've set the staples up in a machine that pushes on them with constant speed, and recorded the reaction force from the staples as they are pushed together. It's basically a rather steep curve that peaks at 400 N, over a distance of about 20 mm and a time of 0.67 seconds (with a constant speed of 30 mm/s).

The background is that I'm trying to decide if I want to go with just a rack and pinion and an electric motor that generates a force of 0,4 kN, or if I should go with having a motor compress a spring that powers a hammer. It doesn't matter much if the compression is slow, but the clamping should happen pretty quickly after the button is pushed, so if I go with the rack and pinion, I have to get a pretty expensive motor to do the pushing. If the motor is permitted to work slower, while compressing the spring and preparing for the next cycle, I might save a bunch of money. All of this is, however, depending on what kind of spring and mass I should go with, and right now I'm completely clueless.

So please, how do I go from knowing the force as a function of time and/or distance to knowing what spring/mass I need to smacking those staples together?

Help would be greatly appreciated!

## Answers and Replies

I'd solve it for the total work needed to push the staple in.. so 400N*0.02M is 8NM... from there, you have to find a spring which requires at least that much work to compress it over however long a distance (I'd suggest 25mm perhaps?)

Don't know on the feasibility of it, but a pneumatic or electromagnetic plunger may be well suited to this too.. even a smaller motor with a flywheel may be able to do it if you have a cam or eccentric.. just some food for thought

Chevreuil
I think the idea should be that I take a much longer distance of compression, like 100 mm, so that the mass can accelerate, and get a high KE without me having to buy a motor that can churn out all of the work in a short distance.

Here's a complicating factor: I don't mind if it takes 5 seconds to get ready for the next stapling, but after a stapling is done, I want the tool to release the staple in less than 1 second, which means that the hammer must retract roughly 20 mm rather quickly. The following 80 mm of retraction (which is most of the work with compressing the spring) can take up to four seconds in total.

If however, the spring is supposed to hold all of its work from just 25 mm of compression, then the motor that can compress it 20 mm in less than one second, is simply powerful enough to do the push by itself, and I can skip the spring.

All input is appreciated! I often get stuck when doing this stuff in an echo chamber.

oh, and electromagnet piston! I haven't thought about that! Is that the same as a "solenoid linear actuator?"

mfb
Mentor
You probably want the equilibrium position of the spring to be at "zero" (staples not pushed yet). Extending the spring 20 mm more will cost some energy, so the sping has to release more energy before. With a longer spring, this contribution is small. As an example, if the spring can be compressed 100 mm and you want 8 J delivered to the staple, the 20 mm just have 1/25 the energy of the 100 mm, so you need 25/24*8J = 25/3 J energy stored for 100 mm compression. Let's round that up to 10 J, then you need a spring constant of 2000 N/m, and a peak motor force of 200 N. Well, doesn't help much so far. But this assumed 400 N over the whole length of 20 mm, you don't need that. What is the total energy you need to push them in? This is the integral of the force over the displacement. It might depend on the pushing speed, something that can (should) be investigated separately.

Chevreuil
I've actually tested the force requirement as a function of the displacement distance with several different (from 0.001 m/s up to 1m/s) constant pushing speeds. They all look about the same. A good approximation would be like this: 10 mm of 200 N, then a triangular peak to 400N, which starts to happen at 10 mm, peaks at 12.5 mm and dies off at 15 mm, followed by another 5 mm of 200N.

That makes 4.75 joules.

If the spring pushes a 250 g mass/hammer head to give it a kinetic energy of 4.75 joules, I need a speed of 4.36 meters per second. This is roughly the impact speed you get if you drop something dense from a height of one meter.

Are you guys positive that as long as the kinetic energy of the spring-accelerated mass equals the push energy? It feels counter-intuitive for some reason.

I'll find some example springs and do some math and see what I come up with. Thanks again, guys!

Nidum
Gold Member
What you are describing is a synchronous spring hammer mechanism . This only works for constant time/cycle applications .

For asynchronous applications you need to latch the spring when fully compressed and then release the latch to fire the mechanism .

Spend some time looking at existing stapler mechanisms and also hammer and hammer drill mechanisms .

Chevreuil
What you are describing is a synchronous spring hammer mechanism . This only works for constant time/cycle applications .

What do you mean by "constant time/cycle applications"? What do you mean by "synchronous"? What's synchronized with what?

For asynchronous applications you need to latch the spring when fully compressed and then release the latch to fire the mechanism .

Spend some time looking at existing stapler mechanisms and also hammer and hammer drill mechanisms .

Latching it up would be great, but I'm still not 100% sure on how to go from my records of the push to picking a spring and hammer head mass.

mfb
Mentor
For asynchronous applications you need to latch the spring when fully compressed and then release the latch to fire the mechanism .
I thought that was the idea.

If we just need 4.75 J, you can half the values in my previous post.

What about some kind of gear system? 5 J in 1 second doesn't need a very powerful motor, and you can even make it nonlinear (like circular motion -> linear motion conversion) to get that extra spike in the required force.

Nidum
Gold Member
In a synchronous spring hammer mechanism the hammer and spring are energised by some external drive so as to oscillate continuously in an approximation to simple harmonic motion . The mechanism is usually energised at a frequency which is just below it's natural frequency .

There can be a lot of stored energy in this simple arrangement .

The basic mechanism has a fixed period or cycle time.

Nidum
Gold Member
@Chevreuil

Draw out a mechanism which you think would work in principle .

Nidum
Gold Member
What about some kind of gear system? 5 J in 1 second doesn't need a very powerful motor, and you can even make it nonlinear (like circular motion -> linear motion conversion) to get that extra spike in the required force.

+1

I am thinking an mechanism a bit like a windshield wiper motor could work coupled with a 'trip' mechanism... the motor stops at top dead center, as soon as it moves past, it releases the spring (Perhaps a cam?) and the spring does the work to drive the staple, the motor continues to turn and then catches the spring and brings it back to top dead center and stops, waiting for the next cycle

Chevreuil
The most simple way to do it would be to just have a big DC motor+ gear transmission to power a rack with a continuous force of 400 N, and a movement speed of about 20 mm/s, to meet the speed requirement. Yes, it would be overpowered for all of the stroke, except for the spike at the end, but it would be safeguarded against unforeseen conditions, such as stapling into a weird material. When the push sequence is complete, the pinion that drives the rack could disconnect from the rack, allowing a weak spring to push it down immediately. The problem is that this require a rather expensive and heavy motor.

What I'd like to understand better, is what would be required if the stapling was done by a hammer head accelerated by a powerful spring. At the end of the cycle, the spring is compressed by a rack, actuated by a pinion which is connected to a smaller, less expensive motor.

Why do I expect the spring+mass to afford me to get a smaller motor? Because although it is important that the push happens quickly, it's perfectly fine is it takes some time to prepare for the next cycle, i.e. the smaller motor can do it's job over a longer period of time, thus having a lower power output. If the motor is directly powering the stapling, then it has to churn out all of that energy in the push phase.

This is my question: How do I pick a spring and mass (or rather: mass and impact velocity) that I know will be able to do the stapling? I'm not sure if I trust the idea that I just have to give the mass the same kinetic energy as the work that is required for the stapling.

Last edited:
mfb
Mentor
If your motor is disconnected (or at least not braking) during the hammering process, apart from the staple and the spring there is nothing else that would take up energy (neglecting the inelastic collision between hammer and staple) until everything stops at the end of the fully pushed staples.

Well, yeah, but if I mounted a couple of staples perfectly on top of each other and then shot the stack with a .177 air gun with a muzzle energy of 10 J, there is no way that the impact of the pellet would cause both staples to fold together perfectly. A whimsy strike with a common hammer however, will work very well (tested). I've also dropped 1kg weights from 1 m and that also works.

Admittedly, I don't have an air rifle where I am now, so I haven't tried that, but there is something about these staples (made out of semi-soft polymer, not sure which kind) that tells me that it's more complicated than to go with complete energy conservation.

Is it perhaps what you just mentioned, the inelasticity of the collision?

mfb
Mentor
I assumed the hammer has significantly more mass than the staples (otherwise the collision will cost some energy), and ignored alignment issues.

the 1kg weight dropped from 1M is 9.8NM
The air rifle would have the problem that the staple will probably buckle from being hit so fast, However, the spring IN the air rifle must be able to hold at least 10J of energy, Perhaps that would make a good test spring?

Here's my idea, using a windshield wiper motor as the drive with a cam... Not fancy drafting work, but you get the idea

#### Attachments

• PF staple driver.jpg
4.1 KB · Views: 396
So I made a simple test rig today. In order to test the kinetic energy required I wanted to use the perfectly predictable acceleration that is gravity. To make sure I hit the staples in the right spot, I mounted them in a very simple wooden vice. To guide the projectile I took a long pipe with a 30 mm OD and cut up a shorter steel pipe with an inner diameter of... well lets say 35 mm. The 0.25 kg piece (about 2.5 inches of pipe) dropped from 1.75 meters consistently failed to make the staples fold up correctly, but from 2.0 meters it did so every time.

A 0.25 kg piece of steel falling from 2.0 meters has about 4.9 Joules of KE when it hits the ground...