Trying to comprehend the Impact Energy

[Martin ConnellyParticipant @martinconnelly55370]

I think impact energy is possibly the energy transferred from the machine to the bit at each collision. In the distant past when I was at school we used to do problems involving collisions of perfectly elastic spheres. It was all based on momentum which I think is mv, mass times velocity. The change in velocity was the important part regarding the energy transfer. So in the case of a hammer contacting a chisel the change in momentum would be because the hammer bounces off the end of the chisel, its velocity is in the approximately reverse direction and probably at a lower value than before contact. At the same time the chisel goes from stationary to moving in the direction the hammer has sent it. If the hammer was in contact for any amount of time then it would be doing work. For a instantaneous contact it is considered to be an impact transfer of energy.

Martin C

[MadMikeParticipant @madmike]

Brian Connelly, an interesting view but it is incorrect. As I said earlier Impact Energy is a value determined by the energy expended to cause the test piece to fail. It's value is, again as I also said earlier, dependent upon the size and composition of the test piece. Whilst your observations may reflect what may be happening when the tool is used the Impact Energy is only relevant to the test.

[MadMikeParticipant @madmike]

Just to clarify a little, the Charpy Test (also known as a Notch Test) would have been on the "chisel tools" to determine their resistance to breaking using a pendulum and measuring the energy required to break the test piece. If you have not been involved in material testing then it is probably worth trying our good friend Mr. Google.

[David AmbroseParticipant @davidambrose86182]

We had a Charpy tester on a training ship I was on as an engineer cadet. Not sure what effect a pitching ship would have had on the accuracy , but it was fun to watch. Certainly less tense than waiting for the break on a tensile tester.

[Kiwi BlokeParticipant @kiwibloke62605]

The waters are getting muddied by imprecise terminology. And we are getting into deeper waters than I know how to manage. Whilst 'impact energy' may be the energy required to break a test piece (as in the Charpy test), any impact upon a material will deliver energy to the material. It may, or may not result in macroscopic damage. Cracks, which lead to wholesale failure (one hopes – in this application) require energy to start, and energy to propagate. (Look up Griffiths, energy and cracks). On this machine's spec. sheet, 'impact energy' (if it means anything at all – spec. sheets can be fanciful, especially if from the Orient…) must mean the energy that each blow delivers.

Another problem is that potential energy is often quoted in units like 'foot pounds', which is dimensionally incorrect (distant memories of Dimension Theory stirred?). Gravitational potential energy = m X g X h, if you remember, so, the inclusion of the acceleration term makes the dimensions correct, as they are for kinetic energy, = 1/2 X m X v X v (sorry, don't know how to do superscripts here…). A Joule = 1 Nm, not any number of kg.m, so is dimensionally consistent, regardless of the type of energy.

However, this doesn't really tell you whether it's going to be an effective breaker, because energy delivered per area is what matters. So a sharp-pointed chisel will inflict different damage to, say a flat, blunt 'chisel' with an end area of a few square whatevers. The system is so full of unknown variables that I think the 'impact energy' figure isn't helpful.

OK, having stirred the muddied waters, I'll slink away.

[Michael GilliganParticipant @michaelgilligan61133]

Thanks [as usual] for your insightful ponderings, Kiwi Bloke

To a large extent, you have re-stated my original lack of comprehension.

However … Just to simplify:

1. The breaker does an excellent job
2. I believe that the quoted Impact Energy must relate to the maximum energy that the machine  can deliver from its  ‘hammer’ to the chisel [because that is quantifiable and, presumably, does not vary according to the shape of the chisel-tip or what it is attempting to break]

MichaelG.

Edited By Michael Gilligan on 13/07/2023 05:58:43

[Michael GilliganParticipant @michaelgilligan61133]

At last …

Although it relates to equipment on a bigger scale, this is informative: **LINK**

https://www.cdrecycler.com/article/getting-a-handle-on-power/

MichaelG.

[Kiwi BlokeParticipant @kiwibloke62605]

er, 'scuse my scepticism, but I think that all the article really says is that the methodology is not standardised, and the figures are therefore essentially meaningless. It's specmanship. Remember the spec war around hi-fi amplifier 'power' ratings? RMS, peak, 'music', 'average', at what level of distortion, etc., etc.? The most important parameter relating to your breaker is that you're satisfied with it…

I find that a 12lb sledge, swung with truly evil intent, is an effective demolition tool. I used to swing a 14lb sledge. But now I'm older… The energy delivered is proportional to the velocity squared, so a considerably faster-moving, but slightly lighter walloper is better – at least for this old bloke!

[Michael GilliganParticipant @michaelgilligan61133]

Posted by Kiwi Bloke on 13/07/2023 07:25:47:

er, 'scuse my scepticism, but I think that all the article really says is that the methodology is not standardised, and the figures are therefore essentially meaningless.

.

But my alternative reading is that they did standardise a methodology which was meaningful and repeatable … but then people started drifting away from it.

It seems increasingly unlikely that I will ever get the comprehension that I was looking for … but I guess it doesn’t really matter.

MichaelG.

[Martin ConnellyParticipant @martinconnelly55370]

The sharp/blunt chisel results in a difference in pressure which is force divided by area. Sharp means a small area so the pressure is higher than blunt with a larger area. Force is mass times acceleration or, in the case of a chisel, deceleration. Force is also the rate of change of momentum. Momentum imparted on the chisel by an impact is mv so if you know the mass you can work out the velocity from the momentum. You can work out the energy imparted at the point of the chisel from loss in kinetic energy which requires mass and velocity to calculate knowing that the chisel stops after a short movement. The one thing that is hard to figure is the deceleration of the chisel as it would take some time to measure the time it took from impact to stationary and enough times to get an average.

The Charpy test, despite being referred as an impact test, is not purely an impact test as the process of bending then breaking a test sample requires that the pendulum does some work on the sample. As I mentioned earlier a true impact does no work as it is considered to be an instantaneous contact that transfers momentum. If you have a hard and brittle sample such as glass then the initial contact is an impact that breaks the sample on contact and so no work is done on it and the change in the momentum of the pendulum is very small so the pendulum swings higher than when a malleable sample that has to be pushed out of the way and has work done on it is tested.

Martin C

Edited By Martin Connelly on 13/07/2023 08:41:50

[Michael GilliganParticipant @michaelgilligan61133]

Posted by Martin Connelly on 13/07/2023 08:41:12:

.

The sharp/blunt chisel results in a difference in pressure which is force divided by area. Sharp means a small area so the pressure is higher than blunt with a larger area.

.

Quite so, Martin …. which Is why I was trying to keep the chisel edge/point out of the discussion

My chisels have the basic cross-section of a 28mm A/F hexagon, and all I am trying to comprehend is how the hammering of that by the machine compares with with what a man hitting it with a lump-hammer would be.

The characteristics of the chisel and its victim would appear to be the same for both candidates, and should therefore cancel-out in the calculation.

MichaelG.

[RobinParticipant @robin]

Kinetic energy is Newton meters Nm in metric, foot poundals ft.pdl in old fashioned and foot pounds force ft.lbf in everyday speak.

A one pound weight presses down with one pound force here on Sol 3 where the lump hammers live.

If you want to do kinetic energy without destroying your brain, I strongly suggest the metric system where Newtons and kilograms are very different beasties

[Michael GilliganParticipant @michaelgilligan61133]

Posted by Michael Gilligan on 13/07/2023 08:57:21:

[…]

My chisels have the basic cross-section of a 28mm A/F hexagon, and all I am trying to comprehend is how the hammering of that by the machine compares with with what a man hitting it with a lump-hammer would be.

The characteristics of the chisel and its victim would appear to be the same for both candidates, and should therefore cancel-out in the calculation.

.

For further clarity … let me re-phrase that statement to remove the potentially ambiguous word ‘it’

My chisels have the basic cross-section of a 28mm A/F hexagon, and all I am trying to comprehend is how the hammering of that by the machine compares with with what a man hitting that with a lump-hammer would be.

The characteristics of both the chisel and its victim would appear to be the same for both candidates, and should both therefore cancel-out in the calculation.

[Kiwi BlokeParticipant @kiwibloke62605]

Well, you can do a bit of guesstimatology…

Estimate the mass of the element in the breaker that flies forwards to strike the chisel's backside; estimate this element's stroke. You know the number of cycles per second, so you can estimate the element's max. forwards velocity (unless the geometry of its driving mechanism is peculiar), so you can get an idea whether the given energy figure is reasonable.

You can also estimate the terminal speed of a lump hammer, and you know its mass, hence you can guesstimate its energy.

What you don't know is whether concrete yields better to very many little taps, or to Thor and his hammer, smiting it mightily. My money would be on Thor, but machines don't get tired.

Edited By Kiwi Bloke on 13/07/2023 10:30:11 (typing with fists again)

Edited By Kiwi Bloke on 13/07/2023 10:30:58

[RobinParticipant @robin]

Posted by Michael Gilligan on 13/07/2023 08:57:21:

My chisels have the basic cross-section of a 28mm A/F hexagon, and all I am trying to comprehend is how the hammering of that by the machine compares with with what a man hitting it with a lump-hammer would be.

Take the weight of the lump hammer in kg

Multiply by 9.81 to get the downward force in Newtons

Add a bit if he is pushing it down rather than just letting it fall.

Multiply by the vertical distance between hammer and chisel in meters.

Put J for Joules after the result.

Compare result with tool specification.

Simples (if I am right)

Robin

[gerry maddenParticipant @gerrymadden53711]

Michael, oh dear, we seem to be going all around the houses on this one! For your straightforward question the straightforward answer is this:

The energy of a lump hammer head falling freely from a height and landing on its target can be found quite simply by: m x g x h For example 2kg x 9.81 x 2metres = 39.24 Joules.

In practice the bloke swinging it will accelerate it a little more than 'g'. A rough estimate could be 30%. So 1.3 x 39.24 = 51 Joules.

Of course it wont be 30% when he's done that a few times though, and more importantly your machine can deliver it's energy probably multiple times a second for as long as you want.

As you say, all the other things are by and large, constant.

Gerry

[gerry maddenParticipant @gerrymadden53711]

Sorry Robin, you sneaked in while I was typing

[duncan webster 1Participant @duncanwebster1]

Michael originally specified lump hammer. To me that is a short shaft 4lb hammer designed to be used in one hand (remember my dad was a stone mason, so I have some familiarity with the terminology) but they do come in different weights. To get Michael's 55N, this hammer would be travelling at 7. 8 m/sec. If we take the working stroke as 600mm (2 ft) the required acceleration is 50.4 m/sec^2, so a force of 91.6N, about 20 lbf. Doesn't sound unreasonable, but at a lot less frequency, and not for very long.

[duncan webster 1Participant @duncanwebster1]

When I worked on gas centrifuges many years ago, they had a calibrated sledge hammer. Machine running at full tilt, many thousands of rpm, in vacuum with quite close clearances, they would hit the casing increasingly hard until they got a rub, followed by complete catastrophic failure. No idea how the calibrated hammer worked, or Why they were doing it, I was very junior back then.

[Michael GilliganParticipant @michaelgilligan61133]

Posted by duncan webster on 13/07/2023 11:15:35:
.

Michael originally specified lump hammer. To me that is a short shaft 4lb hammer designed to be used in one hand (remember my dad was a stone mason, so I have some familiarity with the terminology) …

.

Yes, that was my intent

MichaelG.

[Michael GilliganParticipant @michaelgilligan61133]

Thank you gentlemen … I think we are probably there

Each and every impact from the machine is near enough equivalent to me hitting the chisel with my lump hammer

But the machine runs at 1450 beats per minute and doesn’t tire as quickly as I do.

Q.E.D.

MichaelG.

[duncan webster 1Participant @duncanwebster1]

Posted by David Ambrose on 12/07/2023 22:41:53:

We had a Charpy tester on a training ship I was on as an engineer cadet. Not sure what effect a pitching ship would have had on the accuracy , but it was fun to watch. Certainly less tense than waiting for the break on a tensile tester.

Do you mean ship as in bobbing about on the water? If so why do they need Charpy and tensile test machines? I do realise that Royal Navy shore establishments are also designated 'his majesty's ship', bizarre.

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