Trying to comprehend the Impact Energy

[Michael GilliganParticipant @michaelgilligan61133]

[Michael GilliganParticipant @michaelgilligan61133]

My recently-purchased ‘Electric Demolition Hammer’ [call it what you will] is proving very effective, but I don’t really understand the performance specification, and am hoping that someone versed in such matters [maybe Duncan?] will educate me.

 

It claims to deliver an ‘Impact Energy’ of 55 Joules to a chisel with a 28mm Hex shank, at an impact rate of 1450/min … but how would that translate into familiar units of something like “bloke using a lump hammer”

Grateful for any enlightenment.

Does the weight of the machine enter into the calculation, or is that incidental 

MichaelG.

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Edited By Michael Gilligan on 12/07/2023 08:22:25

[Kiwi BlokeParticipant @kiwibloke62605]

Presumably the 'impact energy' is the kinetic energy of each blow of the internal hammer, delivered to the chisel. The mass (inertia) of the chisel will have an effect on the performance, but the mass of the machine itself, being huge in comparison, isn't really important in practice. How this compares to a bloke + lump hammer system depends on the bloke and the hammer, but your breaker will go on for longer, without having to stop for tea, and will achieve its results by thousands of light strikes, which will cause less collateral damage (usually…).

If beer hasn't addled my maths, 50J is about one quarter of the muzzle energy of a typical .22 round. So, if the breaker doesn't do it, shoot it – lots of times…

On reflection, I don't suppose this helps…

Edited By Kiwi Bloke on 12/07/2023 08:41:55

Edited By Kiwi Bloke on 12/07/2023 08:57:43

[lee websterParticipant @leewebster72680]

I can't tell you anything about impact energy, but I can give a bit of advice for when you use it. Don't put all your weight behind pushing the machine into what you are chiselling. Let the machine do the cutting, you are there just to stop it falling over.

[FulmenParticipant @fulmen]

My math says it's equivalent to dropping a 1kg weight from a height of 5,5m. Not sure if that helps.

[Perko7Participant @perko7]

Interesting question. It's been many years since I did dynamics at uni as part of my engineering studies, but I think the answer would be that the impact energy multiplied by the frequency should approximate the input power minus the various losses. So, 55J impact at 1450 impacts/minute is 79.75kJ/min which is 1329J/sec which is 1329W, which is about 83% of the input power so not an unreasonable figure.

The effectiveness of this input energy in dislodging material would then need to be assessed in relation to the mass of the moving parts, the velocity at the time of impact, and the effort provided by the person holding it. There would seem to be too many variables there to allow a simple comparison with a lump hammer but someone with more recent experience than me might be able to calculate it.

A simple approach suggests that if the machine weighed the same as the lump hammer, and if the velocity at the point of impact was similar, then a similar result would be expected. For how long would be subject to the fitness of the man with the hammer.

Perhaps some empirical evidence is needed for comparison??

[Nicholas FarrParticipant @nicholasfarr14254]

Hi MichaelG, like Lee I can't tell you anything about the impact energy, but I do agree that putting heavy pressure on the machine won't make it cut any faster, in fact it is more likely to deaden the energy. I used to get this with people using industrial hammer drills, where they would sometimes have two blokes pushing on it and getting no where fast. Let the machine do the work.

Regards Nick.

[NealebParticipant @nealeb]

No idea of the equivalent figure for bloke with lump hammer but personal experience based on trying to break up a concrete path with chisel and 7lb club hammer, then going out and hiring an electric breaker of, probably, similar spec to that of the OP, is that the breaker was around 1000 times better than me with hammer. At least. And by a very considerable margin. 10mins with hammer and chisel left a few gouges in the concrete. 10 mins with the breaker left a considerable length of broken concrete fragments. Heaviest bit of the job was carrying the breaker and its bits out of the car, round the house, and up the steps to my back garden and back again an hour or two later. Would have been less but I stopped for coffee…

[RobinParticipant @robin]

I'll give it a try…

Joules is kinetic energy which is measured as force times distance.

Joules is Newtons times meters. (1 Newton force will accelerate a mass of 1 kg at 1 m/s/s)

To work in lump hammers we must convert to English units. Plenty to choose from, I suggest foot pounds

55J = about 40.5 ft.lbf

So a 3 lb lump hammer falling 13.5 feet would arrive with 40 ft.lbf or 55J of energy.

Wow!

[Ady1Participant @ady1]

It's probably like a torque gun for wheelnuts

but instead of going sideyways it goes fronteyways

(No extra charge for the technical report part of this post)

[jaCK HobsonParticipant @jackhobson50760]

I reckon "it does the work of 10 men".

I remember: One average man can work at about one old lightbulb for long periods – 75W

Lets say a little less than double that for someone with a sledge and rests. (I suspect Michael will need rests even using his new tool).

So 10 x ((a little less than 2)x75) = 1329W

Does depend on how big the men are and if you can fit them all in the space.

[FulmenParticipant @fulmen]

Perhaps a more relevant comparison: 55J is the same as dropping the 19kg machine 3 inches.

[Nigel Graham 2Participant @nigelgraham2]

Robin –

Just a small correction.

The Joule is the unit of energy, but all types, not solely mechanical energy.

[Michael GilliganParticipant @michaelgilligan61133]

Thanks for all the replies so far

Just to reassure everyone … I am fully aware [and grateful] that I do not need to apply pressure to the machine

… just holding it in position is plenty for me.

The performance of the machine is excellent … my question is just a matter of academic curiosity.

MichaelG.

[Nigel Graham 2Participant @nigelgraham2]

That allowing the tool to do the work applies to most power-tools, not just these electric chisels.

What though does this mean, on the ratings plate:

"Chiselling", ah Cheq.

Followed by two accelerations, the first of touchingly mathematical niceness. (Sorry, I can't edit the superscript size here). Presumably of the hammer, but are they even relevant to the user? :

9.529m/s², k = 1.5m/s²

[Michael GilliganParticipant @michaelgilligan61133]

Posted by Nigel Graham 2 on 12/07/2023 12:47:23:
.

That allowing the tool to do the work applies to most power-tools, not just these electric chisels.

…

.

It even applies to some hand tools !
I distinctly remember being taught [ in O-level Woodwork] about the heavy Brass-Back on a Dovetail Saw

MichaelG.

[duncan webster 1Participant @duncanwebster1]

It's been covered above in terms of dropped masses, but that's not what the machine is doing. I don't actually know how these machines work, but if we take an analogy a device which compresses an internal spring, then releases this spring to accelerate a striker mass (m) to a velocity (v) to impact the chisel. The force exerted by the spring is equally exerted onto the mass of the machine (M) and so the machine is accelerated in the opposite direction to a velocity (V). Conservation of momentum means that m*v=M*V, so making the machine heavier increases the velocity and hence energy of the striker mass. Trying to type the maths into this thread is challenging. However, any additional mass would need to be very rigidly attached to the machine, human leaning on it has lots of flexibility and damping.

To answer Michael's original 'how does this compare with a man with a lump hammer', I have no idea. I think it depends at least in part on the skill of the man. My dad was a stone mason, he could split a stone exactly where he wanted by looking at it, not quite but he certainly used a lot less effort than I needed, and his split where he wanted it to be.

[SillyOldDufferModerator @sillyoldduffer]

As others have calculated, Michael's Jack Hammer does about 1.3kW of work, which is roughly equivalent to six fit men, or ten weedy ones, and each blow contains as much energy as Robin's "3 lb lump hammer falling 13.5 feet".

Can't be doing with nasty old imperial measure, but that's about 55 Joules hitting the target at about 7 metres per second. The pressure is spread over the area of the lump hammer, rather than concentrated at a chisel point, but enough to deliver a nasty injury.

The Jack Hammer's acceleration figures suggest the speed of it's chisel blows are much the same as Robin's falling lump hammer. Speed makes a big difference. The amount of energy used by a human to push a car slowly won't damage the car. But if the same amount of energy is delivered quickly, the car and human are both damaged. If the human applies the energy with a spear, or hammer, then the car will be damaged badly. Pushed objects have time absorb the energy and maybe move out of the way. When a struck object doesn't have enough time to absorb the energy, the material it is made of is displaced, resulting in cuts, disintegrations and penetrations.

Jack Hammers are designed to deliver sharp blows at the chisel point without wrecking their internals. I believe it's done by the motor slowly storing energy in a flywheel, that operates a faster but still relatively slow moving internal hammer, that delivers all it's energy into the chisel point in a rapid series of short sharp blows. Although most of the energy is dissipated in the target, the chisel point is replaceable, the internals take a beating, and the motor won't last forever!

Dave

[NealebParticipant @nealeb]

If a man can use a hammer at a rate of, say, one blow every 5 sec, i.e. 12 blows per minute, and each blow (see a previous post) is around one-half of a single breaker blow, then we are looking at something of the order of 2000 times more energy dissipated at the chisel tip. No wonder the concrete just gives up.

[Michael GilliganParticipant @michaelgilligan61133]

Posted by duncan webster on 12/07/2023 13:38:28:

[…]

I don't actually know how these machines work, but …

.

It’s not obvious, even with an exploded diagram to study

… but the gist of it seems to be:

1. an electric motor is geared down, and drives a reciprocating piston
2. the chisel is retained within a ‘chuck’ but is free to slide [because of the flat machined on the hexagon]
3. if the machine is run in free space, then the chisel slides down and there is no impact
4. when the weight of the machine is applied, contact is established and it hammers
5. vibration isolation of the handle is included to protect the user.

MichaelG.

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In principle, this is very much like holding the chisel lightly in one hand, and hammering it.

Edited By Michael Gilligan on 12/07/2023 14:45:55

[MadMikeParticipant @madmike]

Impact energy is the energy required to make a standard test piece fail. It is somewhat meaningless in this instance unless you know the size and composition of the test piece used. However if the test piece was a block of cheese it would not be so effective if it required 55 Joules to make the cheese test piece fail. Silly example but I am sure the learned throng will understand.

Edited By MadMike on 12/07/2023 15:37:20

[Martin ConnellyParticipant @martinconnelly55370]

Robin, I think force times distance is the work done. The speed the distance is covered gives the rate of doing the work which is power. Kinetic energy is ½mv² where m is the mass and v is the velocity.

Martin C

[Michael GilliganParticipant @michaelgilligan61133]

An interesting definition, Mike

… I have located it in reference to the Charpy Test

I must admit that I had taken the words more literally i.e. Impact Energy = the energy of the impact

… But yes, I see that ‘perfect’ transfer would require [to borrow a term from elsewhere] ‘impedance matching’

MichaelG.

Edited By Michael Gilligan on 12/07/2023 16:22:41

[Michael GilliganParticipant @michaelgilligan61133]

Useful calculation from a Blacksmith here: **LINK**

http://ffden-2.phys.uaf.edu/webproj/212_spring_2017/Patrick_Woolery/Patrick_Woolery/calculations.html

MichaelG.

[duncan webster 1Participant @duncanwebster1]

Posted by MadMike on 12/07/2023 15:35:40:

Impact energy is the energy required to make a standard test piece fail. It is somewhat meaningless in this instance…….

Edited By MadMike on 12/07/2023 15:37:20

Depends on context, I see nothing wrong with using the words to describe the kinetic energy of the chisel at impact.

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