Power hammer video, Kinyon

Yeah, I wish this was the same for air pressure.

I bought a Whisper Momma and it basically cost me $1500 landed :-( The forge only got hot enough to forge weld when I adjusted it, by adding an extra fire brick base (effectively reducing the size of the fire box.

Apparently this has to do with the altitude where I live (although I can't fathom why), but others with the Whisper Momma swear that it's the best thing since sliced bread.

About a month later I figured out that making a better furnace was a piece of piss. I still have the whisper momma, and when the lining wears out I will "fix" it :-)

Regards Charles

Reply to
Chilla
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(my apologies for re-answering some questions)

Short answer; nope, the formula for calculating the force that's going into hot iron is; energy =3D (how heavy the hammer is) times (how fast it's going).

Long answer; Newton's second law is F=3Dma, but this means; the rate of change of momentum of a body is proportional to the resultant force acting on the body and is in the same direction. In other words, once you stop accelerating something (and it's moving at a constant velocity), it has no more F, or *rate of change* of momentum. Acceleration is not speed squared, either; m/s is speed, and acceleration is m/(s^2), not (m/s)^2. I think what we want is something more like his explanation of impulse, that the impulse granted an object is force F exercised over time. Other posters have stated this with more clarity than I did.

True! I forgot about pivots and levers, sorry. Well, like I said I was picking those numbers out of thin air.

Nope. The amount of force available doesn't change, just where it goes and what its direction is. It's like a wave in a pool; when the wave hits the edge it gets reflected back. Some of its energy is imparted to the wall, and the rest goes back into the water. Your wave is now smaller and heading in the opposite direction, right? but just because it switched doesn't mean it's going to capsize boats. Same thing with a hammer; you swing, it hits, it bounces back. It may bounce back higher than it started, because you gave it more energy than gravity alone, but it doesn't have more energy coming off the anvil than it did coming down. My apologies if you didn't mean that and I'm holding forth on something you already know.

Living nearer the mountains, here, elevation 2000ft. Advantages: easier to find coal mines. Disadvantages; less air to burn it with. I feel your pain.

Reply to
Tozetre

When Tozetre put fingers to keys it was 7/11/08 8:38 AM...

Having just gone back to that wiki page of formulae it looks like we want to be talking about Kinetic Energy, Work, and Power

(ignoring gravitational potential energy) KE = 1/2 x m x v^2 W = /_\KE (trying to make a 'delta') P(avg) = Work / /_\Time

So, getting the hammer going twice as fast does quadruple the Kinetic Energy of the blow, which quadruples Work and approximately quadruples Power (I'm sure contact Time increases slightly for a heavier blow. How much? no clue).

If it doesn't bounce, that means all the energy has left the hammer and gone into the piece (for practical values of 'all'). It it bounces, it's carrying some energy away with it.

The colder the work, the more the hammer bounces, the less energy imparted to the work, the less it is deformed. Hmmm...

Reply to
Carl West

Ah. After all is said and done, it's 1/2 MV^2.

Pete Stanaitis

Reply to
spaco

Blow me down, so it is. My apologies for claiming otherwise!

Reply to
Tozetre

This is probably too obvious, but no power hammer should be running directly off the compressor any way, so why not put in a larger storage tank?

I ran my bull hammer for a short while off of one of those little compressors that roofers use and a larger tank. It was very hard on the little compressor as they have short duty cycles, but the hammer didn't care at all.

For the small one person shop there is usually enough down time between heats for the smaller compressor to refill the tank.

I have since gone to a big honkin' two stage, but that is because I now am set up to run other things off of air and have the potential of having more than one tool in use at a time.

For a short while I considered twinning a smaller compressor into the system and setting the switch at a higher on point than the two stage. Running the numbers it looked like the energy consumption would not have changed much. Compressing a given volume of air to a set pressure just required the same amount of energy.

I do have the smaller unit piped in as a backup at the other end of the piping, but not turned on, strictly a reservoir. This also helps alleviate possible pressure drop issues due to piping size, if I fire up too many pieces of equipment at the far end of the piping.

Mike Graf

Reply to
CGraf

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