Does mass added to an anvil increase efficiency?

[thingmaker3]

Well since we are talking small differences you need a bit more rigor:
1, optical pyrometer to make sure the piece is *exactly* the same temp each time.
2, A weight supported at the same distance over the piece and allowed to free fall in the same orientation to make sure the impact is exactly the same---sort of a guillotine system would work.
3, repeat the test a minimum of 10 times with each anvil mounting.
4, publish results

Also my 515# anvil wasn't dancing in use but it was creeping.

You left out the bit about precision machining test pieces to identical size prior to heating...

[thingmaker3]

More musing... the "clamping force" due to gravity will be six times as great for a 300# anvil as for a 50# anvil.

[Francis Trez Cole]

if it is just used for striking on. the important thing is a solid base attached to the ground firmly. Look at the anvil Brian Brazeal uses it is basically a piece of mild steel attached to legs and pinned to the ground the way he explained it is the connection of a good base makes all the difference when you attach to the earth. The anvil weights about 45 pounds and look at the results he gets. When he came to do a demo in my shop I made one of the anvils it is a great tool to have around I find myself using it all the time.

[Bentiron1946]

I know that my 125# anvil was a lot more enjoyable to work on when I mounted it on an anvil stand made of 10" square box tube filled with sand, 1/2" plate on each end and then dogged down tight to that with 3/8" square mild steel rod. That anvil didn't ring or wiggle on that base but if I got into a good hard swing it would walk on a concrete floor a bit, not a lost but some. I guess that I was a wasting some energy but not near as much as when it was on a eucalyptus stump that was 18" in diameter. That stump sure wasn't very heavy but sure smelled good when hot scale fell in the cracks. :blink:

[bajajoaquin]

I looked at the formula on the Monster Anvils thread, and it looks like a formula for the wrong case. According to the person who came up with this formula from an introductory physics textbook, it describes the energy lost in the rebounding hammer assuming perfectly elastic collision with the anvil. This is not what happens when an experienced blacksmith is forging. According to a mechanical engineering text that I looked at, most industrial forging takes place at nearly zero restitution coefficient. In other words, the collision of hammer and target is nearly inelastic. At zero restitution coefficient, the hammer sticks to the target. At unit restitution coefficient, the hammer rebounds elastic with no energy deposited in deformation of the target. This corresponds more to the case of the beginner wailing away on cold metal. I watched some blacksmiths, including myself, and noticed that the hammer does not rebound much except for finishing blows.

So, the more relevant number is the energy "lost" in deformation for an inelastic collision. The efficiency then becomes this amount divided by the initial hammer energy. Isn't the desirable energy that which goes to do useful work on the target, anyway? Not the energy that sends your hammer rebounding up. I notice that the hammer only rebounds appreciably when I accidentally strike the face of the anvil.

Get out that physics textbook again, and do the calculation for an inelastic collision. This time, the velocity of the hammer and that of the target is equal (sticking) after collision, so there's your extra equation to make up for losing the conservation of energy equation. You have the unknown velocity and the single equation of conservation of momentum. One equation, one unknown, and it can be solved. Or, if the math is too overbearing, immediately join the CBA and get the next California Blacksmith magazine, which will have the answer, I hope.

Of course, this observation seems to fly in the face of the guru's statements at anvilfire.com, but it does seem that blacksmithing is about deforming hot metal rather than seeing how high your ball bearing rebounds.

As you noted, I only put an equation into Excel that was provided by someone else, so I don't have any axe to grind here. But it seems that the equation is valid in understanding the effect of mass on efficiency, even if it doesn't describe the method or amount of energy transferred to the work piece. I did a thought experiment to try to see how the "rebound" equation might be applicable.

What if we take a massless workpiece, and put it on "anvils" of various masses. We could start out with no anvil at all, and because there's no mass in the work piece, when we hit it with a hammer, it just gets pushed aside, and zero energy is transferred to working the piece. Now, if we put in an anvil of infinite mass below it, 100% of the hammer's energy will be transferred to the work piece. Somewhere in between zero and infinity, a fraction of the hammer's energy will be transferred. This is a function of the mass of the anvil (ignoring differences in elasticity in the anvil, otherwise, we'll be talking about wood and concrete vs. steel). Here, the "rebound" equation tells us the percentage of the energy that is available to do work.

The elasticity or plasticity of the workpiece would be the determining function in how that available energy is converted into moving the metal, or doing work.

At least, that's how I see it. I'm open to correction.