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# Double strike hammer

Two hits with one swing
 (+3) [vote for, against]

Carpenter's hammer with movable weight. Spring and back swing keep weight away from main head until split second after main head hits the nail. Movable weight is counter bored so the spring, when compressed, is inside it and weight hits the main head. Total weight and weight distribution are the same as standard hammer. I know you don't get something for nothing from Ma Nature so MFD away!
 — cudgel, Sep 18 2015

http://s1025.photob...ecent]=1&sort=1&o=0 [cudgel, Sep 18 2015]

Low rebound hammer filled wih lead shot [csea, Sep 18 2015]

Elastic vs Inelastic Collisions http://hyperphysics...u/hbase/elacol.html
Rebound is better. [MechE, Sep 28 2015]

like a dead-shot hammer with a solid lump not lead balls. You'd have to figure out how not to have problems with the bounce. You could just double the mass of the hammer for the same effect?
 — bs0u0155, Sep 18 2015

 Make it a triple strike hammer. Or, dare I suggest, a quad strike hammer. Or, let's get real crazy...an infinite strike hammer. Stick a jackhammer mechanism in the hammer head, with a large mass to absorb the recoils.

The bounce of the striking surface is a (surmountable) challenge. The main head would need low mass and very high stiffness. After hitting the nail it actually bounces off and is stopped...it has to be light so the heavy weight can make it strike again.
 — sninctown, Sep 18 2015

I'm with [bs]. If the system works perfectly, you will deliver the same total energy as with a normal hammer of the same mass, but as two small taps rather than one big one. I'm not seeing an advantage; in fact, the opposite.
 — MaxwellBuchanan, Sep 18 2015

It actually doesn't rebound even hitting a steel table at an easy swing. The spring seems to have something to do with it. The two strikes come so close together I can't really hear them.
 — cudgel, Sep 18 2015

You made this? Well, kudos for that.
 — MaxwellBuchanan, Sep 18 2015

[MB] When I had this idea I expected it would deliver lower performance than a regular hammer of the same weight. I was very careful to make it as alike as possible (within .2 oz.) of the same conventional hammer. Not expecting an advantage but a disadvantage, therefore Half Baked. The dead blow effect was unexpected. It is hard for me to tell a difference when driving 12d nails with either hammer into the same piece of wood. The spring started out on the back side of the movable head with the question being would it be compressed enough by the backswing to provide a little extra tap? The double strike sound was just a bit noticeable in that configuration.
 — cudgel, Sep 18 2015

 There are times when multiple lighter blows are desirable, but they are rare. In blacksmithing lighter blows work more on the surface of the piece whereas harder blows have an effect through the piece, and ocassionally this is useful if you are trying to do certain types of detail work. However, it is usually simpler just to switch to a lighter hammer.

 Because, a deadblow hammer (no matter how it is accomplished) is less efficient than a hammer that rebounds. With a traditional soft face mallet, the extra energy goes into deforming the face. With this, I suspect it goes into compressing the spring, but doing so insufficiently to produce a recoil.

So it's interesting, but probably useless.
 — MechE, Sep 18 2015

 I guess any rebound of the first mass is suppressed by the following mass.

 Hmm. There's an interesting physics question here. Suppose you have two hammers of identical mass, swung at the same speed to strike a nail. One hammer has some rebound, the second hammer has no rebound. Which is more effective at driving the nail?

 1st argument: the rebounding hammer leaves with some kinetic energy after the impact; therefore it must deliver less kinetic energy to the nail, and will be the less effective.

2nd argument: the reboundless hammer has all of its downward momentum absorbed by the nail. However, the rebounding hammer not only loses its downward momentum, but acquires some upward momentum on the rebound. This means that the force between the nail and the rebounding hammer must be higher, so the rebounding hammer must be the more efficient.
 — MaxwellBuchanan, Sep 18 2015

[MechE] Yep, just a wall hanger.[MB] The physics is way beyond me but interesting that things don't always fail or succeed the way you think they will. If you can see the photos you know that the face that hits the nail is connected to much more mass than the movable head but less than a normal hammer. I've also wondered if the double, or multiple strikes, if built into the head of a golf driver would be any advantage. Would the club face and ball stay in contact a wee bit longer to transfer more of the energy of the swing to the ball? The downswing of a driver is much faster than that of a hammer so would compress stronger(relatively) springs. Physicists please step forward.
 — cudgel, Sep 18 2015

 [MBhammer] I think your system is too small, hence the conundrum. The mass of the nail and the presumed board , and the stick /slip frictional profile of the nail vs board should be accounted for.

Kind of like the multi-blade razor situation, or Newton's balls. Max. energy transfer in an elastic collision occurs with identical masses.
 — csea, Sep 18 2015

 //The mass of the nail and the presumed board , and the stick /slip frictional profile of the nail vs board should be accounted for.//

 OK, so let's deal with those. Assume that the nail has negligible mass (compared to the hammer head). Assume also that we are driving the nail into an extremely viscous ideal liquid; that overcomes the complexities of "static" versus "dynamic" friction.

So now which hammer displaces the nail furthest: the dead-weight hammer that has no kinetic energy after the impact; or the rebounding hammer whose momentum is not only absorbed but reversed?
 — MaxwellBuchanan, Sep 18 2015

So is the compression of the steel in the nail and the hammer head nullified as to rebound effect in one and not the other?
 — cudgel, Sep 18 2015

Well, in one case that compression is causing the head to rebound, and in the other case it isn't...
 — MaxwellBuchanan, Sep 18 2015

It would seem that the compression and rebound of the steel contributes to driving the nail as well as rebound of the normal hammer. Is this lost with no rebound of the hammer?
 — cudgel, Sep 18 2015

 Re: 1st argument vs 2nd.

 I think the rebounding hammer gets its energy split by the friction of the nail, kind of like two pool balls being hit at the same time by a perpendicular strike from the cue ball. Half goes one way, half goes the other. A strike from the cue ball to both of those balls in line gets all the energy put into it going to one place, wherever both balls hit which would be analogous to the bounce free hammer.

 I'm wondering though if any "shock absorbing" mechanism you put into the hammer to nullify the rebound looses energy in the form of heat or something.

 But definitely, a car crashing into a mud wall vs one bouncing off a brick wall, the mud wall of the same mass is getting pushed further back because it's holding onto the car through the full dissipation of its energy, from the time it hits till the time it comes to a full stop.

 I think the inertia makes the difference. The moving car (or hammer) only has split second to overcome the inertia of the brick wall, then its energy goes off the other way. If there's something to hold onto that moving object for a longer time, the inertia is overcome more effectively. Like running as fast as you can into a heavy refrigerator to move it vs just pushing it.

I think.
 — doctorremulac3, Sep 18 2015

I wonder if any extra advantage is lost in the energy taken to swing it?
 — 2 fries shy of a happy meal, Sep 18 2015

 //So now which hammer displaces the nail furthest: the dead-weight hammer that has no kinetic energy after the impact; or the rebounding hammer whose momentum is not only absorbed but reversed?//

 Quite clearly the dead weight example transferrs the most energy to moving the nail P=mv. The rebounded hammer takes away a portion of the original energy and shows it up as kinetic, then gravitational energy, some of which may be used to pound the nail again.

What hasn't been addressed is oops, looks like [2fries] beat me to it...
 — csea, Sep 19 2015

 // the dead weight example transferrs the most energy to moving the nail P=mv//

 Uh, the energy would be 1/2x mv^2.

The momentum transferred would be mv; but the rebounding hammer must transfer more momentum to the nail, since the velocity is negative on the rebound. Hence the conundrum.
 — MaxwellBuchanan, Sep 19 2015

 I think the rebounding hammer pushes the nail the furthest.

 The dead-weight hammer scenario is an inelastic collision. I.e. some of the initial energy of the system is wasted as heat generated during the inelastic compression of the materials. In the case of the rebounding hammer, a portion of this energy is recovered as kinetic energy, both of the hammer and nail.

 So in other words, I think you want the hammer to bounce as far as possible, and then use this recovered energy to further drive the nail via multiple strikes.

And I've just realised this is basically what [MechE] said.
 — EnochLives, Sep 19 2015

Actually, it might be better to use a screw.
 — MaxwellBuchanan, Sep 19 2015

 I see it this way, in perfect envelope ideals. Action x drives a nail into a material to it's maximum distance. A non rebounding hammer with an action x swing, drives it to this distance = total transfer . A rebounding hammer with action x swing drives to maximum minus completely opposedly wanted rebound action. In both, real cases the hammers would have other losses.

A quick double strike might be an advantage in materials that close up fast and you want to gain on the first hit.
 — wjt, Sep 25 2015

 The problem is that your boundary conditions are an impossible case, so they don't work. It is not possible to have an inelastic collision where kinetic energy is conserved. If you hit something with a hammer that doesn't rebound, a (largish) chunk of the energy is going somewhere other than the nail, typically to (permanently) deforming the head, or heating it.

A rebounding hammer, on the other hand, conserves both momentum and kinetic energy. And since the hammer's momentum is now negative, the nail has a much greater forward momentum.
 — MechE, Sep 25 2015

 To clarify, picture two objects of equal mass, m, in free space. One stationary and one with velocity, Vi.

 In an inelastic collision (non-rebounding), they start with kinetic energy mVi^2, and momentum mVi. When they collide, they stick together, and now have momentum 2mVi/2, or still mVi (twice the mass, half the speed, since momentum is conserved). However, their kinetic energy is now 2m(Vi/2)^2, or mVi^2/2, half of what it was initially. This is because kinetic energy isn't conserved, and the remainder is lost as heat. The more massive the second object, the more energy is lost to heat (and a nail with a wall behind it is fairly massive).

 In the case of a perfectly elastic collision, the ball with the initial velocity is left stationary, and the ball 2 is now moving away at the same velocity the original ball was. The equation for velocity of the first mass in an elastic collision is V1f=V1i*(m1- m2)/(m1+m2). (This may seem counterintuitive, but this is why it's possible for the cue ball to stop dead after hitting another pool ball). Thus all of the kinetic energy is transferred to the second ball.

If the second ball is more massive, it's final speed will be slower, V2F=V1i*2m1/(m1+m2), but the resultant kinetic energy sill still be higher than the inelastic collision mentioned first.
 — MechE, Sep 25 2015

Why is the spring necessary? As long as you're bringing the hammer down at faster than the speed of gravity, inertia should keep the weight separated from the back of the head until the strike.
 — ytk, Sep 25 2015

 [MechE] The two cases have the same masses, the hammers and the (nails forced into material), the same action energy, the swing. One hammer leaves with energy, one doesn't. I would be surprised if the energy from the heat and sound >= the E from the rebound.

The spring would be good for other hammer motion rather than have a weight flopping around.
 — wjt, Sep 26 2015

[ytk] Yes, the spring isn't necessary. My original idea placed the spring behind the floating weight.
 — cudgel, Sep 26 2015

[wjt] Check the math. Momentum is conserved, so no matter what, the kinetic energy of the nail then the hammer comes to a dead stop is 1/2 the kinetic energy of the nail when the hammer rebounds. The energy lost in an inelastic collision is absolutely greater than the rebounding energy.
 — MechE, Sep 28 2015

 [annotate]

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