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Gold Tokens
My question is this: How do I calculated pounds per square inch at impact of a drop hammer on a coining die?

The weight is 50 pounds, the drop is 5 feet. It will impact a coining die that has an area of 1.25 square inches.

Is there a formula I can use to change the weight and height?

Keep it simple. I am 39 years away form high school physics, and didn't need it in college!

Thanks
Richard Hanscom
Fairbanks, Alaska
rpenner
Sadly, this cannot give you pounds per square inch. Even to get average pounds per square inch, you need the time of the impact, or equivalent.

A crude estimate of pounds per square inch is if the features on the coin are x hundredths of an inch, then the 50 pounds drop 6000 hundreds of an inch and stop in a distance of x hundreds of an inch. So the pounds of force average to about 50 times 6000 over x, and you divide that by 1.25 inches to get pounds per square inch. Or, 240000/x pounds-per-square inch -- on average.
Gold Tokens
My 50 pound weight is approximately 8 inches by 6 inches. If it is sitting on the ground, it is imparting about 1 pound per square inch. If it is sitting on top of a die that has an area of 1.25 inch, it is imparting 40 pounds per square inch.

If I raise it up to 5 feet and drop it onto that die, we have 40 pounds per square inch x ????? the force generated by the fall.

Further, let us assume that it stops on impact and does not force the die, sitting on a 2 inch thick plate of steel, any distance into the top of a spruce log.

Further, let us assume that the faces of the dies are flat, with no detail, and that there would be no compression of the material between the dies.

What would the formula be to compute this.

Thanks.
Richard Hanscom
Fairbanks, Alaska
rpenner
If you are assuming that the top surface of the die does not move at all then you are assuming the energy of the weight is entirely dissipated as heating of the weight itself, you need to tell us what the weight is made of and it's shape.

Assuming it is steel or lead, I calculate very weird shapes for it. Is it a wood hammer?

And your assumption that die rests on 2 inches of steel on top of a spruce log (on top of .... ?) doesn't tell us that the top surface of the die doesn't move at all. It mostly tells us that you aren't well setup to have your query answered over the internet.

To figure out pounds of force per square inch, you need 1) the number of square inches (check) and 2) the pounds of force needed to stop the weight. To get (2) you need the a) force of gravity on the weight (check) , (b ) the distance the weight falls (check) and either (i) the effective duration of the collision or (ii) the distance travelled during the collision.

You assert (without measurement) that (ii) is zero, so one way to estimate (i) is via the shape and speed of sound in the hammer. The higher the speed of sound (the harder the hammer) the shorter the time of collision, and the higher the peak pounds per square inches.
Gold Tokens
Sorry, but I did not consider all the variables. To my simple mind, it is pretty straight forward. I really thought there would just be a simple formula.

If a 50# weight is sitting on top of something that is 1.25 in area, then it is exerting 40# per square inch. There is much more energy transferred to the die if it is dropped from 5 feet.

And before you suggest it, yes, I am ignoring the friction from the guides through which the weight drops. I just don't care! I make gold tokens as a hobby. The physics side is just curiosity.

Perhaps there just is no way to convert that energy into pounds per square inch, a measurement my feeble mind would understand.

Quote: "It mostly tells us that you aren't well setup to have your query answered over the Internet." And so I have to ask it differently over the Internet than I would in person?

Seems like this is just too complex a question to be answered in a simple, straight forward manner. Or are you making it too complex?

Richard Hanscom
Fairbanks, Alaska
Enthalpy
It is indeed a complex question with no simple answer.

The pressure depends of the braking time or distance, and these are no consequence of the general hammer's movement. They depend on the behaviour of the deformed part, here of your coin if everything works as it shall.

That is, an impact on hard steel, or on stone, will produce a harder shock (more pounds per square inch during less time) than on a plastic, a rubber...

In the case of a coin, the deformable material will more or less define the pressure, since the hammer can produce any needed pressure - its way would only be shorter. So, the speed and weight of the hammer must suffice to deform the coin for your intent.

The pressure is linked to the coin's material so-called "Yield strength", but not in a simple way, because a coin is not flat, and because most materials get harder as they're deformed.

Subtle software tries to model this behaviour (mainly for plastic injection moulding) but still isn't precise enough.
Gold Tokens
Enthalpy - Thanks. That is an answer I can understand. I was just hoping that there would be a Pounds per square inch at "moment" of impact, without all the other variables thrown in.

Sadly, I have learned about "work hardening" using my manual hydraulic press. It takes three pressings, and annealing between each pressing to bring up the detail in the token.

Which is why I have built a drop hammer. That, and the drop hammer is cooler than the manual hydraulic press!

Thanks again.
Richard Hanscom
Fairbanks, Alaska
Craig
QUOTE (Gold Tokens+Jun 10 2010, 03:06 AM)
My question is this: How do I calculated pounds per square inch at impact of a drop hammer on a coining die?

The weight is 50 pounds, the drop is 5 feet. It will impact a coining die that has an area of 1.25 square inches.

Is there a formula I can use to change the weight and height?

Keep it simple. I am 39 years away form high school physics, and didn't need it in college!

Thanks
Richard Hanscom
Fairbanks, Alaska

You have almost everything you need to find the answer, but the replies I have seen are correct. When you drop the 50 lb hammer onto something from a height of 5 feet it gains kinetic energy while falling. When it impacts the thing it is to strike, the thing takes on the weight of the mass and also must take on the additional force from the mass having energy that must be rid of through some means.

The actual equation to use that would tell you the peak force of impact when it strikes is the relationship between an impulsive force and a change in momentum. You've probably seen F = ma, well re-write it as F = m(deltaV/deltaT) and multiply both sides by deltaT getting: FdeltaT = mdeltaV. This states that the change in momentum of some mass yields the force applied times the time it took to change the momentum from what it was just before the strike to what it is just after. Just before is easy to find, as you can find the velocity the mass has from being dropped from 5 feet easily, and the velocity of the mass after the collision is zero. So you know deltaV it is the velocity just before it strikes the coin minus zero velocity. The mass is also known because you have the weight. That means you need only the time it takes for the hammer to come to a rest from what its maximum velocity was just before the strike to zero velocity.

Given that you are working with hard materials the time it takes for the hammer to come to rest from some speed will be small or very quick. Therefore the force it applies upon the coin is:

F = m (Vf - Vi)/(tf - ti) = m (0 - Vi)/(tf - 0) = -mVi/tf where tf is the final time or the time it takes just at the moment the hammer is striking the mass to come to a rest upon doing so; and Vi is the velocity of the hammer the moment just before it strikes the coin. It always looks as if the hammer simply stops when it hits the coin, but that isn't really true. The coin would have to be beyond the hardest substance known to man, it would have to be a completely rigid body that doesn't flex in anyway. The implications of such an event would be that the force it strikes the coin with would be infinitely large in value for a period of time that is zero seconds long. That doesn't make too much sense; we use the idea a lot in engineering as what we call the impulse function. It is a function that inputs an infinite amount of something into something else, but does so in no time. There is actually a value that can be obtained from this, because the infinitely sized peak is spread over no distance, the area under the curve of that peak is usually an actual value.

In your case, the coin will give a little and therefore the time it takes for the hammer to stop will be non zero. You need that time if you want to know what the force is. Since the time it takes for the hammer to stop will be very small the force it imparts upon striking the coin should be great. With every strike more thermal energy is transferred into the coin and hammer, heating them both up some, but the initial force it strikes with is the same every time if dropped from the same distance and is stopped in the same amount of time.

To mathematically find the time would be a nightmare. You would probably be better off putting a load cell where the coin goes and then drop the hammer upon it. The load cell will read the force it sees upon it and you can set it up such that the maximum force the cell sees is displayed.

I love complex mathematics, but I wouldn't touch trying to find the time it takes that mass upon striking the coin to come to a rest through a mathematical analysis, it would drive you crazy and you'd probably get the wrong value anyway; or even if you got something somewhat close, being it is such a small value, a small difference will relate back to large differences in the calculated force upon the coin.

If not a load cell, a simple scale would work. You can set the scale there in place of the coin, make sure the scale is one that when it feels the maximum force applied upon it; has a way of keeping that value for you to use. If it is a digital scale there is probably some button on there that can be pushed so that it registers and holds the maximum force seen upon it. If it is an analog scale, it can have something like another hand that moves with the first and stays at the maximum displacement the needle moves to.

Just a couple of ideas that may save you some time and effort.

Craig smile.gif
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