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V13 2010 INDEX       E-SYLUM ARCHIVE

The E-Sylum: Volume 13, Number 23, June 6, 2010, Article 24

TECH TALK: COIN STRIKING PRESSURE CALCULATIONS

Dick Hanscom submitted this technical question for the engineers and Minting-technology geeks among our readership. He found some Internet resources to assist him in his calculations, but still has questions. To "Ask Dr. Google" is usually a great first step, but can only take one so far. -Editor

So here is my physics problem. I have built a "new and improved" (read heavier) drop hammer. I am curious how many pounds per square inch will be created upon impact on my die.

The weight is 50 Pounds.
The distance is 5 feet.
The surface area of the die is 1.25 square inches.

Using the Internet I am able to get it down to Kinetic Energy (337.84 J), but don't know if that can be converted into pounds per square inch.

My thinking is that my 50 pound weight is 8x6 inches (48 square inches). Sitting on the ground, that is a little more than 1 Pound per square inch. However, if it is sitting on a 1 square inch die, it is imparting 50 pounds per square inch. The energy is the same, but the area affected is different.

Raise it up and drop the weight 5 feet, the energy is the same whether it lands flat, or on a 1 square inch die. However, all the energy get imparted to a much smaller area when it hits the die.

There's my problem. I must don't know how to ask the Internet the right question to get the right answer.

I don't remember all the search string permutations that I tried, but they include things like "calculate force", "calculate pound per square inch", "convert Kinetic energy," etc. Also included "falling object."

So I got ways to calculate energy in a falling object, but no way to convert that to pounds per square inch.

I will include the answer and the calculations in the second edition! Hopefully, some one will provide the actual calculations and not just the answer.

I am not sure that it will be enough to strike my 1 oz. gold token. I will test it this weekend. I may have to lift it the 5 feet, and then "throw" it down, adding to the acceleration created by gravity. We shall see!

So, can someone assist? -Editor


Wayne Homren, Editor

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