Sorry, Doug, but Mike is right. 1/2 the coin in the dies = twice the usual pressure. That is real simple math and physics to me. Per your examples, Jefferson nickels are noted for weak steps and lack of windows. Yet your 1974 shown looks like s 6 stepper to me and every window showing is solid. Weak strike?
For coins that can't be fully struck (other than transition point) they look that way to me. And as for the transition point thing (which happens to be to low point on the planchet) that shows my point-- weak strike can equal plantchet marks surviving.
I had a Great Aunt with gnarly arthritic fingers and when she'd want to pinch you, she'd sometimes grab to much skin and it would not hurt, but man when she grabbed just a bit of skin, good Lord that hurt. Same principle of area vs pressure, like that a woman's high heel. When weight (pressure being added-force) is placed onto the shoe, the spike of the heel can dent linoleum, but the shoe part will not because the load is spread over a larger area. The same principle can be applied to a coin that is only partially struck. The same force is applied to a much smaller area, creating more distortion in the object being struck. Die set up (stand-off) trial strike, incomplete struck planchet. Note the marks in the unstuck area of the obverse, these marks were there prior to striking and prior to being released from the mint facility.
Thought: reread the sentence, there may be some logic to everyone else. Or at least a reason why everyone disagrees with you.
Only those that have posted dis-agree with Doug. So far neither of you all have convinced me 100%. So keep going...I am just along for the ride. Is this real truth somewhere in the middle? Just not sure - yet.
For Doug's example of a partially struck coins; try this. F = P * A (force = pressure * area). See http://www.school-for-champions.com/science/pressure.htm just to pick one. On the stamping machine, the force is constant. On Doug's example, the coin is struck on only 1/2 the area. I will let you do the math to determine the pressure
I take a three inch piece of quarter inch stainless steel and want to form a 90 degree bend in it, I set the machine to apply a certain amount of pressure and make my part. Then I have to make a 90 degree bend in a 8 foot wide piece of stainless steel, I have to up the pressure immensely in order to make the longer part. When you have only part of the planchet between the dies, you are going to get a much better strike on the part that is struck, because all the pressure is applied to a smaller area. I think that does not mean that the marks you dispute are planchet marks, just that you are thinking that since less of the planchet is in the dies that they somehow do not recieve the full strike, au contraire mon frer, the part that is in the striking area recieves all the pressure. And since part of the planchet is outside the striking area the pressure is concentrated in a smaller area. You could do a simple experiment to prove this, take a 16 ounce hammer and swing it at a cent and then note the cents where you hit it squarely how much it was flattened, compared to the cents you delivered a blow off center to.
The die is putting pressure on the coin and the surface of the die did not change. The coin is not putting pressure on the die directly but indirectly. Therefore the coin is getting 100 psi not 200
"Every action has an equal and opposite reaction." The only thing between the dies is the coin. How much more directly can you be? Aside from that, I site formula and reference. You site .... um?
I don't agree that half the coin is 2x the pressure, if the coin isn't seated correctly that means there is empty space for the metal to be pushed towards (the reason some of the coins are slightly oblong) Instead of the pressure forming a more accurate strike part of that pressure pushes metal into that void. note-I didn't say that they strike is better or worse, just that I don't think it would be twice as much pressure. If there wasn't a blank in place would the dies touch? If not then I would think the pressure would be constant no matter what.
Doug, Please explain how all of these marks are concentrated in the area that represents the highest point of the coin (the jaw) and there are essentially no other contact marks on the entire coin. Additionally, they don't look like contact marks. Most lack the definition and sharpness that is exhibited by a contact mark. Unlike wear which will almost always show on the high points first, contact/surface marks are not drawn to the high points of a coin design. When you have a coin design that shows the same surface flaws in the same areas over years of production, it is perfectly reasonable to conclude that the surface irregularities are the result of planchet flaws in combination with inadequate strike. Just because you don't believe it to be true, doesn't mean it isn't.
Now contrast the coin above with a 1974 with contact marks. Your error example was also a 1974 IIRC. The striking issues of the Jefferson Nickel had been remedied by 1974 and your example has about the best obverse strike characteristics I have ever seen. Notice how defined the contact marks are on the jaw unlike the 1952. I will admit that the TPG's routinely excuse contact marks that are found in this area as planchet flaws but that does not mean we throw the baby out with the bath water. Planchet flaws are real and should be excused during grading IMO.
I think some of the marks are planchet marks, not all-doesn't really matter to me. Why should a coin get a higher grade if it has marks? I think coins should be graded for the condition they are in. I could give a real world example of this but..... won't go there.
It is also worth noting -- and this is EXTREMELY DIFFICULT TO DO IN PHOTOS -- that there is a very subtle difference in planchet defects and contact marks on high grade coins that is visible to the eye and under magnification. What I mean is this: The planchet defects have a less "fresh" or less "sharp" appearance than contact marks. I've always thought it was because of the slight stretching the surface that a planchet goes through during striking, even in an area that doesn't completely strike (i.e. the balloon effect). Perhaps it's because of the heating of the metal itself under such high pressures, but there's a very subtle difference between hits and planchet marks. Once the coin's been circulated, the appearance of these delicate features (essentially, the "edge" and/or "freshness" of the contact/planchet mark) quickly deteriorate. But on extremely high grade specimens -- like those that Lehigh pictured above -- you can begin to see what I'm referring to. In hand, it's much more obvious, but still quite subtle.
p.s. there is also a subtle difference in luster in the not-fully-struck area indicating a planchet defect, versus a contact mark.
Between camlovs and pauls posts - why I am still up in the air on what causes all of it. I think Doug, Paul, Mike and all make valid points. For the two coins posted by Paul I can see why one would have a higher grade than the other - I also could see why someone buying might grade them the same.
I site your formula the die is striking the coin force* area 1" coin * 100 lbs = 100 psi only 1/2 the die struck the coin so .5 inch* 100 lbs = 50 psi do the math before making statements
Great examples, Lehigh! One thing... I disagree. I think they should be "discounted" instead. The thought being they do detract from the eye appeal of the coin. Let's face it -- if you have two coins that are alike in every way, yet one with planchet defects, and the other with none, you're going to choose the one with none. Add in a third coin, this time with hits in the same area of the coin, and you'd likely take the one with planchet defects. Thus "discounted" rather than ignored. Regardless of my petty semantic quibble , thanks for the perfect illustration....Mike
Ice, I'm sorry to say but you are incorrect. Your formula was Pressure= Area * Force. The correct formula is: Pressure = Force / Area or Force = Pressure * Area If you plug the numbers into either of the two above equivalent formulas, they yield an answer of 200 psi. Sometimes I make the same mistake as you've made. To check my work, I look at the units. Pressure is PSI (pounds per square inch) Force is pounds (actually foot-lbs) Area is square inches. Since pressure is measured in POUNDS PER SQUARE INCH, it means that pressure relates to force and area like this: Pressure = (Force is pounds) per (area is square inches) Pressure = Force / Area Hope this helps...Mike
