Very Nice "NEW" lincoln Sets added.
clackamas
Posts: 5,615 ✭
There is 1 new set that although it is at #61 the GPA is 69.02!! with 49% complete. That IMO is probably a short set with all the highest possible grades - simply amazing. Gerry is this you?
61 . 69.02 49% 100% 20.92
There was also a new top 15 set added. I just can't hold my own by adding coins.
61 . 69.02 49% 100% 20.92
There was also a new top 15 set added. I just can't hold my own by adding coins.
0
Comments
My Complete PROOF Lincoln Cent with Major Varieties(1909-2015)Set Registry
to get it's score, it would need 10 points of 68RD's (and I would think that Gerry would have more than that)
My guess it is 29 (only P),30-58 all 67RD except
31-S 66RD
30-P 68RD
34-P 68RD
35-P 68RD
39-D 68RD
40-P 68RD
and one other 1 point 68RD (40-S, 42-S or 55-S)
unless it has a 1919 MS68RD instead of the 1929 and then it doesn't have one of the following 30-P, 34-P,35-P,39-P in MS68RD
but that is just my guess :>
Here is some more info, maybe we can figure it out (I always like a good math problem). The only pre 1930 coins that can give you the 70's (and are NOT spoken for):
1917 MS68RD 4 weight pop 2 (1 in blay's set)
1918-D MS67RD 5 weight pop 1
1919 MS68RD 3 weight pop 14
1919-S MS66RD 5 weight pop 1
Since the set is 49% done he must have 68 or 69 coins. Still too many "unknowns to solve". The set could easily have different "69 point" coins. I just think that he is more likely to have MS68RD's of the short set then the pop 1's above. Since the person needs 117 points of "69" coins, and 7 points of "70 coins" (if they have the 1943 coins, then they would need 10 points of 70), I still think it has MS68RD's. The earlier set I suggested is NOT possible since it does not give you the 49% complete. So I am guessing the person has a:
1919 MS68RD (weight 3 "70 point" coin)
1943 PDS MS68 (weight 1 "68 point" coins)
34-P MS68RD (weight 2 "70 point" coin)
35-P MS68RD (weight 2 "70 point" coin)
39-D MS68RD (weight 2 "70 point" coin)
40-P MS68RD (weight 1 "70 point" coin)
and then 60 coins that have a weight of 114 and each gives you (69 points, mostly MS67RD's in the short set but they will not give you the 114 weight)
Just my thoughts
-SOG
I still think he only needs 2 or 3 points of 70. (If no steel cents, funny the steel cents are the easiest in ultra high grade but drops his/her gpa!) Tell me if I'm doing this correct...
125 ms69's + 2 70's = 127 weighted coins
(69*125 + 70 * 2)/127 = 69.01574
124 ms69's + 3 70's = 127 weighted coins
(69*124 + 70 * 3)/127 = 69.02362
So just one short set ms68 would do. I think Gerry has the ms66 19-s? Boy the 18-d would be interesting coin if he/she had it! What about David Shweitz (sorry if I spelled his name wrong). He posts here, right? Does he have the 65rd 26-s or could this set be his? Hmm. If you have all the short set (71 coins) minus 34-d,35-s,36-s, and all p's 46-56, and minus the steel cents, that's 54 coins weighted at 68. Se he/she'd need 15 more coins weighted at 59 to have 69 coins weighted to 127. Let's see thats about 3.93 per coin, so there has to be a few early S's in ms65... and some of the 3 pt P's in ms67, maybe an ms66 late teen d. Wow, so this set sure does have some nice coins!!!! And this is assuming the worst for the short set (and no steel cents), a few more short set coins in 67 (which would be great ones!) would mean the early coins would have to carry more weight (say more S's!).
I've seen a few ms65 S's sell to people who didn't have registry sets (or at least they were never added to a set) Legend's 19-s and 23-s, one of the 17-s's and one of the 27-s's between Sadler's and Dpoole's sets. Maybe these coins are in this set.