Heritage Jan 11-14, 2024 Auction 3 3

[comix4fun]

On 12/22/2023 at 10:29 AM, delekkerste said:

Also publicly confirmed (275,000 GBP, about $336,000 on the day of the sale):

It's like those contests they used to have at Toys R' Us when we were kids...to get the DREAM SHOPPING SPREE....

I picture David...running down the aisle, filling his cart.... "X-Wing?  Gotta have it" , "Trooper Costume? No brainer"......turns down the James Bond Aisle by accident... "Thunderball? Well, while I'm here, might as well...."

 

[gumbydarnit]

As Dave is running though the store his buddies (I’m looking at you Ryan Condal) are gassing him up “Go Dave Go!!!”

Just like when you were kids, if your bud had the best toys, it was almost as good as you having them!
 

“Dave is it cool if I come over to play with your new X-Wing?”

Edited December 22, 2023 by gumbydarnit

[Bronty]

the image looks terrific so it never occurred to me until now but that flexed scuba fin would be pushing back pretty hard... it would be pretty hard to stand nonchalantly with all that flex.

[batman_fan]

Dave has a piece in his collection I would love to own, hopefully it is auctioned to pay for his recent winnings.

[Brian Peck]

On 12/21/2023 at 7:47 PM, Silver Surfer said:

There must be hundreds of Stormtrooper suits in storage somewhere?

My understanding is this is the best condition complete StormTrooper costume. Greg Jein who used to won it and it was his collection that it and the X-Wing came from. He was THE prop collector, one others come to to get items authenticated. No surprise the StormTrooper was the best example out there.

[Brian Peck]

On 12/22/2023 at 10:37 AM, batman_fan said:

Dave has a piece in his collection I would love to own, hopefully it is auctioned to pay for his recent winnings.

We can all say that. 

[tth2]

On 12/23/2023 at 12:29 AM, delekkerste said:

Also publicly confirmed (275,000 GBP, about $336,000 on the day of the sale):

How could you let that escape your clutches?

[delekkerste]

On 12/22/2023 at 11:25 PM, tth2 said:

How could you let that escape your clutches?

I may have other examples that I personally like more. 

Also, I underbid the main McGinnis image for the Thunderball campaign that was at Heritage last year. I knew that my wife would give me all kinds of grief about the babe content, even though she understood why I had to go for it, and I didn't have any regrets after losing it. It made me realize that I didn't need this version either, and certainly not at an even higher price (though, I had already ruled out a bid early on; IIRC I mentioned that to Dave and others). 

Edited December 23, 2023 by delekkerste

[pestonaccio]

On 12/22/2023 at 4:39 AM, jjonahjameson11 said:

Nope.  It will go for many multiples of that.

I dunno.

i was going to bid on that page, but 5/8k was my price range. For 10k you can buy a HCOL page.

[pestonaccio]

On 12/21/2023 at 7:31 PM, Matches_Malone said:

 

I had it between the $4K-$6K range

Slightly higher, imho, but not over 10k.

small piece, second mini, but iconic.

[jjonahjameson11]

On 12/23/2023 at 9:07 AM, pestonaccio said:

I dunno.

i was going to bid on that page, but 5/8k was my price range. For 10k you can buy a HCOL page.

Let’s revisit after the auction ends.

[pestonaccio]

On 12/23/2023 at 5:06 PM, jjonahjameson11 said:

Let’s revisit after the auction ends.

You are probably right, already at 10k.

WOW!

[tth2]

On 12/24/2023 at 2:27 AM, pestonaccio said:

On 12/24/2023 at 12:06 AM, jjonahjameson11 said:

Let’s revisit after the auction ends.

You are probably right, already at 10k.

WOW!

Formula for OA pricing in the 2020s: (A x 2) + B

where

A = your good faith estimate

B = multiple of $40 (depending on what A is; i.e., if A = double digit, then B = $40, if A = triple digit, then B = $400, and so on)  

[OABearsFan]

The 70's/early 80's John Byrne Marvel art is rare in general, but what seems especially rare these days in the public market are the Byrne MTU pages (and MTIO pages). The Byrne Cap pages (such as the beauty in this HA Signature Auction) come up more often at auction even though there are probably fewer of them. Perhaps a black hole collector or two has them all locked up?

[Matches_Malone]

On 12/21/2023 at 10:39 PM, jjonahjameson11 said:

Nope.  It will go for many multiples of that.

You are certainly right sir. It's already over $10K.

My assessment was based on a previous CL auction. But that was back in 2020. Sandman material has skyrocketed since then. 

[tth2]

On 12/28/2023 at 10:34 PM, Matches_Malone said:

You are certainly right sir. It's already over $10K.

My assessment was based on a previous CL auction. But that was back in 2020. Sandman material has skyrocketed since then. 

That one was a much better image, to boot!  I'd love to see that one come back to the market.

[comix4fun]

On 12/28/2023 at 9:54 PM, tth2 said:

That one was a much better image, to boot!  I'd love to see that one come back to the market.

The clink one was a prelim. 

[alxjhnsn]

On 12/23/2023 at 10:11 PM, tth2 said:

Formula for OA pricing in the 2020s: (A x 2) + B

where

A = your good faith estimate

B = multiple of $40 (depending on what A is; i.e., if A = double digit, then B = $40, if A = triple digit, then B = $400, and so on)  

That would be written as:

=(A*2) + (40*MOD(A, 40))

Mod is a function that divides A by 40 and returns the integer part so MOD (80, 40) gives 2. Just in case you want to put it in Excel.

[Bronty]

Now, I'm not a mathematician, so maybe I'm not understanding but I don't think those are the same.  

I think what Tim is saying (although we are going way way too deep here  ) is that if

A= 10-99 then B = 40; and if

A = 100-999 then B = 400; and if

A = 1000-9999 then B = 4000; and if

A = 10000-99999 then B = 40000; etc.

 

Whereas in your scenario..

if MOD (80,40) is 2 per your explanation, then (40*MOD(A, 40)) = 40*2 = 80 which is not the intention.   

 

 

Tim's intention as I understand it is basically B = 4 x 10^n-1 where n= the number of digits left of the decimal place in A (when A is expressed without extraneous zeroes to the left).

Edited December 29, 2023 by Bronty

[Bronty]

Hmm.    When I write =MOD(80,40) in excel, excel doesn't give me a value of 2, it gives me a value of zero.