Topic: (pics) Physical Bitcoin Bills - For Real World Transactions - Printcoins.com (Read 8824 times)
February 01, 2013, 10:28:58 AM
Due to the increase in valuation printcoins is now offering 0.1 BTC bills which are about equal to a $2 bill. Bills that are greater that 10 BTC have been taken off. Also, Ron Paul bills have been discontinued.
August 11, 2012, 09:03:50 PM
Totally cool! Great idea. Thanks!!!
August 11, 2012, 12:20:56 PM
April 24, 2012, 02:39:13 PM
missed this thread, found it in google images while looking for something.
February 28, 2012, 01:47:15 PM
These will now only be sold in bulk orders of 100. This will likely change back in May, but for now, the small orders are just not worth the time. I would like to sell plastic credit card format printcoins, so let me know if you would like to do bulk orders of these.
January 11, 2012, 09:37:39 PM
Make a QR code sticker/hologram.......WIN
January 11, 2012, 09:21:22 PM
Any plans or thoughts on selling a "checkbook" of xx checks rather than xx loose checks?
January 05, 2012, 05:08:38 PM
Ok we both agree:
According to the docs p is the prime number specifying the finite field used and n is the prime number specifying the order of G.
So I think the number p should be used for all the finite field math operations, for example a = b*c mod p
And n has something to do with G (the base point of the system).
So, I believe that I am correct in that the "overflow" in the finite field math is at p.
I am trying to remember exactly what n means and how it is used. I remember reading it and understanding it at one point but I forget now exactly what it means.
If I find the reference again I will post it here.
We need to nail this down because it affects the all important calculation of the final private key f = a*b mod ?
So you could easily experimentally determine which one it is. Calculate the final private key using each number and let us know which one calculates the correct final private key.
Code:
p = FFFFFFFF FFFFFFFF FFFFFFFF FFFFFFFF FFFFFFFF FFFFFFFF FFFFFFFE FFFFFC2F
n = FFFFFFFF FFFFFFFF FFFFFFFF FFFFFFFE BAAEDCE6 AF48A03B BFD25E8C D0364141
According to the docs p is the prime number specifying the finite field used and n is the prime number specifying the order of G.
So I think the number p should be used for all the finite field math operations, for example a = b*c mod p
And n has something to do with G (the base point of the system).
So, I believe that I am correct in that the "overflow" in the finite field math is at p.
I am trying to remember exactly what n means and how it is used. I remember reading it and understanding it at one point but I forget now exactly what it means.
If I find the reference again I will post it here.
We need to nail this down because it affects the all important calculation of the final private key f = a*b mod ?
So you could easily experimentally determine which one it is. Calculate the final private key using each number and let us know which one calculates the correct final private key.
January 05, 2012, 04:12:37 PM
The number mike calls N is the same number I call p (for "prime") is the prime number:
p = FFFFFFFF FFFFFFFF FFFFFFFF FFFFFFFF FFFFFFFF FFFFFFFF FFFFFFFE FFFFFC2F
All the other parameters for the Bitcoin elliptical curve system can be found here:
https://bitcointalksearch.org/topic/m.635050
and most of the thread at https://bitcointalksearch.org/topic/elliptic-curve-math-question-53177 is interesting and informative.
p = FFFFFFFF FFFFFFFF FFFFFFFF FFFFFFFF FFFFFFFF FFFFFFFF FFFFFFFE FFFFFC2F
All the other parameters for the Bitcoin elliptical curve system can be found here:
https://bitcointalksearch.org/topic/m.635050
and most of the thread at https://bitcointalksearch.org/topic/elliptic-curve-math-question-53177 is interesting and informative.
Between the two of us, one of us has it wrong, because I think the number that needs to be divided by and mod taken is
FFFFFFFF FFFFFFFF FFFFFFFF FFFFFFFE BAAEDCE6 AF48A03B BFD25E8C D0364141.
This is the number I'm calling N. The number above you have called p is the number I would call p as well, but I don't think that's the right number for this purpose.
THe person who has it wrong could very well be me, becuase chances are I implemented it, saw it worked, and forgot about it further. I could have had a brainfart and used the other number instead, but this is the number I always thought i used.
January 05, 2012, 02:58:14 PM
The number mike calls N is the same number I call p (for "prime") is the prime number:
p = FFFFFFFF FFFFFFFF FFFFFFFF FFFFFFFF FFFFFFFF FFFFFFFF FFFFFFFE FFFFFC2F
All the other parameters for the Bitcoin elliptical curve system can be found here:
https://bitcointalksearch.org/topic/m.635050
and most of the thread at https://bitcointalksearch.org/topic/elliptic-curve-math-question-53177 is interesting and informative.
p = FFFFFFFF FFFFFFFF FFFFFFFF FFFFFFFF FFFFFFFF FFFFFFFF FFFFFFFE FFFFFC2F
All the other parameters for the Bitcoin elliptical curve system can be found here:
https://bitcointalksearch.org/topic/m.635050
and most of the thread at https://bitcointalksearch.org/topic/elliptic-curve-math-question-53177 is interesting and informative.
January 05, 2012, 02:31:35 PM
Thanks. I really like this elliptical curve crypto - it is a lot more fun than RSA.
Maybe we should take this to another thread but can you explain this
(F = b*A)
since both keys are 256 bit numbers are we just doing simple multiplication here.
i.e.
5 * 8 = 40 (but obviously much larger numbers). How are we handling 256 bit overflow.
Or maybe I am complete off. Can you provide an example of (F = b*A) using real numbers.
In this case, A is not a 256-bit number. In programming terms, it is an object, a structure with two 256-bit numbers known as X and Y. The definition of "multiplication" has been redefined - in C++ terms, the "multiplication operator" has been overridden with a far more complex definition. So, F = b * A means "Perform the elliptic curve multiplication operation on point object A, using the 256-bit integer b as a parameter"... this operation returns another point object, not a number.
And 256 bit overflow is not a concern once you understand what "finite field" is. Yes, there is overflow, and the overflow is thrown away - but you must critically understand that the overflow doesn't happen at 2^256, it happens at another number known as N, which is a 256-bit number, but smaller than 2^256. So when you do scalar multiplication and the answer exceeds N, you handle the overflow by dividing by N and keeping the remainder (not dividing by 2^256 and keeping the remainder, as "handling 256 bit overflow" strictly means).
N is a constant you can find on the bitcoin wiki, search for "secp256k1".
January 05, 2012, 12:55:01 PM
Update: 0.5 BTC denomination bills now available.
If this price rally keeps up I might even get to roll out my 0.25 bills.
Here is a complete list of all intended colors (going down to 0.01 BTC - which would require bitcoins to be around $100 a piece to make sense):
http://printcoins.com/colors
If this price rally keeps up I might even get to roll out my 0.25 bills.
Here is a complete list of all intended colors (going down to 0.01 BTC - which would require bitcoins to be around $100 a piece to make sense):
http://printcoins.com/colors
January 05, 2012, 10:45:24 AM
Quickly,
First, private keys are just very large numbers but public keys are actually a pair of two very large numbers representing the x, y coordinates of a point on an elliptical curve.
I use lower case letters to represent elements of the selected finite field and capital letters to represent points in the selected elliptical curve group.
So: a = b*c represents simple modulo multiplication for the selected finite field, short for a = ((b*c) mod p)
And: A = B + C would represent the defined point addition operation on the selected elliptical curve group. This is not simple addition, it is the way we define the addition of two points on the elliptical curve.
Then A = a*G is just G+G+G+G+...+G, a times. And in this case the * represents the defined scalar multiplication operation for the selected finite field over the selected elliptical curve group - so no this is not simple multiplication it is a pretty involved function and can be a bit time consuming.
Clear as mud?
First, private keys are just very large numbers but public keys are actually a pair of two very large numbers representing the x, y coordinates of a point on an elliptical curve.
I use lower case letters to represent elements of the selected finite field and capital letters to represent points in the selected elliptical curve group.
So: a = b*c represents simple modulo multiplication for the selected finite field, short for a = ((b*c) mod p)
And: A = B + C would represent the defined point addition operation on the selected elliptical curve group. This is not simple addition, it is the way we define the addition of two points on the elliptical curve.
Then A = a*G is just G+G+G+G+...+G, a times. And in this case the * represents the defined scalar multiplication operation for the selected finite field over the selected elliptical curve group - so no this is not simple multiplication it is a pretty involved function and can be a bit time consuming.
Clear as mud?
January 05, 2012, 10:02:16 AM
Thanks. I really like this elliptical curve crypto - it is a lot more fun than RSA.
Maybe we should take this to another thread but can you explain this
(F = b*A)
since both keys are 256 bit numbers are we just doing simple multiplication here.
i.e.
5 * 8 = 40 (but obviously much larger numbers). How are we handling 256 bit overflow.
Or maybe I am complete off. Can you provide an example of (F = b*A) using real numbers.
January 05, 2012, 09:02:06 AM
Thanks. I really like this elliptical curve crypto - it is a lot more fun than RSA.
January 05, 2012, 07:39:09 AM
Is this the proposal? Should work.
Quote
Person 1:
Create a random private key (a)
From the random private key calculate the standard public key (A = a*G)
Send the public key (A) to person 2
Create a random private key (a)
From the random private key calculate the standard public key (A = a*G)
Send the public key (A) to person 2
Quote
Person 2:
Create a random private key (b)
From the random private key calculate the standard public key (B = b*G)
Send the public key (B) to person 1
Create a random private key (b)
From the random private key calculate the standard public key (B = b*G)
Send the public key (B) to person 1
Quote
Person 1:
From the random private key (a) and the public key created by person 2 (B) calculate the final public key (F = a*B)
[Note a*B = a*b*G which makes the final private key a*b]
Calculate the final public key address of the final public key (F)
From the random private key (a) and the public key created by person 2 (B) calculate the final public key (F = a*B)
[Note a*B = a*b*G which makes the final private key a*b]
Calculate the final public key address of the final public key (F)
Quote
Person 2:
From the random private key (b) and the public key created by person 1 (A) calculate the final public key (F = b*A)
[Note b*A = b*a*G which makes the final private key b*a]
Print the private key (b) on the bill - and cover it with a sticker to hide it
Calculate the public key address of public key F and print it on the bill
Send the bill to person 1
From the random private key (b) and the public key created by person 1 (A) calculate the final public key (F = b*A)
[Note b*A = b*a*G which makes the final private key b*a]
Print the private key (b) on the bill - and cover it with a sticker to hide it
Calculate the public key address of public key F and print it on the bill
Send the bill to person 1
Quote
Person 1
Finally print the private key (a) on the bill
Can match the public key address calcluated above with the public key address on the bill
Finally print the private key (a) on the bill
Can match the public key address calcluated above with the public key address on the bill
Burt you are a great source of clarity. Thanks for the above post!
January 04, 2012, 08:27:22 PM
That's really cool! I suppose as you get more confident with what you're doing the markup fee will drop, or disappear altogether. Also, as demand increases, including for larger amounts, there will be bulk pricing for larger notes. But as it's not too likely that somebody is going to order 100 50 BTC notes right now, it's not really that important. I'm glad you have bulk pricing for the 1 BTC notes. It's still a little too high for me right now. But if you can get your prices down to, say, 106 BTC for 100 1 BTC notes, I'd throw down for that.
You're creating a very important product and I'm glad you're putting a lot of thought into it. This can be something that is truly badass and puts Bitcoin on the map for a much larger segment of society.
You're creating a very important product and I'm glad you're putting a lot of thought into it. This can be something that is truly badass and puts Bitcoin on the map for a much larger segment of society.
Why would markup disappear completely?
As far as 0.06 BTC markup. You are aware there is no central bank which can subsidize printing costs. Things like material, labor, custom holograms, etc do time take and resources. Your expectations may be a little unrealistic.
Well, obviously there is going to be markup. But 3 BTC markup on a 50 BTC note is too steep for me. So is 17 BTC on 100 1 BTC notes.
I'm sure, eventually, there will be competition in the Bit bill market, driving prices down.
I am just giving room for a competitor to step in and eat my lunch. If I charged too little, than what would be the incentive for anyone else to break into the market.
Fair enough.
January 04, 2012, 08:19:32 PM
That's really cool! I suppose as you get more confident with what you're doing the markup fee will drop, or disappear altogether. Also, as demand increases, including for larger amounts, there will be bulk pricing for larger notes. But as it's not too likely that somebody is going to order 100 50 BTC notes right now, it's not really that important. I'm glad you have bulk pricing for the 1 BTC notes. It's still a little too high for me right now. But if you can get your prices down to, say, 106 BTC for 100 1 BTC notes, I'd throw down for that.
You're creating a very important product and I'm glad you're putting a lot of thought into it. This can be something that is truly badass and puts Bitcoin on the map for a much larger segment of society.
You're creating a very important product and I'm glad you're putting a lot of thought into it. This can be something that is truly badass and puts Bitcoin on the map for a much larger segment of society.
Why would markup disappear completely?
As far as 0.06 BTC markup. You are aware there is no central bank which can subsidize printing costs. Things like material, labor, custom holograms, etc do time take and resources. Your expectations may be a little unrealistic.
Well, obviously there is going to be markup. But 3 BTC markup on a 50 BTC note is too steep for me. So is 17 BTC on 100 1 BTC notes.
I'm sure, eventually, there will be competition in the Bit bill market, driving prices down.
I am just giving room for a competitor to step in and eat my lunch. If I charged too little, than what would be the incentive for anyone else to break into the market.
January 04, 2012, 08:07:48 PM
That's really cool! I suppose as you get more confident with what you're doing the markup fee will drop, or disappear altogether. Also, as demand increases, including for larger amounts, there will be bulk pricing for larger notes. But as it's not too likely that somebody is going to order 100 50 BTC notes right now, it's not really that important. I'm glad you have bulk pricing for the 1 BTC notes. It's still a little too high for me right now. But if you can get your prices down to, say, 106 BTC for 100 1 BTC notes, I'd throw down for that.
You're creating a very important product and I'm glad you're putting a lot of thought into it. This can be something that is truly badass and puts Bitcoin on the map for a much larger segment of society.
You're creating a very important product and I'm glad you're putting a lot of thought into it. This can be something that is truly badass and puts Bitcoin on the map for a much larger segment of society.
Why would markup disappear completely?
As far as 0.06 BTC markup. You are aware there is no central bank which can subsidize printing costs. Things like material, labor, custom holograms, etc do time take and resources. Your expectations may be a little unrealistic.
Well, obviously there is going to be markup. But 3 BTC markup on a 50 BTC note is too steep for me. So is 17 BTC on 100 1 BTC notes.
I'm sure, eventually, there will be competition in the Bit bill market, driving prices down.
January 04, 2012, 08:06:49 PM
That's really cool! I suppose as you get more confident with what you're doing the markup fee will drop, or disappear altogether. Also, as demand increases, including for larger amounts, there will be bulk pricing for larger notes. But as it's not too likely that somebody is going to order 100 50 BTC notes right now, it's not really that important. I'm glad you have bulk pricing for the 1 BTC notes. It's still a little too high for me right now. But if you can get your prices down to, say, 106 BTC for 100 1 BTC notes, I'd throw down for that.
You're creating a very important product and I'm glad you're putting a lot of thought into it. This can be something that is truly badass and puts Bitcoin on the map for a much larger segment of society.
You're creating a very important product and I'm glad you're putting a lot of thought into it. This can be something that is truly badass and puts Bitcoin on the map for a much larger segment of society.
Why would markup disappear completely?
As far as 0.06 BTC markup. You are aware there is no central bank which can subsidize printing costs. Things like material, labor, custom holograms, etc do time take and resources. Your expectations may be a little unrealistic.
D&T - Thank you, you are now my official spokesperson for all pricing issues.
Also, anyone is welcome to undercut me if I charge too much. You could probably charge much less and still make a profit if you have access to cheap labor.
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