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jimbocoin 🃏
6140478c9ae12f1d0b540e7c57806649327a91b040b07f7ba3dedc357cab0da5
The SUPERCYCLE guy.

S U P E R C Y C L E nostr:note14nhfeev58tr885prefpg4p0lqa272t239p9a2r2af4w3fs7m68sqjy9nhq

It used to be a thing to use an old Android device for this purpose.

Given the size of the blockchain, and additional space needed for indexing, to run a full archival node plus Electrum server (plus OS) is gonna take at least 1TB space. 2TB would be prudent to accommodate additional growth.

If you just run a pruned node, then space won’t be an issue, but it also means you can’t look up old transactions.

What’s wrong with aluminum foil?

Well, it would actually make the charts less compelling. The current axes were chosen to convey alarm. But consistent axes would reveal something closer to the truth.

Replying to Avatar BitcoinBadger

Y-axis should be uniform 0-100% for all four charts.

nostr:nprofile1qy88wumn8ghj7mn0wvhxcmmv9uq3zamnwvaz7tmwdaehgu3wwa5kuef0qyfhwumn8ghj7ur4wfcxcetsv9njuetn9uq3vamnwvaz7tm9v3jkutnwdaehgu3wd3skuep0qqsxzsz83jdwztcapd2qulzhspnyjvn6jxcypvrl0w3aahp40j4smfgchnfr5 you are right, it is the same

maybe a higher exponent will flatten it further but it's not hard to see it's already getting quite flat now anyway, like, there is a trend that is decelerating by about the same amount each cycle, but it gets closer and closer to a normal y=x*coeff as time goes on... with that power ratio though, so, just to remember in the linear it's gonna be ... well

somewhere between 150-300 this time, and probably 4x as much as that next time, assuming no hyperinflation, at which point the USD value means nothing anymore

Putting aside the math for a second, let’s talk about price.

Price is entirely psychology. The price of something is what one is willing to exchange for it.

The price of #Bitcoin is crossing an uncomfortable chasm. $60k is too large for most people to consider buying crazy internet money, but the price of a sat is too small to reason about. People are not good with decimals.

Once we hit $1M per whole coin, a sat costs $0.01. A penny stock. At that price, we’ll stop pricing in tranches of 100M sats, like a one time 1:100,000,000 stock split. Overnight, Bitcoin will go from $1M, to $0.01, keeping the BTC ticker.

At that moment, everyone’s interests are aligned—the hard money advocates and the degenerate gamblers alike. That’s when the real FOMO moon pump begins.

So I don’t expect us to flatten into a pure exponential growth curve (line in log scale) just yet. Personally, I think that happens somewhere between cent/sat and dollar/sat parity.

> you need to normalize data to its underlying scaling pattern, or you can't see the patterns in the visualisation

Correct. In a data visualization, there is a projection from the domain of the data (time, dollars, etc.) onto the range of values supported by the visual media (pixels, centimeters, etc.). Choosing the function to apply (log/linear) and parameters (base for log, offset, scale) are arbitrary and made for aesthetic reasons.

> patterns appear in a cluster of data when you normalize it

Agreed that the pattern you see depends on the function applied and the parameters you choose. It is my claim that the choice of base and linear scaling parameter are functionally equivalent. The remainder of this post explains how.

Focusing on the Y axis, and assuming you have chosen log scale, here are the parameters you can choose:

- B - base for log function

- m - scale factor for linear projection

- c - offset for linear projection

The linear projection here is from log price to screen pixels. So the total function from price to Y coordinate is:

f(p) = m * logB(p) + c

Let’s consider the algebraic impact of choosing a different base, B’ for a different price projection function, f’:

f’(p) = m * logB’(x) + c

What is the relationship between f and f’, visually? Let’s find out.

It is a rule of logarithms that one can compute a value in a new base according to this formula:

logB(x) = logA(x) / logA(B)

So for us, that means that:

logB’(x) = logB(x) / logB(B’)

Substituting this into our price projection function f:

f’(p) = m * logB(x) / logB(B’) + c

Refactoring:

f’(p) = m * logB(x) / logB(B’) + c * logB(B’) / logB(B’)

f’(p) = (m * logB(x) + c * logB(B’) ) / logB(B’)

f’(p) = (m * logB(x) + c + c * (logB(B’) - 1) ) / logB(B’)

Substituting our original f definition:

f’(p) = (f(p) + c * (logB(B’) - 1) ) / logB(B’)

Refactoring:

f’(p) = f(p) / logB(B’) + c * (logB(B’) - 1) / logB(B’)

Since B, B’ and c are all constants, what this last formula shows is that f’ is a linear projection of f. That is, it fits the form y=mx+b.

I hope none of the above is controversial (unless I’ve made a mistake in the math).

What does this mean for us, in a data visualization context? As noted earlier, visualizing a function requires projecting from the domain of the data into the range of pixel values. Putting log aside, this means, at minimum, picking a scaling factor and an offset. These values are arbitrary and chosen for aesthetic effect.

So irrespective of whether we use f or f’, we’ll end up linearly scaling the values to project them into pixel space using arbitrary, aesthetically chosen parameters. If we have the same aesthetic intent in both cases, we will select projection parameters that yield identical graphs. The parameter values we pick will be different, but the pixel values will be the same (by definition, since we have the same aesthetic intent for both bases).

This is what I mean when I say the graphs are identical. The only way in which they differ is by a linear scaling function, and we control arbitrary linear scaling parameters.

I hope this is clear. The choice of log base and the choice of linear scaling factor are in the same category of arbitrary visualization parameters. Moreover, they have the same effect. If you squish vertically by choosing a higher base, you can stretch vertically to counterbalance that choice by choosing a larger scale factor.

#asknostr nostr:note1znrepuq90typn5njmnughercucdlhxx5jdrhm27h94ejyqc303fq9s2zl8

It’s entirely possible that it’s on my end, but it would take more debugging than I can do at the moment 😅